# The triple Dirac delta-function well: getting KAPPA!

1. Oct 11, 2015

### Lylo

1. The problem statement, all variables and given/known data

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.

2. Relevant 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).

3. 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.

2. Oct 14, 2015

### 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.

3. Oct 15, 2015

### 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 $\frac{\psi_{IV}(s>1)}{ds}.$

Last edited: Oct 15, 2015
4. Oct 16, 2015

### 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 $g=5.$

Last edited: Oct 16, 2015
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