In one-particle, one-dimensional quantum mechanics, if the spectrum of the Hamiltonian is given, can the form of the potential be determined? For instance, can all potentials with spectrum of the form 1/(n+a)2 be determined?
No. There are families of isospectral potentials.
See J. López-Bonilla, J. Morales and G. Ovando, "Darboux Transformations and Isospectral Potentials in Quantum Mechanics" Apeiron 9, 20 (2002), A. Khare and U. P. Sukhatme, "Countable infinity of isospectral potential families" Phys. Rev. A 40, 6185 (1989), M. S. Berger and N. S. Ussembayev, "Isospectral potentials from modified factorization". Phys. Rev. A 82, 022121 (2010)
See also my paper Schmelzer, I. (2009). Why the Hamilton operator alone is not enough, Found. Phys. vol.39, p. 486 – 498, arXiv:0901.3262
The fact that potentials are isospectral, doesn't mean that if one is known the others aren't. How to reconstruct the potential from the spectrum is described in Landau-Lifshitz, for instance and, surely, in many other books. But the procedure is, now, well-known: it suffices to write down the partition function. And for this it's necessary to keep in mind that:
Of course the spectrum means, not only, the eigenvalues of the Hamltonian, but their degeneracies, too!
So for the example mentioned the answer is negative, if the degeneracies aren't provided.
These data are necessary.
The partition function, defined in such a way, must exist. From the correlation functions it defines, the family of potentials is then recovered.
These requirements determine whether eigenvalues and their degeneracies are sufficient.
What is interesting, indeed, is that isospectral potentials are related in a very particular way.
So, apparently, different potentials turn out to be just different ways of describing the same thing.
Dear Dr. Nicolis, thank you for the explanation. I had in mind one-dimensional potentials, where bound-state degeneracy is absent, except some rather extreme cases of potentials.
Dear Dr. Kassner and Dr. Schmelzer , thank you for your answers and the papers.
If the energy eigenvalues are of the form E_n = 1/(n+a)^2, without any degeneracy, then the partition function would be given by the sum of exp(-βE_n) over n. For n->oo, this sum doesn't converge, which means the system isn't consistently closed. That's why the potential can't be recovered in this case: the description is incomplete.
The dimensionality of the system doesn't enter at all, beyond constraining the degeneracy. However one-dimensional systems can have degenerate eigenvalues; for instance, if long range interactions are present.
There are qualitative things that you can say about a confining potential.
For example if you go further up in the spectrum, do the levels further separate or
do they cluster with an upper limit.? Are they randomly spaced or is there an order?
There are isospectral potentials without degeneracy.
To obtain the potential, the eigenvalues are not sufficient, nor are the eigenvalues plus their degeneracy sufficient. You also need the eigenfunctions to reconstruct the potential.
If you have two Hamiltonians H1=Q+Q and H2=QQ+, these are obviously different, if Q is not self-adjoint. But they have the same spectrum under mild conditions, because if |φ> is an eigenvector to H1, Q|φ> is one to H2 with the same eigenvalue:
H1|φ>=Q+Q|φ>=E|φ> => H2 Q |φ>=QQ+Q|φ> = Q E|φ> = E Q |φ>
and if |ψ> is an eigenvector to H2, then Q+|ψ> is one to H1 with the same eigenvalue. So if Q|φ> is defined for any |φ> that is eigenvector to H1, there is a one-to-one correspondence between the eigenvectors. The spectra of H1 and H2 are then the same, but their potentials in position representation need not be the same, because Q is different from identity.
But that's, precisely, the point: the statement that such potentials are ``different'', is, simply, wrong. That they are given by different functions is no different than reflecting a choice of coordinates, in such a case.
However, once more: what does matter for obtaining the potential (more precisely the corresponding ``equivalence class'') from the eigenvalues and the degeneracies of the Hamiltonian is that the partition function exists.
I think that you are forgotten a very important condition in the inverse problem: a complex potential can give a real spectra. As well as the domain of real potentials, the inverse problem shows that a large number of measured eigenvalues is constraining the shape of the potential, i.e, the determination converges towards a unique solution as the number of known energies tends to infinity. However if you consider the initial condition of a domain of complex potentials, it is not difficult to prove that the ambiguity of the chosen a potential cannot be avoided even in 1-D. Thus in such a case the answer to this question is no.
Dear Daniel,
Complex potential can give real and stationary energy states spectra?
Dear Gokaran,
The complex potentials are well known in the inverse problems.
Dear Spiros,
If you restric to real potentials and you want to obtain the shape of the potential from the spectrum, the uniqueness of the potential is given by an even potential V(-x)=V(x), without entering in the boundary conditions.
By the way, I never heard the partition function directly related with a quantum mechanics problem of this kind till this thread. The quantum mechanics that I know is independent of the temperature and only in quantum thermofields is possible to do some correspondences.
Dear Daniel,
Complex potential does exist but corresponding eigenvalue are also complex...
Dear Daniel,
Well there is homework problem given in Griffith with complex potential, where tunneling of Bloch states has been captured using complex potential. Please provide any link where I can lookup. Complex eigenvalue has finite lifetime. It does make sense that complex potential will provide complex energy states.
Dear Gokaran,
My advice is that you work a little bit before making your simple questions, Not only in this thread because I' m starting to be a little tired. Let me to choose a simple example and this is the end with you (at least for today)
The complex potential
V(x)= x 2 −2iax−a 2 + (b2 −1/4)/ (x−ia) 2
Where b >0 (positive) gives the real eigenvalues
En,c= 4n +2 (1-cb)
Being c=(plus, minus) 1.
Have a good day!
Dear Daniel,
The potential that you quoted, which real system have such potential?
Dear Daniel,
thank you for your answers and comments. Yes, I had in mind only real potentials.
Dear Gokaran,
for complex PT-symmetric potentials with real spectra having physical applications, you may refer to the following explanatory paper, by Carl M. Bender,
Article Introduction to PT-Symmetric Quantum Theory
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