Given a many body wave-function for a Fermion system, we can calculate the one-body reduced density matrix straightforwardly. Now suppose we know the one-body reduced density matrix, is there a way to obtain the corresponding wave-function exactly?

  • $\begingroup$ Due Hohenberg-Kohn theorem there is a bijective mapping between the diagonal of the reduced density matrix (just density) and the many body wave function. So one has to "just" solve the inverse problem of finding the external potential which produces the density and then solve the many-body schrödinger equation for that external potential. In practice of course this is never done like that. $\endgroup$ Apr 28, 2016 at 21:36
  • $\begingroup$ Hi Mikael, thanks for your answer. I know there is the Gilbert’s Theorem which basically states the similar thing for the one body reduced density matrix as Hohenberg-Kohn theorem. I want to know if there is a "practical" way to reconstruct the wave-function approximately if not exactly. One approximate way I can think of is to construct a single Slater determinant from the natural orbitals (the Eigenstate of one body reduced density matrix) with large occupancy. I think the question is how to determinate the coefficients if we expand the wave-function in the complete basis. $\endgroup$
    – user115797
    Apr 28, 2016 at 22:33
  • $\begingroup$ You need more information. Having just the one-body reduced density matrices won't uniquely reconstruct the state. You can surely construct "a" state - see for instance mathematik.uni-muenchen.de/~lerdos/WS12/MQM/many.pdf page 57. $\endgroup$
    – Martin
    Apr 28, 2016 at 22:55
  • $\begingroup$ @MikaelKuisma Doesn't the Hohenberg-Kohn theorem require that the state is a ground state of a very special Hamiltonian -- one which is uniquely fixed except for a one-particle potential (which only couples to the one-particle density), such as in the case of interacting electrons in an external potential? $\endgroup$ Apr 28, 2016 at 23:45
  • $\begingroup$ @Martin Thanks very much for pointing out the reference. The proof of theorem 10.2 shows a procedure to reconstruct the state given the one-body reduced density matrix. I have two questions about it. First the author doesn't say whether the way to reconstruct the state is unique or not. Second, the constructed state is a mixed state of single Slater determinant in general. If we know the one body reduced density matrix state is from a pure state, is there a approach to reconstruct the pure state, is it unique? $\endgroup$
    – user115797
    Apr 29, 2016 at 1:56

1 Answer 1


No, in the typical case this is not possible.

Given that the fermions are interacting their combined state will be entangled (or in the language of quantum chemistry, they are correlated). This means that the reduced density matrix for a single electronic coordinate will be a mixed density matrix. Or in other words: The one-body density matrix is a classical probability distribution over a set of (not uniquely defined) wave-functions. Such a set is not reducible to a single wave-function with probability 1.


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