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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:
[tex]\phi'_{0(0)} = \frac{\alpha^{\frac{1}{2}}}{\pi^{\frac{1}{4}}} e^{-\frac{\alpha^2}{2}s^2}[/tex]
To find probability of it being in first excited state, we overlap it with the first excited state:
[tex]\langle \phi'_1|\phi'_{0(0)}\rangle [/tex]
[tex]= \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[/tex]
[tex] = \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[/tex]
Now, we let ##2s = b## and ##a = \frac{\alpha}{\sqrt 2}## and change ##x## to ##x-s## for the second integral:
[tex]=\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][/tex]
[tex]=\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] [/tex]
[tex] = 0[/tex]
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