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Correct way to write the eigenvector in second quantization

  1. Mar 29, 2014 #1
    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.

    I can write this 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 $$ M = \begin{pmatrix}
    A & B\\
    B^* & A^* \\
    \end{pmatrix} $$

    The Hamiltonian is called diagonal when it is expressed as:

    $$ H = \beta^\dagger N \beta - \frac{1}{2} tr(A) $$
    where $$ \beta = \begin{pmatrix}
    b \\
    b^\dagger \\
    \end{pmatrix} $$
    , $$ \beta^\dagger = \begin{pmatrix}
    b & b^\dagger
    \end{pmatrix} $$
    and N is a 2-by-2 matrix .

    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 the eigenvector in second quantization?

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

    Note: where |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
    $$(a_1^\dagger a_2^\dagger ... ) |0> $$
  2. jcsd
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