# Electronic Properties of Materials/Quantum Mechanics for Engineers/Degeneracy

Degeneracy is often talked about in electronics and quantum mechanic in reference to electrons which have the same energy level. In this case, since energy is an eigenvalue, you end up with two electrons with different eigenfunctions which still share the same eigenvalue. We call a quantum state "degenerate" if two or more eigenfunctions have the same eigenvalue as in the case of the electrons, but how does this happen? Well, there are three separate ways; symmetry, exchange, and accidental.

## Degeneracy by Symmetry

This is the form of degeneracy associated with the hybridization of orbitals. Atoms behavior in the x, y and z-directions is the same assuming a spherical potential which generally applies to atoms in isolation.
<FIGURE> "Particle in a 2D Box" (Description)
Imagine a particle in another box, once again with infinite potential on all sides, but this time, in a 2D, giving us the Hamiltonian equation:${\hat {H}}={{{\hat {p}}_{x}}^{2} \over 2m}+{{{\hat {p}}_{y}}^{2} \over 2m}+V(x)+V(y)$
This problem easily breaks into component parts: ${\hat {H}}={\hat {H}}_{x}+{\hat {H}}_{y}$
Substituting in the Schrödinger equation and working through the math finds that:${\begin{aligned}\phi _{n_{1}n_{2}}&=\beta \sin \left({n_{x}\pi \over L_{x}}x\right)\sin \left({n_{y}\pi \over L_{y}}y\right)\\E_{n_{1}n_{2}}&={\hbar ^{2} \over 2m}\left({n_{x}\pi \over L_{x}}\right)^{2}+{\hbar ^{2} \over 2m}\left({n_{y}\pi \over L_{y}}\right)^{2}\end{aligned}}$
Looking at these time-independent eigenfunctions, when $L_{x}=L_{y}$, then we find that $E_{\alpha \beta }=E_{\beta \alpha }$ but $\phi _{\alpha \beta }\neq \phi _{\beta \alpha }$ when $\alpha \neq \beta$. These are different planewave eigenfunctions with the same energy or eigenvalue, which completes the definition of degeneracy by symmetry.
<MATH> The below math has no home :(
When ${\hat {H}}={\hat {H}}(x)+{\hat {H}}(y)$, the solution to the eigenvalue problem is separable. Thus:${\begin{aligned}\phi (x,y)&=\phi (x)\ \phi (y)\\{\hat {H}}\ \phi (x,y)&=\varepsilon \ \phi (x,y)\\H(x)\ \phi (x,y)+H(y)\ \phi (x,y)&=H(x)\ \phi (x)\ \phi (y)+H(y)\ \phi (x)\ \phi (y)\\&=\phi (y)\ \underbrace {H(x)\ \phi (y)} _{\varepsilon _{x}\ \phi (x)}\ +\ \phi (x)\ \underbrace {H(y)\ \phi _{y}} _{\varepsilon _{y}\phi _{y}}\\&=\phi (y)\ \varepsilon _{x}\ \phi (x)+\phi _{x)\ \varepsilon _{y}\ \phi (y)\\&=(\varepsilon _{x}+\varepsilon _{y})\phi (x)\phi (y)\\&=\varepsilon \ \phi (x,y)\end{aligned}}$

## Accidental Degeneracy

Occasionally, two eigenfunctions will just happen to have the same eigenvalue for energy making them "accidentally" degenerate. Accidental degeneracy simply refers to when two planewaves, $\phi$, share the same eigenvalue by accident and not due to symmetry or exchange.

## Degeneracy by Exchange

<FIGURE> "Two Particles in a 1D Box" (Description)
Now let's look at a one dimensional box with two non-interacting particles. This box will once again have infinite potentials outside of the box, as seen in <FIGURE>. As these particles are non-interacting, their potentials never see each other.
Our separable Hamiltonian for this situation will look like:$H=H_{1}+H_{2}=\underbrace {T_{1}+V_{1}(x)} _{Particle\ 1}+\underbrace {T_{2}+V_{2}(x)} _{Particle\ 2}$
The Hamiltonian solution for each particle solves to:${\begin{aligned}\phi _{n}(x)&={\sqrt {2 \over L}}\sin \left({n\pi x \over L}\right)\\E_{n}(x)&={\hbar ^{2} \over 2m}\left({n\pi \over L}\right)^{2}\end{aligned}}$
Combining these equations with our two particle Hamiltonian gives us:${\begin{aligned}\phi _{n_{1}n_{2}}(x_{1},x_{2})&={2 \over L}\sin \left({n_{1}\pi x_{1} \over L}\right)\sin \left({n_{2}\pi x_{2} \over L}\right)\\E_{n_{1}n_{2}}&={\hbar ^{2} \over 2m_{1}}\left({n_{1}\pi \over L}\right)^{2}+{\hbar ^{2} \over 2m_{2}}\left({n_{2}\pi \over L}\right)^{2}\end{aligned}}$<IS x AN EIGENVALUE???>
With this notation, n$_1${\textstyle n_{1}}refers to particle one with quantum number n$_1${\textstyle n_{1}}, while n$_2${\textstyle n_{2}} refers to particle two with quantum number n$_2${\textstyle n_{2}}. Similarly, x$_1${\textstyle x_{1}} and m$_1${\textstyle m_{1}} refer to particle one at position x$_1${\textstyle x_{1}} or mass m$_1${\textstyle m_{1}}, while x$_2${\textstyle x_{2}} and m$_2${\textstyle m_{2}} refer to particle two at position x$_2${\textstyle x_{2}} or mass m$_1${\textstyle m_{1}}. Explicitly, these particles can be at different positions and have different masses, but when the two particles end up having the same eigenvalue for position or mass, m$_1$=m$_2${\textstyle m_{1}=m_{2}} or x$_1$=x$_2${\textstyle x_{1}=x_{2}}, there is degeneracy due to exchange.
<FIGURE> "Title" (Snapshots of classical mechanics)
This has to do with the loss of determinism. In a classical picture, if we take a snap-shot of the system, then wait a second and take another, we can tell which particle is which as the system is deterministic. <FIGURE> Quantum Mechanics is not this way. In the quantum world there is uncertainty which means I can't really tell you that there are "two particles in this box located here, at x$_1${\textstyle x_{1}}, and there, at x$_2${\textstyle x_{2}}." Even if it is known that the wavefunctions are centered on those points, in reality that is just the center of a probabilistic distribution of where the particle might be. <FIGURE>
<FIGURE> "Title" (Note that these are really probability distributions as opposed to positions.)
Applying our classical method to this case, even if by some capacity we could look at the box and define each particle, looking away and looking back means that we can't tell which particle is which anymore, because the wave functions are overlapping. At any point both particles one and two are present in a superposition of particles.
Going back to our equations, this means is that ϕ$_{n1m2}${\textstyle \phi _{n_{1}m_{2}}}, and ϕ$_{n2m1}${\textstyle \phi _{n_{2}m_{1}}} have the exact same energy regardless of if you switch $n_{1}$ and $n_{2}$. This is how exchange leads to degeneracy. We have lost that deterministic world view making it impossible to definitively tell the particles apart if they have the same eigenvalues, as if they had different masses to begin with, we would be able to tell them apart regardless.
Mathematically speaking, E=(ℏπ/L)$^2$ 1/2m (n$_1$$^2$+n$_2$$^2$){\textstyle E=\left({\hbar \pi \over L}\right)^{2}{1 \over 2m}({n_{1}}^{2}+{n_{2}}^{2})}, and switching the n values is the same as switching the particles.

### Implications of Exchange

Comparative to degeneracy by symmetry and accidental degeneracy, degeneracy by exchange has several long reaching implications which should be considered. Imagine we have a system of $N$ non-interactive identical particles. We know wave functions define an interchange operator that operates on our wave function ($\psi$) to switch the variables of two of the particles.$\psi (q_{1},\ q_{2},\ \dots ,\ q_{i},\dots ,\ q_{j},\ \dots ,\ q_{n})$
Each of these q's represent a collection of variables and quantum numbers that represent a given particle. The written order of these variables further corresponds to each particle, precisely representing them and the state that these particles are in. Given this expression we can identify an interchange operator (P$^^$$_{ij}${\textstyle {\hat {P}}_{ij}}) . This interchange operator takes two particles and switches them, essentially switching the $q$ for one particle with the $q$ of another particle. Applying an interchange operator to the wavefunction looks like:
$P_{ij}\ \psi (q_{1},\dots ,\ q_{i},\dots ,\ q_{j},\dots ,\ q_{n})=\psi (q_{1},\dots ,\ q_{i},\dots ,\ q_{j},\dots ,\ q_{n})$
The interchange operator doesn't change much as it simply moves around the parameter which were already in the system. Furthermore, the interchange operator doesn't change the energy of the system which means that our interchange operator and the Hamiltonian operator commute, where $[P_{ij},\ H]=0$. While this was already implied by the non-interacting nature of our particles, this further proves that we can solve for each of these elements independently, thus the wavefunctions ($\psi$) are eigenfunctions of $P_{ij}$ and $H$.
As such, when this operator, the interchange operator, operates on some wavefunction, we know that it has to return an eigenvalue ($P_{ij}\ \psi =\xi \ \psi$). All good operators in Hilbert space will obey this relationship.
As it stands to reason, applying the same interchange operator to the same wavefunction twice will result in the original wave function. The operator switches all the particle parameters, and then switches them back.
What are the values of $\xi$? swapping $i$ and $j$ twice returns to initial state. ${\begin{aligned}P_{ij}(P_{ij}\ \psi )&=P_{ij}\ \xi \ \psi \\&=\xi ^{2}\ \psi =1\ \psi \\\xi ^{2}&=1\ \psi \\\xi &=\pm \ 1\end{aligned}}$In this case, we give these eigenvalues names and when $\xi =+1$ we say that the wavefunction is "symmetric under exchange," and when $\xi =-1$, we say that the wavefunction is "antisymmetric under exchange." It's also worth noting that even though each version of the interchange operator commutes with the Hamiltonian, they don't necessarily commute with each other. It's not universally true that for any exchange that the two are equivalent. This is important because it limits the way that we can express the wavefunctions.
Let's define another operator, ${\hat {P}}$ as a permutation operator which is a series of interchange operators, ${\hat {P}}_{ij}$, that rearrange the variables in the wavefunction ($\psi$). This gives us: $P\psi (q_{1},\ q_{2},\ \dots ,\ q_{n})=\psi (q_{1}',\ q_{2}',\ \dots ,\ q_{n}')$
When $P$ operates, it returns a new wave function with the variables, $q$, with different ordering. If ${\hat {P}}$ contains an even number of swaps, we call it an even permutation, and if has an odd number of swaps, we call it an odd permutation. It's worth nothing that there exists an $N!$ number of total permutations, and these permutation operators will not commute with each other just as the interchange operators don't commute with each other.
Unfortunately, this doesn't make sense. In general, ${\hat {P}}$ operators of different arrangements do not commute because $[P_{ij},P_{lm}]\neq 0$. This means that the wavefunction may be an eigenstate of the Hamiltonian and some permutation, ${\hat {P}}_{1}$, but not be an eigenstate of another permutation, ${\hat {P}}_{2}$. This means that the nature of our wavefunction is now limited. This means that we must express our wavefunction is such a way that is is either totally symmetric or totally antisymmetric. These are two special wavefunctions that commute with the Hamiltonian AND with all $N!$ possible ${\hat {P}}$.
${\begin{aligned}P_{ij}\ \psi _{S}&=\psi _{S}\\P_{ij}\ \psi _{A}&={\begin{cases}\psi _{A}\ &for\ even\ P\\-\psi _{S}\ &for\ odd\ P\end{cases}}\end{aligned}}$ In the case of the symmetric wavefunction the permutation operator acting on the symmetric is equal to the symmetric wavefunction for all permutations. Similarly, in the case of the antisymmetric operator acting on the antisymmetric wavefunction is equal to the antisymmetric wavefunction for even permutation operators and equal to the negative wavefunction for odd permutation operators. This also serves as our definition for symmetric and antisymmetric operators.

#### Postulate Zero

From what we know of the world, $\psi _{S}$ and $\psi _{A}$ are sufficient to describe all systems of identical particles. Particles that are totally symmetric are called Bosons and obey Bose-Einstein statistics (photon, phonon, cooper pair), while particles that are totally are called Fermions and obey Fermi-Dirac statistics (electrons, neutrino, quarks). This is called "Postulate Zero" since we can't actually prove this as fact, it is simply a pattern that every known particle follows, and happens to work out through the relationships of the commutators and our operators.
This means that when we write our solution, we have to make certain that our solution is either totally symmetric or totally antisymmetric. If for example, we wrote a solution to two non-interactive particles in a 1D box that wasn't either totally symmetric or totally antisymmetric, we would need to add different variations of our solution together to achieve either total symmetry or total antisymmetry. This might look like:
${\begin{aligned}&\ \ \left[{\begin{aligned}\psi _{S}&=\phi (x_{1}x_{2})+\phi (x_{2}x_{1})\\\psi _{A}&=\phi (x_{1}x_{2})-\phi (x_{2}x_{1})\end{aligned}}\right]\\\\&P_{12}\psi _{A}=P_{12}(\phi (x_{1}x_{2})-\phi (x_{2}x_{1}))\\&\qquad \ \ \ =\phi (x_{2}x_{1})-\phi (x_{1}x_{2})\\&\qquad \ \ \ =-\phi _{A}\end{aligned}}$As you can see, operating on the wavefunction ($\phi$) returns the negative wavefunction. You can do the same with a symmetric wavefunction. If I tell you that the particles are Fermions or Bosons then you know right away which $\psi _{S}$ or $\psi _{A}$ they reside in. That said, these particles are supposed to be non-interacting and thus the Hamiltonians can't "see" each other, yet they become entangled.${\begin{aligned}\psi _{A}&=\phi (x_{1}x_{2})-\phi (x_{2}x_{1})\\&=\phi (x_{1})\phi (x_{2})-\phi (x_{2})\phi (x_{1})\end{aligned}}$
These are NOT a simple product of single particle states. In quantum mechanics we say that the states are said to be "entangled", and even though the particles do not interact their wavefunctions are entangled. This means that a measurement of one particle has consequences on the other. By looking at the $\psi _{A,S}$ you can see that there is a superposition.

#### Expressing Wave Functions

Imagine $n_{1}=2$ and $n_{2}=3$. Which particle has energy $E_{2}$? We don't know. Each particle is a superposition of being in state $E_{2}$ and $E_{3}$. (Many body physics is neat stuff, but also extremely complicated.)
How do we express a general symmetric or antisymmetric wavefunction for any system? Symmetric wavefunctions are simple, it's just a sum of interchanges ($\phi (\gamma _{i})$). On the other hand, antisymmetric wavefunctions are a little more complicated. To express an antisymmetric wavefunction we use a Slater determinant:
Say $H=h_{1}+h_{2}+\dots +h_{n}$, so $\Phi =\phi _{1}(x_{1})\ \phi _{2}(x_{2})\ \phi _{3}(x_{3})\dots \phi _{n}(x_{n})$.
$h_{j}\ \phi _{j}(x_{j})=E_{j}\phi _{j}(x_{j})$$\psi _{A}(x_{1}\ x_{2}\ x_{3}\dots \ x_{n})={1 \over {\sqrt {N!}}}\left|{\begin{array}{lcl}\phi _{1}(x_{1})&\phi _{2}(x_{1})&\phi _{3}(x_{1})&\cdots &\phi _{n}(x_{1})\\\phi _{1}(x_{2})&\phi _{2}(x_{2})&\phi _{3}(x_{2})&\cdots &\phi _{n}(x_{2})\\\quad \ \vdots &\vdots &\quad \vdots &\ &\quad \vdots \\\phi _{1}(x_{n})&\phi _{2}(x_{n})&\phi _{3}(x_{n})&\cdots &\phi _{n}(x_{n})\end{array}}\right|$
This is an N x N determinant. For an example, look at some three particle system of Fermions:$\psi _{A}(q_{1}\ q_{2}\ q_{3})={1 \over {\sqrt {6}}}[\phi (q_{1}\ q_{2}\ q_{3})-\phi (q_{2}\ q_{1}\ q_{3})+\phi (q_{2}\ q_{3}\ q_{1})-\phi (q_{3}\ q_{2}\ q_{1})+\phi (q_{3}\ q_{1}\ q_{2})-\phi (q_{1}\ q_{3}\ q_{2})]$
A symmetric wavefunction would have the same determinant, but all the terms are added together for the summation.

#### Pauli Exclusion Principle

In the above example, each particle is in a unique state $q_{1},q_{2},q_{3}$. What would happen if $q_{1}=q_{2}$? Then all of the terms would cancel out and the wavefunction would equal zero. This is the crux of the Pauli Exclusion Principle in that Fermions must have unique quantum numbers, because if they share quantum numbers then their antisymmetric wavefunction disappears. This comes from the fact that electrons are indistinguishable particles.
<CITATION> "Quantum Teleportation Between Distant Matter Qubits"