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

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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 $$H = 3+a^\dagger a.$$

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retagged Mar 7, 2014
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.

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THe coefficients are real.

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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.

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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?

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@DavidBarMoshe Any claims about uniqueness would be desirable. However, I don't suspect that to be the case,

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@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.

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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.