# [✓] Join solutions to the Scrödinger Equation for a barrier potential?

Posted 1 year ago
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 I want to obtain the the solutions to the time independent Schrödinger Equation for a barrier potential. I have created three equations: tise1 = -\[HBar]^2/2 m D[\[Psi]1[x], {x, 2}] == e0 \[Psi]1[x] tise2 = -\[HBar]^2/2 m D[\[Psi]2[x], {x, 2}] == (e0 - v0) \[Psi]2[x] tise3 = -\[HBar]^2/2 m D[\[Psi]3[x], {x, 2}] == e0 \[Psi]3[x] and the boundary conditions bc = {\[Psi]1[0] == \[Psi]2[0], \[Psi]2[a] == \[Psi]3[a], D[\[Psi]1[x], x] == D[\[Psi]2[x], x] /. x -> 0, D[\[Psi]2[x], x] == D[\[Psi]3[x], x] /. x -> a}; ic = {} (I intended to use ic to remove "left moving" waves right of the potential (x>a), but I could no figure out how to do this) Next I used sol = DSolve[ Join[{tise1, tise2, tise3}, bc, ic], {\[Psi]1, \[Psi]2, \[Psi]3}, x] But the results are overwhelming and as far as I can tell: wrong. I would like to be able to get the wave function as an piecewise concatenation of psi 1-3 to add the time development and to calculate transmission and reflection rates.Hopefully this makes sense. Any help (or pointers to such) will be appreciated./Mogens
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Posted 1 year ago
 First of all I think you should write tise1 = -\[HBar]^2/(2 m ) D[\[Psi]1[x], {x, 2}] == e0 \[Psi]1[x] tise2 = -\[HBar]^2/(2 m ) D[\[Psi]2[x], {x, 2}] == (e0 - v0) \[Psi]2[x] tise3 = -\[HBar]^2/(2 m) D[\[Psi]3[x], {x, 2}] == e0 \[Psi]3[x] Then, if e0 and v0 are contstants the solutions of the deqs are linear combinations Sin and Cos - functions yielding 6 constants of integration. You should try to fix these to get a solution to your problem.
Posted 1 year ago
 I think it would be better to use DSolve on each piece separately and then apply the boundary conditions, rather than building them into one DSolve call.
Posted 1 year ago
 Frank,Thanks for the advice, but when I solve them individually I get the same constants of integration for each solution (C[1] and C[2]). I there a way to avoid that, or do I need to replace them "by hand". I suppose something like /. C[1] -> c1 etc./Mogens
Posted 1 year ago
 The documentation for DSolve has this example In[1]:= DSolve[y''[x] == y[x], y[x], x, GeneratedParameters -> d] Out[1]= {{y[x] -> E^x d[1] + E^-x d[2]}} 
 This is just general advice. It is a good idea to rescale your problem, so there is no \[HBar] etc. at all. Also do not use Greek letters, code looks quite unreadable with them.