Expectation value of raising/lowering operators

[quasar_4]

Homework Statement

This has been driving me CRAZY:

Show that \langle a(t)\rangle = e^{-i\omega t} \langle a(0) \rangle

and

\langle a^{\dagger}(t)\rangle = e^{i\omega t} \langle a^{\dagger}(0) \rangle

Homework Equations

Raising/lowering eigenvalue equations:

a |n \rangle = \sqrt{n} |n-1 \rangle
a^{\dagger} |n \rangle = \sqrt{n+1} |n+1 \rangle

Time development of stationary states: psi(x)*exp(-i*En*t/hbar)=psi(x)*exp(-i*w_n*t)

The Attempt at a Solution

Suppose we've got the system in some state \psi(0).

Then expanding into the | n(0)\rangle basis (looking just at a, not a dagger here) we have

\langle a(0) \rangle = \langle \psi(0) | a | \psi(0) \rangle = \langle \sum_k c_k n_k | a | \sum_k c_k n_k \rangle = \sum_k \sum_l c_k^* c_l \langle n_k | a | n_l \rangle = \sum_k \sum_l c_k^* c_l \sqrt{n_l} \langle n_k | a | n_{l-1}\rangle = \sum_k \sum_l c_k^* c_l \sqrt{n_l} \delta_{k, l-1}

so for non-trivial <a(0)>, \langle a(0) \rangle = \sum_k |c_k|^2 \sqrt{n_k}

But now

\langle a(t) \rangle = \langle \psi(0) e^{-i \omega t} | a | \psi(0) e^{-i \omega t} \rangle = \langle \psi(0)| e^{i \omega t} a e^{-i \omega t} | \psi(0) \rangle = \langle \psi(0) | a | \psi(0) \rangle \neq e^{-i \omega t} \langle a(0) \rangle

because as far as I know, a does not act on exp(i*w*t) and the two exponential terms cancel out! I get the same sort of problem with a dagger. So... what's the deal?? This is really bugging me, I'd love to know how to do this problem...

 

[diazona]

First of all, thankyouthankyouthankyou for making a decent attempt at the problem

But now

\langle a(t) \rangle = \langle \psi(0) e^{-i \omega t} | a | \psi(0) e^{-i \omega t} \rangle = \langle \psi(0)| e^{i \omega t} a e^{-i \omega t} | \psi(0) \rangle = \langle \psi(0) | a | \psi(0) \rangle \neq e^{-i \omega t} \langle a(0) \rangle

because as far as I know, a does not act on exp(i*w*t) and the two exponential terms cancel out! I get the same sort of problem with a dagger. So... what's the deal?? This is really bugging me, I'd love to know how to do this problem...

Here's the catch: when you did that calculation, you implicitly assumed that |\psi(0)\rangle was an energy eigenstate with a particular eigenvalue E = \hbar \omega. It's not, in general, an eigenstate, and so it doesn't necessarily evolve according to a single exponential factor e^{-i\omega t}. You can't write
|\psi(t)\rangle = |\psi(0) e^{-i\omega t}\rangle
unless you know that the state is an energy eigenstate.

Before you try to generalize that calculation, take another look at this:

Suppose we've got the system in some state \psi(0).

Then expanding into the | n(0)\rangle basis (looking just at a, not a dagger here) we have

\langle a(0) \rangle = \langle \psi(0) | a | \psi(0) \rangle = \langle \sum_k c_k n_k | a | \sum_k c_k n_k \rangle = \sum_k \sum_l c_k^* c_l \langle n_k | a | n_l \rangle = \sum_k \sum_l c_k^* c_l \sqrt{n_l} \langle n_k | a | n_{l-1}\rangle = \sum_k \sum_l c_k^* c_l \sqrt{n_l} \delta_{k, l-1}

Note that you can calculate \langle a(t)\rangle the same way, just make the coefficients c_k functions of time. Specifically,
|\psi(t)\rangle = \sum_k c_k(t) |n_k\rangle = \sum_k c_k e^{-i\omega_k t} |n_k\rangle
Since this is a harmonic oscillator, you know what \omega_k is in terms of k.

so for non-trivial <a(0)>, \langle a(0) \rangle = \sum_k |c_k|^2 \sqrt{n_k}

Be careful there: you started out with a delta function that requires k = l-1. So you shouldn't be winding up with |c_k|^2, since that results from the term where k = l.