# Expectation value of raising/lowering operators

#### quasar_4

1. The problem statement, all variables and given/known data

This has been driving me CRAZY:

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

and

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

2. Relevant 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)

3. 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

Homework Helper
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 $k = l$.

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