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henpenBest ResponseYou've already chosen the best response.0
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 groundstate lavelength (4a). What have I done incorrectly?
 one year ago

JamesJBest ResponseYou've already chosen the best response.1
You haven't written down the correct general solution. You can also have terms \[ ... + B \cos(\omega x) \] with \( \omega = \sqrt{2mE}/\hbar \).
 one year ago

JamesJBest ResponseYou've already chosen the best response.1
This will give you the 4a, as cos(omega.x) = 0 => omega.x = pi/2 + n.pi
 one year ago

henpenBest ResponseYou've already chosen the best response.0
Aren't you assuming A=0 for that solution? \[Asin(\pi/2 \pm n\pi) \ne 0\]
 one year ago

JamesJBest ResponseYou've already chosen the best response.1
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.
 one year ago

JamesJBest ResponseYou've already chosen the best response.1
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
 one year ago

henpenBest ResponseYou've already chosen the best response.0
Ah, that's it! Thanks. What are the boundary conditions? That \[\varphi'(a)=\varphi(a)=0\] Because \[\varphi(x)=0\] for \[x>a, x<a\]?
 one year ago

henpenBest ResponseYou've already chosen the best response.0
*\[\varphi'(a)=\varphi'(a)=0\]
 one year ago

henpenBest ResponseYou've already chosen the best response.0
That guess came from avoiding making the first derivative jump about at + a
 one year ago

JamesJBest ResponseYou've already chosen the best response.1
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.
 one year ago

henpenBest ResponseYou've already chosen the best response.0
\[A \sin( \omega a)+B \cos( \omega a)\] 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.
 one year ago

henpenBest ResponseYou've already chosen the best response.0
@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
 one year ago
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