Indistinguishability in Quantum Mechanics

Question

When describing the defining characteristics of bosons and fermions, I have a problem with the idea of "label switching" - whereby you have the wavefunction for two particles and the particles' labels in the wavefunction are switched and we look at the effect. The wavefunction remains the same: particles are bosons. If not, they are fermions. This is probably a silly question, but if the particles are indistinguishable how can we assign labels to them?

Answer 1

Thats the whole point of dividing the world of particles in bosons and fermions. Because of the fact that they are indistinguishable we could switch two particles and we would NEVER know. So we could switch two particles, let's define the permutation operator $\hat{P}$ that has this effect. Upon applying this operator to a wavefunction that contains two identical particles we get:

$\hat{P}|\psi(x_1,x_2)\rangle = e^{i\delta}|\psi(x_2,x_1)\rangle$.

The phase-factor arises because of the fact that the only thing that we can observe are the probabilities, given by the square of the wave function. This remains unchanged when we add a global phase factor. If we apply the operator $\hat{P}$ again to our wave function we should get our origional wavefunction back:

$\hat{P}^2|\psi(x_1,x_2)\rangle = \hat{P}e^{i\delta}|\psi(x_2,x_1)\rangle = e^{i2\delta}|\psi(x_1,x_2)\rangle$.

This means that our factor $e^{i2\delta}$ should equal1. So the permutation operator yields two possible values (1 or -1) which gives us bosons (+1) and fermions (-1):

$\hat{P}|\psi(x_1,x_2)\rangle = \pm|\psi(x_2,x_1)\rangle$.

Now for the labeling of the particles, that's an artificial labeling we make. This is needed for the calculations, if we want to do interactions between these particles, for example Coulomb interactions, we need these, for Coulomb this gives:

$-\frac{e^2}{r_{i,j}}$

Where $r_{i,j}$ gives the distance between the $i^{th}$ and $j^{th}$.

At the end of all of our calculations we simply add the indistinguishability by summing over all possible permutations of the particles. This way our labeling becomes meaningless for indistinguishable particles, it merely serves as a tool to do the "bookkeeping" within the calculations.

Answer 2

I think your question is best answered by noting that when we model systems of identical particles in quantum mechanics, the notion of labeling the particles is not really necessary and is, in my opinion, a bit misleading because it obscures what is fundamentally going on mathematically.

The main point I am going to make is that when we model systems of particles in general we simply assert that we take the state of the system to live in a Hilbert space that is large enough that it allows for the states of all particles to be present. Then,

for identical particles, we simply assert that the state of the system must live in a certain subspace of this Hilbert space, according to whether the particles are identical bosons or fermions, that captures some well-motivated notion of the particles being identical.

I could describe these constructions very generally, but instead let's consider a two-spin system for the sake of conceptual clarity.

In particular let's consider a system of two identical spin $1/2$ particles. The spin Hilbert space of each individual particle is $\mathcal H_{1/2}$, the two-dimensional spin $1/2$ Hilbert space. We model the Hilbert space of the combined system as a subspace of what's called the tensor product of $\mathcal H_{1/2}$ with itself.

(see this for motivating why we use tensor products for composite systems Should it be obvious that independent quantum states are composed by taking the tensor product?)

In other words, the total Hilbert space is a subspace of $\mathcal H = \mathcal H_{1/2}\otimes\mathcal H_{1/2}$. If you're not familiar with tensor products of Hilbert spaces, then note that what's relevant about this construction for our purposes here is that if $\{|+\rangle, |-\rangle\}$ is a basis for the spin $1/2$ Hilbert space, then the set of all "products" $\{|+\rangle|+\rangle, |+\rangle|-\rangle, |-\rangle|+\rangle, |-\rangle|-\rangle\}$ forms a basis for the tensor product.

We then define the permutation operator $P$ on the tensor product $\mathcal H$ as the unique linear operator whose action on any state of the form $|\psi\rangle|\phi\rangle$ is \begin{align} P|\psi\rangle|\phi\rangle = |\phi\rangle|\psi\rangle \end{align} Finally, we define the bosonic subspace $\mathcal H_+$ of $\mathcal H$ as the subspace of all states $|\psi\rangle$ in $\mathcal H$ for which $$ P|\psi\rangle = +|\psi\rangle $$ and similarly for the fermionic subspace.

At this point, all of this has essentially been pure math. The physical content is then simply that systems of bosons have as their hilbert spaces the bosonic subspace and fermionic systems have as their subspaces the fermionic subspace.