- #1
Lylo
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Homework Statement
I will try to be light on the math as I am just now getting into using LaTex, and I don't want things to get too ugly from me not using it.
Hello there,
I found a thread here on PF concerning a triple delta-function potential well problem, which was a bit informative. However, my quantum professor wants us to solve several problems related to such a potential, introducing the non-dimensionalizing substitution s = x/a to 'scale the variables', resulting in the problem specified by
d2Ψ/ds2 - (ħ2/2ma2) g[∂(s) + ∂(s-1) + ∂(s+1)]Ψ = εΨ ,
where we see our scaled energy in epsilon, and g is a constant integer which describes the delta function's "strength".
I am having some trouble getting to a useful kappa, as I have done for the double potential well. I feel as though I am overlooking something very 'obvious' yet crucial to obtaining the transcendental equation for kappa, so I hope maybe someone else can offer some hints.
Homework Equations
Manifestly there exist four distinct regions of finite potentials; at s=±1 and s=0 we have the delta wells. In a previous assignment we demonstrated that the even solutions can be written as
eks
for s<-1
Beks + Ce-ks
for -1<s<0
Be-ks + Ceks
for 0<s<1
and
e-ks
for 1<s
and that's what I wish to look at here.
We also make use of the continuity equation
## \Delta (\frac{d\psi(s)}{ds})=-g(\psi(s)) ##
and note that our wavefunction must be continuous at the boundaries, and their derivatives are also continuous where their respective wavefunctions are finite (i.e., not at a delta well).
The Attempt at a Solution
Since we expect continuity at s=-1, I will use the first and second wavefunction 'pieces', evaluating them at that point and setting them equal:
## \psi_I(-1) = \psi_I{}_I(-1) ## yields $$ e^k{}^s = Be^k{}^s + Ce^-{}^k{}^s \\ e^-{}^k = Be^-{}^k + Ce^k \rightarrow 1 = B + Ce^2{}^k \rightarrow B = 1 - Ce^2{}^k ;$$
note that we recover this result for s=1, as well. At s=0, we have the unenlightening $$ B+C = B+C .$$
By the by, for good measure we can also write (from our experience with s=-1) $$ C=Be^-{}^2{}^k + e^-{}^2{}^k .$$
We note the discontinuity in ## \psi'(s) ## at, say, s=-1: ## \Delta (\frac{d\psi(1)}{ds})=-g(\psi(1)), ## meaning we need to take derivatives of psi one and psi two, and then subtract the former from the latter: $$ \psi' {}_I{}_I = ke^k{}^s - Ce^2{}^ke^k{}^s(k)-Cke^-{}^k{}^s \\ \psi' {}_I = ke^k{}^s$$ and so we get, after subtracting these two and equating their difference to ## -g\psi(-1) ## $$ -g = -2Cke^2{}^k .$$
It isn't immediately clear to me if this will be useful in getting kappa. I reckon I need to get rid of any B or C dependencies for my final transcendental in kappa, and I have a plethora of what my tyro's eye sees as a "dead-end". Perhaps I have been staring at this stuff for too long. When I did the double delta well problem a year or two ago, I remember some twisty algebra was involved in getting rid of my coefficients in the kappa equation.
Does it look like I am on the right track? Have I missed something, and need a nudge in the right direction? I have always had a bit of trouble playing with groups of equations and trying to construct an equation with only certain components; I am dyslexic so it seems really convoluted to my mind!
Thanks in advance for your help. I'll give this a few re-reads in the hopes that I'll weed out any sneaky typos.
EDIT: I also have tried getting additional equations by exploiting the continuity of the wavefunction's derivatives for finite potentials - i.e., at s=1/2. I am trying to find one of my results doing so, now. If I can find it, I'll add it to the post; if not, I'll do it again when I get time to, provided that it's necessary/useful.