- #1

- 1

- 0

## Main Question or Discussion Point

**Can anyone help me out with von Neumann entropy please??**

I've been wondering if I could be able to calculate the entropy of a system which is part of a bigger system.

For example, let's say that there's a cavity and two atoms(atoms are two-level system).

The hamiltonian of the system would roughly be like this;

[tex] \[

H = \sum_{j} \omega s_{jz} + \omega a^{\dag} a + \sum_{j} g ( s_{j-} a^{\dag} + s_{j+} a ),

\][/tex]

And you know the density matrix. I put the state in this form;

[tex] \[

|\Psi \ket = |\Psi_{atom1} \ket \otimes |\Psi_{atom2} \ket \otimes |\Psi_{cav} \ket = a|1_1, 0_2, 0_3 \ket + b|0_1, 1_2, 0_3 \ket + c|0_1, 0_2, 1_3 \ket

\][/tex]

where 1 is atom1, 2 is atom 2, and 3 is a cavity.

and got this density matrix;

[tex] \[

\rho^{ABC} =

\left( {\begin{array}{ccc}

|a|^2 & ab^\ast & ac^\ast \\

a^\ast b & |b|^2 & bc^\ast \\

a^\ast c & b^\ast c & |c|^2 \\

\end{array} } \right)

\]

[/tex]

Now, I'm trying to see the entropy between the two atoms, and I was wondering about this. I don't know if it's really simple thing or something that's not really been cleared.

First, I thought I had to reduce the matrix to kill the degree of freedom of the cavity. Which gives me this.

[tex] \[

\rho^{AB} =

\left( {\begin{array}{ccc}

|a|^2 & ab^\ast & 0 \\

a^\ast b & |b|^2 & 0 \\

0 & 0 & |c|^2 \\

\end{array} } \right)

\][/tex]

Then I thought now I had to reduce the matrix once more in order to evaluate the entropy. And this gives me a funny result. By choosing different atom to reduce, we get different entropy. The average entropy in a period is the same, but shouldn't it be identical at any time?? I mean, it's the same system. Is there some kind of restriction when you use reduced matrix for entropy like, 'you shouldn't use the reduced matrix to reduce it for the entropy' or something like that?

I really don't know if I'm beeing stupid or anything, but even if I am, let me know.

Thanks in advance!! :D