Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 11.
One consequence of the invariance of the Hamiltonian under translation is that the momentum and Hamiltonian commute:
In quantum mechanics, commuting quantities are simultaneously observable, and we can find a basis for the Hilbert space consisting of eigenstates of both and . We’ve seen that Ehrenfest’s theorem allows us to conclude that for such a system, the average momentum is conserved so that . We can go a step further and state that if a system starts out in an eigenstate of , then it remains in that eigenstate for all time.
First, we need to make a rather subtle observation, which is that
That is, if and commute, then also commutes with the propagator . For a time-independent Hamiltonian, the propagator is
Since this can be expanded in a power series in the Hamiltonian, condition 2 follows easily enough. What if the Hamiltonian is time-dependent? In this case, the propagator comes out to a time-ordered integral
Here the time interval is divided into time slices, each of length . As explained in the earlier post, the reason we can’t just integrate the RHS directly by summing the exponents is that such a procedure works only if the operators in the exponents all commute with each other. If is time-dependent, its forms at different times may not commute, so we can’t get a simple closed form for .
However, if for all times, then commutes with all the exponents on the RHS of 4, so we still get . Another way of looking at this is by imposing the condition we’re saying that if can be expanded in a power series in and , it depends only on , and not on . This follows from the fact that
so that does not commute with any power of .
Given that 2 is valid for all Hamiltonians, then if we start in a eigenstate of , then
Thus remains an eigenstate of with the same eigenvalue for all time. For a single particle moving in one dimension, the state describes a free particle with momentum (and thus a completely undetermined position).