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

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

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

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!