anonymous
  • anonymous
What's wrong with this infinite square well?
Physics
schrodinger
  • schrodinger
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anonymous
  • anonymous
|dw:1360436634776:dw|\[φ''=-\frac{2mE}{ℏ^2} φ\] \[ \varphi= A \sin(\frac{\sqrt{2mE}}{ℏ}x)\] \[\varphi(a)=\varphi(-a)=0\] \[ \pm\frac{\sqrt{2mE}}{ℏ}a=n \pi\] \[ \frac{\sqrt{2mE}}{ℏ}=\pm\frac{n \pi}{a}\] \[\lambda= \frac{2 \pi}{\pm\frac{n \pi}{a}}=\pm\frac{2a}{n}\] Which, for n=1, is shorter than the real ground-state lavelength (4a). What have I done incorrectly?
JamesJ
  • JamesJ
You haven't written down the correct general solution. You can also have terms \[ ... + B \cos(\omega x) \] with \( \omega = \sqrt{2mE}/\hbar \).
JamesJ
  • JamesJ
This will give you the 4a, as cos(omega.x) = 0 => omega.x = pi/2 +- n.pi

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anonymous
  • anonymous
Aren't you assuming A=0 for that solution? \[Asin(\pi/2 \pm n\pi) \ne 0\]
JamesJ
  • JamesJ
The general solution of y '' + w^2.y = 0 is y(x) = A.sin(wx) + B.cos(wx) or y(x) = A.cos(wx + phi0) Either way, there are two parameters. You are applying one of the boundary conditions. You actually are going to apply another boundary condition to y'(x) in order to completely nail down the solution.
JamesJ
  • JamesJ
Alternatively, the two conditions you are using phi(-a) = 0, phi(a) = 0. From this, you get two families of solutions. One with sin, the other with cos. This equivalent to using the second general solution with the phi0 alternating between 0 and pi/2
anonymous
  • anonymous
Ah, that's it! Thanks. What are the boundary conditions? That \[\varphi'(-a)=\varphi(a)=0\] Because \[\varphi(x)=0\] for \[x>a, x<-a\]?
anonymous
  • anonymous
*\[\varphi'(a)=\varphi'(-a)=0\]
anonymous
  • anonymous
That guess came from avoiding making the first derivative jump about at +- a
JamesJ
  • JamesJ
You have two conditions, which solving for the two parameters in the general solution. I recommend you start with A.cos(wx + phi0) and you'll get what you want.
anonymous
  • anonymous
\[A \sin( \omega a)+B \cos( \omega a)\] It is harder to make this expression = 0 in this form than yours, as both summands cannot be made = 0 or the negative of the other by eyeballing it, so yours is simpler. Thanks, I think I've got it.
anonymous
  • anonymous
@JamesJ , sorry to tag you, but I have a (very) brief question: the general solution is \[ \varphi= A e^{\pm i \omega x}\] What is done with \[\Im(\varphi)\]? Is only the real part considered, as is often done in classical mechanics? Thanks again

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