### 4. Gyromagnetic factor of electron

The non-relativistic Schrödinger equation for a charge in a magnetic field reads

$$i\hbar\partial_t\Psi = \left[ \frac{1}{2m} \left( \mathbf{p} - \frac{q}{c_0}\mathbf{A} \right)^2 + q\Phi \right] \psi \qquad (78)$$

In the presence of a homogeneous magnetic field we can set $\mathbf{A} = \frac{1}{2}\mathbf{B} \times \mathbf{r}$ and $\Phi = 0$. In that case the cross-term becomes,

$$-\frac{q}{2mc_0}(\mathbf{A} \cdot \mathbf{p} + \mathbf{p} \cdot \mathbf{A}) = -\frac{q}{mc_0}\mathbf{p} \cdot \frac{1}{2}(\mathbf{B} \times \mathbf{r}) = -\frac{q}{2mc_0}\mathbf{L} \cdot \mathbf{B}$$

with $\mathbf{L} = \mathbf{r} \times \mathbf{p}$ the angular momentum operator. We identify the interaction as the magnetic energy $-\mathbf{m} \cdot \mathbf{B}$ with $\mathbf{m}$ the magnetic moment of the rotating charge. We conclude that its magnetic moment is given by $\mathbf{m} = \mu_B \mathbf{L}$ with $\mu_B = q/2mc_0$ the Bohr magneton (and negative for an electron with $q = -e$). There is no spin in the non-relativistic version but we can add it by hand by making the wave function $\Psi$ a spinor and add the interaction $-g_q \mu_B \mathbf{S} \cdot \mathbf{B}$ knowing that the spin angular momentum operator is $\mathbf{S} = \frac{\hbar}{2}\mathbf{\sigma}$. At this level we can just speculate that the value for the gyromagnetic factor $g_e$ of the electron be equal to 1 as well. However, experimentally it is very close to $g_e = 2$.

The Dirac equation in a magnetic field can be written as,

$$\left[ \gamma^{\mu} \left( i\hbar\partial_{\mu} - \frac{q}{c_0} A_{\mu}(t, \mathbf{x}) \right) - mc_0 \right] \Psi(t, \mathbf{x}) = 0$$

where $\Psi$ is now a bi-spinor. To find $g_e$ we have to map this equation on Eq. (78). To this end we assume $mc_0^2$ to be the dominating energy and set $\Psi(t) = \psi(t) \exp(-imc_0^2t/\hbar)$, with $\psi(t)$ slowly varying with respect to the large energy $mc_0^2$. We will work in the Dirac representation adapted for the electron and set $\Psi = (\Psi_A, \Psi_B)$ in terms of two spinors. We obtain the two coupled equations,

$$\begin{aligned} i\hbar\partial_t\psi_A &= \sigma^k(i\hbar c_0\partial_k - qA_k)\psi_B \\ (i\hbar\partial_t + 2mc_0^2)\psi_B &= \sigma^l(i\hbar c_0\partial_l - qA_l)\psi_A \end{aligned}$$

We can neglect the time-derivative in the second equation and insert the approximation for $\psi_B$ into the first equation. This yields,

$$\begin{aligned} i\hbar\partial_t\psi_A &= \frac{1}{2mc_0^2}\sigma^k\sigma^l(i\hbar c_0\partial_k - qA_k)(i\hbar c_0\partial_l - qA_l) \\ &= \frac{1}{2mc_0^2}(i\hbar c_0\partial_k - qA_k)^2 + \frac{1}{2mc_0^2}i\epsilon_{kln}\sigma^n(i\hbar c_0\partial_k - qA_k)(i\hbar c_0\partial_l - qA_l) \end{aligned}$$
where we have inserted the spin commutator of the Pauli matrices. With the index-conventions the first term can be written as $(\mathbf{p} - q\mathbf{A}/c_0)^2/2m$ denoting orbital angular momentum. In the second term, parts proportional to $\partial_k \partial_l$ and $A_k A_l$ cancel by the anti-symmetry of $\epsilon_{kln}$. The remaining term is proportional to $-i^2\sigma^n \epsilon_{nkl}(\partial_k A_l + A_k \partial_l) = -\sigma^n \epsilon_{nkl}(\partial_k A^l) = -\sigma \cdot \mathbf{B}$. Hence,

$$i\hbar\partial_t\psi_A = \left[ \frac{1}{2m} \left( \mathbf{p} - \frac{q}{c_0}\mathbf{A} \right)^2 - 2\mu_B\mathbf{S} \cdot \mathbf{B} \right] \psi_A$$

This implies that $g_e = 2$. The gyromagnetic factor of the electron is one of the most accurately known constants of nature. The current experimental value is $g_e = 2.00231930436092(36)$. The small deviation stems from the electromagnetic quantum vacuum, the leading QED correction $\Delta g_e = \alpha/\pi = 0.00232...$ was first calculated by Julian Schwinger in 1948 for which he obtained the Nobel Prize.

## H. Irreducible Representations of SU(2) × SU(2)

A major challenge is to find the irreducible representations of the restricted Lorentz group $SO^+(1,3)$. We have seen that boosts are not unitary unless we complexify. Indeed, as was emphasized first by Eugene Wigner [3], the Lorentz group $SO^+(1,3)$ has no unitary irreducible representations of finite dimension. We have seen that the groups $SO(3)$ and $SU(2)$ have finite-dimensional representations labeled by angular momentum. With the unitary trick we have concluded that $SO^+(1,3) \cong SU(2) \times SU(2)$ which implies that representations of $SO^+(1,3)$ can be labeled by the angular momentum as $(j_1, j_2)$ with physical dimension $(2j_1 + 1) \times (2j_2 + 1)$ and with tensor representation $(j_1 \otimes j_2)$. By the addition rule of angular momenta, the z-component in particular, we can split them up in irreducible representations,

$$(j_1 \otimes j_2) \cong (|j_1 - j_2|) \oplus (|j_1 - j_2| + 1) \oplus \cdots \oplus (j_1 + j_2 - 1) \oplus (j_1 + j_2) \quad (79)$$

In literature one often prefers to display the degeneracy $2j + 1$ rather than the angular momentum itself. Then this identity is written as

$$((2j_1 + 1) \otimes (2j_2 + 1)) \cong (2|j_1 - j_2| + 1) \oplus (2|j_1 - j_2| + 3) \oplus \cdots \oplus (2(j_1 + j_2) - 1) \oplus (2(j_1 + j_2) + 1)$$
<table><thead><tr><th>J<sub>z</sub></th><th>j<sub>1</sub> = 5/2, j<sub>2</sub> = 1, possibilities (j<sub>1</sub> + j<sub>2</sub>)<sub>z</sub></th><th>multiplicity</th></tr></thead><tbody><tr><td>7/2</td><td>5/2 + 1</td><td>1</td></tr><tr><td>5/2</td><td>5/2 + 0, 3/2 + 1</td><td>2</td></tr><tr><td>3/2</td><td>5/2 + (-1), 3/2 + 0, 1/2 + 1</td><td>3</td></tr><tr><td>1/2</td><td>3/2 + (-1), 1/2 + 0, -1/2 + 1</td><td>3</td></tr><tr><td>-1/2</td><td>1/2 + (-1), -1/2 + 0, -3/2 + 1</td><td>3</td></tr><tr><td>-3/2</td><td>-1/2 + (-1), -1/2 + (-1), -3/2 + 0</td><td>3</td></tr><tr><td>-5/2</td><td>-5/2 + 0, -3/2 + (-1)</td><td>2</td></tr><tr><td>-7/2</td><td>-5/2 + (-1)</td><td>1</td></tr></tbody></table>

explicitly showing that the total multiplicity (dimension) is preserved. Indeed, for $j_1 > j_2$ we can verify that the righthanded side has the same dimension as the lefthanded side since

$$ \begin{aligned} & [2(j_1 - j_2) + 1] + [2(j_1 - j_2) + 3] + \cdots + [2(j_1 - j_2 + 2j_2) + 1] \\ &= [2(j_1 - j_2) + 1](2j_2 + 1) + \sum_{n=0}^{2j_2} (2n) \\ &= (2j_2 + 1)(2j_1 + 1) \end{aligned} $$

The maximum spin $J = j_1 + j_2$ follows from the unique choice for $J_z = \pm(j_1 + j_2)$, part of the representation $(j_1 + j_2)$ only. Let $j_1 > j_2$. As we go down to $J_z = j_1 + j_2 - 1$ we have already two possibilities for the two spins, $j_1 - 1$, $j_2$ and $j_1$, $j_2 - 1$ covered by the two representations $(j_1 + j_2) \oplus (j_1 + j_2 - 1)$. For instance for two spins $1/2$ this would be the triplet (1) and the singlet (0). As we go down more and more terms in the direct sum become involved but this procedure stops when $J_z = j_1 - j_2$. This implies the relation (79) with all equal weighting factors.

The diagram for the direct product $5/2 \otimes 1 \cong (7/2) \oplus (5/2) \oplus (3/2)$ illustrates this picture. It can be verified explicitly that the direct sum produces the correct multiplicity: $J_z = 7/2$ can only come from the representation $(7/2)$, $J_z = 5/2$ is supported by both $(7/2)$ and $(5/2)$ and thus has double degeneracy, $J_z = 3/2$ by $(7/2)$, $(5/2)$ and $(3/2)$ with triple degeneracy, etc. The total number of possibilities is $6 \times 3 = 18$. In the same way we obtain $(1 \otimes 1) \cong (2) \oplus (1) \oplus (0)$ with dimension 9 on both sides.

The complexification is a mathematical trick to get unitary operators associated with real boosts and rotations. The theorem says that without this trick unitary representations only exist for infinitely many dimensions. This argument is often put forward as a justification for quantum field theory, where particles can be created and destroyed and which is by construction a representation of infinite dimensionality, labeled by the continuous parameter $p^\mu$, generator of translations in space-time. In quantum field theory, particle states are representations of the Poincaré group,
that includes space-time translations and thus has 6 + 4 = 10 generators. In addition to the finite-dimensional angular momentum labels, this brings in the continuous four-vector $p^\mu$ with the Lorentz-invariant property $p_\mu p^\mu = m^2$, defining the mass of the particle, and of course a crucial property of a particle. It makes the group intrinsically infinite dimensional, labeled by spin and mass, and charges when gauge fields are included. See for more details (Chapter 1 of) the book by Steven Weinberg [6], one of the founders (and defenders) of modern quantum field theory.

## 1. Tensor representations

Given two vector spaces $V$ and $W$ with elements $v$ and $w$ for which linear operations $av_1 + bv_2$ are defined, the direct product $V \times W$ is defined as all pairs $(v, w)$. The tensor product is a bilinear map from $V \times W$ to $\mathbb{R}$ or $\mathbb{C}$ often written $V \otimes W$. We can use a basis $\{e_i(v)\}$ and $\{e_j(w)\}$ for both vector spaces and write $(v, w) = \sum_{ij} (e_i(v), e_j(w)) v_i w_j$ and map it to the second-order tensor $v_i w_j$. If $G$ and $H$ are both groups that have a representation in the vector space $V$ and $W$, then $G \times H$ has the obvious tensor representation $(\Pi(G) \otimes \Pi(H)$ in space $V \otimes W$ acting on the tensor $v_i w_j$. For instance the tensor product $\mathbb{R}^3 \otimes \mathbb{R}^3$ consists of all real-valued tensors $B_{ij}$ with $i, j = 1, 2, 3$. Under $SO(3)$ rotations in $\mathbb{R}^3$, the tensor transforms as

$$B'_{ij}(x) = R_i^n R_j^m B_{nm}(R^{-1}x) \qquad (80)$$

We can identify the tensor product in $\mathbb{R}^3 \otimes \mathbb{R}^3$ as a $(j = 1) \otimes (j = 1)$ representation of $SO(3) \times SO(3)$, equivalent to the direct product of two spins 1. By Eq. (79) this group is isomorphic to the irreducible direct sum $(0) \oplus (1) \oplus (2)$. The irreducible subgroups have specific tensor symmetry. The direct sum decomposition of $(1, 1)$ contains the scalar, symmetric representation of spin 0, so that there must be an invariant symmetric tensor in this representation, rapidly identified as the metric tensor $\delta_{ij}$ in $\mathbb{R}^3$. The tensor decomposition is associated with the following identity,

$$B_{ij}(x) = \delta_{ij}T(x) + A_{ij}(x) + S_{ij}(x) \qquad (81)$$

with $T(x)$ a scalar field, $A(x)$ an antisymmetric tensor, and $S$ a symmetric traceless tensor, with respectively 1, 3 and 5 degrees of freedom. The symmetric invariant of the group $(1, 1)$ reveals its bosonic nature.

**Exercise V.18.** $(1 \otimes 1) \cong (0)_S \oplus (1)_A \oplus (2)_S$ and $(\frac{1}{2} \otimes \frac{1}{2}) \cong (0)_A \oplus (1)_S$ The first identity decomposes the product states of two spin 1 into irreducible representations of $SU(2)$ of spin 2, 1 and 0, with each a specific symmetry.
* The spin states $|S = 2, M = \pm 2\rangle$ are uniquely obtained by the product states $|S_1 = 1, m_1 = \pm 1\rangle \times |S_2 = 1, m_2 = \pm 1\rangle$ of two spins 1. Both states are symmetric. Using the lowering operator $S_- = S_{1-} + S_{2-}$ defined earlier, show that

$$ 
\begin{aligned}
|2, 1\rangle &= \frac{1}{\sqrt{2}} [ |1, 0\rangle|1, 1\rangle + |1, 1\rangle|1, 0\rangle ] \\
|2, 0\rangle &= \frac{1}{\sqrt{6}} [ |1, -1\rangle|1, 1\rangle + 2|1, 0\rangle|1, 0\rangle + |1, 1\rangle|1, -1\rangle ] \\
|2, -1\rangle &= \frac{1}{\sqrt{2}} [ |1, 0\rangle|1, -1\rangle + |1, -1\rangle|1, 0\rangle ]
\end{aligned}
$$

so that the entire five-fold multiplet $|S = 2, M\rangle$ is symmetric. The group $(2)_S$ is a representation of $SU(2)$ with spin 2 in a 5D vector space containing spin wave functions that are entirely symmetric in two spin 1 wave functions.

* The total spin state $|S = 1, M = 1\rangle$ must be orthogonal to $|S = 2, M = 1\rangle$. We can choose

$$ |1, 1\rangle = \frac{1}{\sqrt{2}} [ |1, 0\rangle|1, 1\rangle - |1, 1\rangle|1, 0\rangle ] $$

and check that it has $M = 1$. This state is antisymmetric upon exchange of both spins. Verify by lowering operations that

$$ 
\begin{aligned}
|1, 0\rangle &= \frac{1}{\sqrt{2}} [ |1, -1\rangle|1, 1\rangle - |1, 1\rangle|1, -1\rangle ] \\
|1, -1\rangle &= \frac{1}{\sqrt{2}} [ |1, -1\rangle|1, 0\rangle - |1, 0\rangle|1, -1\rangle ]
\end{aligned}
$$

Hence the entire triplet is antisymmetric. The group $(1)_A$ is a representation of $SU(2)$ with spin 1 in a 3D vector space containing spin wave functions that are entirely anti-symmetric in two spin 1 wave functions.

* The singlet state $|S = 0, M = 0\rangle$ must be orthonormal to both $|1, 0\rangle$ and $|2, 0\rangle$ and is necessarily unique up to a phase factor. Check that

$$ |0, 0\rangle = \frac{1}{\sqrt{3}} [ |1, -1\rangle|1, 1\rangle - 2|1, 0\rangle|1, 0\rangle + |1, 1\rangle|1, -1\rangle ] $$

The singlet state of two spins 1 is symmetric. The group $(0)_S$ is a representation of $SU(2)$ with spin 0 in a 1D vector space containing a spin wave function symmetric in two spin 1 wave functions.

* Next consider two spins $\frac{1}{2}$. We write $\frac{1}{2} \otimes \frac{1}{2} \cong (0) \oplus (1)$. Show in the same way that the
triplet with spin 1 is now symmetric and

$$ 
\begin{aligned}
|1,1\rangle &= \left| \frac{1}{2}, \frac{1}{2} \right\rangle \left| \frac{1}{2}, \frac{1}{2} \right\rangle \\
|1,0\rangle &= \frac{1}{\sqrt{2}} \left[ \left| \frac{1}{2}, \frac{1}{2} \right\rangle \left| \frac{1}{2}, -\frac{1}{2} \right\rangle + \left| \frac{1}{2}, -\frac{1}{2} \right\rangle \left| \frac{1}{2}, \frac{1}{2} \right\rangle \right] \\
|1,-1\rangle &= \left| \frac{1}{2}, -\frac{1}{2} \right\rangle \left| \frac{1}{2}, -\frac{1}{2} \right\rangle
\end{aligned}
$$

and since the singlet $|0,0\rangle$ is orthonormal to $|1,0\rangle$, we find the antisymmetric state

$$ |0,0\rangle = \frac{1}{\sqrt{2}} \left[ \left| \frac{1}{2}, \frac{1}{2} \right\rangle \left| \frac{1}{2}, -\frac{1}{2} \right\rangle - \left| \frac{1}{2}, -\frac{1}{2} \right\rangle \left| \frac{1}{2}, \frac{1}{2} \right\rangle \right] $$

The group $(0)_A$ is a representation of $SU(2)$ with spin 0 in a 1D vector space with wave functions anti-symmetric in two spin $\frac{1}{2}$ wave functions. The group $(1)_S$ is a representation of $SU(2)$ in a 3D vector space of symmetric wave functions.

### Example: Tensor representation of three spins 1

For 3 $SU(2)$-spins 1 in $\mathbb{C}^3 \otimes \mathbb{C}^3 \otimes \mathbb{C}^3$, the third-order tensor product decomposes as

$$ (1) \otimes (1) \otimes (1) \cong (0)_A \oplus (1)_S \oplus (1)_M \oplus (1)_{M'} \oplus (2)_M \oplus (2)_{M'} \oplus (3)_S $$

where $M$ and $M'$ stand for a mixture of symmetric and antisymmetric states. The dimension is 27 on both sides. The antisymmetric scalar representation identifies the unique antisymmetric invariant Lévi-Civita tensor $\epsilon_{ijk}$. The associated tensor decomposition of $SU(2) \times SU(2) \times SU(2)$ in $(\mathbb{C}^3)^3$ is thus

$$ C_{ijk}(x) = \epsilon_{ijk}T(x) + S_{ijk}(x) + N_{ijk}(x) $$

with $S_{ijk}(x)$ a tensor that is fully symmetric in the three indices. This tensor has ten independent components (three iii components, $3 \times 2 = 6$ iij components and one ijk component). We can extract the symmetric tensor

$$ V_{ijk}(x) = \delta_{ij}v_k + \delta_{ik}v_j + \delta_{jk}v_i $$

with three degrees of freedom and associated with the vector representation $(1)_S$. The remainder in $S_{ijk}(x)$ has 7 degrees of freedom and corresponds to the octupole representation $(3)_S$.
<table><thead><tr><th>Field</th><th>Representation of SO<sup>+</sup>(1, 3)</th><th>D</th><th>Spin</th><th>Elementary particle</th></tr></thead><tbody><tr><td>scalar (S) or singlet (A)</td><td>(0, 0)</td><td>1</td><td>0</td><td>Higgs boson</td></tr><tr><td>spinor + dual spinor</td><td>(1/2, 0) &oplus; (0, 1/2)</td><td>2 + 2</td><td>1/2</td><td>chiral leptons, quarks</td></tr><tr><td>vector</td><td>(1/2, 1/2)</td><td>4</td><td>1</td><td>four-vector field A<sup>&mu;</sup></td></tr><tr><td>anti-symm. tensor</td><td>(1, 0) &oplus; (0, 1)</td><td>3 + 3 - 2</td><td>1</td><td>massless photon F<sup>&mu;&nu;</sup> + dual with gauge</td></tr><tr><td>symm. traceless tensor</td><td>(1, 1)</td><td>9</td><td>2</td><td>massless graviton with gauge</td></tr></tbody></table>

## 2. Tensor representations of the Lorentz group

Let us generalize this to the Lorentz group $SO^+(1, 3)$ acting on scalar, vector and tensor operators. A scalar field must transform as

$$U(\Delta)\Psi(x)U(\Delta)^{-1} = \Psi(\Delta^{-1}x) \qquad (82)$$

with $U(\Delta) = \exp(-i\omega_{\mu\nu}M^{\mu\nu})$ a unitary transformation with the 6 generators $M^{\mu\nu}$ that represent the Lie algebra of the Lorentz group, or equivalently, the rotations and boosts defined by $M^{ij} = \epsilon^{ijk}J_k$ and $M^{i0} = K_i$ satisfying the commutation relations (55). If we take the derivative on both sides,

$$U(\Delta)\partial^{\mu}\Psi(x)U(\Delta)^{-1} = \partial^{\mu}\Psi(\Delta^{-1}x) = \Delta_{\rho}^{\mu}\partial^{\rho}\Psi(x')|_{x'=\Delta^{-1}x} \qquad (83)$$

Since $\partial^\mu \Psi(x)$ is an example of a contra-variant four-vector operator, we expect that in general four-vector operators must transform as

$$U(\Delta)A^{\mu}(x)U(\Delta)^{-1} = \Delta_{\rho}^{\mu}A^{\rho}(\Delta^{-1}x) \qquad (84)$$

and a second-order four-tensor operator as,

$$U(\Delta)B^{\mu\nu}(x)U(\Delta)^{-1} = \Delta_{\rho}^{\mu}\Delta_{\sigma}^{\nu}B^{\rho\sigma}(\Delta^{-1}x) \qquad (85)$$

By Eq. (2) we immediately conclude that the metric tensor $g_{\mu\nu}$ is an invariant tensor under the representation.

Let us now search for tensor representations of the Lorentz group in some vector space consisting of $n$-component operators $\Psi_a(x)$. To be a representation they must transform according to

$$U(\Delta)\Psi_a(x)U(\Delta)^{-1} = S(\Delta)_a^b\Psi_b(\Delta^{-1}x) \qquad (86)$$

with the requirement of group homomorphism,

$$S(\Delta_1\Delta_2)_a^b = S(\Delta_1)_a^cS(\Delta_2)_c^b \qquad (87)$$
The scalar field is the trivial (0,0) representation of the direct product SU(2) × SU(2) where both groups reduce to the unit operation, and S = 1 without indices. Its (1, 1) representation is equivalent to the one of SO(3) discussed above.

### Example: Tensor representations with spin $\frac{1}{2}$ as building block.

The Dirac field was seen to be a spinor with two components, either lefthanded or righthanded in the Weyl representation. For a small transformation we can write $\Delta^\mu_\nu = \delta^\mu_\nu + \omega^\mu_\nu$ and expand

$$S(\omega)_a^b = \delta_a^b - \frac{i}{4} \omega_{\mu\nu} (\sigma^{\mu\nu})_a^b$$

for some operator $\sigma^{\mu\nu}$ acting on a bispinor. If we restrict to lefthanded chirality we set $S_L = P$, with P defined earlier in Eq. (58) so that $\sigma^{\mu\nu}$ must be related to the Lie algebra of SL(2, C). We have introduced co-variant notation that becomes clear later once we have identified the equivalent of the metric tensor in this representation. The transformation (86) can be expanded to

$$- [\Psi_a, M^{\mu\nu}] = \frac{1}{i} (x^\mu \partial^\nu - x^\nu \partial^\mu) \Psi_a - \frac{1}{2} (\sigma^{\mu\nu})_a^b \Psi_b$$

where the first term is generated by the expansion of the argument $\Delta^{-1}x$ and clearly relates to rotational angular momentum. If we put $x^\mu = 0$ we keep only spin and with $M^{ij} = \epsilon^{ijk}J_k$ and $K_n = M^{n0}$ it follows that,

$$\epsilon^{ijn} [\Psi_a(0), J_n] = \frac{1}{2} (\sigma^{ij})_a^b \Psi_b(0)$$

$$[\Psi_a(0), K_n] = \frac{1}{2} (\sigma^{n0})_a^b \Psi_b(0)$$

At this point we can impose the lefthanded representation $(\frac{1}{2}, 0)$ of SU(2) × SU(2) with $(J_n - iK_n)/2$ in representation $(\frac{1}{2})$ and $(J_n + iK_n)/2$ in representation (0). The commutator of the field with $(J_n \pm iK_n)/2$ stands for $i$ times its small variation in the transformation, which for SU(2) is $-\frac{1}{2}i\sigma^n\Psi_a$ and 0 respectively. Therefore $[\Psi_a(0), J_n] = [\Psi_a(0), -iK_n] = \frac{1}{2}\sigma^n$ so that $(\sigma^{ij})_a^b = \epsilon^{ijn}(\sigma^n)_{ab}$ and $(\sigma^{n0})_a^b = i(\sigma^n)_{ab}$. For the righthanded representation $(0, \frac{1}{2})$ the rotational part $\sigma^{ij}$ remains unaltered but the generator for the boosts changes sign. Thus $(S_R)_a^b = [(S_L^\dagger)^{-1}]_a^b$. This agrees with former results obtained when the mutually dual Weyl representations of the Dirac equation where derived.

We can next consider a tensor $C_{ab}$ with two lefthanded spinor indices $ab$. It must transform as

$$U(\Delta)C_{ab}(x)U(\Delta)^{-1} = S_L(\Delta)_a^c S_L(\Delta)_c^d C_{cd}(\Delta^{-1}x) \quad (88)$$
This is naturally interpreted as a left-handed tensor representation $(\frac{1}{2}, 0) \otimes (\frac{1}{2}, 0) = (\frac{1}{2} \otimes \frac{1}{2}, 0)$. From exercise V.18 we have $(\frac{1}{2} \otimes \frac{1}{2}) = (0)_A \oplus (1)_S$ acting in the vector space of one anti-symmetric singlet and a three symmetric triplet states. It follows

$$ (\frac{1}{2}, 0) \otimes (\frac{1}{2}, 0) = (0, 0)_A \oplus (1, 0)_S \qquad (89) $$

The presence of an antisymmetric singlet implies the existence of an anti-symmetric invariant $\epsilon_{ab}$ of the transformation and indeed it can be checked explicitly (using $\epsilon_{ab} = -i(\sigma^2)_{ab}$) that

$$ S_L(\Delta)_a^c S_L(\Delta)_a^d \epsilon_{cd} = \epsilon_{ab} \qquad (90) $$

The antisymmetric tensor $\epsilon_{ab}$ is invariant in the tensor product space of two left-handed spinors, in much the same way as we have identified the invariance of $\epsilon_{ijk}$ in third-order representation of $SO(3)$ and the tensor $\epsilon_{\alpha\beta\mu\nu}$ in the 4-order tensor representation of the Lorentz group in Minkowski space. Thus anti-symmetric tensors remain anti-symmetric upon transformation and the group identity (89) expresses that

$$ C_{ab} = \epsilon_{ab} D(x) + G_{ab}(x) \qquad (91) $$

with $G_{ab}$ symmetric. Quite remarkably, this anti-symmetric tensor takes over the role of the metric tensor in the spin $\frac{1}{2}$ representation and must be used to lower and raise spinor indices. For instance $\Psi^a = \epsilon^{ab}\Psi_b(x)$. We assume that $(\sigma^\mu)_{ab}$ stands for the usual definition of the Pauli matrices. The definition is

$$ \epsilon_{ab} \equiv -i(\sigma^2)_{ab} = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \qquad (92) $$

It follows that $\epsilon^{ab} = -\epsilon_{ab}$, then $\epsilon_{ab}\epsilon^{bc} = \epsilon_a^c = \epsilon_c^a = \delta_a^c$ again reminiscent of the metric tensor $g_{\mu\nu}$ though unlike $g_{\mu\nu}$ with a subtle minus sign between fully covariant and fully contra-variant version.

Let us now consider tensors that involve the direct product of left- and righthanded spinor operators, that we shall write as $C_{ab}$, the dot implying that this index transforms as a righthanded spinor. Thus,

$$ \begin{aligned} U(\Delta)C_{ab}(x)U(\Delta)^{-1} &= S_L(\Delta)_a^c S_R(\Delta)_b^c C_{cd}^{\dot{d}}(\Delta^{-1}x) = P_{ac}(P^{\dagger,-1})_{\dot{b}\dot{d}} C_{cd}(\Delta^{-1}x) \\ &= P_{ac}C_{cd}(\Delta^{-1}x)(P^{\dagger,-1,T})_{\dot{b}\dot{d}} \end{aligned} \qquad (93) $$

and which is a tensor representation $(\frac{1}{2}, 0) \otimes (0, \frac{1}{2}) = (\frac{1}{2}, \frac{1}{2})$ of the Lorentz group. Its most remarkable property is that we can use it to define a four-vector operator $A_\mu(x)$ according to

$$ C_{ab}(x) := A_\mu(x)(\sigma^\mu)_{ab} \epsilon_{bb'} \equiv A_\mu(x)(\sigma^\mu)_{ab} \qquad (94) $$
The presence of the metric tensor guarantees that the dotted index transforms as a complex-dual spinor field as desired. Hence the representation $(\frac{1}{2}, \frac{1}{2})$ is associated with a four-vector field.

### Exercise V.19. Representation of vector field

We recall that given a spinor with components $\eta_b$ and its Lorentz transformation $P_{ab}\eta_b$, the complex-dual spinor $\xi_a = \epsilon_{ab}\eta_b^*$ transforms as $[P^\dagger]^{-1}_{ab}\xi_b$. We also recall that $\epsilon P^{-1}\epsilon^T = P^T$, equivalent to $\epsilon P^T = P^{-1}\epsilon$, for any $P \in SL(2, \mathbb{C})$ with unit determinant.

* Show that $A_\mu(x)$ Lorentz-transforms as a covariant vector field, defined by Eq. (84).

* Demonstrate the invariance of the antisymmetric tensor $\epsilon_{ab}$ given by Eq. (90).

* Derive the matrix representations of the four matrices $(\sigma^\mu)_{ab}$ and comment.

Second-order Minkowski tensors, defined in Eq. (85), can be constructed from direct products of four-vector fields $C^\mu(x)D^\nu(x)$ and thus, in view of the above, relate to the four-order tensor representation $(\frac{1}{2} \otimes \frac{1}{2}) \otimes (\frac{1}{2} \otimes \frac{1}{2})$ of the group $SU(2)$. Because of the group decomposition $(\frac{1}{2} \otimes \frac{1}{2}) \cong (1)_S + (0)_A$ we obtain the tensor decomposition

$$ (\frac{1}{2}, \frac{1}{2}) \otimes (\frac{1}{2}, \frac{1}{2}) \cong (1, 1)_S \oplus (0, 1)_A \oplus (1, 0)_A \oplus (0, 0)_S $$

with the physical dimension 16 at both sides. The symmetric scalar representation with spin 0 denotes the presence of the metric tensor $g^{\mu\nu}$ that is manifestly invariant in this representation. The symmetric representation $(1, 1)_S$ of two spins 1 must be associated with a symmetric tensor $S^{\mu\nu}$, breaking down into the representation $(0) \oplus (1)$ of spin 0 and 1. The first generates again the metric tensor and can be split off, leaving a traceless matrix. The two representations $(0, 1)_A$ and $(1, 0)_A$, generated by direct products of an anti-symmetric singlet and a symmetric triplet state of two spins $\frac{1}{2}$ and with opposite chirality, stand for antisymmetric matrices $F^{\mu\nu}$ and $G^{\mu\nu}$. Hence we find the decomposition

$$ B^{\mu\nu}(x) = S^{\mu\nu}(x) + F^{\mu\nu}(x) + G^{\mu\nu}(x) + g^{\mu\nu}T(x) \quad (95) $$

According to Eq. (95) the representation allows the separation into symmetric and anti-symmetric tensors, like was seen above for spin $\frac{1}{2}$. We identify the traceless symmetric tensor $S^{\mu\nu}$ with 9
independent components. It can be further decomposed to $(1, 1) \cong (1 \otimes 1) \cong (2) \oplus (1) \oplus (0)$. This representation can therefore describe spin 2, possibly the graviton.

An anti-symmetric 4-tensor has 6 degrees of freedom. The representation decomposition tells that it splits further up into two tensors $F^{\mu\nu}$ and $G^{\mu\nu}$, both with 3 independent components associated with spin 1, and with opposite chirality. On its own, the anti-symmetry facilitates $U(1)$ gauge invariance as we shall see soon, characteristic of zero mass, and reducing the number of degrees of freedom to two. They both look like the photon field. Quite remarkably, *two representations of the photon field exist, with opposite handedness but with equivalent observables*, as we show in the next section. The decomposition above preserves the number of $9+2 \times 3+1 = 16$ degrees of freedom.

### 3. Two Tensor Representations of the Electromagnetic Field

The tensor representation $(\frac{1}{2}, \frac{1}{2}) \otimes (\frac{1}{2}, \frac{1}{2})$ generates the direct sum $(0, 1) \oplus (1, 0)$, suggesting that two different representations must exist for massless spin 1 particles such as the photon. They have opposite chirality but should represent the same photon field since we know of only one.

The fundamental reason behind is the very existence of the Lorentz-invariant 4-pseudo tensor $\epsilon_{\mu\nu}^{\alpha\beta}$. From the proper 4-tensor $F^{\mu\nu}$ we can construct a new tensor,

$$G_{\mu\nu} \equiv \epsilon_{\mu\nu}^{\alpha\beta} F_{\alpha\beta} \qquad (96)$$

This tensor is a pseudo 2-tensor because $\epsilon_{\mu\nu}^{\alpha\beta}$ transforms as pseudo 4-tensor, as was shown in Eq. (53). From the 2-tensor $F^{\mu\nu}$ it is customary to define electric and magnetic field from

$$E^i = -F^{0i} ; \quad B_i = \frac{1}{2} \epsilon_{ijk} F^{jk}$$

that transform as vector and pseudo-vector, respectively. Using the dual pseudo 2-tensor we can do the same and put,

$$B^i = -G^{0i} ; \quad E_i = \frac{1}{2} \epsilon_{ijk} G^{jk}$$

that behave as pseudovector and vector.

### Exercise V.20: Two dual and equivalent representations of the Maxwell Field

* Show that the two definitions of electric and magnetic field coincide. (Hint: use $\epsilon^{0ijk} = -\epsilon^{ijk}$ and $\frac{1}{2}\epsilon^{jk}_{\alpha\beta}F^{\alpha\beta} = \epsilon^{jk}_{0n}F^{0n}$ because for $j, k = 1, 2, 3$ a zero must occur among the indices of $F^{\alpha\beta}$).
* Show that when written as 4 × 4 matrices, $G^{\mu\nu}$ and $F^{\mu\nu}$ take the form

$$F^{\mu\nu} = \begin{pmatrix} 0 & -E_x & -E_y & -E_z \\ E_x & 0 & -B_z & B_y \\ E_x & B_z & 0 & -B_x \\ E_z & -B_y & B_x & 0 \end{pmatrix}, \quad G^{\mu\nu} = \begin{pmatrix} 0 & -B_x & -B_y & -B_z \\ B_x & 0 & E_z & -E_y \\ B_y & -E_z & 0 & E_x \\ B_z & E_y & -E_x & 0 \end{pmatrix}, \qquad (97)$$

which implies that the matrix $G$ is obtained from the matrix $F$ using the transformation $E \to B$ and $B \to -E$. The matrix notation is misleading since it displays vectors and pseudo-vectors on the same footing. This is due to the factor that the tensor $\epsilon_{\mu\nu}^{\alpha\beta}$ is a pseudo-tensor invariant under $P$ and $T$ that is disregarded by the matrix notation.

The pseudo tensor $G^{\mu\nu}$ is associated with the representation $(0,1)$ and is the dual of the proper tensor field $F^{\mu\nu}$ corresponding to $(1,0)$. Clearly, we can construct a tensor representation of the Lorentz group for the Maxwell field using either one of the two tensors, that have unequal transformation under parity, but without any severe consequence for the Lagrangian,

$$\mathcal{L} = -\frac{1}{4} F_{\mu\nu} F^{\mu\nu} = +\frac{1}{4} G_{\mu\nu} G^{\mu\nu}$$

The “dual” symmetry $E \to B$ and $B \to -E$ is broken by the macroscopic Maxwell equations (i.e. in the presence of macroscopic material constants $\epsilon$ and $\mu$), as was first pointed out by Einstein. This issue is not solved and is still a major aspect of current research.
VI. SYMMETRY & NOETHER CHARGES

In this section we investigate the consequences of continuous symmetries in the Lagrangian density. Let us assume that $\mathcal{L} = \mathcal{L}(\Psi_a, \partial_\nu \Psi_a)$ depends on a set of fields and its 4 space-time derivatives and consider small variations generated by a symmetry operation. This operation changes the field by $\Psi'_a = \Psi_a + \delta_s \Psi_a$. We can see that this induces a change in the Lagrangian density

$$ \delta_s \mathcal{L} = \frac{\partial \mathcal{L}}{\partial \Psi_a} \delta_s \Psi_a + \frac{\partial \mathcal{L}}{\partial (\partial_\nu \Psi_a)} \delta_s (\partial_\nu \Psi_a) \quad (98) $$

with implicit operation over the index a. The operation can be said to be a symmetry if the equations of motion are not affected. This is true if the total action (12) is unaffected. Let us set $\delta_s = \epsilon \Delta_s$ with $\epsilon$ small and $\Delta_s$ an operator closely related to the generator of the symmetry group. For the Lagrangian density it is enough that a symmetry operation is characterized by,

$$ \Delta_s \mathcal{L} = \partial_\nu K^\nu \quad (99) $$

knowing that the divergence on the righthand side generates a boundary term in the space-time integral for the action and does not modify the equation of motion. We can work out the small variation, using that $\delta_s(\partial_\nu \Psi_a) = \partial_\nu(\delta_s \Psi_a)$, to get

$$ \Delta_s \mathcal{L} = \partial_\nu \left( \frac{\partial \mathcal{L}}{\partial (\partial_\nu \Psi_a)} \Delta_s \Psi_a \right) + \left[ \frac{\partial \mathcal{L}}{\partial \Psi_a} - \partial_\nu \left( \frac{\partial \mathcal{L}}{\partial (\partial_\nu \Psi_a)} \right) \right] \Delta_s \Psi_a \quad (100) $$

Note that implicit summation over the different fields labeled $\Psi_a$ has been assumed, and that the label a can refer to complex conjugate, treated as an independent variable in the Euler-Lagrange formalism. If $\Psi_a$ obeys the equation of motion, the second term must vanish. Therefore, if the operation is to be a continuous symmetry we can combine Eqs. (99) and (100) to define a continuity equation $\partial_\nu j^\nu = 0$ with

$$ j^\nu = \frac{\partial \mathcal{L}}{\partial (\partial_\nu \Psi_a)} \Delta_s \Psi_a - K^\nu \quad (101) $$

This is Noether's theorem: to every continuous symmetry of the action, a Noether current $j^\nu(x)$ exists that is locally conserved, i.e. satisfies $\partial_\nu j^\nu = \partial_t q + \partial_i j^i = 0$, with $q = j^0/c_0$. If we assume that $j^i$ decays fast enough in space (causality undoubtedly helps if $x \gg c_0t$) we find the Noether charge

$$ Q(t) = \int d^3x q(x) = \frac{1}{c_0} \int d^3x \left( \frac{\partial \mathcal{L}}{\partial (c_0 \partial_t \Psi_a)} \Delta_s \Psi_a - K^0 \right) \quad (102) $$

that is conserved in time.
Symmetry is defined here as an invariance of the Lagrangian. A field $\Psi$ can itself said to be invariant under an operation $U$ if $U\Psi = \exp(i\phi)\Psi$. The two definitions are not equivalent. A symmetry of the Lagrangian may well map an energy eigenfunction of a degenerate energy eigenvalue to another, orthogonal eigenstate. If for instance the ground state of an equation of motion with symmetry $U$ is non-degenerated it will itself be symmetric. Conversely, for every energy eigenstate that is itself not symmetric, there must be a degenerate state, which can be found by applying the symmetry transformation.

A second remark concerns the invariance of wave equations on any transformation $\mathcal{L} \to \alpha\mathcal{L}$. Clearly, the Noether charges and currents transform accordingly. Of course, if $Q(t)$ is conserved than so is $\alpha Q(t)$. This is an invariance, not a real symmetry and introduces no new conserved quantities. The value of $\alpha$ is often chosen (or “built in”) such that we get the familiar equations and energy using the chosen units. A good example is the factor 1/4 in the electromagnetic Lagrangian.

### Exercise VI.1: Global phase invariance

Consider the complex scalar field $\Psi$ with (Schrödinger) Lagrangian

$$ \mathcal{L}(\Psi, \Psi^*, \partial_{\mu}\Psi, \partial_{\mu}\Psi^*) = \frac{i\hbar}{2}\Psi^*(\partial_t\Psi) - \frac{i\hbar}{2}(\partial_t\Psi^*)\Psi - \frac{\hbar^2}{2m}(\partial_i\Psi^*)(\partial_i\Psi) - V(x)\Psi\Psi^* \quad (103) $$

It has no Lorentz-invariance but does exhibit $U(1)$ global phase invariance.

* Show that for this symmetry $\Delta_S\Psi = -i\Psi$ and $\Delta_S\Psi^* = +i\Psi^*$.

* Show that the Noether charge is proportional to

$$ Q = \frac{1}{2} \int d^3x [\Psi^*\Psi + \bar{\Psi}\Psi^*] $$

* Consider the Dirac Lagrangian (66) and the vector and axial $U(1)$ symmetries given in Exercise IV.15. Show that the vector unitary transformation has $\Delta_S\Psi = -i\Psi$ and $\Delta_S\bar{\Psi} = +i\bar{\Psi}$ leading to $\Delta_S\mathcal{L} = 0$. The conserved Dirac charge is

$$ Q_V = \int d^3x \bar{\Psi}\gamma^0\Psi = \int d^3x \Psi^\dagger\Psi $$

Global phase invariance implies conservation of norm (and after second quantization conservation of charge or particle number). Show that the axial unitary transformation has $\Delta_S\Psi = -i\gamma^5\Psi$ and $\Delta_S\bar{\Psi} = -i\bar{\Psi}\gamma^5$ and generates the variation $\Delta_S\mathcal{L} = -2im\bar{\Psi}\gamma^5\Psi$. This is
not a four-divergence so that axial symmetry only exists for a massless “Weyl” fermion. In that case the charge is

$$Q_A = \int d^3 x \bar{\Psi} \gamma^0 \gamma^5 \Psi = \int d^3 x \Psi^\dagger \gamma^5 \Psi$$

and has opposite sign for opposite chiralities.

## Exercise VI.2: SU(2) Noether Charges

Consider the complex (Lorentz-)scalar spinor $\Phi(x) = (\phi_+, \phi_-)/\sqrt{2}$ with Lagrangian

$$\mathcal{L} = \partial_{\mu} \Phi^{\dagger} \partial^{\mu} \Phi - m^2 \Phi^{\dagger} \Phi$$

with symmetry $U(1)$ and $SU(2)$.

1. Argue that $\Phi(x)$ is a $(0, 0)$ representation of the Lorentz group and therefor carries no intrinsic spin, only orbital angular momentum.

2. Show that for $SU(2)$ transformations $\Delta_a \Phi = -\frac{1}{2}i\sigma_a \Phi$ and $\Delta_a \Phi^\dagger = +\frac{1}{2}i\sigma_a \Phi^\dagger$ in terms of the 3 Pauli matrices.

3. Show that the 3 conserved charges are given by

$$Q_a = -\frac{1}{2}i \int d^3 x \left[ \partial_0 \Phi^\dagger \sigma_a \Phi - \partial_0 \Phi \sigma_a \Phi^\dagger \right]$$

4. If we insert an energy eigenfunction (possible since energy is conserved as well) show that

$$Q_z[\mathrm{SU}(2)] = \frac{E}{2} \int d^3 x \left[ \frac{1}{2} \Phi_+^* \Phi_+ - \frac{1}{2} \Phi_-^* \Phi_- + \frac{1}{2} \Phi_+ \Phi_-^* - \frac{1}{2} \Phi_- \Phi_-^* \right]$$

to be compared to the $U(1)$ Noether charge

$$Q_z[U(1)] = \frac{E}{2} \int d^3 x \left[ \Phi_+^* \Phi_+ + \Phi_-^* \Phi_- + \Phi_+ \Phi_-^* + \Phi_- \Phi_-^* \right]$$

We have respected the ordering of fields and complex conjugates to anticipate second quantization. Clearly we can expect the complex fields $\Phi_\pm$ to become bosonic, because with fermionic commutation rules both charges would vanish apart from an infinity.
5. We can impose global SU(2) symmetry to the doublet $\Phi = (\Phi_1, \Phi_2)$ and where both components are independent Dirac bi-spinors. The Lagrangian is then $\mathcal{L} = \bar{\Phi}D(\ell, \ell)\Phi$ with $\ell = i\gamma^\mu\partial_\mu - m$. Thus, $\Delta_n\Phi = -\frac{1}{2}i\sigma_n\Phi$ and $\Delta_n\bar{\Phi} = \frac{1}{2}i\bar{\Phi}\sigma_n$. Show that the associated conserved Noether charge is

$$Q_n^0 = \int d^3x \bar{\Phi}\gamma^0\sigma_n\Phi = \int d^3x \Phi^\dagger\sigma_n\Phi$$

In electroweak theory this charge is associated with the isospin of the doublet $(e, \nu_e)$ of electron and electron-neutrino and of the Higgs doublet. The z-component is proportional to $|\Phi_1|^2 - |\Phi_2|^2$ and is opposite for both field components. Argue that both components of the doublet transform in the same way under $U(1)$ but that the complex-SU(2) dual spinor $\tilde{\Phi} = i\sigma_2\Phi$ transforms under $U(1)$ with opposite sign. The symmetry $U(1)$ generates hypercharge of the spinor $\Phi$ and together with isospin $\sigma_3$ becomes the electromagnetic gauge in electroweak theory.

## Exercise VI.3: Symmetry under space-time translations

A space time translation of a (complex) field $\Psi(x)$ is defined by $\Psi(x + a)$ and $\Psi^*(x + a)$, with $a^\mu$ a four-vector. For infinitively small translations in “direction” $\nu$ it follows that $\Delta_\mu\Psi^{(*)} = \partial_\mu\Psi^{(*)}$.

* Verify that $\Delta_\mu\mathcal{L} = \partial_\mu\mathcal{L}$ independent of the exact action. This means that four translation symmetries can be identified with $K_\mu^\nu = \delta_\mu^\nu\mathcal{L}$ for $\mu = 0, 1, 2, 3$.

* Show that for $\mu = 0$ the conserved charge is

$$Q_0 = \int d^3x \left( \frac{\partial \mathcal{L}}{\partial(\partial_t \Psi_a)} \partial_t \Psi_a - \mathcal{L} \right) \qquad (104)$$

interpreted as energy.

* Show that for $\mu = i$, the conserved charge is

$$Q_i = \int d^3x \frac{\partial \mathcal{L}}{\partial(\partial_t \Psi_a)} \partial_t \Psi_a \qquad (105)$$

interpreted as the component $i$ of total momentum. Energy and momentum are rigorously conserved by an action invariant under space-time translations. The local Noether current is identified as the momentum stress tensor $T_\mu^\nu$ with

$$T_\mu^\nu = \frac{\partial \mathcal{L}}{\partial(\partial_\nu \Psi_a)} \partial_\mu \Psi_a - \delta_\mu^\nu \mathcal{L} \qquad (106)$$
* Show explicitly using the Euler-Lagrange equation that $\partial_{\nu} T^{\nu}_{\mu} = 0$.

* Show that for the Schrödinger Lagrangian (103) time-translation is the only symmetry if the potential is time-independent and has the conserved charge,

$$ 
\begin{align*}
Q_0 &= \int d^3\mathbf{x} \left[ \frac{\hbar^2}{2m} (\partial_i \Psi^*)(\partial_i \Psi) + V(\mathbf{x}) \Psi \Psi^* \right] \\
&= \int d^3\mathbf{x} \left[ -\frac{\hbar^2}{4m} (\partial_i^2 \Psi^*) \Psi - \frac{\hbar^2}{4m} \Psi^* (\partial_i^2 \Psi) + V(\mathbf{x}) \Psi \Psi^* \right] + \frac{\hbar^2}{4m} \int d^3\mathbf{x} \partial_i^2 (\Psi \Psi^*) \\
&= \int d^3\mathbf{x} \frac{1}{2} [\Psi^* H \Psi + \Psi H \Psi^*]
\end{align*}
$$

(Hint: do not forget to recognize field and its complex conjugate as independent fields $\Psi_a$). The term with $\partial_i^2 (\Psi \Psi^*)$ in $Q^0$ integrates to 0 by Gauss' theorem. This shows that we can interpret the conserved charge associated with time translations as total energy (Hamiltonian). Show that the corresponding energy current density can be written as

$$ j^i = \frac{i\hbar}{2m} \int d^3\mathbf{x} [\partial^i \Psi(H\Psi^*) - \partial^i \Psi^*(H\Psi)] $$

* Show that for the free Dirac field the energy is given by

$$ Q_0 = \int d^3\mathbf{x} \overline{\Psi(x)} [-i\hbar\gamma^i\partial_i + mc_0] \Psi(x) $$

* Both Schrödinger and Dirac equations of motion can be written as $i\hbar\partial_t\Psi = H\Psi$ and have conserved energy $Q_0 = \int d^3\mathbf{x} \Psi^\dagger(x,t)H\Psi(x,t)$. Energy governs time evolution. In second quantization, discussed below, $Q_0$ becomes the time-generator.

### Exercise VI.4: Symmetry under Lorentz transformations

Infinitesimal Lorentz transformations are given by $\delta x^{\alpha} = \omega_{\alpha\beta} x^{\beta}$ and determined by 6 independent choices for $\omega_{\mu\nu} \to 0$, with $\omega_{\alpha\beta} = -\omega_{\beta\alpha}$.

* Show that for a scalar real field with Lagrangian $\mathcal{L} = \partial_{\mu}\phi\partial^{\mu}\phi - m^2\phi^2$ we have $\Delta^{\mu\nu}\phi = x^{\mu}\partial^{\nu}\phi - x^{\nu}\partial^{\mu}\phi$ and $\Delta^{\alpha\beta}\mathcal{L} = \partial_{\nu}(K^{\alpha\beta})^{\nu}$ with

$$ (K^{\alpha\beta})^{\nu} = \mathcal{L}(g^{\nu\alpha}x^{\beta} - g^{\nu\beta}x^{\alpha}) $$

Hint: treat $\omega_{\alpha<\beta}$ as 6 independent small parameters.

* Show that the conserved Lorentz charges are given by

$$ Q^{\alpha\beta} = \int d^3\mathbf{x} (x^{\alpha}T^{0\beta} - x^{\beta}T^{0\alpha}) $$
* For $\alpha = i$ and $\beta = j$ this corresponds to total angular momentum. Show that for $\alpha = 0$, $\beta = i$ (boosts) it follows that

$$Q^{0i} = \int d^3x (c_0tP^i(x) - x^iH(x))$$

with $P^i(x) = \partial_0\phi\partial^i\phi$ the momentum density and $H(x) = T_0^0(x)$ the energy density, both following from space-time translational symmetry. We arrive here in the heart of the Poincaré group that also involves translations. The conserved boost charge $Q^{0i}$ brings us to a field formulation of Eq. (4) expressing that center of mass becomes center of energy in special relativity. In second quantization, the boosts operator $Q^{0i}$ becomes unitary.

* The Dirac field transforms under Lorentz transformation as $\Psi'(x) = S(\Delta)\Psi(\Delta^{-1} \cdot x)$ with $S(\Delta) = 1 - \frac{i}{4}\sigma^{\alpha\beta}\omega_{\alpha\beta} + \dots$. Show that $\Delta^{\alpha\beta}\Psi = i(x^\alpha p^\beta - x^\beta p^\alpha - \frac{1}{2}\sigma^{\alpha\beta})\Psi$, with $p^\alpha = -i\partial^\alpha = -i\partial/\partial x_\alpha$ (transforming co-variantly!), so that that the conserved Lorentz charges are

$$Q^{\alpha\beta} = \int d^3x [\Psi^\dagger J^{\alpha\beta} \Psi - \mathcal{L}(g^{0\alpha}x^\beta - g^{0\beta}x^\alpha)]$$

with $J^{\alpha\beta} \equiv x^\beta p^\alpha - x^\alpha p^\beta + \frac{1}{2}\sigma^{\alpha\beta}$. For $\alpha = i$ and $\beta = j$ we recall $\{x^i\} = (x, y, z)$ and $\{p^i\} = \{-p_i\} = (-p_x, -p_y, -p_z)$ and use the orbital angular momentum operator $L = x \times p$ ($x_i p_j - x_j p_i = \epsilon_{ijn}L_n$ abandoning covariant notation) and the spin operator $\sigma^{ij} = \epsilon_{ijn}\sigma_n$ (valid in both Weyl and Dirac representation) so that $J^{ij} = \epsilon_{ijn}(L_n + \frac{1}{2}\sigma_n)I_4$. This identifies the intrinsic spin 1/2 of the Dirac equation as a genuine contribution to angular momentum.

* The Dirac equation can be solved exactly in the presence of a Coulomb potential of the electron. Argue that the energy levels depend on both a principal quantum number $n$ and angular momentum quantum number $j$ associated with $J = L + S$. Show that $j = \ell \pm 1/2$. The standard spectroscopic notation for an energy level is $n^{2s+1}L_J$. The $j$ dependence of energy levels is called fine-structure. Show that the energies of $n^2P_{1/2}$ and $n^2S_{1/2}$ must be the same according to the Dirac equation. In reality they are not, due to the electromagnetic quantum vacuum.

## A. Noether theorem after Second quantization

So far we have considered fields $\Psi_a$ in some Hilbert space. In classical mechanics, an object with degrees of freedom $q_a$ is described by the Lagrangian $L(q_a, \dot{q}_a)$. It is customary to introduce the
canonical momenta $p_a = \partial\mathcal{L}/\partial\dot{q}_a$ and define the Hamiltonian as $H = p_a\dot{q}_a - L$. In the first quantization canonical momentum and $q_a$ become conjugate operators satisfying the canonical commutation relations $[q_a, p_b] = i\hbar\delta_{ab}$ and turning $H$ into the energy operator. Similarly, in field theory, $\Pi_a = \partial\mathcal{L}/\partial(\partial_t\Psi_a)$ can be defined as the canonical momentum of the field. Since no normalization of the Lagrangian has been defined, this fixes the canonical momentum only up to a constant. This factor is usually fixed by the resulting Hamiltonian. In the second quantization, we can propose a real boson field $\Psi_a$ to be quantized according to

$$[\Psi_a(x, t), \Pi_b(y, t)] = i\hbar\delta(x - y)\delta_{ab} \quad , \quad [\Psi_a(x, t), \Psi_b(y, t)] = 0 \qquad (107)$$

and $\Psi_a(x)$ is defined as the destruction operator in Fock space of particle $a$ at space-time $x$. Note that the commutator is evaluated at equal times. Equation (102) expresses the conserved charge $Q$ directly in terms of the (infinitesimal) symmetry operation acting on the field. After second quantization the Noether charge $Q$ becomes an operator that governs the transformation of the quantized field. If we assume that $K^\mu = 0$ we have

$$Q(t) = \int d^3x \Pi_a(x) \Delta_s \Psi_a(x)$$

with $x = (c_0t, x)$. The field transforms under $Q$ as $\Psi' = \exp(-i\epsilon Q)\Psi \exp(i\epsilon Q) \approx \Psi - i\epsilon[Q, \Psi]$ and

$$-i\epsilon[Q, \Psi_b(y)] = -i\epsilon \int d^3x [\Pi_a(x)\Delta_s\Psi_a(x), \Psi_b(y)] = -i\epsilon \int d^3x [\Pi_a(x), \Psi_b(y)]\Delta_s\Psi_a(x) \\ = \epsilon\Delta_s\Psi_b(y) = \delta_s\Psi_b(y)$$

This outcome reconfirms the strong link between continuous symmetries and conserved charges. This is an important property of “consistent” second quantization. For a complex, non-relativistic field the quantization proposed in Eq. (107) does not apply, but the Noether charge will still be promoted to symmetry generator.

In the first quantization, Noether's theorem links conserved charges to a continuous symmetry. In the second quantization, the Noether charge operator becomes the generator of the symmetry.

## 1. Second quantization of massive scalar real field

The free scalar field $\Phi$ with mass $m$ has Lagrangian $\mathcal{L} = \frac{1}{2}\partial_\mu\Phi\partial^\mu\Phi - \frac{1}{2}m^2\Phi^2$ satisfying the Klein-Gordon equation $[\partial_\mu\partial^\mu + m^2]\Phi(x) = 0$. If we insert Fourier components,

$$\Phi(x) = \frac{1}{(2\pi)^3} \int d^4k \, f(k)e^{-ikx}$$
we find that $f(k) = \delta(k^2 - m^2)F(k)$ and $F(-k) = F^*(k)$ for $\Phi \in \mathbb{R}$. Since $\delta(k^2 - m^2) = \delta(E_k - k_0)/2E_k + \delta(E_k + k_0)/2E_k$ with $E_k = \sqrt{m^2 + k^2}$ we can write,

$$ \Phi(x) = \frac{1}{(2\pi)^3} \int \frac{d^3\mathbf{k}}{2E_k} \left[ F(E_k, \mathbf{k})e^{-ikx} + F^*(E_k, \mathbf{k})e^{+ikx} \right] $$

where we performed the operation $\mathbf{k} \rightarrow -\mathbf{k}$ in the second term, associated with negative energies. In the second quantization, the field $\Phi$ becomes an hermitian operator. It is convenient to consider a finite volume $V$ with periodic boundary conditions and discrete wave vectors. This replaces the integral by a discrete sum according to $\int d^3\mathbf{k}/(2\pi)^3 \rightarrow V^{-1}\sum_{\mathbf{k}}$. Upon substituting $F(E_k, \mathbf{k}) = \sqrt{2VE_k}a_k$ and $F^*(E_k, \mathbf{k}) = \sqrt{2VE_k}a_k^\dagger$,

$$ \Phi(x) = \frac{1}{\sqrt{V}} \sum_{\mathbf{k}} \frac{1}{\sqrt{2E_k}} \left[ a_k e^{-ikx} + a_k^\dagger e^{+ikx} \right] \equiv \Phi^+(x) + \Phi^-(x) \quad (108) $$

The quantization rule (107) imposes that $[\Phi(t, \mathbf{x}), \partial_0\Phi(t, \mathbf{y})] = i\delta(\mathbf{x} - \mathbf{y})$ independent on $t$. If we insert the expression for $\Phi(x)$ we can see that this is achieved when

$$ [a_k, a_{k'}] = 0 \quad ; \quad [a_k, a_{k'}^\dagger] = \delta_{kk'} \quad (109) $$

thereby using the lattice sum $V^{-1}\sum_{\mathbf{k}} \exp(ik \cdot \mathbf{x}) = \delta(\mathbf{x})$ for $\mathbf{x} \in V$. In the second quantization $a_k^\dagger$ creates a massive, spinless particle with momentum $\mathbf{k}$ and energy $E_k$ whereas $a_k$ destroys it. Because the field is hermitian, particle and antiparticle are equal and no global phase invariance exists that can be used to gauge the electromagnetic field: the field has electromagnetic charge $Q = 0$.

From the commutation relations (109) it follows straightforwardly that,

$$ \begin{aligned} [\Phi^+(x), \Phi^+(y)] &= 0 \quad ; \quad [\Phi^-(x), \Phi^-(y)] = 0 \\ [\Phi^+(x), \Phi^-(y)] &= \frac{1}{V} \sum_{\mathbf{k}} \frac{e^{-ik(x-y)}}{2E_k} \rightarrow \int \frac{d^3\mathbf{k}}{(2\pi)^3} \frac{e^{-ik(x-y)}}{2E_k} \equiv G_0(x-y) \end{aligned} $$

Note that although the left hand side is an operator identity, the right hand side is just a complex number. The total field commutator in space-time is

$$ [\Phi(x), \Phi(y)] = G_0(x - y) - G_0(y - x) $$

and vanishes at equal times ($x_0 = y_0$) as was imposed earlier in Eq. (107). Because $G_0(x)$ can be shown to be a Lorentz-scalar (see exercise VI.6) we find that $[\Phi(x), \Phi(y)] = 0$ for any space-like four-vector $x - y$ with $(x - y)^2 < 0$. This is a manifestation of the *principle of relativistic causality*: Two operators associated with locally measurable observables at points outside the light cone in space-time must commute.
The total conserved energy, obtained from invariance to time-translations, is given by

$$Q_0 = E = \frac{1}{2} \int_V d^3 x [\partial_t \Phi)^2 + (\partial_x \Phi)^2]$$

When using the commutation relations, this can be written as

$$ 
\begin{aligned}
E &= \frac{1}{2} \sum_{\mathbf{k}} E_k (a_k^\dagger a_k^\dagger + a_k^\dagger a_k) = \sum_{\mathbf{k}} E_k (a_k^\dagger a_k + \frac{1}{2}) \\
&\rightarrow V \int \frac{d^3 \mathbf{k}}{(2\pi)^3} E_k (a_k^\dagger a_k + \frac{1}{2})
\end{aligned} 
\qquad (110) $$

with an infinite positive energy per unit volume from the quantum vacuum.

## 2. Second quantization of Dirac field

We consider the conserved energy $Q^0$ derived above for the Dirac equation (in the Dirac representation of electron-positron) and insert the wave function derived in Eq. (76). The second quantization can be obtained by promoting the Fourier components $\alpha_{ks}$ and $\beta_{ks}^*$ to operators,

$$ \Psi(x) = \frac{1}{\sqrt{V}} \sum_{ks} \frac{1}{\sqrt{2E_k}} \left[ \alpha_{ks} u_{ks} e^{-ik_{\mu}x^{\mu}} + \beta_{ks}^{\dagger} v_{ks} e^{+ik_{\mu}x^{\mu}} \right] \equiv \Psi^{+}(x) + \Psi^{-}(x) \qquad (111) $$

satisfying the Dirac equation $(i\gamma^{\mu}\partial_{\mu} - m)\Psi(x) = 0$. The sum $\sum_{ks}$ is a discrete sum over spin and wave numbers and the functions $u_{ks}$ and $v_{ks}$ are normalized to a plane wave in a finite volume $V$ as in Eq. (108) and we can always take the continuum limit $V \to \infty$ as we have done in the previous paragraph. The term cross-terms $\alpha_{ks}^{\dagger}\alpha_{k's'}$ and $\beta_{ks}\beta_{k's'}^{\dagger}$ that appear in $Q^0$ generate both $V\delta_{kk'}$ upon integrating $x$ over the volume $V$. The cross-terms generate $V\delta_{k,-k'}$ meaning that terms like $u_{ks}^{\dagger}v_{-ks'}$ are generated. Using the orthogonality relations derived in Exercise V.16 (with $c_0 = 1$) we obtain,

$$ Q_0 = \sum_{ks} \frac{1}{2E_k} k_0 2E_k (\alpha_{ks}^{\dagger} \alpha_{ks} - \beta_{ks} \beta_{ks}^{\dagger}) = \sum_{ks} E_k (\alpha_{ks}^{\dagger} \alpha_{ks} - \beta_{ks} \beta_{ks}^{\dagger}) \qquad (112) $$

This can be recognized as the Hamiltonian of electron + positron provided that the operator $\beta_{ks}$ anti-commutes with its adjoint:

$$ \{\beta_{ks}, \beta_{ks}^{\dagger}\} = \beta_{ks} \beta_{ks}^{\dagger} + \beta_{ks}^{\dagger} \beta_{ks} = \delta_{ss'} \delta_{kk'} \qquad (113) $$

This means that changing the order of $\beta_{ks}$ and $\beta_{ks}^{\dagger}$ in Eq. (112) eliminates the minus sign but produces a negative singular energy $-\delta^{(3)}(0)$ that has to be eliminated by other means. It is of course the remnant of the infinitely filled sea of negative energy states that Dirac imagined to
avoid the electron to decay to infinitely negative energies. Some speculate that this singularity
be compensated by the similar positive singularity that we have found earlier for bosons, which
would impose some kind of duality between bosons and fermions (called “supersymmetry”). It is
straightforward to show that if both $\beta_{ks}$ and $\alpha_{ks}$ satisfy anti-commutation rules, and in addition
$\{\alpha_{ks}, \beta_{k's'}^{(\dagger)}\} = 0$ then so do the fields, and $\{\Psi_i(x), \Psi_j^{\dagger}(y)\} = \delta_{ij}\delta(x-y)$ and $\{\Psi_i(x), \Psi_j(y)\} = 0$ at
equal times. This produces the fermionic equivalent of what was proposed in Eq. (107) for bosons
with canonical momentum $\Pi = \partial\mathcal{L}/\partial(\partial_t\Psi) = i\bar{\Psi}\gamma^0 = i\Psi^\dagger$. Note that no conjugate momentum is
associated with $\bar{\Psi}$ since $\partial\mathcal{L}/\partial(\partial_t\bar{\Psi}) = 0$.

## Exercise VI.5: Electromagnetic Charges of Dirac field

The double spinor formulation of the Dirac field does not make it possible to specify the electro-magnetic charges of particle and anti-particle separately since the double spinor transforms as a whole under $U(1)$. Only one charge $q$ features in Eq. (73) that couples photons to the Dirac field.

* Show that the $U(1)$ charge of the Dirac field is given by

$$Q = \int d^3x \Psi^\dagger\Psi = \sum_{ks} \left( \alpha_{ks}^\dagger \alpha_{ks} + \beta_{ks} \beta_{ks}^\dagger \right) = \sum_{ks} \left( \alpha_{ks}^\dagger \alpha_{ks} - \beta_{ks}^\dagger \beta_{ks} + 1 \right)$$

After normal ordering of the operators, particle and anti-particle achieve opposite charge.
The diverging term stands for the Dirac sea.

Many believe that the Dirac equation is only really exploited at this moment after second quantization when the role of antiparticles shows up explicitly, when the problem of negative energies is resolved, and when the anti-commutation of spin 1/2 fermions emerges.

Like we have done for the scalar field in the previous section, we write the quantized Dirac field (111) as $\Psi(x) = \Psi^+(x) + \Psi^-(x)$, separating electron annihilation and positron creation operators, both represented by bispinors.

$$\bar{\Psi}(x) = \frac{1}{\sqrt{V}} \sum_{ks} \frac{1}{\sqrt{2E_k}} \left[ \beta_{ks} \bar{v}_{ks} e^{-ik_{\mu} x^{\mu}} + \alpha_{ks}^{\dagger} \bar{u}_{ks} e^{+ik_{\mu} x^{\mu}} \right] = \bar{\Psi}^{+}(x) + \bar{\Psi}^{-}(x) \quad (114)$$

With the fermionic anti-commutation rules defined above we can verify that $\{\Psi^+(x), \bar{\Psi}^+(y)\} = \{\Psi^-(x), \bar{\Psi}^-(y)\} = 0$. Using the completeness relations for the eigenfunctions $u_{ks}$ and $v_{ks}$ derived in Eq. (75) we can see that

$$\begin{aligned} \{\Psi^+(x), \bar{\Psi}^-(y)\} &= (i\gamma^\mu \partial_\mu + m) G_0(x-y) \\ \{\Psi^-(x), \bar{\Psi}^+(y)\} &= -(i\gamma^\mu \partial_\mu + m) G_0(y-x) \end{aligned}$$
where the derivative is with respect to $x^\mu$ and $G_0$ was defined in the previous section for the scalar field. From this,

$$ \{\Psi(x), \bar{\Psi}(y)\} = (i\gamma^\mu \partial_\mu + m) [G_0(x-y) - G_0(y-x)] \quad (115) $$

This vanishes at equal times ($x_0 = y_0$) and because of the invariance of $G_0(x)$ with respect to a orthochronous Lorentz transformation of $x$ the anti-commutator vanishes for any space-like vector $(x-y)^2 \le 0$. The relativistic causality principle, discussed in the previous section for scalar bosons, here applies to the anti-commutator and thus *involves a subtle minus sign when we swap space-like fermions*. No causality problem exists if we realize that local fermionic observables always involves two fermionic fields. We can check that the observable Lorentz scalar $\rho(x) = \bar{\Psi}(x)\Psi(x)$ commutes if $(x-y)^2 \le 0$,

$$ [\rho(x), \rho(y)] = 0 \quad (116) $$

## B. Wick's theorem for two particles

Wick's theorem is a crucial ingredient to go from equations of motion to scattering amplitudes which are the ones that are usually measured. The latter, when evaluated in perturbation theory, involve time-ordered sequences of the time-evolution operators $H(t_1, x_1)H(t_2, x_2)\dots$ that all depend on time and space. The general theorem states that time-ordered sequences can be reduced to a sum of all possible products of “normally ordered sequences of operators and “contractions”. For a normally ordered operator sequence $N(ABC\dots)$, the destruction operators have moved to the righthand side of the product and the creation operators have moved to the left, and without any other changes. A normal order has the convenient property that its vacuum expectation value vanishes. Contractions are complex numbers (not operators) that stem from the commutator rules that govern the swap of two operators. The time-ordered sequence of an operator corresponds to an order with times increasing time from right to left.

We will illustrate the theorem for two particles. Let us first emphasize that *the time-ordered product can only have two possibilities*:

$$ T[A(x)B(y)] = \begin{cases} +A(x)B(y) & x_0 > y_0 \\ \pm B(y)A(x) & y_0 > x_0 \end{cases} \quad (117) $$

The $+$-sign for $x_0 > y_0$ is obvious, the choice $\pm$ for $x_0 < y_0$ cannot depend on $x$ or $y$ for that would go beyond the spirit of time-ordering and is determined to what happens when space-like vectors
with $(x_0 - y_0)^2 < (x - y)^2$. We shall consider the subtle case $x_0 = y_0$ later and also show that the principle of relativistic causality (Exercise VI.6) guarantees that no discontinuity exists near $x_0 \approx y_0$.

We start with the simple case of a scalar bosonic field $\Phi(x)$ of zero charge. In Eq. (108) this field was split up into a destruction operator $\Phi^+(x)$ with $\Phi^+(x)|0\rangle = 0$, and a creation operator $\Phi^-(y)$, both involving the same particle. If $x - y$ is space-like, $\Phi(x)$ and $\Phi(y)$ commute so that the $+$-option in Eq. (117) must apply. For $x_0 > y_0$ it follows that

$$T[\Phi(x)\Phi(y)] = \Phi^+(x)\Phi^+(y) + \Phi^+(x)\Phi^-(y) + \Phi^-(x)\Phi^+(y) + \Phi^-(x)\Phi^-(y)$$

whereas

$$N[\Phi(x)\Phi(y)] = \Phi^+(x)\Phi^+(y) + \Phi^-(y)\Phi^+(x) + \Phi^-(x)\Phi^+(y) + \Phi^-(x)\Phi^-(y)$$

so that

$$T[\Phi(x)\Phi(y)] - N[\Phi(x)\Phi(y)] = [\Phi^+(x), \Phi^-(y)] = G_0(x-y)$$

If on the other hand $y_0 > x_0$ we find,

$$T[\Phi(x)\Phi(y)] = \Phi^+(y)\Phi^+(x) + \Phi^-(y)\Phi^+(x) + \Phi^+(y)\Phi^-(x) + \Phi^-(y)\Phi^-(x)$$

and with same normal order. Therefore,

$$T[\Phi(x)\Phi(y)] - N[\Phi(x)\Phi(y)] = [\Phi^+(y), \Phi^-(x)] = G_0(y-x)$$

This can be summarized by

$$T[\Phi(x)\Phi(y)] = N[\Phi(x)\Phi(y)] + G_F(x-y) \qquad (118)$$

with the “contraction” of the fields defined by

$$\overline{\Phi(x)\Phi(y)} \equiv G_F(x-y) = i \int \frac{d^4 k}{(2\pi)^4} \frac{e^{-ik(x-y)}}{k^2 - m^2 + i\epsilon} \qquad (119)$$

By construction,

$$\langle 0 | T[\Phi(x)\Phi(y)] | 0 \rangle = G_F(x-y)$$

This is Wick's theorem applied to two scalar bosons. The propagator $G_F(x-y)$ describes the creation of a boson at $x$ and its causal destruction at $y$. The imaginary part of the integrand is proportional to $-2\pi i\delta(k^2 - m^2)$ which refers to a “real” boson. The real part of the integrand also exists away from the mass shell, and the boson is said to be “virtual”.

## Exercise VI.6: Feynman propagator
<figure><img src="image_10.png" /><figcaption>Figure 8. Cauchy contour in the complex plane for $k_0$ that avoids the poles $k_0 = \pm\sqrt{k^2 + m^2 - i\epsilon}$ and that enables to move the integral over the real axis to the entire imaginary axis.</figcaption></figure>

* Show that $G_F(x)$ is invariant under any orthochronous Lorentz transformation $x^\mu \rightarrow \Delta^\mu_\nu x^\nu$.

* Show by contour integration (see Fig. 8) over $k_0$ that $G_F(x)$ introduced in Eq. (119) satisfies $G_F(x) = G_0(x)$ if $x_0 > 0$ and $G_F(x) = G_0(-x)$ if $x_0 < 0$.

* Show that $[p^2 - m^2]iG_F(x) = -\delta^4(x)$, with $p_\mu = -i\partial_\mu$, identifying $G_F$ as the response to a point source in space-time (also called propagator or Green function).

We can do the same operations for the Dirac field. If we are not in the special Majorana representation, the Dirac field is charged and we have the distinct fields $\Psi(x)$ and $\bar{\Psi}(x)$ with their own creation and destruction parts for particle and anti-particle. However, at space-like distances $\Psi(x)$ and $\bar{\Psi}(y)$ anti-commute so that here the minus sign in Eq. (117) must apply for time-ordering. Consequently, we redefine normal order for fermions to generate a minus sign for each swap between creation and destruction operators. For $x_0 > y_0$ we find

$$ 
\begin{aligned}
T[\Psi(x)\bar{\Psi}(y)] &= \Psi^{+}(x)\bar{\Psi}^{+}(y) + \Psi^{-}(x)\bar{\Psi}^{+}(y) + \Psi^{-}(x)\bar{\Psi}^{-}(y) + \Psi^{+}(x)\bar{\Psi}^{-}(y) \\
&= \Psi^{+}(x)\bar{\Psi}^{+}(y) + \Psi^{-}(x)\bar{\Psi}^{+}(y) + \Psi^{-}(x)\bar{\Psi}^{-}(y) - \bar{\Psi}^{-}(y)\Psi^{+}(x) + \{\Psi^{+}(x), \bar{\Psi}^{-}(y)\}
\end{aligned}
$$
whereas normal ordering gives,

$$N[\Psi(x)\bar{\Psi}(y)] = \Psi^{+}(x)\bar{\Psi}^{+}(y) + \Psi^{-}(x)\bar{\Psi}^{+}(y) + \Psi^{-}(x)\bar{\Psi}^{-}(y) - \bar{\Psi}^{-}(y)\Psi^{+}(x)$$

with a subtle minus sign in the last term. We can check that

$$T[\Psi(x)\bar{\Psi}(y)] - N[\Psi(x)\bar{\Psi}(y)] = \{\Psi^{+}(x), \bar{\Psi}^{-}(y)\} = (i\gamma^{\mu}\partial_{\mu} + m)G_{0}(x-y)$$

For $y_0 > x_0$ the time-ordered product changes to

$$T[\Psi(x)\bar{\Psi}(y)] = \bar{\Psi}^{+}(y)\Psi^{+}(x) - \Psi^{+}(y)\bar{\Psi}^{-}(x) + \bar{\Psi}^{-}(y)\Psi^{-}(x) - \Psi^{-}(y)\bar{\Psi}^{+}(x)$$

Because the normal order is unaffected we obtain

$$T[\Psi(x)\bar{\Psi}(y)] - N[\Psi(x)\bar{\Psi}(y)] = -\{\bar{\Psi}^{-}(y)\Psi^{+}(x)\} = (i\gamma^{\mu}\partial_{\mu} + m)G_{0}(y-x)$$

The Wick theorem for two Dirac fermions is therefore

$$T[\Psi(x)\bar{\Psi}(y)] = N[\Psi(x)\bar{\Psi}(y)] + S_{F}(x-y)$$

with the contraction

$$\overline{\Psi(x)\Psi(y)} = (i\gamma^{\mu}\partial_{\mu} + m)G_{F}(x-y) \equiv S_{F}(x-y) = i\int \frac{d^{4}k}{(2\pi)^{4}} \frac{\not k + m}{k^{2} - m^{2} + i\epsilon} e^{-ik(x-y)} \quad (120)$$

Because $(\not k + m)(\not k - m) = k^2 - m^2$ it follows that the 4 × 4 matrix $S_F$ satisfies

$$(i\not\partial - m) i S_F(x - y) = -\delta^4(x - y) \quad (121)$$

and is therefore identified as the propagator of the Dirac field in space-time. This propagator is invariant under any Lorentz transformation of $x-y$ and has no singularity for $x_0 = y_0$ provided that $x \neq y$. Note that $\overline{\Psi(x)\Psi(y)} = -\overline{\Psi(y)\Psi(x)}$.

## Intermezzo: No equal time-contractions in Wick's Theorem

The interaction of a fermion with a photon is described by the Hamiltonian density $H_I = -e\bar{\Psi}(x)\gamma^{\mu}A_{\mu}(x)\Psi(x)$ in which all fields are evaluated at the same time, and even at the same spot in space-time. So far we have avoided equal times in our discussion of time-ordered products. How to “time-order” the product $A(t_1)B(t_1)$ if both operators do not commute? The contraction $G_F(x_0 = 0, \mathbf{x})$ does not vanish and even diverges when also $\mathbf{x} = 0$. We could define $T[A(t_1)B(t_1)] = A(t_1)B(t_1)$ or alternatively as $\pm B(t_1)A(t_1)$ depending on the nature of the commutator.
The normal order $N$ is well-defined for both bosons and fermions and the time-order operation $T$ is supposed to govern the time-evolution according to $U(0,t) = T \exp[-i \int_0^t dt' H(t')]$ and cannot be freely defined either. If the product is already in normal order, say $A^\dagger(t_1)B(t_1)$ the simplest solution would be to say that

$$T[A^\dagger(t_1)B(t_1)] := A^\dagger(t_1)B(t_1) = N[A^\dagger(t_1)B(t_1)]$$

If on the other hand $B(t_1)$ and $A(t_2)$ anti-commute

$$T[B(t_1)A^\dagger(t_1)] := -A^\dagger(t_1)B(t_1) = N[B(t_1)A^\dagger(t_1)]$$

With this approach we set the contractor of fields evaluated at equal times equal to zero. This avoids singularities and produces no contradictions.

### C. Electromagnetic Lagrangian

The observable fields in electromagnetism are the electric and magnetic fields. Yet, they are not part of four-vectors and cannot be the fundamental fields in a Lorentz-invariant theory. The basic field is the four-vector $A^\mu = (\phi/c_0, A^i)$ and transforms as a genuine four-vector in Minkowski space. Because it features as $i\partial_\mu - qA_\mu$ in the Lagrangian of field and matter, it follows that under parity $P$ and time-reversal $T$,

$$\{A_\mu(t, \mathbf{x})\} = \left(\frac{\phi(t, \mathbf{x})}{c_0}, -\mathbf{A}(t, \mathbf{x})\right) \stackrel{P}{\rightarrow} \left(\frac{\phi(t, -\mathbf{x})}{c_0}, +\mathbf{A}(t, -\mathbf{x})\right) \\ \stackrel{T}{\rightarrow} \left(\frac{\phi(-t, \mathbf{x})}{c_0}, +\mathbf{A}(-t, \mathbf{x})\right) \qquad (122)$$

where we recall that time-reversal is also accompanied by complex conjugation. Thus, the vector potential transforms under $SO^+(1,3)$ as a four-vector in Minkowski space but under the non-orthochronous transformation $T$, not part of $SO^+(1,3)$, the vector potential does not transform as a normal Minkowski vector ($c_0t, -\mathbf{x}$).

If we choose the electric field $E = -(1/c_0)\partial_t A - \nabla\Phi$ and magnetic field $B = \nabla \times A$ the two Maxwell equations insensitive to interactions with charges are automatically satisfied: $\nabla \cdot B = 0$ (no magnetic monopoles) and $\nabla \times E = -c_0^{-1}\partial_t B$ (Faraday's induction law). In covariant notation we can easily check that $E^i = -\partial_0 A^i + \partial^i A_0$ and $B_i = \frac{1}{2}\epsilon_{ijk}(\partial^j A^k - \partial^k A^j)$, both terms being equal. Earlier we identified two and only two electromagnetic Lorentz scalars so that the electromagnetic Lagrangian must be of the form

$$\mathcal{L}(E, B) = \frac{1}{2}(E^2 - B^2) + \chi_1(E \cdot B)^2 + \chi_2(E^2 - B^2)^2 + \dots \qquad (123)$$
In particular, despite its Lorentz invariance, a linear term $E \cdot B$ is not allowed since it would break time-reversal and inversion symmetry. The first term is bilinear and gives linear equations of motion. The higher order terms (the “Euler-Heisenberg Lagrangian”) are known to exist when vacuum polarization is taken into account, and giving $\chi_i \sim 2\alpha^2/45m_e^2$. It would predict birefringence of the electromagnetic quantum vacuum in the presence of high magnetic fields such as around neutron stars. It is straightforward to express the first term co-variantly in terms of the vector potential,

$$ \mathcal{L}(\partial_{\mu} A^{\nu}) = -\frac{1}{4} F^{\mu\nu} F_{\mu\nu} \qquad (124) $$

where $F^{\mu\nu} = \partial^{\mu}A^{\nu} - \partial^{\nu}A^{\mu}$ is the antisymmetric electromagnetic field tensor. We can lower and raise indices using the metric tensor. We can verify that $E^i = -F_0^i = -F^{0i} = +F_{0i}$ and $B^i = \frac{1}{2}\epsilon^{ijk}F^{jk} = \frac{1}{2}\epsilon^{ijk}F_{jk}$. These expressions are extremely sensitive to metric convention and different equalities anticipate/favor covariant notation in Minkowski space with metric tensor $D(1, -1, -1, -1)$. We conclude that the free electromagnetic Lagrangian is Lorentz-invariant (of course this is how the symmetry was discovered in the first place) and is described by an anti-symmetric, gauge-invariant second-order four-tensor. We can check that $F_{0i}$ is parity-odd and does not change under time-reversal, for $F_{ij}$ exactly the contrary holds. This means that the electric field $E$ is a normal polar vector in $\mathbb{R}^3$ but that the magnetic field is a pseudo-vector (or axial vector).

### Exercise VI.7: Symmetry and gauge-invariance of classical electromagnetism

* Verify the statement that the electromagnetic field $A^\mu$ is real-valued and that no $U(1)$-symmetry exists that conserves its norm. To put this in more second-quantization language: the photon behaves as a scalar under $U(1)$ and has charge zero. Argue why the existence of a massless particle with finite charge would be impossible to describe in a non-relativistic theory. We will later encounter the Weyl fermions, predicted by Dirac theory, as a representation of the Lorentz group with spin 1/2, zero mass and finite charge and chirality.

* The Lagrangian is invariant under any transformation $(A')_\mu = A_\mu + (1/q)\partial_\mu\theta(x)$. Argue that this transformation is not a physical symmetry but rather a degree of freedom in the theory, called gauge-invariance. There is no conserved quantity related to this invariance either.

* By using the identity $\partial(\partial_\mu A_\alpha)/\partial(\partial_\rho A_\sigma) = \delta_\rho^\mu \delta_\sigma^\alpha$ identifying 16 independent parameters in the electromagnetic Lagrangian, and by writing $\mathcal{L} = -\frac{1}{4}g^{\mu\rho}g^{\nu\sigma}F_{\rho\sigma}F_{\mu\nu}$, establish that the
electromagnetic energy stress tensor (106) is traceless and can be expressed as

$$ T_{\nu}^{\mu} = -F^{\mu\alpha}F_{\nu\alpha} + \frac{1}{4}\delta_{\nu}^{\mu}F^{\alpha\beta}F_{\alpha\beta} \quad (125) $$

Show that $T_0^0 = (\mathbf{E}^2 + \mathbf{B}^2)/2$ and $T^{0i} = (\mathbf{E} \times \mathbf{B})^i$ (Hint: show first that $F^{ij} = \epsilon_{jik}B^k$). Is time evolution governed by energy?

* The classical Maxwell-Lorentz Lagrangian for a charged particle with velocity $v^i = dx^i/dt = \dot{x}^i$ is

$$ \mathcal{L} = \frac{1}{2}m\mathbf{v}^2 + q\mathbf{v} \cdot \mathbf{A} - q\phi = -\frac{1}{2}m\dot{x}_i\dot{x}^i - J^{\mu}A_{\mu}/c_0 $$

with $J^{\mu} = (q, q\mathbf{v})$ a charge current and $A_{\mu} = (\phi, -\mathbf{A})$. Show that gauge-invariance of its equivalent (124) in field theory is maintained if a Lorentz-invariant light-matter interaction term $-J_{\mu}A^{\mu}/c_0$ is added provided that $J^{\mu}$ represents a conserved current, i.e. $\partial_{\mu}J^{\mu} = 0$. (The negative sign of this interaction depends on our choice for the metric and one finds the opposite sign in many references). If we choose $J^{\mu} = (\rho c_0, \mathbf{J})$ show that the equation of motion becomes $\partial_{\nu}F^{\mu\nu} = -J^{\mu}$ and reproduces, choosing $\mu = 0$ and $\mu = i$, the two Maxwell equations with source term: $\nabla \cdot \mathbf{E} = \rho$ and $\nabla \times \mathbf{B} - c_0^{-1}\partial_t \mathbf{E} = \mathbf{J}$. (recall that for our metric $\partial^i = -\partial_i = -\partial/\partial x^i$).

* Richard Feynman added a “gauge-fixing” term $\mathcal{L}_G = -\frac{1}{2}\lambda(\partial_{\mu}A^{\mu})^2$ to the electromagnetic Lagrangian. Show that the equation of motion for $A^{\nu}$ becomes,

$$ \Box A^{\nu} + (\lambda - 1)\partial^{\nu}(\partial_{\mu}A^{\mu}) \equiv M(p)^{\nu}_{\mu}A^{\mu} = J^{\nu} \quad (126) $$

with $\Box = \partial_{\mu}\partial^{\mu}$. Verify that for $\lambda = 0$ gauge invariance exists but that the wave equation is not invertible; 2) for $\lambda = 1$ the equation is $\Box A^{\nu} = J^{\nu}$ and imposes the Lorenz gauge $\partial_{\nu}A^{\nu}$ only in the presence of matter and 3) for $\lambda \neq 0, 1$ the Lorenz gauge is imposed. Demonstrate that for $\lambda \neq 0$ the wave operator $M(k)^{\nu}_{\mu}$ in reciprocal space has an inverse given by

$$ (G_{\gamma}^{0}(k))^{\mu}_{\sigma} \equiv (M(k)^{-1})^{\mu}_{\sigma} = -\frac{1}{k^2} \left( \delta^{\mu}_{\sigma} + \frac{1-\lambda}{\lambda} \frac{k^{\mu}k_{\sigma}}{k^2} \right) $$

We can write $A^{\nu}(k) = (M(k)^{-1})^{\nu}_{\mu}J^{\mu}$ and the longitudinal term proportional to $k^{\nu}k_{\mu}$ drops out whenever $J^{\mu}$ is a conserved current, such as the one in QED, $J^{\mu} = \bar{\Psi}\gamma^{\mu}\Psi$. The choice $\lambda = 1$ made by Feynman gives the electromagnetic propagator

$$ (G_{\gamma}^{0}(k))_{\mu\nu} = -\frac{g_{\mu\nu}}{k^2} \quad (127) $$

The choice $\lambda = \infty$ is the (transverse) Landau gauge.
## Example: Proca Lagrangian and photon mass

The gauge invariance of the electromagnetic field is one of the most precious elements in today's standard model but requires the photon to be massless. Consider the modified electromagnetic Lagrangian, proposed first in 1922 by Louis De Broglie et his student Alexandru Proca,

$$ \mathcal{L}(\partial_{\mu} A_{\nu}, A_{\mu}) = -\frac{1}{4} F^{\mu\nu} F_{\mu\nu} + \frac{1}{2} \left( \frac{m_{\gamma} c_{0}}{\hbar} \right)^{2} A_{\mu} A^{\mu} - \frac{1}{c_{0}} J^{\mu} A_{\mu} \quad (128) $$

We used physical physical units, and $m_γ$ has the dimension of mass.

* Argue that gauge invariance is broken if $m_γ \neq 0$. Derive form the Euler-Lagrange equation, the wave equation

$$ \partial_{\mu} \partial^{\mu} A^{\nu} - \partial^{\nu} \partial_{\mu} A^{\mu} + \left( \frac{m_{\gamma} c_{0}}{\hbar} \right)^{2} A^{\nu} = \frac{1}{c_{0}} J^{\nu} $$

Show that the Lorenz (without “t”!) gauge $\partial_\mu A^\mu = 0$ is imposed for $m_γ \neq 0$. The vacuum solution is $A^\nu = g^\mu \exp(-i\omega t + ik \cdot x)$ with $g^\nu k_\nu = 0$ and dispersion law $\omega^2/c_0^2 - k^2 = (m_\gamma c_0/\hbar)^2$. If we anticipate the energy $\hbar\omega$ and momentum $\hbar k$ of a photon, this law defines $m_γ$ as the photon mass.

* Show that the relation $g^\nu k_\nu = 0$ allows three orthogonal solutions for $A^\mu \sim (g_0, g) \exp(ik_\mu x^\mu)$: two transverse solutions $g \perp k$, $g_0 = 0$ and one longitudinal solution $g \parallel k$ and $g_0 \neq 0$. If the photon mass vanishes we can still use the Lorenz gauge. Show that in that case the longitudinal solution has $E = B = 0$. (Hint: choose the z-axis along $k$ and use $k_\mu = (k_0, 0, 0, -k_0)$). In vacuum, the electromagnetic field with finite mass achieves a longitudinal component and three degrees of freedom exist, against two transverse for massless photons. This conclusion can be argued heuristically by saying that massless photons travel with the speed of light and have no co-moving frame in which we can define a longitudinal spin.

* Show that the Coulomb law for the electric potential is modified to

$$ \partial_t^2 \phi - \nabla^2 \phi + \left( \frac{m_\gamma c_0}{\hbar} \right)^2 \phi = \rho $$

with the solution at low energy for a point charge $q$, $\phi = q \exp(-x/x_\gamma)/(4\pi x)$ with $x_\gamma = \hbar/(m_\gamma c_0)$.
* Show that Faraday's induction law is modified by

$$ \nabla \times \mathbf{B} = \frac{1}{c_0} \partial_t \mathbf{E} + \mathbf{J} - \left( \frac{m_\gamma c_0}{\hbar} \right)^2 \mathbf{A} $$

or equivalently

$$ \frac{1}{c_0^2} \partial_t^2 \mathbf{B} + \nabla \times \nabla \times \mathbf{B} + \left( \frac{m_\gamma c_0}{\hbar} \right)^2 \mathbf{B} = \nabla \times \mathbf{J} $$

De Broglie estimated already $m_\gamma < 10^{-53}$ kg = $5.6 \cdot 10^{-18}$ eV, far less than the supposed mass of neutrinos. With this mass, we find $x_\gamma > \frac{1}{5}$ AU. Laboratory experiments based on Coulomb's law currently impose $m_\gamma < 2 \cdot 10^{-50}$ kg. The most stringent limit $m_\gamma < 2 \cdot 10^{-54}$ kg is obtained by fitting the modified Faraday induction law to measurements of the solar wind and data obtained by the satellite Pioneer flying by Jupiter, known for its large magnetic field.

### Example: Maxwell field as a representation of the Lorentz group

In group theory the vector potential $A^\mu$ is the elementary electromagnetic field with Lagrangian (124). It transforms as a genuine Lorentz 4-vector, $A_\mu(x) = \Delta_\mu^\nu A_\nu(\Delta^{-1}x)$. The transformation $\Delta_\mu^\nu$ is the fundamental representation of the Lorentz group and the vector field will be shown below to have spin 1. The gauge invariance is a crucial property but also implies that observable quantities do not contain $A^\mu$ itself (an important exception is the Aharonov-Bohm effect, due to an observable and gauge invariant phase shift of a quantum particle in a magnetic field).

Can we find a representation involving the observable fields $\mathbf{E}$ and $\mathbf{B}$ themselves? We have defined the electromagnetic vectors $\mathbf{F}_\pm = (\mathbf{E} \pm i\mathbf{B})/\sqrt{2}$ earlier in section III. We can rewrite the free Lagrangian of the electromagnetic field as,

$$ \mathcal{L} = -\frac{1}{4}(\mathbf{F}_+)^2 - \frac{1}{4}(\mathbf{F}_-)^2 \qquad (129) $$

Both vectors $\mathbf{F}_\pm$ transform under $SO(3)$ rotations $\mathbf{R}$ with the 3 angular momentum generators $\mathbf{L}_n$. The magnetic field $\mathbf{B}$ is a pseudovector and unlike $\mathbf{E}$, does not change sign under space inversion $P$ which, we recall, is not part of $SO^+(1,3)$. We have $P\mathbf{F}_\pm = -\mathbf{F}_\mp$, leaving $\mathcal{L}$ invariant. Under time-reversal $T\mathbf{F}_\pm = \mathbf{F}_\mp$ and under charge conjugation, $C\mathbf{F}_\pm = -\mathbf{F}_\pm$. We find $CPT = 1$ but we need both fields to cover all three symmetries. Under the dual transformation $\mathbf{E} \to \mathbf{B}$, $\mathbf{B} \to -\mathbf{E}$ discussed in exercise V.20 both vectors $\mathbf{F}_\pm$ are invariant apart from a phase factor $\mp i$ changing the sign of the Lagrangian without consequences for the equations of motion. Under boost transformations, it was established in section III that $\mathbf{F}'_\pm(x') = \mathbf{M}(\pm)\mathbf{F}_\pm(x) = \mathbf{M}(\pm)\mathbf{F}_\pm(\Delta^{-1}x')$
with $M(\pm)^T M(\pm) = 1 = M(\mp)M(\pm)$ and $\det M(\pm) = 1$. Since $M_{\pm}^{\dagger} = M_{\mp}$ the transformation matrix is hermitian and not unitary. We can verify that,

$$M(\pm, \xi) = \exp(\pm i\epsilon \cdot \xi)$$

with $\xi \equiv \xi \hat{\beta}$ and $\tanh \xi \equiv \beta$ as before. However, the product of two boosts matrices $M(\pm, \xi)$ is not necessarily another boost: No closed Lie algebra is generating the set. Since $-i\epsilon \cdot \hat{x}$ is rotation along the x-axis we find $[\epsilon \cdot \hat{x}, \epsilon \cdot \hat{y}] = -1 \times -i\epsilon \cdot \hat{z}$ which is a negative rotation along the z-axis. The set $\{M(\pm, \xi)\}$ is therefore not a group. To get a group we must include the unitary rotations and write for a general Lorentz transformation,

$$M(\pm, \Delta) = \exp(\pm i\epsilon \cdot \xi - i\alpha \cdot L) \qquad (130)$$

This defines two equivalent non-unitary sets of complex 3 × 3 matrices with unit determinant part of $SL(3, \mathbb{C})$. They satisfy the group structure of $SO^+(1, 3)$. We have $M(\pm, \Delta)F_{\pm} = F'_{\pm}$ and this matrix Lorentz-transforms $F_{\pm}$ from one reference frame $\mathcal{R}$ to another frame $\mathcal{R}'$ via a mixture of rotations and boosts. The electromagnetic field $F_{\pm}$ is uniquely defined in a given reference frame but we can go from one reference to another either directly or indirectly. We can reach the $\mathcal{R}_2$ by first going from $\mathcal{R}$ to $\mathcal{R}_1$ with the transformation $\Delta_1$, followed by going to $\mathcal{R}_2$ with the transformation $\Delta_2$. We can also go directly from $\mathcal{R}$ to $\mathcal{R}_2$ using $\Delta_2 \cdot \Delta_1$. Both must be equal and we conclude that

$$M(\pm, \Delta_2 \cdot \Delta_1) = M(\pm, \Delta_2) \cdot M(\pm, \Delta_1) \qquad (131)$$

We have found two equivalent representations of the Lorentz group in $SL(3, \mathbb{C})$. If the field $F_+$ transforms as $M(+, \Delta)$ its dual field transforms as $(M(+, \Delta)^{\dagger})^{-1} = M(-, \Delta)$, similar to what we have found in Exercise IV.9 for the chiral representation of the Dirac field. This implies that in this representation the dual field of $F_+$ is $F_-$. Together they have 6 degrees of freedom, reminiscent of two (complex) spins 1, but 2 disappear by gauge invariance that applies for zero mass.

### Exercise VI.8: Chiral representations of the electromagnetic field

In Dirac theory the notion of chirality was seen to be Lorentz invariant, giving rise to the direct sum of the two representations $(0, \frac{1}{2}) \oplus (\frac{1}{2}, 0)$ with opposite chirality and with axial global phase symmetry if the mass vanishes. From the above we conclude that the complex electromagnetic field with $E = \pm iB$ with $F_{\pm} = 0$ is Lorentz-invariant and preserves this form in any reference
frame obtained by either rotations or boosts. Show that this is satisfied for the plane wave $E = (\hat{x} \pm i\hat{y}) \exp(ikz)$ which is the complex notation for a plane wave with circular polarization $\pm$.

### Exercise VI.9: Quantization of the electromagnetic field

In the Coulomb gauge $\nabla \cdot A = 0$, the equation for the electric potential separates from the one for $A$ to the electrostatic Poisson equation $-\nabla^2 \phi = \rho$. In vacuum no charges exist and we can put $\phi = A_0/c_0 = 0$. This gauge is possible but is not Lorentz-invariant. The electromagnetic field in vacuum is entirely given by the vector potential

$$A(\mathbf{x}, t) = \sum_{jk\mathbf{k}} \frac{g_k}{\sqrt{V}} \left[ a_{jk\mathbf{k}} \hat{e}_{jk} e^{-i\omega_k t + i\mathbf{k}\cdot\mathbf{x}} + a_{jk\mathbf{k}}^{\dagger} \hat{e}_{jk} e^{i\omega_k t - i\mathbf{k}\cdot\mathbf{x}} \right] \quad (132)$$

in terms of the two normalized polarization vectors $\hat{e}_{1,2k}$ orthogonal to $k$ and assumed real-valued. Hence, only two degrees of freedom exist for each wave vector. Traditionally we adopt a finite quantization volume $V$ with discrete wave numbers. Quantization is performed by making $a_{\mu k}$ and $a_{\mu k}^{\dagger}$ canonical bosonic operators with $[a_{jk}, a_{j'k'}] = \delta_{jj'}\delta_{kk'}$.

* Argue qualitatively why it is always possible to impose the Coulomb gauge.

* Explain why $\omega_k > 0$ and why $A(\mathbf{x}, t) = A^{\dagger}(\mathbf{x}, t)$ implies that the photon is equal to its anti-particle.

* Electric and magnetic fields are given by $E = -c_0^{-1} \partial_t A$ and $B = \nabla \times A$. Show that the energy is given by

$$H \equiv \int d^3\mathbf{x} \left( \frac{1}{2}\mathbf{E}^2 + \frac{1}{2}\mathbf{B}^2 \right) = \sum_{jk\mathbf{k}} g_k^2 \omega_k^2 \left[ a_{jk\mathbf{k}} a_{j'k}^{\dagger} + a_{j'k}^{\dagger} a_{jk\mathbf{k}} \right] = \sum_{jk\mathbf{k}} \hbar \omega_k \left[ a_{jk\mathbf{k}}^{\dagger} a_{jk\mathbf{k}} + \frac{1}{2} \right] \quad (133)$$

with the choice $g_k = \sqrt{\hbar c_0 / 2\omega_k}$. In particular, demonstrate that the terms $a_{jk\mathbf{k}}a_{j'k}^{\dagger}$ and $a_{jk\mathbf{k}}^{\dagger}a_{j'k}^{\dagger}$ cancel.

* Show that

$$i\hbar \partial_t A(\mathbf{x}, t) = [A(\mathbf{x}, t), H]$$

After quantization, energy governs again time-evolution.

* Thermal averages are calculated using the density matrix of the thermal ensemble, diagonal in the Fock space of photon occupation numbers of wave numbers states, $\rho(\{n_k\}) =$
$Z^{-1} \exp(-\beta H)|\{n_k\}\rangle\langle\{n_k\}|$, with an occupation $n_k$ of the state $k$ (and polarization $i$) and $Z = \text{Tr} \exp(-\beta H)$. Show that (keep volume $V$ finite),

$$ \langle a_{j\mathbf{k}}^{\dagger} a_{j'\mathbf{k}'} \rangle \equiv \text{Tr} \rho a_{j\mathbf{k}}^{\dagger} a_{j'\mathbf{k}'} = f_{\mathbf{k}} \delta_{jj'} \delta_{\mathbf{k}\mathbf{k}'} $$

with $f_{\mathbf{k}} = 1/[\exp(\beta\hbar\omega_k) - 1]$.

* Show from this that as $V \to \infty$,

$$ \begin{aligned} \langle A_i(0,0)A_j(t, \mathbf{x}) \rangle &= \hbar c_0 \delta_{ij} \int \frac{d^3\mathbf{k}}{(2\pi)^3} \frac{1}{2\omega_k} \left[ (f_{\mathbf{k}} + 1)e^{i\mathbf{k}\cdot\mathbf{x}-i\omega_k t} + f_{\mathbf{k}}e^{-i\mathbf{k}\cdot\mathbf{x}+i\omega_k t} \right] \\ &= \hbar c_0 \delta_{ij} \int \frac{d^3\mathbf{k}}{(2\pi)^3} \frac{1}{2\omega_k} (2f_{\mathbf{k}} + 1)e^{i\mathbf{k}\cdot\mathbf{x}-i\omega_k t} \end{aligned} \quad (134) $$

This formula was used earlier to calculate the Unruh effect.

* The Coulomb gauge used above is not Lorentz-invariant. For $T = 0$ we can fix this by translating Eq. (134) to

$$ \langle A_{\mu}(0,0)A_{\nu}(t, \mathbf{x}) \rangle = -\hbar g_{\mu\nu} \int \frac{d^3\mathbf{k}}{(2\pi)^3} \frac{1}{2\omega_k/c_0} e^{-i\omega_k t + i\mathbf{k}\cdot\mathbf{x}} $$

Show that for both negative and positive $t$,

$$ \langle T[A_{\mu}(0)A_{\nu}(x)] \rangle = -i\hbar g_{\mu\nu} \int \frac{d^4k}{(2\pi)^4} \frac{e^{-ikx}}{k^2 + i\epsilon} \equiv (G_{\gamma}^0(x-y))_{\mu\nu} \quad (135) $$

with $x^{\mu} = (c_0t, \mathbf{x})$. We recover the Feynman propagator derived earlier in Eq. (127). This is the electromagnetic equivalent of Eq. (119) derived for a scalar massive boson.
<figure><img src="image_11.png" /><figcaption>Figure 9. The Casimir effect: Low energetic photons created by quantum vacuum fluctuations cannot obey the boundary conditions between two ideal metallic plates so that a net pressure form the outside exists that attracts the plates.</figcaption></figure>

### Example: The Casimir effect (Casimir, 1948)

The sum $\sum_{\mu k} \frac{1}{2} \hbar \omega_k$ implies an infinite electromagnetic energy per unit volume in the quantum vacuum. One may think that it is compensated by a yet unknown mechanism. This may be true, nevertheless this energy changes when constraints are imposed. Let us simplify to scalar massless bosons in 1D and compare the sum $\sum_k \frac{1}{2} \hbar \omega_k$ for a hole with Dirichlet boundary conditions for the field at $z = 0$ and $z = L$ so that $k_n = n\pi/L$, $n = 1, 2, \dots$ to the value if the boundary condition did not exist. Both are infinite, but surprisingly the difference is finite and measurable. Hendrik Casimir had the clever idea to add a small absorption increasing with frequency to make both finite. Show that

$$ \Delta E_0(L) = \lim_{\gamma \downarrow 0} \left[ \frac{\hbar c_0}{2} \sum_{n=1}^{\infty} k_n e^{-k_n \gamma} - \frac{\hbar c_0 L}{2 \pi} \int_0^{\infty} dk \, k e^{-k \gamma} \right] = -\frac{\pi \hbar c_0}{24L} \quad (136) $$

The change in vacuum energy is finite and negative, intuitively because some modes are excluded by the boundary conditions. The Casimir force between the boundaries $F = -dE_0/dL$ is attractive and varies as $1/L^2$. The generalization to 2D metallic plates with large surface A and electromagnetic vacuum is straightforward and gives $E_0(L) = -(\pi^2/720)\hbar c_0 A/L^3$. It was measured for the first time in 1958 by Casimir and Sparnaay in the Philips Research laboratory. Casimir argued later that for a spherical geometry $A \sim R^2$ and $L = 2R$ this energy $E_0(R) \sim -C/R$ could compensate the Coulomb repulsion $E_C \sim +1/R$ and explain why the electron is stable, with a bonus since the value for C would fix the fine-structure constant $\alpha \sim e^2/\hbar c_0$. Unfortunately he was wrong since C turned out to be negative because the curvature of the sphere costs vacuum energy.

By Eq. (112) the zero-point energy for fermions is negatively diverging. Nevertheless, the Casimir energy difference for massless fermions in the Casimir geometry is hardly different from the one
found for photons, and the force is also attractive [16].

### Example: The Lamb shift (Willis Lamb, 1947, Nobel Prize 1955)

The Lamb shift refers to a small separation in energy of about 1000 MHz between the energy levels $2S_{1/2}$ and $2P_{1/2}$ in atomic hydrogen that have different orbital angular momentum but due to the spin 1/2 of the electron have equal total angular momentum $J = 1/2$. According to the Dirac equation they must therefore be degenerated. The Lamb shift was observed first in 1947 by Willis Lamb and his student Robert Retherford. The same year Hans Bethe demonstrated it to be caused by the coupling of the electron to the electromagnetic quantum vacuum. In 2020 the Lamb shift was measured at CERN in the anti-hydrogen atom, a severe test for the equivalence of matter and anti-matter.

The Lamb shift can be understood as a manifestation of Casimir energy. The following semi-classical argument is due to Richard Feynman and taken from Ref. [16]. Similar to Eq. (136) we can calculate the change in energy due to the presence of a dilute gas of atoms. The dielectric constant of such a gas depends on frequency and their number density $\rho$. For a dilute gas it is approximately given by $\varepsilon(\omega) = 1 + \rho\alpha_n$ with the atomic polarizability given by

$$ \alpha_n(\omega) = \frac{4\pi e^2}{m_e} \sum_m \frac{f_{nm}}{\omega_{mn}^2 - (\omega + iA_n/2)^2} $$

for the atoms in state $n$, and with $m_e$ the electron mass, $\hbar\omega_{mn} = \hbar(\omega_m - \omega_n)$ the energy difference between two quantum states and $f_{nm}$ the so-called oscillator strength associated with the dipole transition $n \to m$. Mathematically, we require a small positive imaginary part in frequency not only to avoid a singularity at resonant frequencies $\omega_{mn}$, physically it implies most of all causality, more precisely in the time-domain the polarizability is a convolution kernel,

$$ \alpha_n(t - t') = \int_{-\infty}^{\infty} \frac{d\omega}{2\pi} \alpha_n(\omega) e^{-i\omega(t-t')} $$

This kernel vanishes for $t - t' < 0$ and decays exponentially as $\exp[-A_n(t - t')/2]$ for $t - t' > 0$ due to radiation damping. The imaginary part thus also implies that the atom strongly radiates near a resonance frequency $\omega_{nm}$ so that the polarization decays exponentially after it has been excited. At sufficiently high frequencies atomic forces are negligible and we expect Newton's law $m_e\ddot{x} = eE(t)$. With the atomic polarization $d_n = er_n = \alpha_n E$ (hence polarization density $P_n(x) = \delta(x - x_a)d_n$) and the relation $D = E + 4\pi P$ in Gaussian units, we should find the polarizability $\alpha_n = -4\pi e^2/m_e\omega^2$ so that we must have the sum rule $\sum_m f_{nm} = 1$. Given a wave
vector $\mathbf{k} = (n_x, n_y, n_z)\pi/L$ imposed by boundary conditions on the sides of the volume $V = L^3$, the associated frequency is $\omega_k = kc_0/\sqrt{\varepsilon(\omega_k)}$. For low number density the change in Casimir energy thus becomes

$$ \Delta E_n = \frac{\hbar c_0}{2} \sum_{\mathbf{k},g} \left[ \frac{k}{\sqrt{\mathrm{Re}\,\varepsilon(\omega_k)}} - k \right] \rightarrow -\frac{\hbar}{\pi} \frac{e^2}{m_e} \frac{N}{c_0^3} \sum_m \mathrm{Re} \int_0^\infty d\omega \frac{\omega^3 f_{nm}}{\omega_{mn}^2 - (\omega + iA_n/2)^2} $$

where we have converted the discrete sum $\sum_k$ over wave numbers into the integral $V/(2\pi)^3 \times \int d^3k$, summed over 2 polarizations and have set $\rho V = N$, the number of atoms. Even after subtraction of the infinite Casimir energy of the vacuum we find that the integral brutally diverges, but we can isolate the divergence associated with a free electron, obtained by putting $\omega_{nm} = 0$. The self-energy of the electrons is,

$$ \Delta E(0) = N \times \frac{1}{\pi} \frac{e^2}{m_e c_0^3} \hbar \int_0^{\omega_c} d\omega \omega \approx N \times \frac{1}{2\pi} \frac{e^2}{\hbar c_0} \times m_e c_0^2 \quad (137) $$

if we use a cut-off $\omega_c \approx mc_0^2/\hbar$ to cope with the divergence, assuming that at this relativistic scale the classical picture should break down. If we subtract this energy, we find for the change in energy per atom

$$ \frac{\Delta E_n - \Delta E(0)}{N} = -\frac{1}{\pi} \frac{e^2}{m_e c_0^3} \hbar \mathrm{Re} \sum_m \int_0^\infty d\omega \frac{f_{nm} \omega_{nm}^2 \omega}{\omega_{nm}^2 - (\omega + iA_n)^2} $$

The frequency integral is taken between $\omega = 0$ and $\omega = \infty$ and does not vanish due to causality arguments, but still diverges only logarithmically. If we use the cut-off $\omega_c$, we can move the integral to the imaginary axis using Cauchy's theorem, with the quarter circle $\omega = \omega_c \exp(i\phi)$ with $0 \le \phi \le \pi/2$ not contributing to the real part,

$$ \frac{\Delta E_n - \Delta E(0)}{N} = \frac{1}{2\pi} \frac{\hbar e^2}{m_e c_0^3} \sum_m \int_0^{\omega_c} \frac{ds}{\omega_{nm}^2 + s^2} = \frac{1}{2\pi} \frac{\hbar e^2}{m_e c_0^3} \sum_m f_{nm} \omega_{nm}^2 \log \left( \frac{m_e c_0^2}{\hbar \omega_{nm}} \right) \quad (138) $$

This is exactly the expression first obtained by Bethe for the Lamb shift. Using the quantum-mechanical theory for the oscillation strengths, a second useful sum rule can be derived for the hydrogen atom with Coulomb potential, $\sum_m \omega_{nm}^2 f_{nm} = (4\pi e^2/3m_e)|\Psi_n(0)|^2$ in terms of the wave function at the origin. This means that if we replace the relatively large and slowly varying logarithmic factor by some proper average value, the Lamb shift is proportional to the probability to find the electron at the origin. This vanishes for the $2P_{1/2}$ level (and all levels with $\ell > 0$) but not for the $2S_{1/2}$ level. We will see later in the quantum field theory that the electron self-energy truly diverges, and that no known mechanism is currently known that makes it finite.
## Example: Angular momentum and spin 1 of the Maxwell field

The vector potential $A^\mu$ of the Maxwell field transforms under Lorentz transformations as a four-vector field, i.e.

$$ (A')^\mu(x^\gamma) = \Delta_\nu^\mu A^\nu \left[ (\Delta^{-1})^\gamma_\delta x^\delta \right] = \Delta_\nu^\mu A^\nu \left( \Delta_\delta^\gamma x^\delta \right) $$

The transformation of space-time $x$ generates - via Noether's theorem - orbital angular momentum as was seen for the scalar field. The transformation as a Lorentz four-vector generates the intrinsic spin 1. From the transformation of $\Delta_\mu^\nu$ given by Eqs. (49) and (50) we find that

$$ \delta \Delta_\mu^\nu = -\frac{i}{2} (M^{\alpha\beta})_\mu^\nu \omega_{\alpha\beta} = \frac{1}{2} (g_\mu^\alpha g^{\beta\nu} - g^{\alpha\nu} g_\mu^\beta) \omega_{\alpha\beta} $$

so that the infinitesimal group transformation of the field reads,

$$ \Delta^{\alpha\beta} A_{\mu} = \frac{1}{2} (g_{\mu}^{\alpha} A^{\beta} - g_{\mu}^{\beta} A^{\alpha} + x^{\beta} \partial^{\alpha} A_{\mu} - x^{\alpha} \partial^{\beta} A_{\mu}) $$

To find the Noether charge we also need $\partial\mathcal{L}/\partial(\partial_0 A_\mu) = -F^{0\mu} = (0, E)$. For $\alpha, \beta \neq 0$ the K term drops out (see analysis of scalar field) so that the Noether charge is

$$ \begin{aligned} Q^{ij} &= \frac{1}{c_0} \int d^3 x \frac{\partial \mathcal{L}}{\partial(\partial_0 A_\mu)} \Delta^{ij} A_\mu \\ &= -\frac{1}{2c_0} \int d^3 x \left[ F^{0i} A^j - A^i F^{0j} + F^{0\mu} (x^j \partial^i A_\mu - x^i \partial^j A_\mu) \right] \end{aligned} \quad (139) $$

The angular momentum pseudo-vector can be defined as $J_n = \epsilon_{ijn}Q^{ij}$. This can be rearranged to

$$ J_n = \frac{1}{c_0} \int d^3 x \left[ -F^{0i} \epsilon_{ijn} A^j + F^{0l} \epsilon_{ijn} x^i \partial^j A_l \right] \equiv S_n + L_n \quad (140) $$

It is possible to show that $J_n$ is proportional to the space integral of $x \times (E \times B)$ (i.e. the vector product of position and electromagnetic momentum density, see e.g. Chapter I of [15]). To proceed we insert the expression (132) for the quantized field in vacuum and choose a basis of circularly polarized vectors,

$$ A(x, t) = \sum_{\sigma k} \frac{g_k}{\sqrt{V}} \left[ a_{\sigma k} \hat{e}_{\sigma k} e^{-i\omega_k t + i\mathbf{k} \cdot \mathbf{x}} + a_{\sigma k}^{\dagger} \hat{e}_{\sigma k}^{*} e^{i\omega_k t - i\mathbf{k} \cdot \mathbf{x}} \right] \quad (141) $$

with $\sigma = \sigma(\mathbf{k})$ standing for circular polarization with associated transverse polarization vectors $\hat{e}_{\sigma k} = (\hat{x} + i\sigma\hat{y})/\sqrt{2}$ that are orthonormal for $\sigma = \pm$, with $\mathbf{k}$ chosen along the $z$ direction of a righthanded $xyz$ frame. We want to define left- and right-handedness as $\sigma(-\mathbf{k}) = -\sigma(\mathbf{k})$. If we
define $\sigma$ and $\sigma'$ both with respect to $\mathbf{k}$, we get

$$S_n = \frac{V}{c_0} \sum_{\mathbf{k}} \sum_{\sigma\sigma'} \frac{i\omega_k g_{\mathbf{k}}^2}{V} \left[ a_{\sigma k} a_{-\sigma'-k} e^{-2i\omega_k t} (\hat{e}_{\sigma\mathbf{k}})^i \epsilon_{ijn} (\hat{e}_{-\sigma'-\mathbf{k}}^*)^j - \text{h.c.} \right] \\ + \frac{V}{c_0} \sum_{\mathbf{k}} \sum_{\sigma\sigma'} \frac{i\omega_k g_{\mathbf{k}}^2}{V} \left[ a_{\sigma k} a_{\sigma'k}^\dagger (\hat{e}_{\sigma\mathbf{k}})^i \epsilon_{ijn} (\hat{e}_{\sigma'\mathbf{k}}^*)^j - a_{\sigma k}^\dagger a_{\sigma'k} (\hat{e}_{\sigma\mathbf{k}}^*)^i \epsilon_{ijn} (\hat{e}_{\sigma'\mathbf{k}})^j \right]$$

with an extra factor $V$ stemming from in the volume integral in Eq. (140) that defines total angular momentum. We have

$$(\hat{e}_{\sigma \mathbf{k}})^i \epsilon_{ijn} (\hat{e}_{-\sigma' \mathbf{k}}^*)^j = \left( \frac{\hat{x} + i\sigma \hat{y}}{\sqrt{2}} \times \frac{\hat{x} - i\sigma' \hat{y}}{\sqrt{2}} \right)_n = -\frac{i}{2}(\sigma + \sigma')\hat{k}_n = -i\sigma\delta_{\sigma\sigma'}\hat{k}_n$$

The transformation $\mathbf{k} \to -\mathbf{k}$ changes the sign of the time-dependent terms in $S_n$ so that they both vanish. With $g_{\mathbf{k}}^2 = \hbar c_0 / 2\omega_k$ and $V \to \infty$ we obtain

$$S = \sum_{\mathbf{k}\sigma} \frac{1}{2} \left[ a_{\sigma k} a_{\sigma \mathbf{k}}^{\dagger} + a_{\sigma \mathbf{k}}^{\dagger} a_{\sigma k} \right] \hbar \sigma \hat{\mathbf{k}} = \sum_{\mathbf{k}\sigma} a_{\sigma k}^{\dagger} a_{\sigma \mathbf{k}} \hbar \sigma \hat{\mathbf{k}}$$

This assigns the intrinsic spin $\hbar\sigma\hat{\mathbf{k}} = \pm\hbar\hat{\mathbf{k}}$ (a pseudovector!) to a free, circularly polarized photon with wave number $\mathbf{k}$.

## Exercise VI.10: orbital angular momentum of the quantized electromagnetic field

Show by insertion of Eq. (141) that the orbital angular momentum of the quantized electromagnetic field is given by

$$L = V \int \frac{d^3 \mathbf{k}}{(2\pi)^3} \sum_{\sigma} \frac{1}{2} \left[ a_{\sigma k} \hat{e}_{\sigma \mathbf{k}} \left( -i\hbar \mathbf{k} \times \nabla_{\mathbf{k}} \right) \cdot a_{\sigma k}^{\dagger} \hat{e}_{\sigma \mathbf{k}}^{*} + \text{h.c.} \right]$$

We recognize the universal orbital angular momentum operator $\mathbf{r} \times \mathbf{p}$ that becomes $-i\nabla_{\mathbf{k}} \times \hbar\mathbf{k}$ in Fourier space.

## D. Quantum Electrodynamics

After second quantization of the photon field and the Dirac field it becomes possible to convert photons to fermions and vice-versa. For instance the Dirac-Maxwell interaction Hamiltonian is given by

$$H_I(t) = \frac{q}{c_0} \int_V d^3 \mathbf{x} \bar{\Psi}(x) \gamma^\mu A_\mu(x) \Psi(x) \equiv \int_V d^3 \mathbf{x} H_I(\mathbf{t}, \mathbf{x}) \quad (142)$$
where we can insert the quantized Dirac field (111) and the quantized Maxwell field (132). The interaction identifies $q\gamma^\mu$ as the interaction vertex between fermions and photon, and allows the destruction of a photon with the combined creation of a pair of electron and positron, and vice versa. In second-order perturbation theory this offers the possibility to create an electron-positron pair temporarily from a photon.

<figure><img src="image_12.png" /><figcaption>Figure 10. The Feynman rules in the coupling of Dirac fermions and the electromagnetic gauge field with the convention that time runs from bottom to top. The lines can go either from left to right (shown here) or from right to left depending on the direction of the (three-)momentum; $A^+$ stands for the destruction of a photon, $A^-$ for the creation of a photon, usually visualized without arrow since photon and anti-photon are equal; $\Psi^+(\bar{\Psi}^-)$ refers to the destruction (creation) of an electron; $\Psi^-(\bar{\Psi}^+)$ represents the creation (destruction) of a positron and time running backwards. Horizontal (or sometimes vertical) lines stand for the propagators of photons and fermions, the dot indicates the fermion-photon interaction with as coupling constant the electromagnetic charge. At this interacting node always two fermion lines (one arriving one leaving) occur and a single photon line.</figcaption></figure>

This course is not intended to go into the rich and subtle details of perturbation theory. The specific example of quantum vacuum polarization illustrates 1) how equations of motions can be converted into physical scattering events; 2) how built-in symmetry guarantees conservation laws to be obeyed in scattering events and 3) that technical problems do remain such as divergencies as large energies. In scattering theory the scattering (S)-matrix describes all events that can happen between an “incident” configuration and an “outgoing” configuration, given the constraints imposed by symmetries. Given the time-dependent interaction $H_I(t)$, standard quantum mechanics tells us
that the S-matrix is a unitary operator given by the time-ordered product (T)

$$ 
\begin{aligned}
S(t_i \rightarrow t_f) &= T \exp \left( -\frac{i}{\hbar} \int_{t_i}^{t_f} dt' H_I(t') \right) \\
&= 1 - \frac{i}{\hbar} \int d^4 x' H_I(t', x') + \frac{(-i)^2}{2\hbar^2} T \int d^4 x' \int d^4 x'' H_I(t', x') H_I(t'', x'') + \dots (143)
\end{aligned}
$$

For simplicity, we shall use units such that $c_0$ and $\hbar$ are equal to one. The first term in this expansion generates 8 contributions from $\Psi^{\pm}$ (destruction of electron or creation of positron), $\overline{\Psi}^{\pm}$ (creation of electron or destruction of positron) and $A_{\mu}^{\pm}$ (creation or destruction of photon) among which for instance the scattering of an electron by an incident photon. However, none of them survives in a quantum vacuum with only virtual particles.

To calculate Feynman diagrams it is convenient to let $V \to \infty$ and to go over to the continuum description with $\alpha_{ks} = \alpha_s(k)/\sqrt{V}$ etc. with anti-commutation rule $\{\alpha_s(k), \alpha_s^\dagger(k')\} = (2\pi)^3\delta(k-k')$, with the Dirac delta-distribution $\sqrt{V}\delta_{kk'} \to (2\pi)^3\delta(k-k')$ as $V \to \infty$ (and idem *mutadis mutandis* for the photon field). This generates energy-momentum relations easily.

### Exercise VI.11: No first-order photon emission

<figure><img src="image_13.png" /><figcaption>Figure 11. The Feynman diagram for the emission of a photon by a charged fermion that changes momentum.</figcaption></figure>

We consider the possible process $e(q) \to e(q') + \gamma(k)$ that emits a photon with four-momentum $k$ by a fermion with four-momentum $q$ that is scattered to $q'$. Show that the first-order matrix element in Eq. (143) contains a term equal to

$$ 
\begin{aligned}
-i \int d^4 x' H_I(x') &= -ie \int \frac{d^4 q}{(2\pi)^4} \int \frac{d^4 k}{(2\pi)^4} 2\pi\delta((q-k)^2 - m^2) \times 2\pi\delta(k^2) \times 2\pi\delta(q^2 - m^2) \\
&\quad \times \sum_{sjs'} \alpha_{(q-k)s}^{\dagger} a_{kj}^{\dagger} \alpha_{qs'} \times \overline{u}_{(q-k)s'} \cdot \gamma^{\mu} (\hat{e}_{kj})_{\mu} \cdot u_{qs}
\end{aligned}
$$
that destroys a fermion with four-momentum $q$, and creates a photon with four-momentum $q-k$
as well as a fermion with four-momentum $q$. The integral over $x'$ imposes $q-q'-k=0$. Show that
the conservation laws cannot be satisfied for $m > 0$ (hint: evaluate the condition $(q-k)^2-m^2=0)$.

**Example: Electron self-energy (Victor Weisskopf, 1939) (taken from Ref. [16])**

The electron-self energy $\Sigma(k)$ can be defined as a correction to the fermionic propagator $S_F$ due to the interaction of the electron with a virtual photon in the quantum vacuum. This process is shown in Fig. (12). In Eq. (137) we provided a non-relativistic calculation showing that this energy diverges quadratically at high optical frequencies. We here provide the relativistic calculation. In four-momentum space we can write it as

$$S_F(k) = S_F^0(k) + S_F^0(k)(-i)\Sigma(k)S_F^0(k) \quad (144)$$

with $S_F^0(k)$ that can be read off from Eq. (120),

$$S_F^0(k) = i \frac{\not k + m}{k^2 - m^2 + i\epsilon} = \frac{i}{\not k - m + i\epsilon}$$

where we used $(\not k + m)(\not k - m) = k^2 - m^2$. If we iterate Eq. (144) we can formally write

$$S_F(k) = \frac{i}{\not k - m - \Sigma(k) + i\epsilon} \quad (145)$$

which implies that the self-energy of the electron can be interpreted as renormalization of the electron mass (and explains why we included the factor $-i$ above in its definition), though dependent on the four-momentum $k$ and also recalling that $\Sigma(k)$ is in principle a $4 \times 4$ complex matrix. Using the Feynman gauge for the photon we can write down the expression and Wick's theorem that generates a contraction on both fermion field and photon field with conservation of four-momentum,

$$-i\Sigma(k) = (-ie)^2 \int \frac{d^4 q}{(2\pi)^4} \gamma^\mu (G_\gamma^0)_{\mu\nu}(q) S_F^0(k-q) \gamma^\nu = -e^2 \int \frac{d^4 q}{(2\pi)^4} \frac{1}{q^2 + i\epsilon} \gamma_\nu \frac{\not k - \not q + m}{(k-q)^2 - m^2 + i\epsilon} \gamma^\mu$$

From the anti-commutator $\{\gamma^\mu, \gamma^\nu\} = 2g^{\mu\nu}$ we can verify that $\gamma_\nu\gamma^\mu = 4I_4$ and $\gamma_\nu\not k\gamma^\nu = -2\not k$.
Hence,

$$\Sigma(k) = -2ie^2 \int \frac{d^4 q}{(2\pi)^4} \frac{1}{q^2 + i\epsilon} \frac{\not q - \not k + 2m}{(k-q)^2 - m^2 + i\epsilon}$$

We can perform a Wick rotation to the imaginary axis, avoiding all the poles for $q_0$ in the complex plane (see Ex. VI.6, the extra poles at $q = \pm\sqrt{q^2 - i\epsilon}$ are still avoided) and define the Euclidian
4D vector $p = (p_0 = iq_0, q)$ to get

$$ \Sigma(k) = -2ie^2i \int \frac{d^4p}{(2\pi)^4} \frac{1}{p^2} \frac{ip_0\gamma^0 - \gamma^i p_i - \not k + 2m}{p^2 - k^2 + m^2 + 2ik_0p_0 - 2\mathbf{p} \cdot \mathbf{k}} \\ \approx 2e^2 \int_{m \ll p \le \Lambda} \frac{d^4p}{(2\pi)^4} \frac{2m - \not k}{(p^2)^2} \sim e^2m \log\left(\frac{\Lambda}{m}\right) $$

We kept only the integral for $p \gg m$ and removed odd parts that then integrate to zero. We see that the integral is positive and diverges, yet only logarithmically. The relativistic approach demonstrates that the divergence is genuine: No momentum higher than $mc_0$ exists and therefore the integral is not compensated by any known contribution. When we go back to physical units, the mass correction is proportional to the product of $e^2/4\pi\epsilon_0\hbar c_0 \equiv \alpha$, an unknown logarithmic factor and $m$. Even if the logarithm would be finite and say of the order of 10, this correction is significant! Furthermore when the self-energy acts on an electron with Dirac wave function $u_{ks}$ we have $(\hbar\not k - mc_0)u_{ks} = 0$, so that the anisotropic self-energy proportional to $\not k$ drops out. The conclusion is that $\delta m = +\alpha m \log(\Lambda/mc_0)$ has a logarithmical divergence that is put in the "observed mass" of the electron, with the hope of finding one day a better theory that removes the divergence.

<figure><img src="image_14.png" /><figcaption>Figure 12. The electron self energy: a photon is temporarily created from the vacuum by its coupling to the electron. This diagram diverges and renormalizes the electron mass. Inside an atom this diagram generates the Lamb shift.</figcaption></figure>
Example: Quantum Vacuum Polarization (Paul Dirac, Werner Heisenberg, 1934)

<figure><img src="image_15.png" /><figcaption>Figure 13. Left: The Feynman diagram for quantum vacuum polarization (sometimes also called photon self energy) where an electron-positron pair is virtually created from a propagating photon. Right: Bubble diagram that renormalizes the vacuum state.</figcaption></figure>

We can consider the vacuum matrix element $\langle 0 | S(-\infty, \infty) | 0 \rangle$ and investigate what processes occur in the quantum vacuum. For vacuum polarization we can ask how the photon propagator is affected by the existence of the quantum vacuum. To this end we consider

$$ (G_{\gamma}(x-y))_{\mu\nu} = \langle 0 | T[A_{\mu}(x)SA_{\nu}(y)] | 0 \rangle $$

For $S = 1$ this reduces to the free photon propagator obtained in Eq. (135). The time-ordered product of operators can be evaluated using Wick's theorem. The second-order perturbation can be written as

$$ (\delta G_{\gamma}(x-y))_{\mu\nu} = (-ie)^2 \frac{1}{2} \langle 0 | T \{ A_{\mu}(x) \int d^4x' d^4x'' [\bar{\Psi}(x')\gamma^{\alpha}A_{\alpha}(x')\Psi(x')\bar{\Psi}(x'')\gamma^{\beta}A_{\beta}(x'')\Psi(x'')] A_{\nu}(y) \} |0\rangle $$

Wick's theorem excludes contractions of fields at equal times and generates products of contractions, $A_{\mu}(x)A_{\alpha}(x') \times A_{\beta}(x'')A_{\nu}(y)$ for the photon and $\bar{\Psi}(x')\Psi(x'') \times \bar{\Psi}(x')\Psi(x'')$ as the first possibility. The last two just differ in sign so that an extra minus-sign is generated. A second possibility exists to contract $A_{\mu}(x)$ with $A_{\beta}(x'')$ and $A_{\nu}(y)$ with $A_{\alpha}(x')$ but makes the same contribution. Hence,

$$ (\delta G_{\gamma}(x-y))_{\mu\nu} = (-ie)^2 \int d^4x' d^4x'' (G_{\gamma}^0(x-x'))_{\mu\alpha} \text{Tr} [\gamma^{\alpha} \cdot S_F(x' - x'') \cdot \gamma^{\beta} \cdot S_F(x'' - x')] (G_{\gamma}^0(x'' - y))_{\beta\nu} $$
and thus in reciprocal space,

$$ (G_{\gamma}(q))_{\mu\nu} = \frac{-ig_{\mu\nu}}{q^2} + \frac{-ig_{\mu\alpha}}{q^2} \Pi^{\alpha\beta}(q^2) \frac{-ig_{\beta\nu}}{q^2} \quad (146) $$

with

$$ \Pi^{\alpha\beta}(q^2) = -(-ie)^2 \int \frac{d^4 k}{(2\pi)^4} \text{Tr} \left( \gamma^{\alpha} \frac{i(k+m)}{k^2 - m^2 + i\epsilon} \gamma^{\beta} \frac{i(k+q+m)}{(k+q)^2 - m^2 + i\epsilon} \right) \quad (147) $$

Electron-positron pairs with arbitrary large energies are created with conservation of momentum at both vertices. The integral over diverges as $k^3dk^2/k^4 = kdk$, even for $q = 0$ when the propagator $\Pi^{\alpha\beta}(q^2 = 0)$ can be seen to be proportional to $g^{\alpha\beta}$, which would leave us with a highly singular $1/q^4$ vacuum polarization. We therefore expect that regularization produces a finite expression that vanishes for $q = 0$. In addition, because this operator is independent on the gauge used for the photon field $A^{\mu}$, any gauge other than the Feynman gauge (see Exercise VI.7) imposes the Lorenz gauge so that we must have $q_{\alpha}\Pi^{\alpha\beta}(q^2) = 0$. We thus anticipate that $\Pi^{\alpha\beta}(q^2) \sim if_1(q)[q^2g^{\alpha\beta} - q^{\alpha}q^{\beta}]$. Wick's theorem facilitates a third option where the fields $A^{\mu}$ and $A^{\nu}$ also contract to produce $A_{\mu}(x)A_{\nu}(y) = (G_{\gamma}^{(0)}(x-y))_{\mu\nu}$. The remainder is independent on $x$ and $y$ and is visualized by a so-called bubble-diagram. It can be seen to be generated by $\langle 0|S|0 \rangle$ without explicit reference to the photon propagator. For a good choice of the vacuum state $|0 \rangle$ we expect this to be equal to unity, assuming that the vacuum state is stable and eigenfunction of the total hamiltonian. So, bubble diagrams should either cancel or renormalize coupling constants. They will not be discussed here, and will simply be ignored.

## Exercise VI.12: Charge Regularization in QED

* For $q^2 = 0$, show that expression (147) reduces to,

$$ \Pi^{\alpha\beta}(q^2 = 0) = -4e^2 \int \frac{d^4 k}{(2\pi)^4} \frac{(m^2 - k^2)g^{\alpha\beta} + 2k^\alpha k^\beta}{(k^2 - m^2 + i\epsilon)^2} $$

Use that $\text{Tr}\gamma^{\mu}\gamma^{\nu} = 4g^{\mu\nu}$, $\text{Tr}\gamma^{\mu}\gamma^{\nu}\gamma^{\alpha}\gamma^{\beta} = 4(g^{\mu\nu}g^{\alpha\beta} - g^{\mu\alpha}g^{\nu\beta} + g^{\mu\beta}g^{\nu\alpha})$ and that traces of odd numbers of Dirac matrices always vanish.

* By replacing the four-dimensional Minkowski integral by an $1+(n-1)$ dimensional Minkowski integral and by generalizing the denominator to a power $p > 2$, show that

$$ \Pi^{\alpha\beta}(0) = -4e^2 g^{\alpha\beta} \int \frac{d^nk}{(2\pi)^n} \left[ \frac{2/n-1}{(k^2 - m^2 + i\epsilon)^{p-1}} + \frac{2m^2/n}{(k^2 - m^2 + i\epsilon)^p} \right] $$

Hint: show first that the average over the surface “$k^2 = \text{constant}$” is $\langle k^{\alpha}k^{\beta} \rangle = k^2g^{\alpha\beta}/n$.
• Show that after Wick rotation of $k_0$ to the imaginary axis (see Fig. 8), this takes the form

$$ \begin{aligned} \Pi^{\alpha\beta}(0) &= -4e^2 i (-1)^p g^{\alpha\beta} \int \frac{d^n x}{(2\pi)^n} \left[ \frac{1-2/n}{(x^2+m^2)^{p-1}} + \frac{2m^2/n}{(x^2+m^2)^p} \right] \\ &= -\frac{4e^2}{(2\pi)^4 \hbar^2} i (-1)^p g^{\alpha\beta} \left[ \left(1-\frac{2}{n}\right) I(n,p-1) + \frac{2m^2}{n} I(n,p) \right] \\ &= -\frac{4e^2}{(2\pi)^4 \hbar^2} i (-1)^p m^2 g^{\alpha\beta} \frac{p-2}{p-1-n/2} I(n,p) \end{aligned} $$

where we have introduced the integral

$$ I(n,p) = \int \frac{d^n x}{(x^2 + m^2)^p} = \frac{S_n}{2m^{2p-n}} \frac{\Gamma(n/2)\Gamma(p-n/2)}{\Gamma(p)} = \frac{\pi^{n/2}}{m^{2p-n}} \frac{\Gamma(p-n/2)}{\Gamma(p)} $$

valid for $p > n/2$. The second equality for $\Pi^{\alpha\beta}(0)$ applies $\Gamma(x+1) = x\Gamma(x)$ so that $I(n, p-1) = m^2(p-1)I(n, p)/(p-1-n/2)$. The dimensional regularization sets the value back to $p = 2$ and puts $n = 4 + \epsilon$. For $\epsilon > 0$ is $\Gamma(-\epsilon/2) \approx -2/\epsilon$ finite with the result that $\Pi^{\alpha\beta}(q^2 = 0) = 0$.

• For finite four-momentum $q$ of the photon this procedure can be repeated. For $p = 2$ the singular value for $I(n, 2)$ remains and the outcome is

$$ \Pi^{\alpha\beta}(q^2) = \frac{-8ie^2}{(2\pi)^4} \pi^{2+\epsilon/2} m^\epsilon \Gamma(-\epsilon/2) (q^\alpha q^\beta - q^2 g^{\alpha\beta}) \int_0^1 dx (x^2 - x) \left[ \frac{q^2}{m^2} (x^2 - x) + 1 \right]^{\epsilon/2} $$

<figure><img src="image_16.png" /><figcaption>Figure 14. The Feynman diagrams for electron-electron scattering (in the reference frame of zero momentum, mediated by the electromagnetic gauge boson, left) and Compton scattering of an electron by a photon in the X-ray (right). Both diagrams have a second version with equal result. A second Compton diagram exists where the two photon lines are interchanged. On each vertex of a Feynman diagram four-momentum conservation is satisfied.</figcaption></figure>
With $\Gamma(-\epsilon/2) = -2/\epsilon + c$ and collecting only $1/\epsilon$ divergent and constant order zero terms, this becomes

$$ \Pi^{\alpha\beta}(q^2) = \frac{-8ie^2}{(2\pi)^4} \pi^2 (q^\alpha q^\beta - q^2 g^{\alpha\beta}) \left\{ -\left(\frac{2}{\epsilon} - \log \pi - \log m - c\right) \int_0^1 dx(x^2 - x) \right. \\ \left. - \int_0^1 dx(x^2 - x) \log \left[1 + \frac{q^2}{m^2}(x^2 - x)\right] \right\} $$

The first positive divergent term between brackets is removed by redefining the electron charge $e$. The final result for the vacuum polarization vertex, bringing back the physical units, becomes

$$ \Pi^{\alpha\beta}(q^2) = i \frac{2\alpha}{\pi} (q^\alpha q^\beta - q^2 g^{\alpha\beta}) \int_0^1 dx(x^2 - x) \log \left[ 1 + \lambda_C^2 q^2 (x^2 - x) \right] \quad (148) $$

The vacuum polarization creates a spatial charge structure at the (reduced) Compton wavelength $\lambda_C = \hbar/mc_0 = 3.8 \cdot 10^{-13}$ m, much smaller than the size of atoms. It is proportional to $\alpha = e_R^2/4\pi\epsilon_0\hbar c_0 \approx 1/137$, recognized as the QED fine structure constant.

* Vacuum polarization generates a small correction to the Coulomb interaction in e.g. the scattering of two electrons with low energies. Show that the S-matrix element for the process $e(k) + e(k') \to \gamma(q) \to e(k-q) + e(k'+q)$ can be written as

$$ f_{ee}(k, q, \{s\}) = (-ie)^2 \left[ \bar{u}_{k-q,s} \gamma^{\mu} u_{ks'} \right] \left( G_{\gamma}(q) \right)_{\mu\nu} \left[ \bar{u}_{k'+q,s''} \gamma^{\mu} u_{k's''} \right] $$

Show that $q_\mu \bar{u}_{k-q,s} \gamma^\mu u_{ks'} = 0$. Hint: recall that $(\gamma^\mu k_\mu - m)u_{ks} = 0$ and $\bar{u}_{ks}(\gamma^\mu k_\mu + m) = 0$. This means that any longitudinal term in the photon propagator can exist but cancels in the observable scattering amplitude. Gauge invariance is behind this cancelation.

* If the momenta of both electrons are non-relativistic, i.e. $\hbar k_i \ll mc_0$ in physical units, the Dirac equation reduces to $(i\gamma^0 k_0 - m)u = 0$, with solution $u_{ks} = \sqrt{2m}(e_{ks}, 0)$ and $E_k = k_0 = m$ independent of momentum. Recall that $\Psi^+ \sim u_{ks}/\sqrt{2E_k}$. Let us choose the two incident electrons at rest so that $k = k' = (k_0, 0)$. For low energies, the exchange of four-momentum between the two electrons only concerns momentum, not energy, so that $q = (0, q)$. Verify that $\bar{u}_{ks}\gamma^0 u_{ks'} = 2m\delta_{ss'}$ and $\bar{u}_{ks}\gamma^i u_{ks'} = 0$ and show that the scattering amplitude $f_{ee}$ is given,

$$ f_{ee}(k, q, s, s') = \frac{(-ie)^2}{(2m)^2} (G_{\gamma}(q))_{00} \left[ \bar{u}_{ks} \gamma^0 u_{ks} \right]^2 = -i \left( \frac{e^2}{q^2} + \frac{\alpha e^2}{15\pi} \lambda_C^2 \right) $$

independent on spin. Since in the Born approximation $S(k \to k') \approx 1 - i\langle k | V(r) | k' \rangle$, this can be recognized as the scattering amplitude for a potential $V$ that in real space takes the
form,

$$V_{ee}(r) = \frac{e^2}{4\pi r} + \frac{\alpha}{15\pi} e^2 \lambda_C^2 \times \delta^3(\mathbf{r}) \quad (149)$$

The quantum vacuum correction also modifies the Coulomb law between electron and proton at scales on the order of the Compton wavelength of the electron, though with opposite sign and generates a small shift in the energy $(\Delta E)_{QV} = \langle 1S | \delta V | 1S \rangle \sim -(\alpha e^2 / 15\pi) \lambda_C^2 |\Psi_s(0)|^2$ of an s-state of Hydrogen and this gives a contribution to the Lamb shift discussed earlier. Using Eq. (138), it is given by $(\Delta E)_L = +(4\alpha e^2 L / 3) \lambda_C^2 |\Psi_s(0)|^2$ with $L \approx 12$ a logarithmic factor that describes virtual transitions of the electron to high energy states. The Lamb shift is much larger and positive (around 1050 MHz), but the vacuum polarization is not negligible since the shift is known with large precision.

## E. Local symmetry

So far we have considered global continuous symmetries and have linked them to conserved charges. Let us consider local transformations. For instance, a local $U(1)$ transformation would transform the wave function as $\Psi'(x) = \exp(-iq\theta(x))\Psi(x)$. Clearly this will act on derivatives and produce terms like $\partial_\mu \Psi \partial^\mu \Psi^*$ that are not gauge-invariant,

$$\partial_\mu \Psi' = e^{-iq\theta} (\partial_\mu \Psi - iq\partial_\mu \theta \Psi)$$

The minus sign and the inclusion of the charge $q$ in the $U(1)$ transformation is convenient since it produces the right sign for the Noether current, proportional to the charge. Local invariance can be repaired by replacing the space-time derivative by the so-called covariant derivative involving a gauge field $A_\mu$,

$$\partial_\mu \rightarrow D_\mu = \partial_\mu + iqA_\mu \quad (150)$$

Note that $A_\mu$ must transform as a contra-variant vector for this expression to be fully contra-variant invariant. We can check that $D'_\mu e^{-iq\theta} \Psi = e^{-iq\theta} D_\mu \Psi$ provided that the gauge field transforms as

$$A'_\mu = A_\mu + \partial_\mu \theta \quad (151)$$

We can recognize the gauge freedom we found for the electromagnetic four-vector potential. The transformation keeps the vector field real-valued. Therefore this transformation introduces de facto
an interaction between the field $\Psi$ and the electromagnetic field, and links the $U(1)$ symmetry that guaranteed probability and charge conservation to the gauge freedom of the Maxwell field. The gauge-invariant Lagrangian describing matter and electromagnetic field thus becomes

$$ \mathcal{L} = \mathcal{L}[\Psi_a, (\partial_\mu + iq_a A_\mu)\Psi_a] - \frac{1}{4}F^{\mu\nu}F_{\mu\nu} \quad (152) $$

For the Dirac field in particular we see that,

$$ \mathcal{L} = \bar{\Psi} (i\gamma^\mu \partial_\mu - m) \Psi - A_\mu J^\mu - \frac{1}{4} F^{\mu\nu} F_{\mu\nu} \quad (153) $$

with the Dirac charge-current $J^\mu = q\bar{\Psi}\gamma^\mu\Psi$ associated with global $U(1)$ symmetry. This current is still conserved by global phase invariance because this symmetry leaves $A^\mu$ unaffected. We see that the source term in Maxwell's equations (see Exercise V.2) is automatically generated by the minimal coupling procedure. Minimal coupling extends global to local phase invariance but no new conserved currents arise.

### Example: Dirac-Maxwell Hamiltonian

Show that the Dirac equation for a charge $q$ coupled to the electromagnetic field in physical units takes the form

$$ \left[ \gamma^{\mu} \left( -i\hbar \partial_{\mu} + \frac{q}{c_0} A_{\mu} \right) + mc_0 \right] \Psi = 0 $$

which is equivalent to the Hamiltonian

$$ H = c_0 \boldsymbol{\alpha} \cdot \left( \boldsymbol{p} - \frac{q}{c_0} \boldsymbol{A} \right) + q\Phi + \gamma^0 mc_0^2 $$

in terms of the previously defined matrices $\alpha^i = \gamma^0 \gamma^i$.

We can repeat the same procedure for spinors and try to impose a local $SU(2)$ symmetry. The difference is that this group in non-Abelian. Let us consider a complex doublet $\phi = (\phi_1, \phi_2)$ with Lagrangian

$$ \mathcal{L}(\phi, \phi^*, \partial_\mu \phi, \partial_\mu \phi^*) = \partial_\mu \phi^\dagger \partial^\mu \phi - m^2 \phi^\dagger \phi \quad (154) $$

This Lagrangian is invariant under global $U(1)$ symmetry as well as under global $SU(2)$ symmetry. This operator rotates the spinor in complex $2D$ space and was seen to have 3 generators $T_a = \frac{1}{2}\sigma_a$
with Lie algebra $[T_a, T_b] = i\epsilon_{abc}T_c$ and orthonormalization $\mathrm{Tr} T_a^\dagger T_b = \frac{1}{2}\delta_{ab}$. This last property is not unimportant since it leaves the charge as the only free parameter and any modification will also affect the Lagrangian and the Lie algebra. A global $SU(2)$ transformation is given by $U = \exp(-ig\alpha_a T_a)$, with 3 real-valued coefficients $\alpha_a$, the charge $g$ and the generators $T_a$ that replace the Abelian $U(1)$ and the electromagnetic charge $q$. It becomes local when the $\alpha_n$ depend on $x$. In that case

$$ \partial_{\mu}\phi' = \partial_{\mu}(U\phi) = U\partial_{\mu}\phi + (\partial_{\mu}U)\phi $$

This time we introduce 3 real-valued gauge field $A_\mu^a(x)$, equal to the number of generators, and write $A_\mu = A_\mu^a(x)T_a$ (with implicit sum over $a$) to restore the local symmetry. This makes the vector potentials part of the Lie algebra, a vector space, of the underlying group, here $SU(2)$, and the 4-vector $A_\mu$ is a $2 \times 2$ matrix for each component.

If we introduce $D_\mu = \partial_\mu + igA_\mu = \partial_\mu + igT_a A_\mu^a$, we want to impose that $D_\mu(U\phi) = U D_\mu\phi$. We get

$$ U^\dagger D'_\mu(U\phi) = U^\dagger(\partial_\mu + igA'_\mu)U\phi = \partial_\mu\phi + U^\dagger(\partial_\mu U)\phi + U^\dagger igA'_\mu U\phi := (\partial_\mu + igA_\mu)\phi $$

so that the gauge field must transform as

$$ A'_{\mu} = U A_{\mu} U^{\dagger} + \frac{i}{g} (\partial_{\mu} U) U^{\dagger} \qquad (155) $$

If we consider small transformations for which the Lie algebra comes in, this becomes

$$ A'_{\mu} \approx T_a A_{\mu}^a - ig\alpha_b [T_a, T_b] A_{\mu}^a + (\partial_{\mu}\alpha_b)T_b \qquad (156) $$

or,

$$ (A'_{\mu})^a = A_{\mu}^a - g\epsilon_{abc}\alpha_b A_{\mu}^c + \partial_{\mu}\alpha_a + \mathcal{O}(\alpha^2) \qquad (157) $$

Thus, the Lie commutator determines how the gauge field transforms. Equation (156) reduces to the $U(1)$ gauge field transformation if the generators commute, that is if the symmetry transformation is Abelian.

The next step is to look out for a Lorentz-invariant Lagrangian for the gauge field itself. The four-tensor $F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$ is invariant only under a local $U(1)$ gauge.

### Exercise VI.13: Non-Abelian field Tensor

By construction, the commutator $F_{\mu\nu} \equiv [D_\mu, D_\nu]$ transforms as a second-order contravariant 4-tensor and under gauge-transformation as $U^\dagger F'_{\mu\nu} U = F_{\mu\nu}$.
* Show that $F_{\mu\nu} = igT_aF_{\mu\nu}^a$, with

$$F_{\mu\nu}^a = \partial_\mu A_\nu^a - \partial_\nu A_\mu^a + g\epsilon_{abc}A_\mu^b A_\nu^c \quad (158)$$

Use that $[\partial_\mu, A_\nu^a] = (\partial_\mu A_\nu^a)$.

* Show that

$$-\frac{1}{2g^2} \mathrm{Tr} F^{\mu\nu} F_{\mu\nu}^{\dagger} = -\frac{1}{4} (F^a)^{\mu\nu} (F^a)_{\mu\nu} \quad (159)$$

which is both a Lorentz-invariant and locally gauge-invariant real-valued scalar.

The non-commutativity of the underlying symmetry gauge-field introduces the gauge field itself into the Lagrangian, rather than just its derivatives, so to guarantee the invariance under the local gauge transformation.. In second quantization this implies that the gauge-field interacts with itself. The total Lagrangian of the spinor, Lorentz-invariant and with local SU(2) symmetry, becomes

$$\mathcal{L} = D_{\mu} \phi^{\dagger} D^{\mu} \phi - m^2 \phi^{\dagger} \phi - \frac{1}{4} (F^a)^{\mu\nu} (F^a)_{\mu\nu} \quad (160)$$
VII. SPONTANEOUS SYMMETRY BREAKING

Spontaneous symmetry Breaking (SSB) refers to the situation that the ground state of an object (in particle physics this notion refers to the “quantum vacuum”) does not have the same symmetry as the Hamiltonian. In a “simple” system this is never the case: the ground state of the hydrogen atom exhibits all the symmetries of the Hamiltonian. SSB is a relatively recent concept that was first introduced in condensed matter physics by P.W. Anderson around 1964. He argued that the superconducting state of a conductor can be associated with the broken $U(1)$ symmetry associated with particle conservation. This gives rise to the Anderson-Higgs mechanism that gives mass to the intermediate boson that carries the force, here the photon. The mechanism was independently discovered in particle physics (Higgs et al) using a relativistic description in which the Higgs particle also appears. It has now become the essential ingredient in the description of both strong and weak force and explains why particles associated with gauge fields achieve mass and as such violate the Goldstone theorem described below. Philip Anderson argued in a classical paper on the philosophy of science that SSB is an example of a concept that could never have been discovered by trying to solve microscopic equations on a macroscopic level: *More is different*.

Before we address the notion of broken symmetry it is important to be clear what we call symmetry operations and what is more related to degree of freedom. These arguments are taken from the excellent review on broken symmetries in [4]. Symmetry imposes physical laws and conservations laws, whereas degrees of freedom refers to a liberty to describe physics. We have seen in electromagnetism that gauge fields can be transformed without affecting any physical law. In classical mechanics we have the freedom to label our particles any way we want but no physical law can ever depend on this. In quantum mechanics this principle is more visible: the wave function of *N* identical particles must be symmetric or antisymmetric in any change of labels. We can rotate the axes with respect to which we measure the orientation of objects, the physical laws cannot not change (this is the Euclidean Principle discussed earlier). Yet, symmetry and broken symmetry need a reference frame to be visible. We give two well-known examples of broken symmetry.

The physics underlying a ferromagnet is rotationally invariant, yet the ground state exhibits a magnetization $M$ that has some well-defined direction in space. In the case of a ferromagnet textbooks usually introduce a small magnetic field $B$ and impose an interaction $E = -M \cdot B$ which determines the reference. If the magnetization remains as the external field vanishes we can say that the rotational symmetry of the ferromagnet is spontaneously broken. This kind of limit often does not commute with other limits such as the thermodynamic limit or the limit that the
temperature goes to zero.

Another example is the obvious fact that macroscopic objects, like the pencil on your desk, tend to be in an approximate eigenstate of the position operator, rather than in one of the momentum operator which is the one that commutes with the Hamiltonian. Clearly the object is large, but not large enough and translational symmetry is broken. We can imagine our pencil to be confined by an external trap that does the symmetry breaking. As we make the object more macroscopic, the trap potential can be smaller. The limit of vanishing trap and the thermodynamic limit do not commute and for a finite large system this results in broken translational symmetry.

In both examples we can express the ground state of the object as a superposition of many states that obey the symmetry of the Hamiltonian. As the object comes larger the energy difference between these states becomes smaller, typically proportional to the volume. In the thermodynamic limit they are hard to discriminate, and a “tower” of many more or less equivalent ground states appears that individually respect the global symmetry. Without external interaction, the system will evolve over all possible approximate ground states and settle in the real ground state after a long time, typically the inverse level spacing. The presence of a small energy barrier, though larger than the level spacing, will avoid this and create a superposition of states that is not an eigenstate of the Hamiltonian.

These arguments illustrate the possibility to break global symmetries. What about broken local symmetry? We have seen that local symmetry requires the presence of gauge fields to repair the loss of local symmetry. A real mathematical theorem, called *Elitzur's theorem* and demonstrated in 1975 hence only recently, states that *local symmetries cannot be spontaneously broken*. In particular, gauge invariance cannot be spontaneously broken, which is consistent with their interpretation as degrees of freedom. Apparently, local symmetries are much more robust against perturbations and are not easily broken.

### Example: Local and global translational symmetry

Consider the Hamiltonian of the ideal, non-relativistic gas consisting of $N$ identical particles, $H = \Sigma_n(p_n)^2/2m$. It has local translational symmetries expressed by $U_n = \exp(ip_n \cdot a_n)$ which moves particle $n$ around over a distance $a_n$. The ground state with energy $E=0$ has all particles in their lowest momentum state $k=0$ which is infinitely extended. We can confine particle $n$ by a harmonic trap $V = \frac{1}{2}m\omega_0^2 x_n^2$ which breaks the local symmetry. We will assume $N$ to be “large” and $\omega_0$ to be “small”. The ground state energy is now $E = \frac{3}{2}\hbar\omega_0$. We also have that
$\langle \Delta x_n^2 \rangle = 3\hbar/2m\omega_0$. As the trapping vanishes and the number of particles increases, we see that $E \to 0$ and $\langle \Delta x_n \rangle^2 \to 0$ without any confusion about order of limits: The local symmetry is not broken because for any large $N$ the symmetry is not destroyed for any small $\omega_0$.

Next consider a non-ideal gas with all $N$ particles trapped, described by the Hamiltonian

$$H = \sum_n \frac{p_n^2}{2m} + \sum_{(n,l)} V(\mathbf{x}_n - \mathbf{x}_l) + \sum_n \frac{1}{2} m \omega_0^2 \mathbf{x}_n^2$$

* Introduce total momentum $\mathbf{P} = \sum_n \mathbf{p}_n$, center of mass $\mathbf{X} = \sum_n \mathbf{x}_n/N$. The internal position vector of the particle is $\mathbf{y}_n = \mathbf{x}_n - \mathbf{X}$. Show that $[X^a, P^b] = i\hbar\delta_{abs}$. For $\omega_0 = 0$ global translational invariance guarantees total momentum to be conserved. We can set $\mathbf{p}_n = \mathbf{P}/N + \hbar\mathbf{k}_n$ with $\sum_n \mathbf{k}_n = \sum_n \mathbf{y}_n = 0$. Show that for large $N$ $\mathbf{y}_n$ and $\mathbf{k}_n$ are approximately canonical, i.e. $[y_n^a, k_l^b] = i\delta_{ab}\delta_{nl} + \mathcal{O}(1/N)$ and that the Hamiltonian can be written as

$$H = \frac{\mathbf{P}^2}{2M} + \frac{1}{2} M \omega_0^2 \mathbf{X}^2 + H_{\text{int}}(\mathbf{y}_n, \mathbf{k}_n)$$

with $M = Nm$ total mass and $H_{\text{int}}(\mathbf{y}_n, \mathbf{k}_n) = \sum_n \hbar^2 \mathbf{k}_n^2 / 2m + \sum_{nl} V(\mathbf{y}_n - \mathbf{y}_l)$.

* Show that the ground state energy and lowest excited states are given by $E \sim n\hbar\omega_0 + E_0(N)$ and that $\langle X^2 \rangle \sim n\hbar/M\omega_0$, where $E_0(N)$ is the ground state of $H_{\text{int}}$, independent on $\omega_0$ when sufficiently small, and expected to be proportional to $N$. For $\omega_0 < \mathcal{O}(1/N)$ a tower of states exists consisting of very many approximate ground states, each with approximate translational invariance on scales $d \ll \sqrt{\hbar/mN\omega_0}$.

* Argue that global translational symmetry is spontaneously broken: The limits $\omega_0 \to 0$ and $N \to \infty$ do not commute. A macroscopic object with very large $N$ will not be stable against a perturbation with very small $\omega_0$ and the true ground state will be localized in space

<figure><img src="image_17.png" /><figcaption>Figure 15. The tower of (ground) states for particles trapped in a harmonic potential and the Goldstone excitations. In the thermodynamic limit, the first are approximately translationally invariant, the second become gapless.</figcaption></figure>
and equal to a superposition of many approximate ground states that is not a momentum eigenstate.

* Goldstone's Theorem. We want to know the lowest excitations of $H_{\text{int}}$. To this end we imagine small perturbations around the equilibrium positions for which we can take a cubic 1D lattice with lattice spacing $a$, volume $L$ and periodic boundary conditions (the generalization to a cubic 3D sample is straightforward). Let us consider isotropic harmonic forces between the particles ($\omega \gg \omega_0$ !):

$$H_{\text{int}}(y_n, k_n) = \sum_n \frac{\hbar^2 k_n^2}{2m} + \frac{1}{2} m \omega^2 (y_n - y_{n-1})^2 \quad (161)$$

This Hamiltonian can be diagonalized exactly by first going to the Fourier space. For each coordinate is

$$y_n = \frac{1}{\sqrt{N}} \sum_q e^{iqna} y_q \quad ; \quad y_q = \frac{1}{\sqrt{N}} \sum_n e^{-iqna} y_n$$

$$k_n = \frac{1}{\sqrt{N}} \sum_q e^{iqna} k_q \quad ; \quad k_q = \frac{1}{\sqrt{N}} \sum_n e^{-iqna} k_n$$

where $\sum_q$ is a sum over discrete quasi-momentum in the first Brillouin zone of the lattice $q = j(2\pi/L)$ with $j = -N/2 + 1, \dots, N/2$, $N = L/a$. Show that $[y_q, k_{q'}] = i\delta_{q+q'}$ and $y_{-q} = y_q^\dagger$ etc. Show that the Hamiltonian takes the form

$$H_{\text{int}} = \frac{\hbar^2 k_{q=0}^2}{2m} + \sum_{q \neq 0} \left[ \frac{\hbar^2}{2m} k_q k_{-q} + \frac{1}{2} m \omega_q^2 y_q y_{-q} \right]$$

with $\omega_q = 2\omega|\sin(qa/2)|$. Since $k_{q=0} = P/\sqrt{N}$ the first term should be ignored since it was already split off above and generates a tower of total momentum states near $k=0$. Using a well-known Bogoliubov transformation, we define a new operator $b_q = (Ak_q + iBy_q)/\sqrt{2}$ with $A, B$ appropriately chosen to get rod of units ($A^2 = \hbar/2m\omega_q$, $B^2 = m\omega_q/2\hbar$). Demonstrate the bosonic commutator $[b_q^\dagger, b_{q'}] = \delta_{qq'}$ and show that this Hamiltonian can be further simplified to

$$H_{\text{int}} = \sum_{q \neq 0} \hbar \omega_q \left( b_q^\dagger b_q + \frac{1}{2} \right)$$

Gapless excitations $n\hbar\omega_q$ exist that we associate with phonons. The broken translational symmetry is accompanied by gapless (in particle physics "massless") excitations at low energies. This is a manifestation of Goldstone's theorem. The ground state energy $E_0$ of $H_{\text{int}}$ is proportional to $(L/a)\hbar\omega$ (to see this use $\sum_q \rightarrow (L/2\pi) \int_0^{\pi/a} dq$ as $L \rightarrow \infty$) and is thus
proportional to the total number of particles N as was argued earlier. The lowest excitations have $q \sim 1/L$ with energy $\hbar\omega_q \sim \hbar\omega a/L \sim \hbar\omega/N$. If we assume that $N \times \langle y_q y_q^\dagger \rangle < \langle X^2 \rangle$, i.e. the cumulative particle displacement induced by the phonons is smaller than the size of the ground state, we find that $\omega_0 < \omega/N^2$, so that $\hbar\omega/N \gg \hbar\omega_0$. Thus even the Goldstone modes have energies much larger than the approximate degeneracy of the ground state.

### Exercise VII.1: Classical magnetism

Consider N classical magnets with magnetic moment $M$ of equal size $\mu$ but with arbitrary direction. They interact with nearest neighbors with Hamiltonian,

$$H = -J \sum_{(i,j)} M_i \cdot M_j$$

For $J > 0$ energy is minimized by aligning spins which would create a macroscopic magnetization. However, by global rotational $SO(3)$ symmetry, this ground state is infinitely degenerated.

* At temperature $T$ the average magnetization is given by ($\beta = 1/kT$)

$$\langle M \rangle = \frac{1}{Z} \sum_{i} e^{-\beta E_{i}} M_{i}$$

Show that $\langle M \rangle = 0$ at $T = 0$ and at any temperature.

* As usual we add a small global perturbation $-\sum_i B \cdot M_i$. Show that at zero temperature $\langle M \rangle = N\mu\hat{B}$ i.e. in the direction of $B$ but independent on its value and $\langle M \rangle = 0$ for $kT \gg J\mu^2$. Argue that the limits $B \to 0$ and $N \to \infty$ do not commute at zero temperature: The $SO(3)$ symmetry is spontaneously broken at low enough temperatures and for large enough N an average magnetization $M = N\mu\hat{B}$ emerges.

### Example: Néel state in anti-ferromagnetism: the perfect SSB

Louis Néel, Nobel Prize 1970; Founding paper of SSB: P.W. Anderson, 1952 [7].

Consider the Heisenberg Hamiltonian

$$H = -J \sum_{(i,j)} S(i) \cdot S(j) \qquad (162)$$

this time formulated in terms of quantum spins on a cubic lattice, i.e. $S(i) = \frac{\hbar}{2}\sigma(i)$ on each site, spins on different sites commute, and the spin components of a spin at a given site follows the commutation algebra of the 3 Pauli matrices. The Hamiltonian, working on product states of spins on different sites has global $SU(2)$ symmetry.
• Show that

$$[\mathbf{S}(1) \cdot \mathbf{S}(2), \mathbf{S}(1) + \mathbf{S}(2)] = 0$$

and generalize this to the statement that total spin $\mathbf{S} = \sum_i \mathbf{S}(i)$ commutes with the Heisenberg Hamiltonian. Hence $[H, \mathbf{S}^2] = 0$.

If $J < 0$ a ferromagnetic state is not a favorable ground state since it would maximize the energy. Because the sum is done over nearest neighbors, classically, the energy is minimized by a configuration with opposite spins neighbor spins in the z-direction. This classical anti-ferromagnetic “Néel” state is still strongly degenerated because any axis would do. Quantum-mechanically however, this state is not even an eigenstate of the Hamiltonian. To see this, we define $S_±(i) = S_x(i) \pm S_y(i)$. This operator raises or lowers the quantum number $m = \pm \frac{1}{2}$ of $S_z(i)$. Thus,

$$H = -J \sum_{(i,j)} \left[ S_z(i)S_z(j) + \frac{1}{2}(S_+(i)S_-(j) + S_-(i)S_+(j)) \right]$$

We write $|s = \frac{1}{2}, m = \frac{1}{2}\rangle = |\uparrow\rangle$ and $|s = \frac{1}{2}, m = -\frac{1}{2}\rangle = |\downarrow\rangle$.

• Show that $S_+(i)|\downarrow\rangle = |\uparrow\rangle$ and $S_-(i)|\uparrow\rangle = |\downarrow\rangle$ and all others combinations equal to 0. Hence, for the product state of two neighboring spins,

$$[S_+(i)S_-(i+1) + S_-(i)S_+(i+1)] |\uparrow\rangle|\downarrow\rangle = |\downarrow\rangle|\uparrow\rangle$$

Conclude that the Néel state is not an eigenstate of $H$.

We can define two equivalent sublattices A and B of the cubic lattice that regroup nearest neighbors. Their total spin is denoted by $\mathbf{S}(A)$ and $\mathbf{S}(B)$.

<figure><img src="image_18.png" /><figcaption>Figure 16. The tower of states for the anti-ferromagnetic Heisenberg model.</figcaption></figure>
* Show that global SU(2) symmetry is generated by the 3 components of total spin $\mathbf{S} = \sum_n \mathbf{S}(n)$, with $[S_n, S_m] = i\hbar\epsilon_{nml}S_l$.

* Show that like total spin $\mathbf{S}$, both sublattice spins $\mathbf{S}(A), \mathbf{S}(B)$ obey the usual commutation relations $[S_n, S_m] = i\hbar\epsilon_{nml}S_l$. Show that this implies that all spin components always commute with total norm: $[\mathbf{S}, \mathbf{S}^2] = 0$. Conclude that $\mathbf{S}^2, \mathbf{S}(A)^2$ and $\mathbf{S}(B)^2$ mutually commute.

Like was done for the gas above we can define for each dimension of the cubic lattice with (here) $2N = L/a$ sites and periodic boundary conditions, with $q = j(2\pi/L)$ with $j = -N + 1, \dots, N$, or $q = (-1 + 1/N, \dots, 1)\pi/a$,

$$ \mathbf{S}(2n) = \frac{1}{\sqrt{N}} \sum_q e^{2niqa} \mathbf{S}_q \quad ; \quad \mathbf{S}(2n+1) = \frac{1}{\sqrt{N}} \sum_q e^{(2n+1)iqa} \mathbf{S}_q $$

i.e. we associate even and odd lattice sides with A and B, both lattices with sites $n = 1, \dots, N$, and assume for simplicity that $N$ is even so that both sublattices have integer spin and can have total spin 0. Note that $\sum_n \exp(2niqa) = N\delta(q = \mod(\pi/a))$ is only non-zero for $q = 0$ and $q = \pi/a$ in the allowed first Brillouin zone, we find $\sum_n \mathbf{S}(2n) = \mathbf{S}(A) = \sqrt{N}(\mathbf{S}_{q=0} + \mathbf{S}_{q=\pi/2})$ and $\sum_n \mathbf{S}(2n+1) = \mathbf{S}(B) = \sqrt{N}(\mathbf{S}_{q=0} - \mathbf{S}_{q=\pi/2})$

* Show that the Hamiltonian can be written as

$$ H = -\frac{J}{N}\mathbf{S}(A) \cdot \mathbf{S}(B) + \sum_{\mathbf{q} \neq 0, \pi/a} H(\mathbf{q}) $$

to be identified with a “collective” Hamiltonian $H_{\text{coll}}$ plus internal excitations. (Hint: Write $H = -J/2 \sum_n \mathbf{S}(2n) \cdot \mathbf{S}(2n+1) - J/2 \sum_n \mathbf{S}(2n) \cdot \mathbf{S}(2n-1)$ and use that $\sum_n \exp[2ni(q+q')a] = N\delta(q+q'=\mod(\pi/a))$. The wave number $q+q'$ can take the values $0, \pm\pi/a, 2\pi/a$ for the allowed $q$ and $q'$ in the discrete first Brillouin zone).

* Show that $H_{\text{coll}}$ has eigenvalues

$$ E = -\frac{J\hbar^2}{2N} (S(S+1) - S_A(S_A+1) - S_B(S_B+1)) $$

with $S = 0, 1, \dots, 2N$, and $S_{A,B} = 0, 1, \dots, N$, i.e. all have integer spin. The unique ground state of $H_{\text{coll}}$ occurs for $S = 0$ and $S_A = S_B = N$ which may look like a Néel state but is in fact a perfectly rotationally symmetric singlet state and thus necessarily a superposition of all possible Néel states having all possible axes. This is a highly entangled state, arguably extremely fragile against small perturbations from the environment and this extreme Schrödinger-cat state is not likely to survive the macroscopic limit.
* Show that the lowest excitations of $H_{\text{coll}}$ are of order $|J|\hbar^2/N$ occurs at fixed $S_A = S_B = N$ and $S = 0, 1, \dots$. These energies are very small when $N$ is large. This is again Anderson's tower of states and corresponds to varying orientations of the spins $S(A)$ and $S(B)$ at constant norm. Any change in the spin amplitudes $S^2(A)$ or $S^2(B)$ of the sublattices requires an energy $|J|\hbar^2$ that is much larger.

We anticipate the spontaneous symmetry breaking to a Néel state with a well-defined axis $z$. To this end we investigate the influence of a perturbation that favors such state and write

$$H_{\text{coll}} = -\frac{J}{N} S(A) \cdot S(B) - \mu [S_z(A) - S_z(B)]$$

where we can assume that $S(A)^2 = S^2(B) = N^2$ take a constant value and only change in quantized orientation. Finding the ground state of two anti-ferromagnetic super-spins is a standard, though technical exercise in quantum mechanics (see [4] for more details). The result is

$$\Delta E_0(\mu) = \frac{3\hbar}{4} \sqrt{2\mu\hbar|J|} - \mu\hbar N$$

We see that for any $\mu > 0$ a the energy of a macroscopic lattice is lowered by $\Delta E_0 = -\mu\hbar N$ corresponding to two sublattices with opposite spin $S_z(A) = -S_z(B) = N\hbar/2$ associated with the classical Néel state. This state is a superposition of states in Anderson's tower. The symmetries $S_x$ and $S_y$ in $SU(2)$ (any global rotation of all spins over $x$ and $y$ axis) are broken and the global symmetry reduced to rotation symmetry around the $z$-axis with generator $S_z$, isomorphic to $U(1)$.

## A. Order parameter and cosets of broken symmetry states

In complex many-body systems we expect in general no degeneracies although, for a finite volume, eigenstates may have energies differences that are of order $1/V$ and thus close. The conventional picture is that a ground state is an eigenstate $\Psi_0$, stable in time, that corresponds to the unique minimum of the energy (or maybe of some more general free energy). If the energy is symmetric under a set of transformations $U(g)$ belonging to some group $g \in G$ the uniqueness guarantees that $U(g)\Psi_0 = \exp(i\phi_g)\Psi_0$, i.e. the state itself is symmetric under all transformations $U(g)$.

In the case of spontaneously broken symmetry a stable state $\Psi$ emerges that does not have the symmetry expressed by the group $G$ and is no longer an energy eigenstate. As a result, $U(g)|\Psi\rangle =$
$|\Psi'\rangle \neq \exp(i\phi_g)|\Psi'\rangle$ so that $|\langle\Psi'|\Psi\rangle| < 1$ by the Cauchy inequality. Nevertheless, there may be a residual symmetry left: Some subset $H$ of transformations in $G$ may still exist such that $h|\Psi\rangle = \exp(i\phi_h)|\Psi\rangle$. This set $H \subset G$ can be shown to have group structure although if $H = 1$ can happen so that no real symmetry remains. All states $g \cdot H|\Psi\rangle$ are equivalent to $g|\Psi\rangle$ which is one coset in the quotient space $G/H$. Different cosets are associated with different choices for the broken symmetry. For the superfluid we find $H = 1$ and $G/H = G$ and all states in $G$ are different. For the Néel state we found that $H = U(1)$ and $SU(2)/U(1) = S^2$ and different broken symmetries correspond to different orientations of the quantification axis on the unit sphere $S^2$ in $\mathbb{R}^3$. This is not a group. We recall that only if the group $H$ is normal ($gH = Hg$ for all $g \in G$), than the quotient set $G/H$ is a group.

With this picture we can investigate the orthogonality between different states in the thermodynamic limit. A many-particle quantum state is typically written as a direct product $|\Psi\rangle = \otimes_i|\Psi_i\rangle$ on $N$ particles or spins distributed in space. We can apply the following hand-waving argument for the scalar product of two different macroscopic wave functions,

$$ \langle\Psi'|\Psi\rangle = \langle\otimes_i\Psi'_i|\otimes_j\Psi_j\rangle = \prod_{i=1}^{N} \langle\Psi'_i|\Psi_i\rangle \approx \langle\Psi'_i|\Psi_i\rangle^N $$

The last inequality assumes approximate translational invariance, not broken by a global symmetry operation. For states $|\Psi'_i\rangle = h|\Psi_i\rangle = \exp(i\rho_h)|\Psi_i\rangle$ with $h \in H \subset G$, the global symmetry guarantees the scalar product of $|\Psi'\rangle$ and $|\Psi\rangle$ to survive the thermodynamic limit and not to converge to zero. For any other set of wave functions related by operations $g \notin H$ the scalar product is no longer symmetry-protected. Since $|\langle\Psi'|\Psi\rangle| < 1$ it follows that $\langle\Psi'|\Psi\rangle \to 0$ as $N \to \infty$.

Given a spontaneously broken symmetry from group $G$ to its subgroup $H$, the group $H$ is the residual symmetry group that relates equivalently broken symmetry states and is one coset element in $G/H$. All other cosets in the quotient set $G/H$ are non-equivalent broken states and orthogonal in the thermodynamic limit.

With these elements we can now characterize more quantitatively the occurrence of a broken symmetry. The definition is clearly inspired by the Landau theory of phase transitions in statistical physics.

**Definition of broken symmetry.** Let $U = \exp\{(i\alpha_a Q_a)\}$ be a group $G$ of symmetry operations that commutes with the Hamiltonian. A state $|\Psi\rangle$ is said to have broken $G$-symmetry if there exists an
operator $\Phi(x)$ such that

$$ \langle \Psi | [Q_a, \Phi(x)] | \Psi \rangle \neq 0 \qquad (163) $$

for at least one generator $Q_a$. The operator $O(x) = [Q_a, \Phi(x)]$ is called the order parameter, and vanishes in the absence of G-broken symmetry. The operator $\Phi$ is called the interpolating field.

We can check that the criterion is the infinitesimal version of,

$$ \langle \Psi | U^\dagger \Phi(x) | U \Psi \rangle \neq \langle \Psi | \Phi(x) | \Psi \rangle \qquad (164) $$

From this we see that if $|\Psi\rangle$ respects the symmetry, $U|\Psi\rangle = e^{i\phi}|\Psi\rangle$, the expectation value of the order parameter vanishes: $\langle\Psi|O(x)|\Psi\rangle = 0$. The choice of the order parameter is not unique and in most cases it can be chosen such that the broken symmetry state is an eigenfunction of the order parameter.

### Exercise VII.2: Order parameter of Néel state

We have shown above that the Heisenberg antiferromagnet spontaneously breaks the collective spin-rotations along $x-$ and $y-$axis generated by the two unitary operators $S_{x,y} = \sum_n S_{x,y}(n)$. Define the staggered magnetization $M(n) = S(n \in A)$ and $-M(n) = S(n \in B)$. Show that $[S_x, M_y(n)] = iM_z(n)$ and $[S_y, M_x(n)] = iM_z(n)$ so that we can propose the local order parameter $O(n) = M_z(n)$. Show that the Néel state is an eigenfunction of this order parameter.

### Example: Interacting Bose-Einstein Condensate (superfluid)

The Lagrangian density of a Schrödinger field in an external potential $V$ and with pointlike repulsive self-interaction is given by

$$ \mathcal{L}(\Psi, \Psi^*, \partial_\mu \Psi, \partial_\mu \Psi^*) = \frac{i\hbar}{2} \Psi^*(\partial_t \Psi) - \frac{i\hbar}{2} (\partial_t \Psi^*) \Psi - \frac{\hbar^2}{2m} (\partial_i \Psi^*)(\partial_i \Psi) - V(x) \Psi^* \Psi - \frac{\lambda}{2} (\Psi^* \Psi)^2, \quad (165) $$

with $\lambda > 0$.

* Show that the equation of motion is the non-linear Schrödinger equation

$$ i\hbar\partial_t\Psi(x,t) = \left( -\frac{\hbar^2\nabla^2}{2m} + V(x) + \lambda|\Psi|^2 \right) \Psi(x,t) \qquad (166) $$
Both Lagrangian and equation of motion have global $U(1)$ symmetry and the conserved charge $Q$ derived earlier in Eq. (103) without self-interaction remains valid, that we shall call $N$ here because it becomes the particle-number operator later,

$$N = \frac{1}{2} \int d^3x [\Psi^\dagger(x)\Psi(x) + \Psi(x)\Psi^\dagger(x)]$$

* Show that the conserved energy (104) associated with invariance under time-translations is given by the Hamiltonian,

$$H = \int d^3x \left[ -\frac{\hbar^2}{4m}(\partial_i^2 \Psi^\dagger)\Psi - \frac{\hbar^2}{4m}\Psi^\dagger(\partial_i^2 \Psi) + V(x)\Psi^\dagger\Psi + \frac{\lambda}{2}\Psi^\dagger\Psi^\dagger\Psi\Psi \right]$$

The second quantification of the non-relativistic field is proposed by making $\Psi$ and $\Psi^\dagger$ bosonic conjugate operators in Fock space with $[\Psi(x), \Psi^\dagger(x')] = \delta(x - x')$. To find an order parameter associated with broken $U(1)$ symmetry we can propose as interpolating field $\Phi(x) = \Psi(x)$.

* Verify that $[N, \Psi(x)] = -\Psi(x)$. The order parameter for broken $U(1)$ symmetry is the annihilation operator for a particle at position $x$. One needs a state with high quantum coherence to avoid that the expectation value of $\Psi(x)$ vanishes. For a statistical mixture of Fock states it vanishes: $\langle n_i | \Psi(x) | n_i \rangle = 0$.

* The two independent fields in the Lagrangian (165) are $\Psi_1 = \Psi$ and $\Psi_2 = \Psi^*$. Show that the commutators (107) for second quantization are not satisfied but that nevertheless

<figure><img src="image_19.png" /><figcaption>Figure 17. Interference revealed by phase contrast between two Bose Einstein Condensates with varying distance. On the left a single condensate for comparison. On the right only one of the interfering condensates is shown, the other has been eliminated from the image. Taken from Andrews et al [12]. See also Ref. [13]</figcaption></figure>
the $U(1)$ charge becomes the generator for local $U(1)$ transformations, as expressed by
$$-i[N, \Psi_a] = \Delta\Psi_a.$$

In second quantization, broken $U(1)$ symmetry emerges if the physical state has non-zero expectation value. We can use the complete normalized set of wave functions $\psi_i(x)$ of kinetic energy plus trap (due for instance to a cold atom trap) with the energy $\epsilon_i$ and write $\Psi(x) = \sum_i \psi_i(x)a_i$ and the creation operator $\Psi^\dagger(x) = \sum_i \psi_i^*(x)a_i^\dagger$ for a particle at position $x$. It can be checked that as usual $[a_i, a_j^\dagger] = \delta_{ij}$, and $[a_i, a_j] = [a_i^\dagger, a_j^\dagger] = 0$. Fock states (in state $\psi_i$) are eigenstates of the number operator: $a_i^\dagger a_i|n_i\rangle = n_i|n_i\rangle$. One derives from the commutation rules that $a_i^\dagger|n_i\rangle = \sqrt{n_i+1}|n_i+1\rangle$ and $a_i|n_i\rangle = \sqrt{n_i}|n_i-1\rangle$. We easily find $N = \sum_i a_i^\dagger a_i$.

If we separate the ground state $\Psi(x) = a_0\psi_0(x) + \sum_{i \neq 0} \psi_i(x)a_i$ into a global part plus excitations that we shall ignore initially (because of their relatively high energies), we infer that broken symmetry implies that the (unperturbed) wave function in Fock space is equal to an eigenfunction of the annihilation operator $a_0$ for a particle in the ground state.

* Show that $a_0|\alpha\rangle = \alpha|\alpha\rangle$ for any $\alpha \in \mathbb{C}$, with the coherent state

$$|\alpha\rangle = e^{-|\alpha|^2/2} \sum_{n=0}^{\infty} \frac{\alpha^n}{\sqrt{n!}} |n\rangle \quad (167)$$

Coherent states have many elegant properties that can be verified. First, they are not energy eigenstates. Secondly, they are normalized but only “approximately orthogonal” : $\langle\alpha|\beta\rangle = \exp(-|\alpha - \beta|^2/2) \exp(i\text{Im}\,\alpha^*\beta)$. Finally, they constitute an (over-) complete set,

$$\frac{1}{\pi} \int_{\mathbb{C}} d\alpha |\alpha\rangle\langle\alpha| = 1$$

and

$$e^{i\theta N_0}|\alpha\rangle = |e^{i\theta}\alpha\rangle$$

which confirms the role of $N_0 = a_0^\dagger a_0$ as the generator of $U(1)$ applied to coherent states.

* Show that $\langle\alpha|\Psi(x)|\alpha\rangle = \alpha\psi_0(x)$ if excitations are neglected, and that the number of particle $N_0$ in the ground state fluctuates as $(\Delta N_0)^2 = N_0$.

The quantum expectation value of the wave function has broken symmetry $U(1)$ symmetry with well-defined phase yet with indefinite number of particles. This notion prompts us to go the grand-canonical energy $H - \mu N$,

$$H - \mu N = \int d^3x \left[ -\frac{\hbar^2}{4m} (\partial_i^2 \Psi^\dagger) \Psi - \frac{\hbar^2}{4m} \Psi^\dagger (\partial_i^2 \Psi) + V(x) \Psi^\dagger \Psi - \mu \Psi^\dagger \Psi + \frac{\lambda}{2} \Psi^\dagger \Psi^\dagger \Psi \Psi \right]$$
Except for the trapping potential, this free energy is a non-relativistic version of Higgs Lagrangian (16).

* Show that again by restricting to the first level $i = 0$ of the trap, we find

$$ \langle \alpha | H_0 - \mu N_0 | \alpha \rangle = (\varepsilon_0 - \mu) |\alpha|^2 + \frac{\lambda}{2V_0} |\alpha|^4 \quad (168) $$

with $1/V_0 = \int d^3x |\psi_0|^4$, and for $\mu > \varepsilon_0$ it has a local minimum at $|\alpha|^2 = V_0(\mu - \varepsilon_0)/\lambda$. Hint: use $\langle \alpha | a_0^\dagger | \alpha \rangle = \langle \alpha | a_0 | \alpha \rangle^* = \alpha^*$. The minimum only fixes the amplitude of $\alpha$ and not its phase so that $U(1)$ symmetry is still preserved and the ground state is highly degenerated.

* If the energy of $n$ particles in the ground state is $E_0(n)$ show that $E_0(n) = \varepsilon_0 n + (\lambda/2V_0)n(n-1)$. From this show that the expectation of the order parameter evolves in time as

$$ \langle \alpha | e^{iHt/\hbar} \Psi(x) e^{-iHt/\hbar} | \alpha \rangle = \alpha \psi_0(x) e^{-i\varepsilon_0 t/\hbar} e^{i|\alpha|^2(\exp(-i\lambda t/V_0\hbar) - 1)} \rightarrow \alpha \psi_0(x) e^{-i\mu t/\hbar} \quad (169) $$

with the chemical potential $\mu = E_0(N_0 + 1) - E_0(N_0) = \varepsilon_0 + \rho\lambda$. In the approximation we take the thermodynamic limit where the average number of particles in the ground state becomes large at constant density $\rho$. In this limit $\varepsilon_0 \sim 1/V_0^{2/3} \rightarrow 0$. As the number of atoms becomes macroscopic the chemical potential is much smaller than the energies $E(N_0)$ of the Fock states. *Time-evolution is thus extremely slow and only affects the phase.*

In the thermodynamic limit fluctuations $(\delta N_0)/\langle N_0 \rangle \sim 1/\sqrt{\langle N_0 \rangle}$ are small and can be neglected. We have $N_0 = |\alpha|^2$ and $|\psi_0|^2 = 1/V$. The total energy is $E_0 = \varepsilon_0 N_0 + (\lambda/2)\langle N_0 \rangle^2/V$. The chemical potential is $\mu = \partial E_0/\partial N_0 = \varepsilon_0 + \lambda(N_0/V) = \varepsilon_0 + \lambda\rho$. The relation $N_0 = V(\mu - \varepsilon_0)/\lambda$ implies that the field has settled in a highly degenerated minimum of the grand-canonical energy. Choosing the coherent state with phase Arg($\alpha$) is a manifestation of spontaneously broken $U(1)$ symmetry.

* Given the level spacing $\mu$ between states of different energy $E_0(n)$ around the approximate ground state energy $E_0$, argue that Anderson's scenario with tower of states applies.

* Consider a perturbation $\delta H = \varepsilon_p \int_{[\theta_0 \pm \delta\theta]} d\theta |\beta\rangle\langle\beta|$, with $\beta = \sqrt{N_0} \exp(i\theta)$ that breaks phase invariance. Show that the energies of nearby coherent states are perturbed by the amount $\delta E \approx (\varepsilon_p/N_0)\text{Err}(N_0\delta\theta)$. Show that the limits $\delta\theta \rightarrow 0$ and $N_0 \rightarrow \infty$ do not commute. For any phase difference $\delta\theta \neq 0$ two coherent states are orthogonal in the thermodynamic limit $\langle N_0 \rangle \rightarrow \infty$. Phase-invariance is broken, particle number is only approximately conserved and all states related by $U(1)$ transformations are inequivalent so that no real symmetry subgroup persists.
* Show that the non-relativistic energy operator $\frac{i\hbar}{2}\Psi^{\dagger}(\partial_t\Psi) - \frac{i\hbar}{2}(\partial_t\Psi^{\dagger})\Psi$ is not affected by amplitude changes of the wave function $\Psi = |\Psi|\exp(i\theta)$ - equal to $\phi_H$ in Eq. (18) - but only by phase fluctuations, contrary to the relativistic energy operator $\partial_t\Psi^{\dagger}\partial_t\Psi$ that produces both. *The massive Higgs boson emerges only in a relativistic model, and does not survive in the non-relativistic approximation.* The phase fluctuations do survive and correspond to the massless Goldstone boson.

We can finally study small fluctuations by linearizing Eq. (166), and set $\Psi(x, t) = \alpha\psi_0\exp(-i\mu t/\hbar) + \delta\Psi$. We can look for eigenfunctions and eigenenergies but it is easier to look for traveling waves. By assuming that approximate translational invariance, the fluctuations must be plane waves with wave number $q$. We write

$$ \Psi(x, t) = \alpha\psi_0 e^{-i\mu t/\hbar} \left[ 1 + \sum_{q \neq 0} \left\{ \alpha_q e^{-i\phi_q/2} e^{i\mathbf{q}\cdot\mathbf{x} - i\omega t} + \beta_q e^{i\phi_q/2} e^{-i\mathbf{q}\cdot\mathbf{x} + i\omega t} \right\} \right] $$

with real-valued coefficients and a phase factor.

* Show that the linearization of the nonlinear wave equation produces the equation

$$ \left( \begin{array}{cc} \hbar\omega - \mu - E_q & -\mu \\ -\mu & -\hbar\omega - \mu - E_q \end{array} \right) \left( \begin{array}{c} \alpha_q \\ \beta_q \end{array} \right) = 0 $$

with $E_q = q^2/2m$. This leads to the dispersion law for the quasi-particles

$$ \hbar\omega_q = \pm\sqrt{E_q(E_q + 2\mu)} $$

For $q < 1/\xi$ this is phonon-like, for $q > 1/\xi$ this is particle like. Here $\xi = \sqrt{\hbar/m\mu}$ is the healing length of the condensate. Their 'gapless' energies are again characteristic of the Goldstone modes that appear after a continuous symmetry is broken.

## Example: the Josephson effect (Brian Josephson, Nobel Prize 1973)

The prediction and observation of the Josephson effect has been of historic importance because it settled the debate about the presence or not of broken global phase symmetry in superconductors. The theoretical prediction by Josephson came after having been inspired about Anderson's ideas of broken symmetry in superconductors to explain the Meissner effect, and was awarded the Nobel prize in 1973: the tunneling of Cooper pairs between two nearby superconductors creates a "super" current that needs no voltage drop to exist but only a phase difference between Cooper condensates.
The following toy model is due to Richard Feynman and demonstrates that the Josephson effect must be more generally valid in the presence of symmetry breaking.

* Consider two superfluids to which we assign quantized fields $\Psi_1$ and $\Psi_2$. The Josephson Hamiltonian is given by

$$H = \varepsilon_0 \Psi_1^\dagger \Psi_1 + \varepsilon_0 \Psi_2^\dagger \Psi_2 + \varepsilon_J (\Psi_1^\dagger \Psi_2 + \Psi_2^\dagger \Psi_1)$$

with $[\Psi_i, \Psi_j^\dagger] = \delta_{ij}$ the usual commutation rule in second quantification. The interaction term annihilates a boson in one condensate and creates one in the other, without specification of details (which is here the proximity effect due to overlapping wave functions in real space that provides a value for the Josephson energy $\varepsilon_J$). Show that

$$[H, \Psi_1^\dagger \Psi_1] = -[H, \Psi_2^\dagger \Psi_2] = \varepsilon_J (\Psi_2^\dagger \Psi_1 - \Psi_1^\dagger \Psi_2)$$

* Show that particle number is conserved and that

$$\left\langle \frac{d}{dt} \Psi_1^\dagger \Psi_1 \right\rangle = - \left\langle \frac{d}{dt} \Psi_2^\dagger \Psi_2 \right\rangle = \frac{\varepsilon_J}{\hbar} \sqrt{N_1 N_2} \sin(\phi_0 - \phi_1)$$

where $\phi_i$ are the global phases for both condensates. This reveals a super current driven by the global phase difference induced by broken $U(1)$ symmetry.

### Exercise VII.3: Order parameter of Heisenberg ferromagnet: the great exception

Let us this time put $J > 0$ in the Heisenberg Hamiltonian (162) with $SU(2)$ global symmetry. Classically, the state with lowest energy is any configuration with all spins aligned, but all directions are possibly and quantum-mechanically the ground state could be a superposition of all states with arbitrary directions to restore symmetry. As in the case of antiferromagnetism with $J < 0$, both symmetries $S_x$ and $S_y$ are spontaneously broken and act as each other's interpolating field. We can here propose the order parameter $S_z$. The special feature here is that this ferromagnetic order parameter is also a generator of the symmetry group and commutes with the Hamiltonian.

* We can follow the treatment that we have used for the anti-ferromagnetic case and decompose the spin into $N$ Fourier components,

$$S(n) = \frac{1}{\sqrt{N}} \sum_q e^{niqa} S_q$$
We can see that total spin is $\mathbf{S} = \sqrt{N}S_{q=0}$. Show that the Hamiltonian takes the form

$$H = -J \sum_{q} e^{-iqa} \mathbf{S}_{q} \cdot \mathbf{S}_{-q} = -\frac{J}{N} S^{2} + H_{\text{magnon}}(q \neq 0)$$

The total spin $\mathbf{S}$ has z-components $S_z = (-N/2, -N/2+1, \dots, N/2)\hbar$. Since it obeys a Lie algebra $S^2$ has the value $\lambda = s(s+1)\hbar^2$ with $s = 0, 1, \dots, N/2$. The states with lowest energy $E_0 = -J\hbar^2 N/4$ happens for $S^2 = \frac{N}{2}(\frac{N}{2}+1)\hbar^2 \approx N^2\hbar^2/4$ and is $N+1$ times degenerated and with first excitations with finite energy gap $J\hbar^2$. Since $N$ is large the degenerated ground state corresponds to an almost free classical orientation of the total spin. The ground states obey the $SU(2)$ symmetry. Most of them are highly entangled Schrödinger Cat states. Argue that no tower of states exists and that the above scenario of spontaneous symmetry breaking does not apply.

* By adding a Zeeman interaction $H_Z = -\mu S_z$ we break the symmetry and the privileged ground state becomes $|j = \frac{N}{2}, m = \frac{N}{2}\rangle = |\uparrow\uparrow\cdots\rangle$, whose energy is reduced maximally to $E_0 = -J\hbar^2 N/4 - \mu\hbar N/2$.

* We can reduce the symmetry and consider the solvable Ising model where the spin is forced to be aligned along $z$: $H = -J \sum_{(i,j)} S_z(i)S_z(j)$. The global symmetry is now reduced from continuous group $SU(2)$ to the discrete group $\mathbb{Z}_2$ and local fluctuations have become “gapped” by the energy $J$. Argue that the ground states at energy $E_0 = -N\hbar^2 J/4$ are given by (with $|\alpha|^2 + |\beta|^2 = 1$),

$$|\Psi_0\rangle = \alpha|\uparrow\uparrow\cdots\rangle + \beta|\downarrow\downarrow\cdots\rangle$$

most of them being highly entangled and the degeneracy exhibiting the $\mathbb{Z}_2$ symmetry. For the same reasons as above the $\mathbb{Z}_2$ symmetry is expected to be broken in favor of a ferromagnetic state that is still an eigenstate of the Hamiltonian with finite magnetization along the $z$-axis.

* **Transverse 1D Ising model.** We can add a transverse magnetic field along the $x$-axis which introduces an energy $-\mu_x S_x$ with $\mu_x = gB > 0$. Show that for $\mu_x \gg J\hbar$ the ground state is $|ggg\cdots\rangle$ with the single spin state $|g\rangle = (|\uparrow\rangle + |\downarrow\rangle)/\sqrt{2}$ and $M_x = -\mu N$ proportional and opposite to magnetic field, and with $M_z = 0$. An exact calculation reveals the existence of a (second-order) quantum phase transition from ferromagnetic to paramagnetic at $J = \mu_x$ with $M_x \neq 0$: $M_z = 0$ for $J < \mu_x$ and $M_x = 0$ and $M_z \neq 0$ for $J > \mu_x$.
## Example: Time crystals I

Crystalline structure of matter in space is an ubiquitous feature at low enough temperatures and deeply rooted in the spontaneous symmetry breaking of translational symmetry. Does a similar feature exist associated with time translations? After all, time and space are strongly connected via Lorentz symmetry, so why could one be broken and not the other? This would be a *time crystal*, originally defined as a ground state that has periodic, eternal motion in time, the equivalent of long-range order in crystals. It will not use energy, not perform work, neither produce entropy. The question was posed first in 2011 by Frank Wilczek, Nobel Prize laureate for the discovery of asymptotic freedom of quarks in SU(3) gauge theory. A lively debate was initiated whether this such *perpetuum mobile* could really exist without violating first principles such as the Second Law. Comments exist in literature that Wilczek's proposition was actually not the ground state and thus not stable.

The major complication with this argument is that time translations are generated by the energy operator. For time translation to be broken for the ground state $|\Psi_0\rangle$, we need to find an interpolating field $\Phi$ such that $\langle\Psi_0|[H, \Phi]|\Psi_0\rangle \neq 0$. Thus we need $\partial_t\Phi \neq 0$. However, the thermal average of any operator also involves the Boltzmann factor $\exp(-\beta H)$ and cannot depend on time:

$$ \langle\Phi(t)\rangle = \mathrm{Tr} \left[ e^{-\beta H} e^{iHt} \Phi(0) e^{-iHt} \right] = \mathrm{Tr} \left[ e^{-\beta H} \Phi \right] $$

This suggests that time crystals cannot exist in thermal equilibrium. Actually this conclusion can be reconsidered with another definition that we will come back to later, yet it proves that spontaneously broken time-invariance does not come for free. We have to explore media that do not thermalize (due to what is called *many-body localization* in phase space). In spatial crystals long-range spatial order is described by a correlation function that does not decay to zero at large distances. We will describe long-range time correlations below.

## B. Goldstone's Theorem

In the previous section we have noticed that the broken, continuous symmetry leads to new excitations that are massless (or gapless). This is a manifestation of undoubtedly the most fundamental theorem in the theory of SSB, that we shall new prove.
## Goldstone's Theorem

If a global, continuous symmetry is spontaneously broken in the absence of long-ranged interactions, and leaving some (discrete) translational symmetry intact, then there exists a mode in the spectrum whose energy vanishes as its wave number approaches zero.

### Demonstration (adapted from Ref. [4])

The demonstration of this theorem is subtle since in the thermodynamic limit a tower of states is expected at $q = 0$ that we want to avoid because the theorem focusses on the excitations at small $q$. To cope with this, we take a quantization volume $\Omega$ for the $q$-vectors and consider $Q(V) = \int_V d^3x j^0$ in terms of the Noether charge density $j^0$ in Eq. (102), with $V \ll \Omega$ in order not to project everything right away on $q = 0$. Because the tower of states is located at $q = 0$ and thus has weight throughout the volume $\Omega$, the procedure $V \to \infty$ after $\Omega \to \infty$ excludes de facto the mode $q = 0$. More precisely, if the interpolating field is sufficiently local (thus excluding long-range interactions), we expect that $[Q(V), \Phi(y)] = 0$ for $y \notin V$ so that its expectation value vanishes for a wave function at $q = 0$ normalized on the much larger volume $\Omega$. We shall illustrate this below explicitly for the superfluid. After second quantization, our Hilbert space is the product space of all Fock spaces of the states with wave number $q_i$ and we use the complete set $|\{n_q\}\rangle = |n_0, n_{q_1}, \dots\rangle$ of Fock states with occupation $n_q$ in the state with wave number $q$. We insert this into the definition of the order parameter in Eq. (163),

$$ \begin{aligned} \langle\Psi|[Q(V), \Phi]|\Psi\rangle &= \sum_{\{n_q'\}} [\langle\Psi|Q(V)|\{n_q\}\rangle\langle\{n_q\}|\Phi|\Psi\rangle - \text{c.c.}] \\ &= \int_V d^3x \sum_{\{n_q'\}} [\langle\Psi|j^0(x,t)|\{n_q\}\rangle\langle\{n_q\}|\Phi|\Psi\rangle - \text{c.c.}] \end{aligned} $$

The accent in $\{n_q'\}$ means any state $|n_0, 0, \dots\rangle$ in the tower of states is excluded as $\Omega \to \infty$ so that the sum only contains excitations. In the thermodynamic limit $\Omega \to \infty, n_0 \to \infty$ the ground state $|\Psi\rangle$ becomes an eigenstate at energy $E(0)$ and $q = 0$. Since the expression above is not to vanish in the presence of broken symmetry, we expect that at least a single excitation $|n_0, 0, \dots, n_q = 1, 0, \dots\rangle$ must be created by either the operator $\Phi$ or $j^0$. A single excitation $|n_0, 0, \dots, n_q, 0, \dots\rangle$ is created by the operator $\Phi$. We write,

$$ j^0(x, t) = e^{-iHt - i\mathbf{P} \cdot \mathbf{x}} j^0(0, 0) e^{iHt + i\mathbf{P} \cdot \mathbf{x}} $$
and obtain,

$$
\begin{align*}
\langle \Psi | [Q(V), \Phi] | \Psi \rangle &= \int_V d^3 x \sum_{\{n_q\}'} e^{i[E(q)-E(0)]t + i q \cdot x} [\langle \Psi | j^0(0,0) | \{n_q\} \rangle \langle \{n_q\} | \Phi | \Psi \rangle - c.c.] \\
&= \sum_{\{n_q\}'} \delta_V(q) [e^{i(E_n(q)-E(0))t} \langle \Psi | j^0(0,0) | \{n_q\} \rangle \langle \{n_q\} | \Phi | \Psi \rangle - c.c.]
\end{align*}
$$

where $\delta_V(\mathbf{q})$ is a function strongly peaked at $\mathbf{q} = 0$ because we have kept $V$ finite but large. This expression must be non-zero for $\Psi$ to have broken symmetry so that there must be at least (and typically) one state excitation with $\mathbf{q} \in V^{-1}$ that is produced by the joint action of $j^0(0,0)$ and $\Phi$ on $|\Psi\rangle$. This gives a recipe to identify and classify Goldstone modes but we will not go into this technical subject here. Long-rang interactions like Coulomb potentials generate singularities at $\mathbf{q} \to 0$ and must therefore be excluded for these arguments to hold. Note that by construction the ground state is orthogonal to any excitation, i.e. $\langle \Psi | n_0, 0, \cdot n_q = 1, 0, \cdots \rangle = 0$, so that the presence of both operators is essential. The interpolating function $\Phi$ may still depend on $x$, but because $\mathbf{q} \approx 0$ this dependence disappears as $V \to \infty$.

According to Noether's theorem, the charge $Q$ is conserved. If the interpolating field $\Phi$ has no explicit time-independence either (it may still not commute with the Hamiltonian since the time-dependence $\exp(-iHt)$ has already been acknowledged) then $d/dt \langle \Psi | [Q, \Phi] | \Psi \rangle = 0$. Thus,

$$
\sum_{\{n_q\}'} \delta_V(\mathbf{q}) i [E(\mathbf{q}) - E(0)] \left[ e^{i[E_n(\mathbf{q}) - E(0)]t} \langle \Psi | j^0(0,0) | \{n_q\} \rangle \langle \{n_q\} | \Phi | \Psi \rangle - c.c. \right] = 0
$$

This means that the mode or modes $|\{n_q\}'\rangle$ that have been identified earlier must be gapless: $E_n(\mathbf{q}) - E_0 \to 0$ as $\mathbf{q} \to 0$. This proves Goldstone's theorem. Note that the theorem does not exclude the existence of gapped modes, but guarantees that they are not excited by e.g. $\Phi$ from the ground state.

Example: the superfluid

We shall illustrate these arguments more explicitly for the Bose-Einstein Condensate discussed earlier. The broken symmetry is $U(1)$ with generator $Q(V) = \int_V d^3x \Psi^\dagger(x)\Psi(x)$. The interpolating field is $\Phi(y) = \Psi(y) = \sum_q a_q \exp(iq \cdot y)/\sqrt{\Omega}$ and does not commute with $Q(V)$ when $y \in V$. The ground state is a coherent state $|\Psi_0\rangle = |\alpha_0, 0, 0, \cdots\rangle$. We see immediately that $\Phi \cdot |\Psi_0\rangle$ does not create an excitation out of the tower of states and should not contribute to the commutator for
the order parameter. Thus, if we ignore time-dependence and set $t = 0$,

$$ \begin{aligned} \langle \Psi_0 | [Q(V), \Phi(\mathbf{y})] | \Psi_0 \rangle &= - \int_V d^3\mathbf{x} \sum_{\{n_q\}} \langle \Psi_0 | \sum_{\mathbf{q}} a_{\mathbf{q}} \frac{e^{i\mathbf{q}\cdot\mathbf{y}}}{\sqrt{\Omega}} | \{n_q\} \rangle \langle \{n_q\} | \frac{1}{\Omega} \sum_{\mathbf{q}'} \sum_{\mathbf{q}''} e^{i(\mathbf{q}''-\mathbf{q}')\cdot\mathbf{x}} a_{\mathbf{q}''}^{\dagger} a_{\mathbf{q}'} | \Psi_0 \rangle \\ &= - \sum_{n_0=0}^{\infty} \frac{1}{\Omega} \sum_{\mathbf{q}} \sum_{\mathbf{q}'} \sum_{\mathbf{q}''} \delta_V(\mathbf{q}'' - \mathbf{q}') \times \\ & \qquad \langle \Psi_0 | \frac{e^{i\mathbf{q}\cdot\mathbf{y}}}{\sqrt{\Omega}} a_{\mathbf{q}} | n_0, n_q = 1, 0, \dots \rangle \langle n_0, n_q = 1, 0, \dots | a_{\mathbf{q}''}^{\dagger} a_{\mathbf{q}'} | \Psi_0 \rangle \end{aligned} $$

where we have selected the single excitation at wave number $\mathbf{q}$ that is the only Fock state that can make a non-zero contribution in the first matrix element, and kept the sum over the particle occupation $n_0$ of the condensate. We see that necessarily $\mathbf{q}' = 0$ since $a_{\mathbf{q}'}$ acts on the condensate and that $\mathbf{q}'' = \mathbf{q}$ to create the same excitation as is annihilated by the interpolating field in the first factor. What remains is

$$ \begin{aligned} \langle \Psi_0 | [Q(V), \Phi(\mathbf{y})] | \Psi_0 \rangle &= - \sum_{n_0=0}^{\infty} \frac{1}{\Omega} \sum_{\mathbf{q}} \delta_V(\mathbf{q}) \frac{e^{i\mathbf{q}\cdot\mathbf{y}}}{\sqrt{\Omega}} \alpha_0 \langle \alpha_0 | n_0 \rangle \langle n_0 | \alpha_0 \rangle \rightarrow - \frac{\alpha_0}{\sqrt{\Omega}} \int \frac{d^3\mathbf{q}}{(2\pi)^3} \delta_V(\mathbf{q}) e^{i\mathbf{q}\cdot\mathbf{y}} \\ &= - \frac{\alpha_0}{\sqrt{\Omega}} \end{aligned} $$

for $\mathbf{y} \in V$. We have taken the thermodynamic limit with $|\alpha_0|^2 = N_0 = \rho_0\Omega$. As $V \to \infty$, we also have $\delta_V(\mathbf{q}) \to \delta^3(\mathbf{q})$. We find the same result as directly from the commutator $[Q, \Phi(\mathbf{y})] = -\Psi(\mathbf{y})$ but this analysis identifies explicitly that the Goldstone modes at small momenta are deeply related to the broken symmetry.

* Show that any intermediate state $|\{n_q\}\rangle = |n, 0, \dots\rangle$ inside the tower of states at $\mathbf{q} = 0$ has zero contribution in the commutator above as $\Omega \to \infty, n_0 \to \infty$ at fixed $V$. More precisely,
$$ \sum_{n=0}^{\infty} \langle \Psi_0 | Q(V) | n, 0, \dots \rangle \langle n, 0, \dots | \Phi(\mathbf{y}) | \Psi_0 \rangle - \sum_{n=0}^{\infty} \langle \Psi_0 | \Phi(\mathbf{y}) | n, 0, \dots \rangle \langle n, 0, \dots | Q(V) | \Psi_0 \rangle = - \frac{V}{\Omega} \frac{\alpha_0}{\sqrt{\Omega}} $$

### C. Long-range Correlations

Spontaneously broken symmetry implies the presence of long-range correlations, for instance all spins aligned or counter-aligned in magnetic materials. Nn quantum systems we can even encounter the onset of off-diagonal long-range order. For $N$ particles, the many-body wave function (in first quantization) in real space is $\Psi(\mathbf{x}_1, \mathbf{x}_2, \dots, \mathbf{x}_N)$ and we can define a diagonal pair-correlation function

$$ \rho_D(\mathbf{x}, \mathbf{y}) = \int d^3\mathbf{x}_3 \cdots d^3\mathbf{x}_N |\Psi(\mathbf{x}, \mathbf{y}, \mathbf{x}_3, \cdots, \mathbf{x}_N)|^2 \quad (170) $$
This function can have long-range or infinite range order, for instance in a crystal. We can also define the off-diagonal pair-correlation function,

$$ \rho_O(\mathbf{x}, \mathbf{y}) = \int d^3\mathbf{x}_2 \cdots d^3\mathbf{x}_N \Psi^*(\mathbf{x}, \mathbf{x}_2, \mathbf{x}_3, \cdots, \mathbf{x}_N) \Psi^*(\mathbf{y}, \mathbf{x}_2, \mathbf{x}_3, \cdots, \mathbf{x}_N) \quad (171) $$

and say that off-diagonal long-range order occurs if this function does not decay or hardly decays for distant $\mathbf{x}$ and $\mathbf{y}$. For instance in a superfluid with broken $U(1)$-symmetry we have completely lost track of the position of the individual bosons. All that we can tell is how many bosons are on average at position $\mathbf{x}$. We can define the spatial correlation as

$$ \rho_O(\mathbf{x}, \mathbf{y}) = \langle \Psi_0 | \Psi^\dagger(\mathbf{x}) \Psi(\mathbf{y}) | \Psi_0 \rangle = |\alpha_0|^2 \psi_0(\mathbf{x})^* \psi_0(\mathbf{y}) \quad (172) $$

In the second equation we assumed the presence of a coherent state in the ground state. In the thermodynamic limit $\psi_0 = 1/\sqrt{V}$ and $|\alpha_0|^2 = N_0$. We see that the superfluid has infinite off-diagonal order. “Supersolid” phases have been discovered that possess both diagonal and off-diagonal long range order.

### Example: Long-range order as a signature for SSB

It is generally possible to define a local order parameter $O(\mathbf{x}, t)$ such that

$$ O = \int_V d^d\mathbf{x} O(\mathbf{x}, t) $$

This can be local spin, any local deviation, local phase etc. A system is said to have long-range order (LRO) if

$$ \lim_{V \to \infty} \langle O(\mathbf{x}, t) O(\mathbf{x}', t) \rangle \to \sigma^2 \quad (173) $$

for $|\mathbf{x} - \mathbf{x}'|$ large, say of order $V^{1-\epsilon}$ so to stay inside the box.

* Consider a Hamiltonian $H = \sum_i p_i^2 + \sum_{(i,j)} V(r_i - r_j)$ that commutes with total momentum $\mathbf{P}$ and has eigenstates $|\Psi_n\rangle$ at energy $E_n$. Show that, in the presence of thermal equilibrium and without SSB, the thermal expectation $\langle O(\mathbf{x}, t) \rangle$ depends nor on $\mathbf{x}$, neither on $t$. (Hint: write $O(\mathbf{x}, t) = \exp(-i\mathbf{P} \cdot \mathbf{x}) \exp(iHt) O(0, 0) \exp(i\mathbf{P} \cdot \mathbf{x}) \exp(-iHt)$.)

* Show that in the presence of LRO,

$$ \lim_{V \to \infty} \frac{1}{V^2} \langle O^2 \rangle = \sigma^2 \quad (174) $$
Long-range order signals the occurrence of SSB, while nothing is perceived in the average. We can generalize Eq. (173) to long-range crystalline order,

$$ \lim_{V \to \infty} \langle \mathcal{O}(\mathbf{x}, t) \mathcal{O}(\mathbf{x}', t) \rangle \to \sigma^2(\mathbf{x} - \mathbf{x}') \quad (175) $$

again for $|\mathbf{x} - \mathbf{x}'|$ large, where now $\sigma^2(\mathbf{x} - \mathbf{x}')$ is a periodic function in $d$ dimensions. With the Fourier transformation

$$ \mathcal{O}_q(t) = \int d^d \mathbf{x} e^{i\mathbf{q} \cdot \mathbf{x}} \mathcal{O}(\mathbf{x}, t) $$

this is equivalent to

$$ \lim_{V \to \infty} \frac{1}{V^2} \langle \mathcal{O}_q \mathcal{O}_{-\mathbf{q}'} \rangle = \delta_{\mathbf{q}\mathbf{q}'} \sigma_{\mathbf{q}}^2 \quad (176) $$

with $\sigma_q^2$ nonzero for a few lattice vectors $\mathbf{q}$. For $\mathbf{q} = \mathbf{q}' = 0$ we are back to the case (174) discussed above above provided $\sigma_0^2 \neq 0$.

### Example: No quantum time crystal in a many-body ground state (Adapted from [9])

We can try to formulate long-range time order similar to Eq. (175) and define a time crystal to have the property

$$ \lim_{V \to \infty} \langle \mathcal{O}(\mathbf{x}, t) \mathcal{O}(\mathbf{x}, 0) \rangle \to \sigma^2(t) \quad (177) $$

for $t$ large and with $\sigma^2(t)$ some periodic function. We want this function to be more or less independent on $\mathbf{x}$ so that the qualification makes time crystals a collective effect. Then the global order parameter would satisfy

$$ \lim_{V \to \infty} \frac{1}{V^2} \langle \mathcal{O}(t) \mathcal{O}(0) \rangle \to \sigma^2(t) \quad (178) $$

It is quite easy to get eternal oscillations in a stable ground state locally. Consider the two-level atom with $H = -\hbar\Omega\sigma_z$ with ground state $|\Psi_0\rangle = (1, 0)$ ("spin up") and excited state $|\Psi_1\rangle = (0, 1)$ ("spin down"). We can verify that in the ground state

$$ \langle \Psi_0 | \sigma_x(t) \sigma_x(0) | \Psi_0 \rangle = \exp(-i\Omega t) $$

We do not want to call this a time crystal but we do have identified a local ground state with an eternally oscillating spin component. We will show that the quantum expectation value in the many-body ground state $\langle \Psi_0 | \mathcal{O}(t) \mathcal{O}(0) | \Psi_0 \rangle$ cannot oscillate in time as $V \to \infty$. The proof
uses the assumption that the order parameter $\mathcal{O}$, like the Hamiltonian (without too much long-range interactions) is the integral of a local operator. Consider the operator $\mathcal{O}(t) = \int_V d^d\mathbf{x} \mathcal{O}(\mathbf{x}, t)$ and write

$$ \begin{aligned} \langle \Psi_0 | \mathcal{O}(t) \mathcal{O}(0) - \mathcal{O}(0)^2 | \Psi_0 \rangle &= \langle \Psi_0 | e^{iHt} \mathcal{O}(0) e^{-iHt} \mathcal{O}(0) - \mathcal{O}(0)^2 | \Psi_0 \rangle = \langle \Psi_0 | e^{iE_0 t} \mathcal{O}(0) e^{-iHt} \mathcal{O}(0) - \mathcal{O}(0)^2 | \Psi_0 \rangle \\ &= i \int_0^t ds \langle \Psi_0 | \frac{d}{ds} \mathcal{O}(0) e^{-i(H-E_0)s} \mathcal{O}(0) | \Psi_0 \rangle \end{aligned} $$

By Cauchy's inequality is $|\int dx f(x)| \le \int dx |f(x)|$ so,

$$ |\langle \Psi_0 | \mathcal{O}(t) \mathcal{O}(0) - \mathcal{O}(0)^2 | \Psi_0 \rangle| \le \int_0^t ds |\langle \Psi_0 | \mathcal{O}(0) (H - E_0)^{1/2} e^{-i(H-E_0)s} (H - E_0)^{1/2} \mathcal{O}(0) | \Psi_0 \rangle| $$

Next we apply the Schwarz inequality $|\langle \Psi | \Phi \rangle| \le \sqrt{\langle \Psi | \Psi \rangle |\langle \Phi | \Phi \rangle}$,

$$ \frac{1}{V^2} |\langle \Psi_0 | \mathcal{O}(t) \mathcal{O}(0) - \mathcal{O}(0)^2 | \Psi_0 \rangle| \le \frac{1}{V^2} \int_0^t ds \langle \Psi_0 | \mathcal{O}(0) (H - E_0) \mathcal{O}(0) | \Psi_0 \rangle = \langle \Psi_0 | [\mathcal{O}(0), [H, \mathcal{O}(0)]] | \Psi_0 \rangle \frac{t}{V^2} $$

If we assume that also $H$ is approximately an integral $H = \int_V d^d\mathbf{x} h(\mathbf{x})$ of local contributions (thus excluding too long-range interactions) with local commutation rules $[h(\mathbf{x}), \mathcal{O}(\mathbf{x}')] \sim g(\mathbf{x}) \delta_{\mathbf{x}\mathbf{x}'}$ and idem for $g(\mathbf{x})$, it follows that only one volume integral remains in the double commutator so that

$$ \frac{1}{V^2} |\langle \Psi_0 | \mathcal{O}(t) \mathcal{O}(0) - \mathcal{O}(0)^2 | \Psi_0 \rangle| \le K \times V \times \frac{t}{V^2} $$

In the thermodynamic limit $V \to \infty$ and for arbitrarily large $t$ the righthand side vanishes. The expectation value of $\mathcal{O}(t)\mathcal{O}(0)$ in the ground state cannot be a periodic function of time for large times. The proof only applies to the ground state in which case the order parameter $\mathcal{O}(t=0)$ induces excitations to higher levels due to local fluctuations, so that $\langle \mathcal{O}(0)(H - E_0)\mathcal{O}(0) \rangle > 0$. This outcome confirms that time crystals do not exist in the way they were conceived originally. With some more effort the proof can be extended to finite temperatures.
## D. Hohenberg-Mermin-Wagner Theorem

Is the broken symmetry state stable against fluctuations, either quantum or thermal? We already showed that the broken-symmetry state is accompanied by Goldstone modes that have energies that are small for small wave numbers. Therefore they can easily be excited thermally. Are they able to destroy the long range order? The answer is yes in low dimensions.

### Hohenberg-Mermin-Wagner Theorem

*There is no phase with spontaneous breaking of a continuous symmetry for $T > 0$, in $d = 2$ dimensions for an infinite system, and no long-range order on arbitrary scales can exist.*

We will not give a proof of this important theorem since it would involve the introduction of many technical tools. Nevertheless we give the following more handwaving argument relying on a well-known theorem: the fluctuation dissipation theorem (FDT). Goldstone's theorem demonstrates the existence of bosonic excitations $g$ (of magnetization, of phonons, ...) at low energies $E(q)$. The FDT takes the form already seen in Eq. (7) for quantum-vacuum fluctuations of the electric field. It relates the correlations in space-time to the propagation of the excitations and hence to its dispersion law. For a relation $\hbar\omega \sim q$ it takes the form

$$ \langle g(x)g(0) \rangle \sim -\mathrm{Im} \int_0^\infty \frac{d\omega}{2\pi} \int \frac{d^d q}{(2\pi)^d} \frac{e^{i\mathbf{q}\cdot\mathbf{x}}}{\omega^2 - q^2 c^2 + i0} \left[ \frac{1}{e^{\beta\hbar\omega} - 1} + \frac{1}{2} \right] \\ \sim \int_0^\infty dq \, q^{d-2} e^{i\mathbf{q}\cdot\mathbf{x}} \left[ \frac{1}{e^{\beta\hbar q c} - 1} + \frac{1}{2} \right] \quad (179) $$

The first factor in the first line is the propagation of the excitation, the second factor is the thermal weight with which it has been excited, with the $1/2$ its value at $T = 0$. For $T = 0$ no problems occur for $d = 2$, but the integral diverges in one dimension. At finite temperature we consider excitations $\hbar q c \ll 1/\beta$ where the thermal factor diverges as $1/\beta\hbar q c$. This time the integral diverges as $dq \, q^{d-3}$. This is no problem in 3D but in 2D a logarithmic divergence exists for an infinite medium. In a finite medium we can use a lower cut-off $q = \pi/L$ and we see that $\langle g(0)^2 \rangle$ diverges slowly as $\log L$ with the size of the medium. For large $x$ the correlation function is the Fourier transform of $1/q^2$ in reciprocal space, hence in real space it grows as $\log x$, like the Coulomb law in 2D. We conclude that in low dimensions fluctuations are large and break the long-range order that would have been created by the spontaneous symmetry breaking. A variant of the theorem shows that perfect and infinite 2D crystals cannot exist, but deviations are too small to observe for finite lattices.
### Exercise VII.4: HMW theorem for quadratic dispersion

When $E \sim q^2$ the FDT takes the form,

$$ \langle g(\mathbf{x})g(0) \rangle \sim -\mathrm{Im} \int_0^\infty dE \int \frac{d^d\mathbf{q}}{(2\pi)^d} \frac{e^{i\mathbf{q}\cdot\mathbf{x}}}{E - q^2/2m + i0} \left[ \frac{1}{e^{\beta E} - 1} + \frac{1}{2} \right] \quad (180) $$

* Show that at $T = 0$, the thermal fluctuations $\langle g(0)^2 \rangle$ scale as $dq \bar{q}^{d-1}$ for small wave numbers $q$ and at finite temperature contribute $dq \bar{q}^{d-3}$, hence logarithmically divergent in $2D$.

### Exercise VII.5: One dimensional lattice

The Hamiltonian for the 1D lattice with harmonic coupling between neighbors is given by

$$ H = \kappa \sum_{(n,m)} (x_n - x_m)^2 $$

with $x_m = ma + \Delta x_m$. In a classical description we can drop the kinetic energies as long as we are interested in thermal expectation values for positions only. The minimum energy is reached when all $\Delta x_n = 0$ which would correspond to perfect crystalline order. We investigate the thermal fluctuations of $\Delta x_m$ in a crystal of length $L = Na$. We can decompose the displacement in $N$ discrete Fourier components $\{q_j\} = (2\pi/L)\{-N/2 + 1, \dots, N/2\}$,

$$ \Delta x_n = \frac{1}{\sqrt{N}} \sum_q e^{iqna} \Delta x_q \quad ; \quad \Delta x_q = \frac{1}{\sqrt{N}} \sum_{n=0}^{N-1} e^{-iqna} \Delta x_n $$

where we can use $\sum_{n=0}^{N-1} e^{i(q+q')na} = N\delta_{q+q'}$ to link the two equations. We have assumed periodic boundary conditions with $x_0 = x_N$ so that $\sum_n \Delta x_n = 0$ (and thus $\Delta x_{q=0} = 0$) meaning that the center of mass of the crystal as a whole does not move. The different $\Delta x_n$ are not independent.

* Show that the Hamiltonian can be written as

$$ H = E_0 + \kappa \sum_{q \neq 0} h(q) \Delta x_q \Delta x_{-q} $$

with $h(q) = 2[1 - \cos(qa)]$. Conclude from this that in thermal equilibrium different Fourier modes are independent parameters and that,

$$ \langle \Delta x_q \Delta x_{q'} \rangle = \delta_{q+q'} \frac{\frac{1}{2}kT}{\kappa h(q)} $$
• Show that

$$ (\Delta x_n - \Delta x_0)^2 = \frac{1}{N} \sum_{qq'} (e^{iqna} - 1) (e^{iq'na} - 1) \Delta x_q \Delta x_{q'} $$

with thermal average

$$ \langle (\Delta x_n - \Delta x_0)^2 \rangle = \frac{\frac{1}{2}kT}{\kappa N} \sum_{q \neq 0} \frac{4 \sin^2(qna/2)}{h(q)} $$

• In the thermodynamic limit $\sum_q \to L \int_{-\pi/a}^{\pi/a} dq/2\pi$. Show that

$$ \langle (\Delta x_n - \Delta x_0)^2 \rangle = \frac{1}{2} \frac{kT}{\kappa} \int_{-\pi}^{\pi} \frac{dy}{2\pi} \frac{\sin^2(yn/2)}{\sin^2(y/2)} = \frac{n}{2} \frac{kT}{\kappa} $$

because the integral over $y$ equals $n$. For $n = 1$ this gives $\langle \kappa (\Delta x_n - \Delta x_{n-1})^2 \rangle = \frac{1}{2}kT$ independent of $n$. Hence, all local phonon oscillations are equilibrated and the total energy is $N \times \frac{1}{2}kT$ and thus extensive (because $N$ independent modes exist). The typical difference between displacements separated by $n$ sites grows with $\sqrt{nkT}$ and for arbitrary small temperature always becomes larger than the lattice spacing $a$. The conclusion is that thermal fluctuations destroy long-range crystalline order in 1D.

• Show that the mean square displacement at any site $n$ is given by

$$ \langle (\Delta x_n)^2 \rangle = \frac{1}{2} \frac{kT}{N\kappa} \sum_{q \neq 0} \frac{1}{4 \sin^2(qa/2)} \to \frac{kT}{N\kappa} \sum_{j=1}^{\infty} \frac{1}{4(\pi j/N)^2} = \frac{N kT}{24 \kappa} $$

which is an extensive quantity for $N$ large essentially because the lowest Goldstone mode $j = 1$ dominates. Conclude from the different results that

$$ \langle \Delta x_n \Delta x_0 \rangle = \frac{N kT}{24 \kappa} \left(1 - 6 \frac{n}{N}\right) $$

for $n \ll N$ (of course for $n = N$ this correlation should vanish). Thus, site oscillations have long-range correlation over many lattice sites due to phonons with long wavelengths.

### Exercise VII.6: the classical XY-model

The classical XY-model couples nearest neighbor spins on a cubic 2D lattice aligned in some direction $\phi$ in the $xy$ plane with interaction

$$ H = -J \sum_{(i,j)} \mathbf{s}_i \cdot \mathbf{s}_j = -J \sum_{(i,j)} \cos(\phi_i - \phi_j) $$
with $J > 0$. The model has the global continuous symmetry of spin rotation in the plane. For $kT \gg J$ we expect all spins to rotate freely with no net magnetization. For $kT \ll J$ the energy is minimized if all spins are aligned in the same direction and we could expect a ferromagnetic state with “small” thermal fluctuations of the phases. We can therefore expand $\cos(\phi_i - \phi_j) \approx 1 - \frac{1}{2}(\phi_i - \phi_j)^2$ and use the same procedure as in the first example of this section. In this harmonic approximation the phase differences between neighbor spins are not i.i.d. (independent identical and even Gaussian distributed) because the sum $\sum_{i=0}^N (\phi_{i+1} - \phi_i) = \phi_N - \phi_0 = 0(\mod 2\pi)$ along one axis, and long wavelength Goldstone modes impose correlations. We insert discrete Fourier modes $q = -\pi/a, -\pi/a+2\pi/L, \dots, \pi/a-2\pi/L, \pi/a$ along both axis. Hence $N^2 = (L/a)^2$ discrete Fourier modes exist.

* Show that the Hamiltonian becomes,

$$H = E_0 - \sum_{\mathbf{q} \neq 0} E(\mathbf{q}) \phi_{\mathbf{q}} \phi_{-\mathbf{q}}$$

with $E(\mathbf{q}) \sim Jq^2a^2/2$ for $q \ll a$. Hence Goldstone modes exist with energies $E(\mathbf{q}) \sim q^2$.

* Show that the classical thermal expectation value of the phase correlation is

$$\langle \phi_{\mathbf{q}} \phi_{\mathbf{q}'} \rangle = \delta_{\mathbf{q}+\mathbf{q}'} \frac{kT}{E(\mathbf{q})}$$

so that $\langle |\phi_{\mathbf{q}}|^2 \rangle \sim 1/q^2$ diverges at finite temperature for small $q$.

* Show that, upon taking the continuous limit $L \to \infty$, phase correlation between different sites on the x-axis is given by

$$\begin{aligned} \langle (\phi_n - \phi_0)^2 \rangle &= \frac{2}{N^2} \sum_{\mathbf{q}} \frac{1 - \cos(aq_x n)}{\beta E(\mathbf{q})} \rightarrow 2a^2 \int_{q \le \frac{\pi}{2a}} \frac{d^2\mathbf{q}}{(2\pi)^2} \frac{1 - \cos(aqn \cos \theta)}{\beta E(\mathbf{q})} \\ &\rightarrow \frac{2kT}{\pi J} \int_0^{n\pi/2} dx \frac{1 - J_0(x)}{x} \end{aligned}$$

which grows slowly as $K + \log n$ for large $n$. Due to the large weight of excitations with very long wave length coherence is maintained over a long distance. When $kT \gg J$ phase fluctuations between neighbors are large and our harmonic approximation breaks down completely. For $kT \ll J$ phase is randomized if $\langle (\phi_n - \phi_0)^2 \rangle \ge (2\pi)^2$ which happens when $n > \exp(4\pi^3 J/kT)$. This is large but finite. Before that happens we can estimate that

$$\langle \mathbf{s}_n \cdot \mathbf{s}_0 \rangle = \langle e^{i(\phi_n - \phi_0)} \rangle \approx e^{-\frac{1}{2} \langle (\phi_n - \phi_0)^2 \rangle} \sim n^{-kT/\pi J}$$
Because $\phi_n - \phi_0$ is a sum of many thermally uncorrelated Fourier modes it has Gaussian statistics, which has been applied in the second equality. Spin correlations decay as a weak power-law, sign of quasi-long-range order, nevertheless at large distances the ferromagnetic phase is completely destroyed by thermal fluctuations.

## Example: The Peierls argument for the Ising model

The classical Lenz-Ising model (in the absence of an external magnetic field) is given by the Hamiltonian

$$H = -\frac{J}{2} \sum_{(i,j)} \sigma_i \cdot \sigma_j$$

with $J > 0$ and the sum running over nearest neighbors on a (say) cubic lattice with $\sigma_i = \pm 1$. The model was proposed and solved in 1D at any temperature by Walter Lenz and his student Ernst Ising in 1924. It did not reveal a phase transition to a ferromagnetic state with finite magnetization. The solution in 2D was obtained analytically by Lars Onsager in 1944 [22], and reveals spontaneous magnetization below a critical temperature. This demonstration revealed that original arguments, based on analytic behavior (with $\beta$) of the Gibbs ensembles, that no phase transition should occur in any dimension, were flawed. The subtlety is the thermodynamic limit. In the Lenz-Ising model configurations in a finite volume with all spins “up” ($\sigma_i \equiv 1$) or “down” ($\sigma_i \equiv -1$) have equal energy $-JN$ and are thus thermally equally probable. Hence the total magnetization $M = N^{-1} \sum_{i=1}^{N} \sigma_i$ is on average zero for all temperatures, end of story. Note that the limits $T \to 0$ and $M \to \infty$ do not commute, reminiscent of the scenario of broken symmetry, here $\mathbb{Z}^2$.

Unlike in the XY model, the symmetry of the Lenz-Ising model is not continuous, gapless Goldstone modes do not exist (a minimum gap $\Delta E = J$ exists for $q \neq 0$ excitations) and the HMW theorem does not apply: A spontaneous magnetization occurs in 2D at low enough temperatures.

<figure><img src="image_20.png" /><figcaption>Figure 18. Ising model in 1D with single domain wall</figcaption></figure>
Just after the introduction of the Ising model, Rudolf Peierls made an historical argument why there should be a phase transition in 2 and higher dimensions at finite temperature. Consider first the model in 1D and the ground state of all spins “up”. We can imagine the creation of a domain with $M < N$ spins “down” around a given spin defined at the origin. The cost in energy is $J + J$, independent of $M$. However, such a chain has a finite entropy $S = k \log W$ because $W = M$ possibilities exist to have it include the origin. The change in free energy is thus $\Delta F = \Delta E - T\Delta S = 2J - kT \log M$. In 1D and at any finite temperature, thermal fluctuations can create domains with opposite spin. The state in thermal equilibrium is disordered and no net magnetization exist. The exact solution shows that spin correlations $\langle \sigma_i \sigma_j \rangle$ decay exponentially with $|i - j|$.

The argument changes significantly in 2D. On a square lattice four nearest neighbors exist and a domain wall involving $M$ consecutive spins around the origin costs an energy $(4-2) \times J \times (M-2) + (4-1) \times J = 2J(M+1)$ since except for the 2 edges two neighbor spins stay aligned. The number of non-intersecting paths that cross the origin can be estimated as $W = M^2 3^M$ since the path can start typically on a lattice of size $M^2$ and have at each step $4-1=3$ possibilities to choose the next step. This gives $\Delta F = 2J(M+1) - kT M \log 3 - 2kT \log M$. If $kT/J < 2/\log 3 = 0.367$ thermal fluctuations cannot support the formation of domains walls and the thermal equilibrium is “frozen” in one of the two perfectly ferromagnetic ground states. For $kT/J > 2/\log 3$ thermal fluctuations destroy the long-range order and $M = 0$. The exact Onsager solution has $kT_c = 2.27J$ (which means that we have overestimated $W$ and counted self-intersecting paths).

## Example: The BKT transition

In the 2D XY model it now known that below a certain temperature a BKT (Berezinskii-Kosterlitz-Thouless) phase transition occurs from a disordered to another disordered state. This transition is driven by the formation of vortices. Vortices are topological geometries of spin orientations. This was the first discovery of a transition of topological nature, where spins align in vortex states or not, depending on what is thermodynamically preferred. Such a transition is not covered by the HMW theorem. It was rewarded by the Nobel Prize in 2016.

There is no non-zero order parameter that shows up but some correlation length exists that diverges on the transition. Below the critical temperature the correlation of the field $\langle \Psi^{\dagger}(x) \Psi(0) \rangle$ decays algebraically as in the classical XY model, and not exponentially as is the case above the critical