Chapter 12: Identical Particles

Note

Quantum mechanics is like Belgian endives: it is an acquired taste - Anonymous, 2021

Summary

Attention

In this chapter, we examine indistinguishable particles and accomplish two objectives. First, we discuss the possible symmetries a many-body quantum state can take upon the application of the exchange operator. Second, we discuss how those symmetries affect the ground and first excited states of the He atom, which we treat using a perturbative approach (See chapter Chapter 11: Time-independent perturbation theory).

Imagine a two-particle state, obtained by the tensor product of two single-particle states (see Chapter 5: Combining two spin-1/2 particles):

$|a\rangle_{1} \otimes|b\rangle_{2}$

This state is a combination of particle “1” in state “a” and particle “2” in state “b”. The exchange operator is defined as:

$\hat{P}_{12}\left(|a\rangle_{1} \otimes|b\rangle_{2}\right)=|b\rangle_{1} \otimes|a\rangle_{2}$

We find that because we are dealing with indistinguishable particles, the outcome of an observation does not change upon application of the exchange operator, and as a result, states must be either symmetric or anti-symmetric upon application of the exchange operator. Even more important, we realize that any such state cannot be a superposition of a symmetric and antisymmetric states. In physics, we called particles that have a symmetric states: bosons and those that have an antisymmetric one: fermions.

So, the take-home message is that if we deal with electrons (which are fermions), then their quantum state must be anti-symmetric upon exchange.

Using this important result, we examine the ground state of the He atom. The He atom is treated using perturbation theory since its Hamiltonian:

$\hat{H}=\frac{\hat{\mathbf{p}}_{1}^{2}}{2 m_{e}}+\frac{\hat{\mathbf{p}}_{2}^{2}}{2 m_{e}}-\frac{Z e^{2}}{\left|\hat{\mathbf{r}}_{1}\right|}-\frac{Z e^{2}}{\left|\hat{\mathbf{r}}_{2}\right|}+\frac{e^{2}}{\left|\hat{\mathbf{r}}_{1}-\hat{\mathbf{r}}_{2}\right|}$

can be seen as two hydrogen-like (with $Z=2$ ) systems (i.e., each electron separately experiences the Coulombic attraction of the He protons), along with a perturbation (last term on the right) corresponding to the electrostatic interaction (repulsion) between the two electrons of He. We solve this problem at first order and find an approximate value of the ground state energy of He. More importantly, we build the state (at zero-th order) using two $n=1, l=0, m=0$ states but realize that such a combination is symmetric under exchange! This forces us to remember the spin-angular momentum of the electrons and build the spin part of the state as an antisymmetric combination of spin-1/2 states as we have done in Chapter 5: Combining two spin-1/2 particles. Doing so allows us to create a anti-symmetric state since the product of a symmetric (spatial coordinates) and anti-symmetric (spin) functions is anti-symmetric.

Further, we perform the same type of analysis for the first excited state of He, which turns out to have a fourfold degeneracy. By doing so, we come to the conclusion that in order for the four zero-th order excited states to be antisymmetric, we must make various combinations of position- and spin- parts of the quantum states. This leads to an energy term that has no classical correspondence and is solely due to the necessity of creating anti-symmetric states.

All in all, the take home message is that the indistinguishability of particles in quantum mechanics leads to effects that have no classical equivalent. This is not completely surprising since, in classical physics, even identical particles can be distinguished (since we can “keep an eye on them” at all time since there is no Heisenberg principle that forbids that, classically).

Note

The variational principle

There are a number of other methods that can be used to obtain the energy-eigenstates of physical systems for which an analytical solution does not exist. We have dicussed the time-independent perturbation approach in Chapter 11: Time-independent perturbation theory. Another such method is coined the variational principle.

It stems from the following observation: for any state $\ket{\psi}$ , we have:

$E_{0} \leq\langle\psi|\hat{H}| \psi\rangle$

where $E_{0}$ is the ground state energy. The equality in the equation above can only be achieved for the true ground state $\ket{\psi_0}$ . Thus, if one can devise a sampling method, we can minimize the expectation value of the Hamiltonian and find the true ground state (or, at least an approximation) numerically.

Once we know the ground state, we can find excited states by working in the space of states that are orthogonal to $\ket{\psi_0}$ , etc.

Learning Material

Copy of Slides

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Key Learning Points

TBA

Test your knowledge

The exchange operator has eigenvalues $\pm 1$ , therefore I know that it is…
1. Hermitian
2. Unitary
3. Hermitian and unitary
Consider the quantum state of two indistinguishable particles. A measurement indicates that this state is an equal superposition of a fully symmetric and fully anti-symmetric state under exchange of the particle. What do you conclude?
1. These particles are bosons.
2. These particles are fermions
3. These particles do not exist.
We combine two electrons into a single state. A measurement indicates that the spatial part of the state is anti-symmetric upon exchange of the particles. What do you conclude?
1. This state has a total spin zero.
2. This state has a total spin of one.
3. There isn’t enough information to answer this question conclusively.
What is the variational principle?
1. It is an approach based on the fact the lowest possible expectation value of the Hamiltonian is the ground state energy of the system.
2. It is an perturbative approach consisting in varying the interaction parameters until the electron density is normalized.
3. It a theoretical approach based on the fact that the variation in density leads to quantum oscillations.
In the He atom, the reason why the spin-part of the state is well-defined is due to…
1. The fact there is an external magnetic field.
2. The quantum state describing the combination of two electrons must be symmetric upon exchange.
3. The quantum state describing the combination of two electrons must be anti-symmetric upon exchange.
4. The spin part of the state is not well-define in general. This is a trick question.

Hint

Find the answer keys on this page: Answers to selected test your knowledge questions. Don’t cheat! Try solving the problems on your own first!

Homework Assignment

Solve the following problems from the textbook:

Recitation Assignment

Solve the following problems from the textbook: