Translational invariance and conservation of momentum

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:

\displaystyle \left[P,H\right]=0 \ \ \ \ \ (1)

In quantum mechanics, commuting quantities are simultaneously observable, and we can find a basis for the Hilbert space consisting of eigenstates of both {P} and {H}. We’ve seen that Ehrenfest’s theorem allows us to conclude that for such a system, the average momentum is conserved so that {\left\langle \dot{P}\right\rangle =0}. We can go a step further and state that if a system starts out in an eigenstate of {P}, then it remains in that eigenstate for all time.

First, we need to make a rather subtle observation, which is that

\displaystyle \left[P,H\right]=0\rightarrow\left[P,U\left(t\right)\right]=0 \ \ \ \ \ (2)


That is, if {P} and {H} commute, then {P} also commutes with the propagator {U\left(t\right)}. For a time-independent Hamiltonian, the propagator is

\displaystyle U\left(t\right)=e^{-iHt/\hbar} \ \ \ \ \ (3)

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

\displaystyle U\left(t\right)=T\left\{ \exp\left[-\frac{i}{\hbar}\int_{0}^{t}H\left(t^{\prime}\right)dt^{\prime}\right]\right\} \equiv\lim_{N\rightarrow\infty}\prod_{n=0}^{N-1}e^{-i\Delta H\left(n\Delta\right)/\hbar} \ \ \ \ \ (4)


Here the time interval {\left[0,t\right]} is divided into {N} time slices, each of length {\Delta=t/N}. 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 {H} is time-dependent, its forms at different times may not commute, so we can’t get a simple closed form for {U\left(t\right)}.

However, if {\left[P,H\left(t\right)\right]=0} for all times, then {P} commutes with all the exponents on the RHS of 4, so we still get {\left[P,U\left(t\right)\right]=0}. Another way of looking at this is by imposing the condition {\left[P,H\left(t\right)\right]=0} we’re saying that if {H\left(t\right)} can be expanded in a power series in {X} and {P}, it depends only on {P}, and not on {X}. This follows from the fact that

\displaystyle \left[X^{n},P\right]=i\hbar nX^{n-1} \ \ \ \ \ (5)

so that {P} does not commute with any power of {X}.

Given that 2 is valid for all Hamiltonians, then if we start in a eigenstate {\left|p\right\rangle } of {P}, then

\displaystyle P\left|p\right\rangle \displaystyle = \displaystyle p\left|p\right\rangle \ \ \ \ \ (6)
\displaystyle PU\left(t\right)\left|p\right\rangle \displaystyle = \displaystyle U\left(t\right)P\left|p\right\rangle \ \ \ \ \ (7)
\displaystyle \displaystyle = \displaystyle U\left(t\right)p\left|p\right\rangle \ \ \ \ \ (8)
\displaystyle \displaystyle = \displaystyle pU\left(t\right)\left|p\right\rangle \ \ \ \ \ (9)

Thus {U\left(t\right)\left|p\right\rangle } remains an eigenstate of {P} with the same eigenvalue {p} for all time. For a single particle moving in one dimension, the state {\left|p\right\rangle } describes a free particle with momentum {p} (and thus a completely undetermined position).

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