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Correct way to write the eigenvector of a diagonalized hamiltonian in second quantization

+ 7 like - 0 dislike
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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 $$p \,a|0> + q \, a^\dagger|0> $$ (where the column $\alpha$ has been used as the basis) or $$p \,a^\dagger|0> + q \, a|0> $$ (where $\alpha^\dagger$ has been used as the basis)?

Note: Here $|0>$ 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 $$(a_1^\dagger a_2^\dagger ... ) |0> $$

This post imported from StackExchange Physics at 2014-04-05 11:03 (UCT), posted by SE-user user1906035
asked Mar 29, 2014 in Theoretical Physics by user1906035 (35 points) [ no revision ]
retagged Apr 5, 2014
I forgot to mention that the Hamiltonian is Hermitian. I will edit the question to add the conditions on A and B.

This post imported from StackExchange Physics at 2014-04-05 11:04 (UCT), posted by SE-user user1906035

2 Answers

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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.

This post imported from StackExchange Physics at 2014-04-05 11:04 (UCT), posted by SE-user webb
answered Apr 3, 2014 by webb (20 points) [ no revision ]
+ 1 like - 0 dislike

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):

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

with:

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

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.

This post imported from StackExchange Physics at 2014-04-05 11:04 (UCT), posted by SE-user luming
answered Apr 5, 2014 by BaBQ (95 points) [ no revision ]

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