# Rotation operator (quantum mechanics)

This article concerns the rotation operator, as it appears in quantum mechanics.

## Quantum mechanical rotations

With every physical rotation $R$, we postulate a quantum mechanical rotation operator ${\widehat {D}}(R):H\to H$ that is the rule that assigns to each vector in the space $H$ the vector
$|\alpha \rangle _{R}={\widehat {D}}(R)|\alpha \rangle$ that is also in $H$. We will show that, in terms of the generators of rotation, ${\widehat {D}}(\mathbf {\hat {n}} ,\phi )=\exp \left(-i\phi {\frac {\mathbf {\hat {n}} \cdot {\widehat {\mathbf {J} }}}{\hbar }}\right),$ where $\mathbf {\hat {n}}$ is the rotation axis, ${\widehat {\mathbf {J} }}$ is angular momentum operator, and $\hbar$ is the reduced Planck constant.

## The translation operator

The rotation operator $\operatorname {R} (z,\theta )$, with the first argument $z$ indicating the rotation axis and the second $\theta$ the rotation angle, can operate through the translation operator $\operatorname {T} (a)$ for infinitesimal rotations as explained below. This is why, it is first shown how the translation operator is acting on a particle at position x (the particle is then in the state $|x\rangle$ according to Quantum Mechanics).
Translation of the particle at position $x$ to position $x+a$: $\operatorname {T} (a)|x\rangle =|x+a\rangle$
Because a translation of 0 does not change the position of the particle, we have (with 1 meaning the identity operator, which does nothing): $\operatorname {T} (0)=1$ $\operatorname {T} (a)\operatorname {T} (da)|x\rangle =\operatorname {T} (a)|x+da\rangle =|x+a+da\rangle =\operatorname {T} (a+da)|x\rangle \Rightarrow \operatorname {T} (a)\operatorname {T} (da)=\operatorname {T} (a+da)$
Taylor development gives: $\operatorname {T} (da)=\operatorname {T} (0)+{\frac {d\operatorname {T} (0)}{da}}da+\cdots =1-{\frac {i}{\hbar }}p_{x}da$ with $p_{x}=i\hbar {\frac {d\operatorname {T} (0)}{da}}$
From that follows: $\operatorname {T} (a+da)=\operatorname {T} (a)\operatorname {T} (da)=\operatorname {T} (a)\left(1-{\frac {i}{\hbar }}p_{x}da\right)\Rightarrow {\frac {\operatorname {T} (a+da)-\operatorname {T} (a)}{da}}={\frac {d\operatorname {T} }{da}}=-{\frac {i}{\hbar }}p_{x}\operatorname {T} (a)$
This is a differential equation with the solution
$\operatorname {T} (a)=\exp \left(-{\frac {i}{\hbar }}p_{x}a\right).$
Additionally, suppose a Hamiltonian $H$ is independent of the $x$ position. Because the translation operator can be written in terms of $p_{x}$, and $[p_{x},H]=0$, we know that $[H,\operatorname {T} (a)]=0.$ This result means that linear momentum for the system is conserved.

## In relation to the orbital angular momentum

Classically we have for the angular momentum $\mathbf {L} =\mathbf {r} \times \mathbf {p} .$ This is the same in quantum mechanics considering $\mathbf {r}$ and $\mathbf {p}$ as operators. Classically, an infinitesimal rotation $dt$ of the vector $\mathbf {r} =(x,y,z)$ about the $z$-axis to $\mathbf {r} '=(x',y',z)$ leaving $z$ unchanged can be expressed by the following infinitesimal translations (using Taylor approximation):
${\begin{aligned}x'&=r\cos(t+dt)=x-y\,dt+\cdots \\y'&=r\sin(t+dt)=y+x\,dt+\cdots \end{aligned}}$
From that follows for states: $\operatorname {R} (z,dt)|r\rangle =\operatorname {R} (z,dt)|x,y,z\rangle =|x-y\,dt,y+x\,dt,z\rangle =\operatorname {T} _{x}(-y\,dt)\operatorname {T} _{y}(x\,dt)|x,y,z\rangle =\operatorname {T} _{x}(-y\,dt)\operatorname {T} _{y}(x\,dt)|r\rangle$
And consequently: $\operatorname {R} (z,dt)=\operatorname {T} _{x}(-y\,dt)\operatorname {T} _{y}(x\,dt)$
Using $T_{k}(a)=\exp \left(-{\frac {i}{\hbar }}p_{k}a\right)$ from above with $k=x,y$ and Taylor expansion we get: $\operatorname {R} (z,dt)=\exp \left[-{\frac {i}{\hbar }}\left(xp_{y}-yp_{x}\right)dt\right]=\exp \left(-{\frac {i}{\hbar }}L_{z}dt\right)=1-{\frac {i}{\hbar }}L_{z}dt+\cdots$ with $L_{z}=xp_{y}-yp_{x}$ the $z$-component of the angular momentum according to the classical cross product.
To get a rotation for the angle $t$, we construct the following differential equation using the condition $\operatorname {R} (z,0)=1$:
${\begin{aligned}&\operatorname {R} (z,t+dt)=\operatorname {R} (z,t)\operatorname {R} (z,dt)\\[1.1ex]\Rightarrow&{\frac {d\operatorname {R} }{dt}}={\frac {\operatorname {R} (z,t+dt)-\operatorname {R} (z,t)}{dt}}=\operatorname {R} (z,t){\frac {\operatorname {R} (z,dt)-1}{dt}}=-{\frac {i}{\hbar }}L_{z}\operatorname {R} (z,t)\\[1.1ex]\Rightarrow&\operatorname {R} (z,t)=\exp \left(-{\frac {i}{\hbar }}\,t\,L_{z}\right)\end{aligned}}$
Similar to the translation operator, if we are given a Hamiltonian $H$ which rotationally symmetric about the $z$-axis, $[L_{z},H]=0$ implies $[\operatorname {R} (z,t),H]=0$. This result means that angular momentum is conserved.
For the spin angular momentum about for example the $y$-axis we just replace $L_{z}$ with S$_y$=ℏ/2 σ$_y${\textstyle S_{y}={\frac {\hbar }{2}}\sigma _{y}} (where $\sigma _{y}$ is the Pauli Y matrix) and we get the spin rotation operator $\operatorname {D} (y,t)=\exp \left(-i{\frac {t}{2}}\sigma _{y}\right).$

## Effect on the spin operator and quantum states

Operators can be represented by matrices. From linear algebra one knows that a certain matrix $A$ can be represented in another basis through the transformation $A'=PAP^{-1}$ where $P$ is the basis transformation matrix. If the vectors $b$ respectively $c$ are the z-axis in one basis respectively another, they are perpendicular to the y-axis with a certain angle $t$ between them. The spin operator $S_{b}$ in the first basis can then be transformed into the spin operator $S_{c}$ of the other basis through the following transformation: $S_{c}=\operatorname {D} (y,t)S_{b}\operatorname {D} ^{-1}(y,t)$
From standard quantum mechanics we have the known results S$_b$|b+⟩=ℏ/2 |b+⟩{\textstyle S_{b}|b+\rangle ={\frac {\hbar }{2}}|b+\rangle } and S$_c$|c+⟩=ℏ/2 |c+⟩{\textstyle S_{c}|c+\rangle ={\frac {\hbar }{2}}|c+\rangle } where $|b+\rangle$ and $|c+\rangle$ are the top spins in their corresponding bases. So we have: ${\frac {\hbar }{2}}|c+\rangle =S_{c}|c+\rangle =\operatorname {D} (y,t)S_{b}\operatorname {D} ^{-1}(y,t)|c+\rangle \Rightarrow$ $S_{b}\operatorname {D} ^{-1}(y,t)|c+\rangle ={\frac {\hbar }{2}}\operatorname {D} ^{-1}(y,t)|c+\rangle$
Comparison with S$_b$|b+⟩=ℏ/2 |b+⟩{\textstyle S_{b}|b+\rangle ={\frac {\hbar }{2}}|b+\rangle } yields $|b+\rangle =D^{-1}(y,t)|c+\rangle$.
This means that if the state $|c+\rangle$ is rotated about the $y$-axis by an angle $t$, it becomes the state $|b+\rangle$, a result that can be generalized to arbitrary axes.