Show that wave packet is an eigenstate to operator

[Bapelsin]

Show that wave packet is an eigenstate to operator [SOLVED]

Homework Statement

For a harmonic oscillator we can define the step up and down operators \hat{a} and \hat{a}^{\dagger} and their action as

\hat{a}=\sqrt{\frac{m\omega}{2\hbar}}(\hat{x}+\frac{\imath}{m\omega}\hat{p}) \quad \hat{a}|n\rangle = \sqrt{n}|n-1\rangle

\hat{a}^{\dagger}=\sqrt{\frac{m\omega}{2\hbar}}(\hat{x}-\frac{\imath}{m\omega}\hat{p}) \quad \hat{a}^{\dagger}|n\rangle = \sqrt{n+1}|n+1\rangle

Show that the Gaussian wave-packet

\Psi(x)=\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}e^{-\frac{m\omega}{2\hbar}(x-x_t)^2+\frac{\imath}{\hbar}p_t(x-x_t)}

is an eigenstate to \hat{a} with eigenvalue \alpha.
\hat{a}|a\rangle = \alpha | a\rangle

Express the eigenvalue in terms of x_t and p_t.

Homework Equations

See above.

The Attempt at a Solution

\hat{p} \rightarrow -i\hbar\frac{\partial}{\partial x}

gives the expression for

\hat{a}=\sqrt{\frac{m\omega}{2\hbar}}(\hat{x}+\frac{\hbar}{m\omega}\frac{\partial}{\partial x})

\hat{a}|\Psi\rangle = \sqrt{\frac{m\omega}{2\hbar}} \left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\left(e^{-\frac{m\omega}{2\hbar}(x-x_t)^2+\frac{\imath}{\hbar}p_t(x-x_t)} + \frac{\hbar}{m\omega}(-\frac{m\omega}{\hbar}(x-x_t)+\frac{\imath}{\hbar}p_t(e^{-\frac{m\omega}{2\hbar}(x-x_t)^2+\frac{\imath}{\hbar}p_t(x-x_t)})\right) = <br /> \Psi\sqrt{\frac{m\omega}{2\hbar}}\left(1+(x-x_t)+\frac{\imath}{m\omega}p_t\right)

which is obviously not a scalar times \Psi.

What am I doing wrong here?

 

[cepheid]

You forgot the x in the first term in parentheses (which comes from the x operator multiplied by Psi).

Here, cut out the clutter (h is supposed to be hbar in what follows):

C = (mω/h)^1/2

A = (h/mω)

Then:

a|Ψ> = C(x + A ∂/∂x)Ψ

= C(xΨ + A∂Ψ/∂x)

= C(xΨ + A[(-mω/h)(x-x_t) + (i/h)p_t]Ψ)

=C(xΨ + A[-A^-1xΨ + A^-1x_tΨ + (i/h)p_tΨ])

See anything that might cancel?

 

[Bapelsin]

Yes. Thank you!