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 $$\begin{pmatrix}
p\\
q \\
\end{pmatrix} $$ , 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: 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
$$(a_1^\dagger a_2^\dagger ... ) |0> $$
$$ H = \displaystyle\sum_{<i,j>} A_{ij} a_i^\dagger a_j + \frac{1}{2}\displaystyle\sum_{<i,j>} [B_{ij} a_i^\dagger a_j^\dagger + B_{ij}^* a_j a_i ]
$$
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 $$\begin{pmatrix}
p\\
q \\
\end{pmatrix} $$ , 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: 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
$$(a_1^\dagger a_2^\dagger ... ) |0> $$
0 commentaires:
Enregistrer un commentaire