The triple Dirac delta-function well: getting KAPPA

[Lylo]

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

e^ks
for s<-1

Be^ks + Ce^-ks
for -1<s<0

Be^-ks + Ce^ks
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.

 

[Lylo]

Hello,

I was going to edit my first post again, but it seems as though I cannot. I cannot find the edit option again, at least.

While walking around campus the day after I posted this, a few ideas hit me which I will be trying out tonight. I also realized some issues with my post, which I will clear up here:

The Schrödinger equation is incorrect, because of an oversight on my part. The potential V(s) is ## \frac{\hbar^2g}{2ma} (η), ## where η denotes the scaled delta-functions; however, the Schrödinger equation becomes $$ \epsilon\psi = \frac{-\partial^2\psi}{\partial s^2} - g[η]\psi $$ and we note that ## \partial(sa) → \frac{1}{|a|}\partial s .## Of course, g ∈ ℤ and is constant.

 

[Lylo]

Hello,

I have made some more additions to my list of equations in my efforts for obtaining a transcendental in κ.

Given the wavefunction 'pieces' for the even solutions, we also have: $$ \psi_e = \begin{cases}
κe^{κs} & s < -1\\
κBe^{κs} - κCe^{-κs} & -1 < s < 0\\
-κBe^{-κs} + κCe^{κs} & 0 < s < 1 \\
-κe^{-κs} & 1 <s \\
\end{cases}. $$
Applying the continuity equation about the discontinuity in ## \frac{d\psi}{ds} ## at ## s=0 ## gives:

$$ \begin{cases}
\frac{d\psi}{ds}|_+ =& -κB + κC \\
\frac{d\psi}{ds}|_- =& κB - κC\\

\end{cases} \\ Δ\frac{d\psi}{ds} = (-κB + κC) - (κB - κC) = -2κB + 2κC; \\ \text{or, using } B = 1 - Ce^{2k}: \\ -2κ + 2κCe^{2κ} + 2κC.$$
This implies that $$ -2κ + 2κCe^{2κ} + 2κC = -g(1 - Ce^{2κ} + 2κC), $$
which I obtain as leading to: $$ \frac{2κ}{g} = \frac{1-C(e^{2κ}+1)}{1-C(e^{2κ}-1)}. $$
This is close to an equation involving a ## \tanh 2k ,## but I haven't been able to produce such an equation.

Any tips?

EDIT: I also decided to exploit the continuity of the derivatives of ##\psi## at finite potentials, as well as the symmetry in the wavefunctions about ##s=0:## $$ κBe^{-k/2} - κCe^{k/2} = -(-κBe^{-k/2} + κCe^{κ/2}) → 0 = 0,$$
with an equivalent result for ##\frac{d\psi_I(s<-1)}{ds}## and ##\frac{\psi_{IV}(s>1)}{ds}. ##

 

[Lylo]

Hello,

For those who are interested, I finally pieced together something which seems okay for ##\kappa.## Recall from post # 1 that ## g = 2C\kappa e^{2\kappa}; ## using this fact we can also obtain a value for ##C## upon inspection: ## C=\frac{g}{2\kappa}e^{-2\kappa} ##. I have already proven that $$ \frac{2\kappa}{g} = \frac{1-C(e^{2\kappa}+1)}{1-C(e^{2\kappa}-1)}, $$ so implementing our new expression for ##C## into our equation of ##\frac{2\kappa}{g}## yields $$ 2\kappa [1-\frac{g}{2\kappa}(1+e^{-2\kappa})] = g[1-\frac{g}{2\kappa}(1-e^{-2\kappa})].$$

Now, for ##g=5##, Wolfram returns three roots for this transcendental in ##\kappa##:## \begin{align}
\kappa &≈ -2.42392 \\
\kappa &≈ 2.0756 \\
\kappa &≈2.73481
\end{align}.##

We know that ## \kappa = \frac{\sqrt{-2mE}}{\hbar};## so for our case in epsilon (## \epsilon=\frac{2mEa^2}{\hbar^2} ##): ## \kappa = \frac{\sqrt{\frac{-2m \hbar^2 \epsilon}{2ma^2}}}{a} \rightarrow \kappa = \frac{\sqrt{- \epsilon}}{a}.##

The only restriction for ##\kappa## is ##\kappa ∈ ℝ.## Our ##\epsilon ## is assumed to be ## < 0.## It is trivial to show that ## - \epsilon = a^2\kappa^2 \rightarrow \epsilon = -a^2\kappa^2.## The negative sign ensures we always recover a negative value for ## \epsilon ## notwithstanding the sign of ## \kappa.## So, it looks like our three ##\kappa##'s are in favor of three bound states at ##g=5.##