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chris_avfc
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Homework Statement
Already defined that for a 1D well with one finite wall the eigenvalue solutions are given by
k cot(kl) = -α
Show the eigenvalue solutions to well with both walls finite is given by
tan(kl) = 2αk / (k^2 - α^2)
Well is width L (goes from 0 to L) with height V_0
Homework Equations
k cot(kl) = -α
tan(kl) = 2αk / (k^2 - α^2)
Time Independent Schrödinger Equation.
The Attempt at a Solution
I will explain briefly as there is a lot of equations and it will look a mess, I can always upload a picture of my work if needs be.
General Solution:
u(x) = Ce^-αx + De^αx
For x < 0
C = 0, so exponential doesn't go to ∞.
For x > L
D = 0 so exponential doesn't go to ∞.
Using boundary conditions where the function and it's derivative must be continuous
With the solution between 0 and L being
u(x) = A sin(kx)
Substituting in L for x
A sin(kL) = Ce^-αL
Ak cos(kL) = -αCe^-αL
Dividing the second by the first
k cot(kL) = -α (As given in question)
This breaks down when substituting in 0 for x.
A sin(k0) = De^α0
Ak cos(k0) = αDe^α0
0 = αD
I thought about changing the coordinates so the well runs from -L/2 to L/2, but that just appears to give
k cot(kL/2) = -α
And
k cot(kl/2) = α
I was/am looking in one textbook, and that manages to get it down to
k tan (kL/2)= α
k cot (kL/2)= -α
Still not what I need, but might be closer?