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Qm, infinite sq well doubles in width

  • Thread starter 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 dont 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]
 
Last edited:

Answers and Replies

Have you visualised the solution as a linear superposition of the eigen states of the the new well?

I agree, as you seem to state, that all the sine states have no components in this solution. I'm not sure how you approximate the TDSE as the full one seems to give you everything you want fairly quickly and easily.
 
See http://www-atm.physics.ox.ac.uk/user/jpaulin/Applet.html [Broken] from some fun with a quantum wave packet
 
Last edited by a moderator:
6
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Still having a little problem with part (a), anyone any ideas?
 
when you say you are having a problem with part (a) does that mean that you have done part (b)?
 

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