Harmonic Oscillator, overlap in states

[unscientific]

Homework Statement

Particle originally sits in ground state about x=0. Equilibrium is suddenly shifted to x=s. Find probability of particle being in new first excited state.

Homework Equations

The Attempt at a Solution

Shifted wavefunctions are for ground state: $\phi'_0 = \frac{\alpha^{\frac{1}{2}}}{\pi^{\frac{1}{4}}} e^{ -\frac{\alpha^2}{2}(x-s)^2}$ and for first excited state: $\phi'_1 = \frac{2\alpha^{\frac{3}{2}}}{\pi^{\frac{1}{4}}} (x-s) e^{-\frac{\alpha^2}{2} (x-s)^2}$.

When the equilibrium is shifted to x=s, the particle's ground state at x=0 wavefunction becomes:

$$\phi'_{0(0)} = \frac{\alpha^{\frac{1}{2}}}{\pi^{\frac{1}{4}}} e^{-\frac{\alpha^2}{2}s^2}$$

To find probability of it being in first excited state, we overlap it with the first excited state:

$$\langle \phi'_1|\phi'_{0(0)}\rangle$$
$$= \frac{2\alpha^2}{\pi^{\frac{1}{2}}} e^{-\frac{(\alpha s)^2}{2}} \int_{-\infty}^{\infty}(x-s) e^{-\frac{\alpha^2}{2}(x-s)^2 } dx$$
$$= \frac{2\alpha^2}{\pi^{\frac{1}{2}}} e^{-\frac{(\alpha s)^2}{2}} \left[ \int_{-\infty}^{\infty} (x-2s)e^{-\frac{\alpha^2}{2}(x-s)^2} dx + \int_{-\infty}^{\infty}s \space e^{-\frac{\alpha^2}{2}(x-s)^2} \right] dx$$

Now, we let $2s = b$ and $a = \frac{\alpha}{\sqrt 2}$ and change $x$ to $x-s$ for the second integral:

$$=\frac{2\alpha^2}{\pi^{\frac{1}{2}}} e^{-\frac{(\alpha s)^2}{2}} \left[ -\frac{(2s)\pi^{\frac{1}{2}}}{\sqrt 2 \alpha} dx + s \int_{-\infty}^{\infty} e^{-\frac{\alpha^2}{2}x'^2} dx' \right]$$

$$=\frac{2\alpha^2}{\pi^{\frac{1}{2}}} e^{-\frac{(\alpha s)^2}{2}} \left[ -\frac{\sqrt 2 \pi^{\frac{1}{2}}s}{\alpha} + s\pi^{\frac{1}{2}} \frac{\sqrt 2}{\alpha} \right]$$

$$= 0$$

 

[dauto]

What you're looking for is $\langle \phi'_1|\phi_{0}\rangle^2$

 

[unscientific]

What you're looking for is $\langle \phi'_1|\phi_{0}\rangle^2$

Why is it $\phi_0$ and not $\phi_0'$?

 

[dauto]

Because the particle was in the state $\phi_0$, not $\phi_0'$

 

[unscientific]

Because the particle was in the state $\phi_0$, not $\phi_0'$

Got it, thanks!

 

[unscientific]

What you're looking for is $\langle \phi'_1|\phi_{0}\rangle^2$

I did that, and I got an answer of $2(\alpha s)^2$..