Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 10, Exercise 10.3.5.
In a system with two particles, the state in the basis is given by where is the position of particle . We can define the exchange operator as an operator that swaps the two particles, so that
To find the eigenvalues and eigenvectors of we have
where is the eigenvalue and is the eigenvector. Using the same argument as before, we can write
Equating coefficients in the first and third lines, we arrive at
which gives the same symmetric and antisymmetric eigenfunctions that we had before:
We can derive a couple of other properties of the exchange operator by noting that if it is applied twice in succession, we get the original state back, so that
Thus the operator is its own inverse.
Consider also the two states and . Then
However, the last line is just equal to the inner product of the original states, that is
This means that
Thus is both Hermitian and unitary.
Shankar asks us to show that, for a general basis vector , . One argument could be that, since the basis spans the space, we can express any other vector such as as a linear combination of the vectors, so that applying to means applying it to a sum of vectors, which swaps the two particles in every term. I’m not sure if this is a rigorous result. In any case, if we accept this result it shows that if we start in a state that is totally symmetric (that is, a boson state), this state is an eigenvector of with eigenvalue . Similarly, if we start in an antisymmetric (fermion) state, this state is an eigenvector of with eigenvalue .
Now we can look at some other properties of . Consider
This follows because the operator operates on the first particle in the state which on the RHS of the first line is at position . Thus , that is, returns the numerical value of the position of the first particle, which is . This means that in terms of the operators alone
In the last two lines, the operator is the momentum of particle , and the result follows by applying the operators to the momentum basis state .
For some general operator which can be expanded in a power series of terms containing powers of and/or , we can use 10 to insert between every factor of or . For example
That is, for any operator we have
The Hamiltonian for a system of two identical particles must be symmetric under exchange of the particles, since it represents an observable (the energy), and this observable must remain unchanged if we swap the particles. (In the case of two fermions, the wave function is antisymmetric, but the wave function itself is not an observable. The wave function gets multiplied by if we swap the particles, but the square modulus of the wave function, which contains the physics, remains the same.) Thus we have
[Note that this condition doesn’t necessarily follow if the two particles are not identical, since exchanging them in this case leads to an observably different system. For example, exchanging the proton and electron in a hydrogen atom leads to a different system.]
The propagator is defined as
and the propagator dictates how a state evolves according to
Since the only operator on which depends is , then is also invariant, so that
Multiplying from the left by and subtracting, we get the commutator
For a symmetric state or antisymmetric state , we have
This means that states that begin as symmetric or antisymmetric remain symmetric or antisymmetric for all time. In other words, a system that starts in an eigenstate of remains in the same eigenstate as time passes.