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Delta potential barrier in a box

  • Thread starter tommy01
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  • #1
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hi,...
i unfortunately couldn't find a solution to this problem although it seems like a classical textbook problem...

how can i solve the (time independent) schroedingerequation for the following potential

[tex]V(x) = \infty[/tex] for x<=-1
[tex]V(x) = a\delta(x)[/tex] for -1<x<1
[tex]V(x) = \infty[/tex] for x>=1

so at x=0 there is a barrier.

i thought i can set the solutions to the left and right of x=0 for a free particle but i don't know how to handle x=0.

thanks,...
tommy
 
Last edited:

Answers and Replies

  • #2
2,111
16
I haven't done this, but I think I can see how it can be done. Basically you need to define the wave function [tex]\psi(x)[/tex] in parts, for intervals -1<x<0 and 0<x<1 separately (they will be trigonometric functions), so that they vanish at the end points x=-1 and x=1, and that you get certain kind of sharp angle at the x=0. You should somehow solve how big jump there has to be in the derivative [tex]\psi'(x)[/tex] at the origin (this could be the most understanding demanding part. You can ask more questions if you don't know how to calculate with equations like [tex]D_x\theta(x)=\delta(x)[/tex] and so on). After this, substitute in some attempt of trigonometric functions with some constants, and try to figure out what set of equations you get for the constants.
 
  • #3
Gokul43201
Staff Emeritus
Science Advisor
Gold Member
7,051
16
hi,...
i unfortunately couldn't find a solution to this problem although it seems like a classical textbook problem...

how can i solve the (time independent) schroedingerequation for the following potential

[tex]V(x) = \infty[/tex] for x<=-1
[tex]V(x) = a\delta(x)[/tex] for -1<x<1
[tex]V(x) = \infty[/tex] for x>=1

so at x=0 there is a barrier.

i thought i can set the solutions to the left and right of x=0 for a free particle but i don't know how to handle x=0.

thanks,...
tommy
You need to apply BCs to determine the coefficients of the plane wave states. But there's one trick here: while the wave function itself is continuous at x=0, its derivative is not. The discontinuity in [itex]\partial \psi/\partial x [/itex] at x=0 (due to the delta function potential there) can be found by integrating the SE over a tiny window about x=0 and taking the limit that the width of this window goes to zero.
 
  • #4
3
0
i think the idea here is this: you have and incident wave, and reflected wave and a transmitted wave. split the problem into two parts. in region (1) (ie before the potential) the wave could be of the form q(x) = Aexp(ikx) + bexp(-ikx) (+ikx for he incident, -ikx for reflected). and in region (2) (after the potential) q(x) = Cexp(ikx) (transmitted wave).
one boundary conditon obviously is continuity of q(x), but the bc of the first derivative is found as said by Golkul43201.
 
  • #5
2,111
16
It is not necessary to start with plane waves. You would then have to solve suitable superpositions to get the boundary conditions at the x=-1 and x=1 correctly. But we already know that the wave function is going to be trigonometric functions that vanish at these end points. So the attempt

[tex]
\psi(x) = A\sin(B(x+1)),\quad -1<x<0
[/tex]

[tex]
\psi(x) = C\sin(D(x-1)),\quad 0<x<1
[/tex]

gives a good start. The task that remains to be carried out, is that one must solve what constants A,B,C,D satisfy the correct boundary condition at the origin x=0.
 
  • #6
3
0
oh yeah of course, they aren't normalizable. i completely spaced on the fact that this is in a box. obviously we need bound states... sorry to the op.
 
  • #7
Gokul43201
Staff Emeritus
Science Advisor
Gold Member
7,051
16
Oops. I did too. Nevertheless, the procedure I outlined ought to work.
 
  • #8
40
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thanks for your answers.

i'll outline the solution method i found for later viewers:

this potential can be handled exactly the same way as the infinite square well with the exception of having 2 conditions at x=0. one to have a continuous wave function so [tex]\Psi_{left}(0)=\Psi_{right}(0)[/tex] and the other one to manage the delta potential which demands for a discontinuous first derivation of [tex]\Psi(x)[/tex] at x=0. this discontinuity condition is obtained by integrating over the schroedingerequation.

thanks again and greetings.
tommy
 

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