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  Constructing a Hamiltonian (as a polynomial of $q_i$ and $p_i$) from its spectrum

+ 6 like - 0 dislike

For a countable sequence of positive numbers $S=\{\lambda_i\}_{i\in N}$ is there a construction producing a Hamiltonian with spectrum $S$ (or at least having the same eigenvalues for $i\leq s$ for some $s$)?

Here by 'Hamiltonian' I understand a polynomial of $p_i$ and $q_i$ (or equivalently - $a_i$ and $a_i^\dagger$) of $k$ pairs of variables and of order $2n$. Both $k$ and $n$ can be functions of $S$ and $s$.

For example, for the spectrum $$S =\{3,5,7,9,\ldots\}$$ one of Hamiltonians working for any $s$ is $$H = 3+a^\dagger a.$$

This post has been migrated from (A51.SE)
asked Feb 17, 2012 in Theoretical Physics by Piotr Migdal (1,260 points) [ no revision ]
retagged Mar 7, 2014 by dimension10
Are the coefficients in this polynomial supposed to be real, or integer...? (in the latter case, you only have countably many such polynomials and uncountably many possible spectra) If the coefficients are real, your polynomial would have to somehow encode a subset of $N$ in the coefficients, which might be difficult to do) An interesting question, BTW.

This post has been migrated from (A51.SE)
THe coefficients are real.

This post has been migrated from (A51.SE)
Not sure that it is relevant to your question, but there was a discussion on related topic at physics.stackexchange: http://physics.stackexchange.com/questions/13480/in-quantum-mechanics-given-certain-energy-spectrum-can-one-generate-the-corresp . Probably some references from there might be useful for this problem too.

This post has been migrated from (A51.SE)
Surely, the solution can be not unique, for example, the adjoint action by a unitary transformation would not change the spectrum. Will you be satisfied with any Hamiltonian having the given spectrum?

This post has been migrated from (A51.SE)
@DavidBarMoshe Any claims about uniqueness would be desirable. However, I don't suspect that to be the case,

This post has been migrated from (A51.SE)
@PiotrMigdal: For a finite spectrum there is no unique solution. I suspect this is also true for the infinite case, but I am not sure.

This post has been migrated from (A51.SE)

3 Answers

+ 6 like - 0 dislike

It seems there is a simply way to do this for polynomials with finite degree $d$. Since $a^\dagger a$ is the number operator, we can take $a^\dagger a = N$, where $N$ is the number of excitations corresponding to a particular level. Then if the Hamiltonian has the general form $H = \sum_{k=0}^d c_k (a^\dagger a)^k$, the energy corresponding to a particular state is $E_N = \sum_{k=0}^d c_k N^k$. Since $N$ is constant for a given eigenstate of the Hamiltonian, the equation for $E_N$ is just a linear equation in the variables $\{c_k\}_k$. Since you have the spectrum, you have $E_N$, and hence you can solve using Guassian elimination (or whatever your preferred technique is for solving linear systems of equations). Even if the spectrum is infinite, you will only need $d+1$ equations to fix the values of $c_k$, so this is a simple calculation.

This post has been migrated from (A51.SE)
answered Feb 18, 2012 by Joe Fitzsimons (3,575 points) [ no revision ]
I think, the [Newton interpolation formula](http://en.wikipedia.org/wiki/Finite_difference#Newton.27s_series) may be used instead of Gaussian elimination in this case

This post has been migrated from (A51.SE)
Joe, thanks for your solution; I overcomplicated it by looking at more modes. It is even easier to deal with it for $\sum_k c_k a^{\dagger k} a^k$ as then it can be solved step by step (i.e. adding $n$-th eigenvalue does not modify $c_k$ for $k

This post has been migrated from (A51.SE)
+ 4 like - 0 dislike

Here's a new paper which may be related this problem, and appeared today on the archive:

Recovering the Hamiltonian from spectral data

Cyrille Heriveaux, Thierry Paul


This post has been migrated from (A51.SE)
answered Feb 24, 2012 by Cristi Stoica (275 points) [ no revision ]
+ 3 like - 0 dislike

It was already mentioned in a comment above about a topic in physics.SE, that it may be related with inverse scattering problem. I only may add, that there is precise method of construction of Shroedinger operator $p^2 + V(q)$ with arbitrary finite spectrum for one-dimensional case. It is very well developed due to application to soliton theory. The polynomial operators $V(q)$ have some problems due to infinities for $q \to \pm \infty$, but for $V(q)$ with fast convergence to zero there are hundreds papers. Especially simple construction may be found, e.g., in M. Ablowitz, H. Segur, Solitons and the Inverse Scattering Transform, SIAM, Philadelphia, 1981.

The idea, that if you have some $L_1 = p^2 + u(q)$ with given spectrum, there is an elegant method to add yet another eigenvalue $\zeta$. You should consider equation $L_1 g = \zeta g$, find the solution $g(q)$. Then $L_2 = p^2 + u + 2 (\ln (g))''$ saves all eigenvalues of $L_1$, but also has new eigenvalue $\zeta$ with eigenfunction $1/g(q)$. So you may start with $u=0$ and add eigenvalues one after another.

Yet, as I mentioned here is a problem with polynomial case, it is one-dimensional and may be only useful thing is idea to search for trick like $g \to 1/g$ with adding eigenvalues...

[ADD] (1). I think, it is possible also to consider products $L_k(q_k)$, etc. (2). It is not unique, e.g., isospectral deformation of $L$ is described by KdV equation (yet, it was also mentioned in a reference in Physics.SE thread)

This post has been migrated from (A51.SE)
answered Feb 18, 2012 by Alex V (300 points) [ no revision ]

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