Bound states for sech-squared potential

[broegger]

Hi,

I'm working on an introductory qm project, hope somebody has the time to help me (despite the length of this post), it will be highly appreciated.

My goal is to determine the bound states and their energies for the potential

$$V_j(x) = -\frac{\hslash^2a^2}{2m}\frac{j(j+1)}{\cosh^2(ax)}$$​

for any positive value of j (I think I'm supposed to show that j has to be integer at some point, but I don't know). By a change of variable, $$ax\rightarrow y$$, I have rewritten the Schrödinger equation for this potential to:

$$-u''(y)-\frac{j(j+1)}{\cosh^2(y)}u(y) = Eu(y).$$​

I will denote the corresponding Hamiltonian H_j. I have shown that the ground state for V_j(x) is $$\psi_0(x) = A\cosh(ax)^{-j}$$ (this is all correct, has been verified). Now here is where serious problems begin. I am given these ladder operators:

$$a_{j+} = p + ij\tanh(y)$$
$$a_{j-} = p - ij\tanh(y)$$
​

where $$p = -i\frac{d}{dy}$$ (no h-bar). I have shown that $$H_j = a_{j+}a_{j-} - j^2$$ and $$H_{j-1} = a_{j-}a_{j+} - j^2$$ and I have used this to show that if $$|\psi>$$ is an eigenstate for H_j then $$a_{j-}|\psi>$$ is an eigenstate for H_(j-1) and $$a_{(j+1)+}|\psi>$$ is an eigenstate for H_(j+1), both with the same eigenvalue (energy). Thus the purpose of the a-operators is to carry an eigenfunction for the potential characterized by the value j (H_j) over to an eigenfunction for the potential characterized by the value j-1 or j+1 (H_(j-1), H_(j+1)).

Now I should be able to find all bound states and their energies (I have been told that there is only one bound state for every (integer?) value of j, so this amounts to showing that the ground state I found earlier is the only bound state there is). I'm clueless on this last step.

 

[arivero]

Hi,

I'm working on an introductory qm project, hope somebody has the time to help me (despite the length of this post), it will be highly appreciated.

My goal is to determine the bound states and their energies for the potential

$$V_j(x) = -\frac{\hslash^2a^2}{2m}\frac{j(j+1)}{\cosh^2(ax)}$$​

Transparent potential. You solve it by supersymmetry. Honestly.

 

[broegger]

Don't know what you're talking about. Honestly.

 

[arivero]

There are five potentials called "shape invariants" in one dimension: 1/r^2, sec^2, sech^2, cosec^2, cosech^2 Yours is one of them. They can be solved by finding a superpotential, ie a function W(x) and an operator $$Q=-d/dx +W(x)$$ such that $$QQ^+$$ is your hamiltonian, and then $$Q^+Q$$ is a hamiltonian having the same eigenvalues except one. Bu repeating this trick, it is possible to attach your problem to the free one. It is a very nice trick.

 

[arivero]

Now I read the rest of your post, I see it is the same technique! So google around searching for "transparent potentials", "isospectral", "Gengenshtein", "Boya", "SUSY Quantum Mechanics", and variants of it.

Also this
http://www.iop.org/EJ/abstract/0305-4470/26/21/020

Who told you about this?

 

[arivero]

Now I should be able to find all bound states and their energies (I have been told that there is only one bound state for every (integer?) value of j, so this amounts to showing that the ground state I found earlier is the only bound state there is). I'm clueless on this last step.

Perhaps the step you are looking for lies around the fact that the zero eigenvalue does no have a partner, because the ladder operator hits a nullspace. Then one of the first order operators $$a_\pm$$ can we used to solve for the eigenvector, it should be
$$\psi^0=e^{\int \pm W(x) dx}$$
where W (your tanh) is sometimes called the "superpotential" due to an old article from Witten. Obviously only one of the two signs +, -, must give a valid eigenfunction, which in turn is conscistent with the observation that only one of the two paired equations has a zero eigenvector. Is this the step you were looking for?

I prefer to see the ladder mechanism as a iteration of superpotential pairings. In each pairing, you have two hamiltonians having exactly the same spectrum except for one eigenvalue, which happens to be the zero eigenvalue. Say the (-) potential has this extra eigenvector. Then you produce it explicitly with the exponential formula above, you get the rest of the eigenvectors from the (+) via the ladder $$a$$, and you add a constant term to the hamiltonian so you are ready to pair it with the next one, and so on.

As far as I know, there are five families of superpotentials you can use for the pairing trick: 1/x, tan x, tanh x, cot x, coth x. Besides, you can use the Heaviside funtion and you can pair delta-function potentials!

 

[broegger]

Hmm, thanks for helping, but I don't get it. It's an intro-course, remember :) As you say, it's something about the zero eigenvalue, but I don't know how to reason...

 

[dextercioby]

There's only one thing to be said.Classically,that potential is integrable...

Write the Tindep SE and see if you can identify a change of variable.

Daniel.

 

[dextercioby]

May i reccomend chapter 2 of this arXiv:hep-th/9405029 v2 5 May 94 ...?


Daniel.

 

[arivero]

May i reccomend chapter 2 of this arXiv:hep-th/9405029 v2 5 May 94 ...?
Daniel.

Very long article. The cosec^2 is, although, in page 16, but regularised for a box of size L. A shorter, but unnormalised, presentation, appears in page 2 of http://xxx.unizar.es/ps/quant-ph/0311139 I guess that reference 4 in this paper has more detail, but it is a old one, offline.

Also page 11 of
http://xxx.unizar.es/ps/math-ph/9810022
and of course references therein

 

[arivero]

Hmm, thanks for helping, but I don't get it. It's an intro-course, remember :) As you say, it's something about the zero eigenvalue, but I don't know how to reason...

Sorry I have been too extense. I will try again. Take your pair
$$a_{j+} = p + ij\tanh(y)$$
$$a_{j-} = p - ij\tanh(y)$$
The hamiltonians
$$H^+=a_{j+}a_{j-}=H_j+j^2$$
$$H^-=a_{j-}a_{j+}=H_{j-1}+j^2$$

are really the ones having the same set of eigenvalues, because if $$a_{j+}a_{j-} f= K f$$, then $$a_{j-}a_{j+} (a_{j-} f)= K (a_{j-} f)$$ and reciprocally.

But now consider the vector |f> solution of $$a_{j-} f(y)=0$$. It is an eigenvector of $$H^+$$, with eigenvalue 0. But the mapping via $$a_{j-}$$ into $$H^-$$ fails here: it goes to the null vector, it does not produce a valid, normalisable, eigenvector of $$H^-$$. So, surprise, the hamiltonian (+) has an eigenvector which is not present in the hamiltonian (-).

We can calculate this eigenvector very trivially, because after all it is the solution of $$-f'(y) - j \tanh(y) f(y)=0$$. Thus

$$f_0(y)= e^{-j \int \tanh(y)}= e^{-j \ln \cosh y}$$

is the "new" eigenvector exclusive of $$H^+$$

 

[arivero]

Appendix: note there do not exist any normalisable solution of $$a_{j+} f(y)=0$$. Thus the problem does not appear in the reverse direction.

 

[broegger]

I think I finally get what you're saying, arivero. I'm not at home right now, so I can't work it out thoroughly, but I hope your still with me in case I don't get it. Also, how do I open those links you gave me?

Thank you so much for taking the time!

 

[arivero]

Sorry I gave you the postscript. Use instead
http://arxiv.org/pdf/quant-ph/0311139
http://arxiv.org/pdf/math-ph/9810022
http://arxiv.org/abs/hep-th/9405029

You can first to work out your problem and then to use the links as extra documentation.

 

[broegger]

I totally get it now :) Very elegant, indeed. Thank you, arivero.

I found out that $$H_j$$ has int(j) bound states (int refers to the integer part of j), so I was wrong about there being exactly one bound state for every value of j (j need not be integer neither).

Now, the only thing left is to derive a general formula for all the eigenstates of $$H_j$$. I have shown that for $$0<j\leq1$$ there is exactly one bound state, the ground state $$|\psi_0\rangle=\cosh(y)^{-j}$$ (apart from normalization). I know I should apply the $$a_{j+}$$-operator continuously to these groundstates to generate the eigenstates for higher values of j, but I can't get a neat, general formula. Any hints? :)

 

[arivero]

I know I should apply the $$a_{j+}$$-operator continuously to these groundstates to generate the eigenstates for higher values of j, but I can't get a neat, general formula. Any hints? :)

No neat thing, just iterate and iterate, AFAIK. Perhaps in the thesis of Arturo Ramos or here http://wigner.unizar.es/personal_p/jfc/pspubs/pub96.ps or here http://wigner.unizar.es/personal_p/jfc/pspubs/pub93.ps (sorry no PDF, only postcript files).

Still, there are some neat formulae for the asymtotics, ie for the phase shifts of the scattering across these potentials and their Jost functions, because each new eigenvalue is just a pole in the S-matrix.

 

[broegger]

Ok, thanks.

I've thought about something else. What if the ground state for $$H_{l+1}$$ is degenerate (so that there are two groundstates), let's call this other ground state $$|\psi_0'\rangle$$; then it could be that $$a_{(l+1)-}\psi_0'\rangle$$ yields an acceptable wavefunction (not the 0-function) and then the spectra for $$H_l$$ and $$H_{l+1}$$ would in fact be identical. How come this isn't the case?

Also, how do we know that $$a_{l+}\psi\rangle=0$$ has no solution?

 

[arivero]

Ok, thanks.

I've thought about something else. What if the ground state for $$H_{l+1}$$ is degenerate (so that there are two groundstates), let's call this other ground state $$|\psi_0'\rangle$$; then it could be that $$a_{(l+1)-}\psi_0'\rangle$$ yields an acceptable wavefunction (not the 0-function) and then the spectra for $$H_l$$ and $$H_{l+1}$$ would in fact be identical. How come this isn't the case?

Also, how do we know that $$a_{l+}\psi\rangle=0$$ has no solution?

Applause! I silenced this (which I myself learn from a friend, Casahorran) because you told your problem was set up as only an introductory QM exercise.

The point is that the eigenfunctions for the, er, vacuums, are, as we saw,
$$f_0(y)= e^{\pm j \int W(y)}$$
ie, one comes with the sign +, another one with the sign -. Conclusion: only one of the two vacuums is normalisable. The other one blows itself away.

Perhaps here was the very point of Witten when suggesting to see this kind of models as "Supersymmetric Quantum Mechanics": to show that there was a (new?) way of spontaneusly breaking SUSY only for a discrete set of states, particularly for the low lying state, the vacuum.

In some sense, this gives a condition to W(y): it must be such that at least one of the vacuum eigenfunctions is not normalisable so the symmetry "spontaneusly" break. I think I have already mentioned that I only know of five valid superpotentials, but of course there is the additional condition of having the free particle as the original partner.

 

[broegger]

You lost me again :)

 

[broegger]

I figured it out... Just one more question:

Appendix: note there do not exist any normalisable solution of $$a_{j+} f(y)=0$$. Thus the problem does not appear in the reverse direction.

How can we prove that?

 

[arivero]

I figured it out... Just one more question:
How can we prove that?

It is just a rephrasal of the last message. For the trick to function, either Exp(+ \int W(x)) or Exp(- \int W(x) ) must be normalisable but no both. Given that we have an exponential, it is very easy to say *given a W(x)* if the trick is going to work or not

 

[broegger]

Yep, I finished my project now. Thank you very much for helping me :-)