# Double-Well Delta Potentials - Schrödinger Equation

Page 177 on Davies' book- Spectral theory of diff operatrs contains the following computation problem: Calculate the negative eigenvalues and the corresponding eigenfunctions of the following operator: $H:= -\frac{d^2 }{dx^2 } -\delta_{-r} -2\delta_{r}$ .

The book gives the calculation for the operator $$-\frac{d^2 }{dx^2 } -2\delta_{-r} -2\delta_{r} \tag{*},$$ but there few things I really need to understand before trying to solve the exercise:

1) In the symmetric potential case, it's valid to assume the eigenfunctions are even or odd... But can we assume the same thing in the asymmetric case?

2) Can we deduce something from writing our new operator as: $H = (-\frac{d^2 }{dx^2 } -2\delta_{-r} -2\delta_{r} )+ \delta_{-r}$ ?

3) In his calculation, Davies' says that the operator (*) has excatly two negative eigenvalues ... How can he see that? Can someone explain me how did he get the boundary conditions- $f'(r+)-f'(r-)=-2f(r)$ , $f'(-r+ ) - f'(-r- )=-2f(-r)$ ? (I understand we have a jump discontinuity of the first derivative (and the second), but how it implies these boundary conditions?

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Note that omitting a dieresis is usually just as much of a misspelling as replacing a vowel by a completely different vowel. In many languages with diacritical marks, including German in this case, the marked vowels have only a historical connection to the unmarked ones and have an unrelated pronunciation. Thus writing "Schrodinger" is about as bad as writing "Schridinger". If you don't have vowels with diaresis on your keyboard, you can always copy them from the Web, e.g. from the corresponding Wikipedia articles. –  joriki Jul 17 '12 at 16:23
OK... Sorry... I fixed it...Thanks! –  joshua Jul 17 '12 at 16:30

1) No, you can't assume that the eigenfunctions are even or odd. You can only do that when the Hamiltonian commutes with the parity operator.

2) I guess you could try deducing something from that, e.g. using perturbation theory with $\delta_{-r}$ as the perturbation, but I wouldn't go down that road; it doesn't seem very promising.

3) I don't know whether he "saw" that the operator has exactly two negative eigenvalues – I presume he says that because he performed the calculation. Regarding the boundary conditions: The jump discontinuity is in the first derivative, not the second; the second derivative has a delta peak at that point, since it has to cancel the delta peak from the potential in the Schrödinger equation. Integrating the second derivative yields the first derivative, and integrating over $-2\delta f$ yields a jump of height $-2f$; since there's no jump in $f$ itself, these two have to be equal, so the first derivative must have a jump of height $-2f$.

[Edit in response to the comment:]

Away from the delta peaks, for negative eigenvalues the solution is a superposition of two exponentials decaying towards positive and negative $x$ values, respectively. There are three regions, left, right and centre. In the left region there can be no leftward increasing component, and in the right region there can be no rightward increasing component. That leaves four unknown amplitudes, of which we can arbitrarily set one to $1$ since the wavefunction will be normalized:

$$f(x)= \begin{cases} \mathrm e^{\lambda x}&x \le -r\;,\\ b_+\mathrm e^{\lambda x}+b_-\mathrm e^{-\lambda x}&-r \lt x \le r\;,\\ c\mathrm e^{-\lambda x}&r \lt x\;.\\ \end{cases}$$

As you wrote, in the symmetric case, we can assume that the wavefunction has definite parity. For positive parity, we get

$$f_+(x)= \begin{cases} \mathrm e^{\lambda x}&x \le -r\;,\\ b\mathrm e^{\lambda x}+b\mathrm e^{-\lambda x}&-r \lt x \le r\;,\\ \mathrm e^{-\lambda x}&r \lt x\;.\\ \end{cases}$$

The continuity condition is

$$\mathrm e^{-\lambda r} = b\mathrm e^{\lambda r}+b\mathrm e^{-\lambda r}\;,$$

and the jump condition that you wrote is

$$-\lambda\mathrm e^{-\lambda r}-\left(\lambda b\mathrm e^{\lambda r}-\lambda b\mathrm e^{-\lambda r}\right)=-2\mathrm e^{-\lambda r}\;.$$

(There's only one of each now because of the symmetry.)

We can solve the first condition for $b$ and substitute it into the second:

$$b=\frac{\mathrm e^{-\lambda r}}{\mathrm e^{\lambda r}+\mathrm e^{-\lambda r}}\;,$$

$$-\lambda\mathrm e^{-\lambda r}-\lambda\frac{\mathrm e^{-\lambda r}}{\mathrm e^{\lambda r}+\mathrm e^{-\lambda r}}\left(\mathrm e^{\lambda r}-\mathrm e^{-\lambda r}\right)=-2\mathrm e^{-\lambda r}\;.$$

Dividing by $-\lambda\mathrm e^{-\lambda r}$ yields

\begin{align} 1+\frac{\mathrm e^{\lambda r}-\mathrm e^{-\lambda r}}{\mathrm e^{\lambda r}+\mathrm e^{-\lambda r}} &= \frac2\lambda\;,\\\\ \frac{\mathrm e^{\lambda r}}{\mathrm e^{\lambda r}+\mathrm e^{-\lambda r}} &= \frac1\lambda\;,\\\\ 1+\mathrm e^{-2\lambda r}&=\lambda\;. \end{align}

Since the left-hand side is strictly decreasing and the right-hand side is strictly increasing, this equation has exactly one solution, which for $r=1$ Wolfram|Alpha locates at $\lambda \approx1.10886$.

You can do the same thing for negative parity to find the second eigenvalue. In the asymmetric case, you'll have to do a little more work, since you have to keep all three constants if you can't use symmetry to simplify.

P.S.: If you're wondering how come $\lambda$ occurs without $r$ in the equation even though $r$ seems to be the only length scale in the problem: There's a hidden length scale because the jump in $f'$ should have units of inverse length, so the delta strength $2$ introduces a characteristic length $1/2$.

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Can you please exaplin me what is the calculation one should do in order to "see" that there excatly two negative eigenvalues? I think I'll manage to do everything else...Thanks –  joshua Jul 17 '12 at 16:31
@joshua: I guess somewhere in your book there is the statement that a $\delta$ binds exactly a single bound state. So two $\delta$s bind ...? –  Fabian Jul 17 '12 at 19:16
@Fabian: Does it really work like that? Do $n$ deltas bind $n$ bound states? –  joriki Jul 17 '12 at 19:26
OK... I think I got it, except for the parity thing... If I can't use it in the assymetric case, so I'll have 4 eigenfunctions and 4 eigenvalues? Is there any book or paper that covers the calculation of such a thing? Thanks a lot again ! –  joshua Jul 17 '12 at 19:39
@joriki: yes, if they are sufficiently far (further than the decay length of the bound state) apart from each other. Obviously, if you bring the two $\delta$'s on the same spot, you get a $\delta$ with double strength which still has a single bound state (the antibonding state merges at some point with the continuum). –  Fabian Jul 17 '12 at 21:27

OK... Here is my attemp to solve this question for the asymmetric case: So also in this case we are given three regions: $I) x<-r \qquad II) -r\leq x \leq r \qquad III) x>r$ . In both regions $I,III$ , we can regard the potential as being $0$ , and hence we get a regular ode which can be solved directly, and as you said, the solution will be a superposition of two exponential functions. After ommiting the "blow-up" cases, we are left with solutions of the form: $f(x)= \begin{cases} \mathrm e^{\lambda x}&x \lt -r\;,\\ b_+\mathrm e^{\lambda x}+b_-\mathrm e^{-\lambda x}&-r \le x \le r\;,\\ c\mathrm e^{-\lambda x}&r \lt x\;.\\ \end{cases}$ .

We now have continuity conditions-

(1) $e^{-\lambda r } = b_+e^{-\lambda r } + b_- e^{ \lambda r }$

(2) $b_+ e^{ \lambda r}+ b_- e^{-\lambda r}= ce^{- \lambda r }$

As for the third and fourth condition, we get (by integrating over small interval):

(3) $f' (-r+ ) - f'(-r-) = -f(-r)$. Hence: $\lambda b_+ e^{-\lambda r} - \lambda b_- e^{\lambda r} - \lambda e^{-\lambda r} = - e^{-\lambda r}$

(4) $f'(r+ ) -f'(r-) = -2f(r)$ , Hence: $- \lambda c e^{-\lambda r} -\lambda b_+ e^{ \lambda r} + \lambda b_- e^{-\lambda r } = -2ce^{-\lambda r}$ .

We received four equations with 4 variables ($\lambda, b_- , b_+ , c$) , which implies we have a unique solution...

Am I right? Does this imply we have only one eigenvalue in this asymmetric case? Is there any elegant trick I can do in order to solve these equations?

Hope you're still here and can verify my answer

Thanks a lot both of you !

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It all looks correct to me -- up to the part where you count solutions. You can't count that way because $\lambda$ doesn't occur linearly. You have to eliminate $b_-$,$b_+$ and $c$ (which you can since they occur linearly), and the result will be an equation for $\lambda$ that may have any number of solutions. We know that it would have two solutions if the strength of the second delta were $-2$ instead of $-1$, and one solution if it were $0$ instead, so I'd guess that you'll find either one or two solutions. There's only one way to find out :-) –  joriki Jul 18 '12 at 17:26
Regarding how to solve the equations: The important thing is not to get distracted by all the $\mathrm e$s and $r$s and $\lambda$s; consider those as constants and eliminate $b_-$, $b_+$ and $c$ like you would in any set of linear equations. Those can be solved quite mechanically, but if you want to do it elegantly, I think something like $\lambda(1)\pm(3)$ and/or $\lambda(2)\pm(4)$ should be a good start. –  joriki Jul 18 '12 at 17:31
@joriki - Thanks a lot...I apreciate your kind help. I'll write the full answer to myself tomoroow morning in order to get a final answer (and then to see that I understand the steps). Hope I'll succeed. Thanks a lot again ! Btw- Is there any physical interpretation of such an equation? –  joshua Jul 18 '12 at 17:35