## Double Delta Potential

Hello folks,

I am doing Griffiths myself, and I ran into a problem involving the double delta potential. The problem and my solution is described in the pdf file (see below).

If you cannot see the pdf file, please click here: http://spinor.sitesled.com/doubledelta_pf.pdf.

I think there should be 3 bound states, but how to do the algebra?
Attached Files
 doubledelta_pf.pdf (58.5 KB, 411 views)

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 Please help someone...
 The pdf has some small mistake(s). I have corrected them and will upload the corrected version in a few hours.

Mentor

## Double Delta Potential

The results of Problem 2.1 (c) (top of page 30) simplify things a bit.

To start, take $\psi (x)$ to be even. This gives a couple of relationships between B, C, D, and F.

This is my (extended) working (attached pdf).
Attached Files
 doubledelta2.pdf (46.6 KB, 631 views)

 Have solved the problem with some help, thanks If I have anything else to ask I'll post here.
 Now I need to work out the transmission coefficient for the scattering problem version...I have set up all the equations, now is there any way I can do the elimination using a computer-algebra-system (CAS) like Mathematica or Matlab? Any ideas?
 Hello, I am working on this problem too. I found the same wave functions as the one posted in the above pdf file. I also found the energy for the even bound state in terms of the LambertW function. For the odd bound state I found the energy as zero, and I think the calculations in the pdf file posted in this thread also lead to that. Is this acceptable? Does this mean such a state does not exist? Should there be two bound states in a double delta potential?
 If the two delta functions are very close together then they behave as a single delta function and there is one bound state. If they are very far apart then they each have a bound state: but this situation can also be analyzed as even and odd coupled states, with the coupling equal to zero. As you bring the delta functions closer together from infinity, the coupling forces the bound states apart in energy, raising the energy of the odd state. At a critical distance, the energy of the odd state goes to zero and from there on only the even state is bound. If you have done the math correctly and the energy of the odd state is zero, then the textbook has cooked up the problem to give you the critical separation.
 Wow, its nice to see an old thread resurrected after over 3 years I received an email a few months ago from someone who mistook me for someone else, but gave me a link to a paper he co-authored on the analytical form of the wavefunctions for the double delta potential well. The reference is T.C. Scott, J.F. Babb, A. Dalgarno and John D. Morgan III, "The Calculation of Exchange Forces: General Results and Specific Models", J. Chem. Phys., 99, pp. 2841-2854, (1993). I haven't looked into this paper because I couldn't get hold of it, and eventually I forgot about it (the semester is a wonderful thing...). And back in '06, I didn't quite analyze it in as much detail...certainly not up to the Lambert W functions. With zero coupling, you are basically solving for two different problems and superposing the (shifted) wavefunctions. The sum isn't a solution to any Schrodinger equation since you'd have to throw away cross terms like $V_{i}\psi_{j}(1-\delta_{ij})$ (i, j = 1, 2), which is of course wrong. So, its an approximation, albeit a poor one. Personally, I have reservations against mathematically looking at the situation from the point of view of even and odd states (sum and difference) with zero coupling for the simple reason that the wavefunction for the potential $V_{1}+V_{2}$ is not equal to the sum of wavefunctions for $V_{1}$ and $V_{2}$. Sure we use this approximation in atomic physics and in chemistry for MO theory, but its better to say that in the weak coupling regime, the eigenfunction has a similar behaviour as do the even or odd states. Plus, this is a Delta 'function' potential, so I'd be careful with the approximations. I think its a better idea to cook up a problem with two rectangular wells, study them in weak coupling and then blow up their height (or depth in this case) and shrink the width to make them delta functions. Nevertheless, your physical conclusions seem correct, and since you've worked out the wavefunctions analytically in such detail, you no doubt have done the math correctly. But your conclusions in the 0724 post are correct.

 Quote by maverick280857 With zero coupling, you are basically solving for two different problems and superposing the (shifted) wavefunctions. The sum isn't a solution to any Schrodinger equation since you'd have to throw away cross terms like $V_{i}\psi_{j}(1-\delta_{ij})$ (i, j = 1, 2), which is of course wrong. So, its an approximation, albeit a poor one.
It's a bad approximation if you have two delta functions and two electrons. But if you only have one electron, isn't it exactly correct?

 Quote by conway It's a bad approximation if you have two delta functions and two electrons. But if you only have one electron, isn't it exactly correct?
$$H_1 = \frac{p^2}{2m} - k\delta(x-a) \qquad \mbox{Wavefunction }: \psi_1$$

$$H_2 = \frac{p^2}{2m} - k\delta(x+a) \qquad \mbox{Wavefunction }: \psi_2$$

$$H = \frac{p^2}{2m} -k(\delta(x-a)+\delta(x+a)) \qquad \qquad \mbox{Wavefunction }: \psi$$

For one particle,

$$H\left(\psi_1 \pm \psi_2\right) = H_1\psi_1 \pm H_2\psi_2 + (H_2\psi_1 \pm H_1\psi_2) = E_1\psi_1 \pm E_2\psi_2 + \mbox{ cross terms }$$

For it to be exactly correct, you'd want

$$H\left(\psi_1 \pm \psi_2\right) = E_{\pm}\left(\psi_1 \pm \psi_2\right)$$

wouldn't you? (You can think of 'a' is a measure of the 'coupling distance'.)

 I think we're going to agree with each other in the end. Instead of two delta function take two hydrogen nucleii (protons). The ground state of each one is the familiar s orbital. But as a system it is correct to take two identical ground states, symmetrical and antisymmetrical. The electron can be in either one, or partly in both. By taking appropriate superpositions of these two degenerate ground states, you can place the electron on this proton, that proton, or shared between them. As you move the protons closer together, the coupling forces the degenerate states apart, the symmetric one being the true ground state. If an electron is attached to one proton, then you must analyze it as a superposition of the symmetric and antisymmetric states, which evolve at different frequencies. So after a time the electron will be on the other proton. It is true that as the protons approach each other you can no longer write the wave function as a simple +/- combination of the familiar s orbitals. In the limiting case the protons become a helium nucleus and the symmetric and antisymmetric combinations reappear as the familar s orbital squeezed by a factor of 2, and the p orbital oriented along the line by which the protons converged. The dihydrogen molecular ion is an interesting special case which appears along the way.

 Quote by conway If the two delta functions are very close together then they behave as a single delta function and there is one bound state. If they are very far apart then they each have a bound state: but this situation can also be analyzed as even and odd coupled states, with the coupling equal to zero. As you bring the delta functions closer together from infinity, the coupling forces the bound states apart in energy, raising the energy of the odd state. At a critical distance, the energy of the odd state goes to zero and from there on only the even state is bound. If you have done the math correctly and the energy of the odd state is zero, then the textbook has cooked up the problem to give you the critical separation.
I've worked out the problem myself (from problem 3 chapter 6 in Merzbacher) and got a transcendental equation like the one
in the pdf atached file. I sketched a graphic solution and don't see that there is a critical separation for the 2 delta
functions at which one of the energies states is zero. For The odd state energy, k goes from zero to $gm/\hbar^2$ and for the even state energy, k goes from $2gm/\hbar^2$ to $gm/\hbar^2$ as the separation of the two delta functions varies from 0 to infinite, so both energies converge as k converges to $k=gm/\hbar^2$ at infinite. Can someone confirm this?

 I wonder if you're working out one-dimensional solutions and I'm talking about the three-dimensional case? Even in one dimension it's hard for me to see how you get an odd solution when the two delta functions are much closer together than the decay constant of the wave function. But I haven't actually done the calculation. It just seems like the flipping over is too drastic.

 Quote by conway I wonder if you're working out one-dimensional solutions and I'm talking about the three-dimensional case? Even in one dimension it's hard for me to see how you get an odd solution when the two delta functions are much closer together than the decay constant of the wave function. But I haven't actually done the calculation. It just seems like the flipping over is too drastic.
Yes, I'm talking about the one dimensional problem. It seems easy to skech the equation and see what I'm saying. Anyway I would like a confirmation on this.

 OK, you are correct for the one dimensional case. You can always take a standing wave solution, consisting of two travelling waves in opposite directions, and put a delta potential at any nodal point without changing anything, because the charge is always zero at the only point where the potential exists. The limiting case is the zero-frequency standing wave which is the odd state you are talking about. I don't know if it's the same in three dimension.
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