I am studying diagonalization of a quadratic bosonic Hamiltonian of the type:

$$ H = \displaystyle\sum_{<i,j>} A_{ij} a_i^\dagger a_j + \frac{1}{2}\displaystyle\sum_{<i,j>} [B_{ij} a_i^\dagger a_j^\dagger + B_{ij}^* a_j a_i ] $$

in second quantization.
Also, $A = A^\dagger$ and $B = B^T$, so that the Hamitonian is Hermitian.

This can be written in matrix form as $$ H = \frac{1}{2} \alpha^\dagger M \alpha - \frac{1}{2} tr(A)$$ where $ \alpha = \begin{pmatrix} a \\ a^\dagger \\ \end{pmatrix} $ , $ \alpha^\dagger = \begin{pmatrix} a & a^\dagger \end{pmatrix} $ and M is $ M = \begin{pmatrix} A & B\\ B^* & A^* \\ \end{pmatrix} $

Question: Can I numerically diagonalize the matrix M to get eigenvalues and eigenvectors of the Hamiltonian? If yes, then what would be the right way to write those eigenvectors in second quantization?

e.g. If for 2$\times$2 matrix M, one of the numerically calculated eigenvectors is $\begin{pmatrix} p\\ q \\ \end{pmatrix} $ , then, should it be written as $$p \,a|0\rangle + q \, a^\dagger|0\rangle $$ (where the column $\alpha$ has been used as the basis) or $$p \,a^\dagger|0\rangle + q \, a|0\rangle $$ (where $\alpha^\dagger$ has been used as the basis)?

Note: Here $|0\rangle$ is the vacuum state for 'a' type (bosonic) particles. End of Question.

Note : Consider a simpler Hamiltonian $$ H = \displaystyle\sum_{<i,j>} A_{ij} a_i^\dagger a_j $$ and note that its eigenvectors are of the form $$(a_1^\dagger a_2^\dagger ... ) |0\rangle $$

  • $\begingroup$ I forgot to mention that the Hamiltonian is Hermitian. I will edit the question to add the conditions on A and B. $\endgroup$ Mar 31, 2014 at 13:56
  • $\begingroup$ Is your rewritten hamitonian correct? somthing seems missing $\endgroup$
    – unsym
    Apr 6, 2014 at 7:48
  • $\begingroup$ It seems correct. $\endgroup$ Apr 6, 2014 at 9:10
  • $\begingroup$ More on Bogoliubov transformation: physics.stackexchange.com/… $\endgroup$
    – Qmechanic
    Apr 6, 2014 at 11:48

3 Answers 3


Take the Hamiltonian you write at end of your question as the example(where your statement about the eigenvector is not right), the procedure to obtain the eigenvalue and eigenvectors of other Hamiltonian is nearly the same.

$$H=\sum_{ij}A_{ij}a_i^\dagger a_j$$

Writing in matrix form you have(suppose it is a $n\times n$ matrix):

$$H=\alpha^\dagger A \alpha$$


$$ \alpha^\dagger=(a_1^\dagger, a_2^\dagger,\dotsb,a_n^\dagger )$$

Now you are ready to diagonalize Matrix analytically or numerically:

$$A=X^\dagger D X$$

Where $D=[E_1,E_2,\dotsb,E_n]$ is the eigenvalue you want, and $X$ is just a Unitary matrix that diagonalize $A$. Indeed, it is the normalized mathematical eigenvectors of $A$ write in column way(which is same in Fortran if you do it numerically).

Now substitute this to the original Hamiltonian:

$$H=\alpha^\dagger A \alpha=\alpha^\dagger X^\dagger DX\alpha=\beta^\dagger D \beta$$

where $\beta$ is $\beta =X\alpha=(\beta_1,\beta_2,\dotsb,\beta_n)^T$. Because this is a unitary transformation, you can easily check that the commute or anti-commute relation still holds for $\beta_i,\beta_2,\dotsb$.

Now you can consider the system is consisting of non-interacting qusi-particles $\beta_i,\beta_2,\dotsb$, then the eigenstate of the system is easily obtained: $$|\psi \rangle=(\beta_1^\dagger)^{n_1}(\beta_2^\dagger)^{n_2}\dotsb(\beta_n^\dagger)^{n_n}|0\rangle$$

$n_i=0,1$ for fermions and $n_i=0,1,2,3\dotsb$ for bosons.

Also, see this thread with the very similar question.

  • $\begingroup$ Thanks for the reply. But, I don't understand how to get the exponents $n_1$, $n_2$, $n_3$,.... for a particular eigenvector from the matrix X. $\endgroup$ Apr 14, 2014 at 19:46

What you're talking about looks a lot like a Bogoliubov transformation. The idea would be that you would decompose your Hamiltonian into something along the lines of \begin{equation} H = \sum_{i, j} \mathcal{C}_i ~c_i^\dagger c_i \end{equation} so that the eigenvalues are the $\mathcal{C}$ and the eigenvectors are the eigenvectors of $c_i^\dagger c_i$.

Specifically, if you consider $U$ to be the diagonalizing unitary matrix that takes your original matrix $M \mapsto U \mathcal{C} U^\dagger$, then by directly inserting this into your definition for your Hamiltonian you would get \begin{equation} H = \alpha^\dagger \left (U \mathcal{C} U^\dagger \right ) \alpha \end{equation} and, hence, the individual elements of $U^\dagger \alpha$ are the new "eigen-operators" of the Hamiltonian. The spectral properties of those operators, combined with the diagonal $\mathcal{C}$, will give the new eigenvalues and eigenstates in terms of the old $a$-basis.

  • 1
    $\begingroup$ This is unfortunately not correct, but this is a common mistake. The creation and annihilation operators associated with the eigenvectors of $M$ would not be bosonic. It would be correct for fermions, where the transformations preserve the euclidean norm, but for bosons you are working on a symplectic space (as in classical mechanics). The right way to do it is to look at the eigenvectors of $\Sigma M$, where $\Sigma=\mbox{diag}(1,-1)$ and you furthermore have to normalize them with respect to the norm induced by $\Sigma$. $\endgroup$
    – ulf
    Dec 4, 2014 at 14:31
  • $\begingroup$ Please explain where the logical error is, and how it contrasts with the other answer given to this question. I've just made a change of basis, nowhere do I require that the $\alpha$ commute or anti-commute. $\endgroup$
    – webb
    Dec 4, 2014 at 18:45

@webb's comment (since the answer is longer than a comment allows for).

Firstly, you have to distinguish between diagonalizing a matrix representation of an operator in a given basis and diagonalizing the coefficient matrix of an operator in a set of creation and annihilation operators.

We are dealing with the latter here. If (as you point out correctly) we need a Bogoliubov transformation, the transformation of the form $U^\dagger \alpha$ is not sufficient. You need to allow for mixing of creation and annihilation operators.

But then you can't just diagonalize the coefficient matrix. The logical error is that you imply that $c_i^\dagger c_i$ have the properties of a number operator, but why should this be true? The best you can do is bring the coefficient matrix into Jordan normal form, which leads to additional terms of different type.

For bosons it turns out that you cannot generally write the resulting Hamiltonian in the form $$ \mathcal H= \sum_i \mathcal C_i \, c_i^\dagger c_i. $$ This would assume that the matrix is generally diagonalizable. This is for instance not true when Goldstone modes exist.


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