Expectation value of raising/lowering operators

In summary, the conversation discusses a problem involving the calculation of <a(t)> and <a^dagger(t)> in a harmonic oscillator system. The solution involves expanding the state into the basis of energy eigenstates and taking into account the time evolution of the system. It is also important to note that the state must be an energy eigenstate in order for the calculation to be valid.
  • #1
quasar_4
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Homework Statement



This has been driving me CRAZY:

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

and

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

Homework Equations



Raising/lowering eigenvalue equations:

[tex] a |n \rangle = \sqrt{n} |n-1 \rangle [/tex]
[tex] a^{\dagger} |n \rangle = \sqrt{n+1} |n+1 \rangle [/tex]

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 [tex] \psi(0) [/tex].

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

[tex] \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} [/tex]

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

But now

[tex] \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 [/tex]

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...
 
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  • #2
First of all, thankyouthankyouthankyou for making a decent attempt at the problem o:)

quasar_4 said:
But now

[tex] \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 [/tex]

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

Before you try to generalize that calculation, take another look at this:
quasar_4 said:
Suppose we've got the system in some state [tex] \psi(0) [/tex].

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

[tex] \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} [/tex]
Note that you can calculate [itex]\langle a(t)\rangle[/itex] the same way, just make the coefficients [itex]c_k[/itex] functions of time. Specifically,
[tex]|\psi(t)\rangle = \sum_k c_k(t) |n_k\rangle = \sum_k c_k e^{-i\omega_k t} |n_k\rangle[/tex]
Since this is a harmonic oscillator, you know what [itex]\omega_k[/itex] is in terms of [itex]k[/itex].
quasar_4 said:
so for non-trivial <a(0)>, [tex] \langle a(0) \rangle = \sum_k |c_k|^2 \sqrt{n_k} [/tex]
Be careful there: you started out with a delta function that requires [itex]k = l-1[/itex]. So you shouldn't be winding up with [itex]|c_k|^2[/itex], since that results from the term where [itex]k = l[/itex].
 

FAQ: Expectation value of raising/lowering operators

What is the expectation value of a raising/lowering operator?

The expectation value of a raising/lowering operator is a measure of the average value that the operator would yield when applied to a quantum state. It represents the most probable outcome of a measurement of the operator.

How is the expectation value of a raising/lowering operator calculated?

The expectation value of a raising/lowering operator is calculated by taking the inner product of the operator with the quantum state, and then multiplying it by the complex conjugate of the inner product. This value is then normalized by the square root of the inner product of the state with itself.

What is the physical significance of the expectation value of a raising/lowering operator?

The expectation value of a raising/lowering operator is physically significant because it gives an idea of the average energy level of the quantum system. It can also be used to determine the probability of a particular energy level being measured.

Can the expectation value of a raising/lowering operator be negative?

Yes, the expectation value of a raising/lowering operator can be negative. This indicates that the quantum state has a higher probability of being in a lower energy level than the average energy level represented by the expectation value.

How does the expectation value of a raising/lowering operator change with time?

The expectation value of a raising/lowering operator can change with time as the quantum state evolves. This evolution is governed by the Schrödinger equation, and the expectation value can be calculated at any given time by using the time-dependent wavefunction of the system.

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