Assume we start with a generic many-body Hamiltonian: $$ H=\sum_{ij} t_{ij} a_i^\dagger a_j+\sum_{mnlk}U_{mnkl}a_{m}^{\dagger}a_{n}^{\dagger}a_la_k. $$ Now if there is only the one-body part, which has a matrix form we can perform a series of canonical transformations to diagonalize it using a naive but effective way like Jacob's method.

My question is for the two-body term, the operator is kind of like a (rank-4)tensor, so I am wondering whether there is a general mathematical way to diagonalize the problem like Jacob's method. After that, the Hamtonlian will have the following form: $$ H'= \sum_{I} \epsilon_I a_I^\dagger a_I+\sum_{IJ}\epsilon_{IJ}a_{I}^{\dagger}a_{J}^{\dagger}a_Ja_I+\text{higher body term} $$

  • $\begingroup$ To be clear, you are asking for whether one can diagonalize a Hamiltonian of the form of your first equation but where one body terms are all zero, right $t_{ij}=0$? And that you expect the answer to have the form of your second equation? $\endgroup$
    – KF Gauss
    Commented Nov 7, 2023 at 10:39
  • $\begingroup$ Actually, you can consider we first diagonalize the one body part, which corresponds to a Fourier transformation(change of basis), then start to "diagonalize" the two-body part. The problem is there is a general method to diagonalize the one-body part since it resembles a lot like a matrix with only two indices, and the two-body part can then be considered as a rank-4 tensor, However, I don't know any existing method that can systematically help me to get the form I describe in the second equation. $\endgroup$
    – ZhiYu Fan
    Commented Nov 7, 2023 at 12:49
  • $\begingroup$ @ZhiYuFan So in essence, you are asking if there is a standardized (and efficient) way to bring a Hamiltonian of the form of the first equation into the form of the second equation, where the "higher body terms" are interactions with at least 5 creation/annihilation operators? -- Do you have any constraints on the transformation? In which way to $H$ and $H'$ have to be related? $\endgroup$ Commented Nov 7, 2023 at 15:31
  • $\begingroup$ @NorbertSchuch A standard way would be using the BCH formula. $\endgroup$
    – ZhiYu Fan
    Commented Nov 8, 2023 at 1:57
  • $\begingroup$ @ZhiYuFan In don't understand your comment. My question is: Should H and H' be related by a unitary? Is it sufficient if H' has the same low-energy spectrum as H? Or what does H' have to satisfy (i.e. how does it have to relate to H) to be a valid answer to your question? $\endgroup$ Commented Nov 8, 2023 at 2:10

1 Answer 1


In general case this Hamiltonian is not exactly solvable - in fact, there are many methods and approximations developed for dealing with such Hamiltonians in various situations, just to give a few examples:

  • various mean-field approaches - like Hartree-Fock or Habbard-Stratonovich
  • perturbation methods - e.g., Dyson expansion
  • non-perturbative techniques, like renormalization group

In fact, finding the ground state of this Hamilronian is already an achievement - see Why is the ground state important in condensed matter physics?

However, in some special cases (i.e., for particular choices of $U_{mnkl}$) it can be solved exactly, using bosonization or Bethe-Ansatz.

I suggest looking for inspiration (for more precise questions) in the following threads (I didn't put them in any particular order):
Is there a way to guess/find unitary transformations for Hamiltonians? Why is Hartree-Fock considered a mean-field approach?
Is Hartree-Fock a standard mean field approximation?
Higher-order perturbation in Kondo problem
What Are Some Examples of Physics Problems With Many Different Approaches That Give the Same Answer?
In what ways does quantum field theory (QFT) extend quantum mechanics (QM)? Can Fermi liquid be obtained by a canonical transformation?

Update (in response to the OP edit)
The OP suggests looking more specifically at Hamiltonian $$ H'= \sum_{I} \epsilon_I a_I^\dagger a_I+\sum_{IJ}\epsilon_{IJ}a_{I}^{\dagger}a_{J}^{\dagger}a_Ja_I+\text{higher body term} $$ Note that $$\sum_{IJ}\epsilon_{IJ}a_{I}^{\dagger}a_{J}^{\dagger}a_Ja_I= \sum_{IJ}\epsilon_{IJ}n_{I}n_{J},$$ where $n_I=a_{I}^{\dagger}a_I$ is the occupation number. The Hamiltonian without "higher body terms" (which are really still two-body, but representing the exchange interaction) is significantly simpler, and enters a number of well-known models, depending on $\epsilon_I, \epsilon_{IJ}$ and the meaning of indices $I,J$. Notably: Hubbard model, Anderson model, various models of Coulomb blockade, Tomonaga-Luttinger model, etc. Some of these are exactly solvable (Tomonaga-Luttinger, Anderson), for others there exist well developed approximate methods and non-perturbative results.

  • 1
    $\begingroup$ The term "Bogoliubov" should probably be in there too (to make it easier to search for the associated approximate diagonalization techniques), although I suppose that's all wrapped up in your first bullet point. $\endgroup$
    – march
    Commented Nov 6, 2023 at 17:13
  • $\begingroup$ @march I thought about including it... but I had doubts, since it is really applied to a bilinear Hamiltonian, after mean field approximation. $\endgroup$
    – Roger V.
    Commented Nov 6, 2023 at 17:31
  • 1
    $\begingroup$ Well, that makes sense, but there is, after all, Hartree-Fock Bogoliubov. $\endgroup$
    – march
    Commented Nov 6, 2023 at 19:50
  • $\begingroup$ Thanks for your explanation, but I guess all the methods here are essentially a perturbative point of view. I guess I am asking this problem from a different perspective. $\endgroup$
    – ZhiYu Fan
    Commented Nov 7, 2023 at 2:22
  • $\begingroup$ @ZhiYuFan see the update in response to your edit. $\endgroup$
    – Roger V.
    Commented Nov 7, 2023 at 10:04

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