Delta well + infinite barrier -> bound state

  • #26
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How shall I say? For a potential of infinite range, the statement only holds asymptotically. But for a potential of finite range d, if A exp (ikx)+B exp(ikx) =C sin (kx+phi) holds asymptotically (i.e. abs(A)=abs(B)) then it holds down to x=d.
Thanks for your reply DrDu. That does clear things up a bit. But how does one get a transcendental equation for [itex]\phi[/itex] which 'cannot be solved exactly'? I seem to have found one which can be solved exactly for [itex]\phi[/itex]. They seem to want something like [itex]\phi = f(\phi)[/itex]. Do you see something wrong in my reasoning?
 
  • #27
DrDu
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As you keep complaining I just tried to solve the problem for the bound state.
so for x<d psi is A sinh kx and for x>d psi is B exp(-kx).
The condition from the delta function is (with the limit epsilon to 0 understood)
[tex] \frac{\psi'(d+\epsilon)}{\psi(d)}-\frac{\psi'(d-\epsilon)}{\psi(d)}=\frac{2m\lambda}{\hbar^2}[/tex] or
[tex] -k - k \coth(kd)=\frac{2m\lambda}{\hbar^2}[/tex].
That looks quite transcendental and different from what you ever proposed.
 
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  • #28
DrDu
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Your post #20 seems to be correct. So the only transcendental content is the arctan.
 
  • #29
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As you keep complaining I just tried to solve the problem for the bound state.
so for x<d psi is A sinh kx and for x>d psi is B exp(-kx).
The condition from the delta function is (with the limit epsilon to 0 understood)
[tex] \frac{\psi'(d+\epsilon)}{\psi(d)}-\frac{\psi'(d-\epsilon)}{\psi(d)}=\frac{2m\lambda}{\hbar^2}[/tex] or
[tex] -k - k \coth(kd)=\frac{2m\lambda}{\hbar^2}[/tex].
That looks quite transcendental and different from what you ever proposed.
I wasn't complaining DrDu, appreciate your help :-). But this is what I did to get that condition for the delta function:

[tex]-\frac{\hbar^2}{2m}\int_{d-\epsilon}^{d+\epsilon}\frac{d^2\psi}{dx^2}dx - \lambda\int_{d-\epsilon}^{d+\epsilon}\psi(x)\delta(x-d) dx = E\int_{d-\epsilon}^{d+\epsilon}\psi(x) dx[/tex]

the rhs = 0 in the limit of epsilon -> 0. This gives the equation I got.

By the way, the equation you wrote in your post does not contain any phase [itex]\phi[/itex]. I assume you have solved only the bound state problem, which I have solved as well. I am now referring to the problem I wrote in post #18 below.
 
  • #30
reilly
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Think of the delta function, with positive lambda, as a very narrow, deep potential well. Use your common sense and experience with the potential well to get the appropriate general form of the solution.
Regards,
Reilly Atkinson
 
  • #31
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Think of the delta function, with positive lambda, as a very narrow, deep potential well. Use your common sense and experience with the potential well to get the appropriate general form of the solution.
Regards,
Reilly Atkinson
Reilly, thanks for your reply, but which part are you referring to?

As I stated in an earlier post, I was able to get the bound state solution. But in post #18, I posed the second part of the problem, which deals with the asymptotic nature of the wavefunction in case of a scattering problem for the same V(x). I solved that and got an equation for the phase shift.

But as far as I can tell, the phase shift can be determined exactly from that equation and unless I have made a mistake in the computation, there is no need to assume that [itex]\phi \approx ak[/itex] and determine [itex]a[/itex] as the question demands.
 

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