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t1mbro
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Qm, infinite sq well doubles in width (reposted more clearly)
Edit: posting a little clearer. I was asleep when I did this last time
I don't have any notes and have to sit an exam tomorrow so i would appreciate a little help understanding this. I have the answers so i don't need them just a description of how to get to them so I can apply it hopefully to other questions! Thanks for any help
A particle is in the ground state
[itex]
u_1(x)=\left\{\begin{array}{cc}\sqrt{(2/w)}cos[\frac{(\pi)x}{w}],&\mbox{ if }
\frac{-w}{2}<x< \frac{w}{2}\\0, & \mbox{ if } x\leq \frac {-w}{2}, x \geq \frac {w}{2}\end{array}\right.
[/itex]
of a 1D square infinite potential well. The wall Separation is suddenly doubled to [itex]2w[/itex]. The expansion takes palce symetrically so that the centre remains around [itex]x = 0[/itex]
a)explain briefly why the wavefunction immidiatelyafter the wall has moved is [itex]u_1(x)[/itex].
hint: consider the approximate form of the TDSE [itex]i \hbar \Delta \psi \simeq (\hat{H} \psi) \Delta t[/itex]
b)Express [itex]u_1(x)[/itex] as a sum of the eigenfunctions in the new potential well
c)By calculating the appropriate overlap integral determine the probabliltiy that the particle will be found in the new groundstate of the new box.
[ans: [itex]p_1 = \frac{64}{9 \pi^2}[/itex]
Edit: posting a little clearer. I was asleep when I did this last time
I don't have any notes and have to sit an exam tomorrow so i would appreciate a little help understanding this. I have the answers so i don't need them just a description of how to get to them so I can apply it hopefully to other questions! Thanks for any help
A particle is in the ground state
[itex]
u_1(x)=\left\{\begin{array}{cc}\sqrt{(2/w)}cos[\frac{(\pi)x}{w}],&\mbox{ if }
\frac{-w}{2}<x< \frac{w}{2}\\0, & \mbox{ if } x\leq \frac {-w}{2}, x \geq \frac {w}{2}\end{array}\right.
[/itex]
of a 1D square infinite potential well. The wall Separation is suddenly doubled to [itex]2w[/itex]. The expansion takes palce symetrically so that the centre remains around [itex]x = 0[/itex]
a)explain briefly why the wavefunction immidiatelyafter the wall has moved is [itex]u_1(x)[/itex].
hint: consider the approximate form of the TDSE [itex]i \hbar \Delta \psi \simeq (\hat{H} \psi) \Delta t[/itex]
b)Express [itex]u_1(x)[/itex] as a sum of the eigenfunctions in the new potential well
c)By calculating the appropriate overlap integral determine the probabliltiy that the particle will be found in the new groundstate of the new box.
[ans: [itex]p_1 = \frac{64}{9 \pi^2}[/itex]
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