Calculating Many-Body Correlator: c$_{L}^{\dagger}$d

[gonadas91]

Hi all, I am trying to calculate the correlator between two different creation and annihilation operators on a many body system. The system is composed by two wires (left and right) and an energy level between them $\epsilon_{0}$. So we have the creation and annihilation operators on the wires, $c_{L,R}^{\dagger},c_{L,R}$, and the ones with the dot level $d^{\dagger}, d$. Then, how could I find out the correlator (equal time correlation):

\begin{eqnarray*}
\langle c_{L}^{\dagger}(x)d\rangle
\end{eqnarray*}

Any ideas?

 

[eljose79]

Hello,

To find the correlator between two different creation and annihilation operators on a many body system, we can use the following formula:

\begin{eqnarray*}
\langle c_{L}^{\dagger}(x)d\rangle = \frac{\langle \Psi | c_{L}^{\dagger}(x)d |\Psi \rangle}{\langle \Psi | \Psi \rangle}
\end{eqnarray*}

where $|\Psi \rangle$ is the state vector of the many body system.

To calculate this correlator, we first need to determine the state vector $|\Psi \rangle$. This can be done by solving the Schrödinger equation for the system, which will give us the energy eigenstates and their corresponding eigenvalues.

Next, we can express the creation and annihilation operators in terms of the energy eigenstates. For example, we can write $c_{L}^{\dagger}(x)$ as a linear combination of the energy eigenstates $|n \rangle$:

\begin{eqnarray*}
c_{L}^{\dagger}(x) = \sum_{n} c_{L}^{\dagger}(x)|n \rangle
\end{eqnarray*}

We can do the same for the other operators $c_{L}, d^{\dagger}, d$. Then, we can substitute these expressions into the formula above to calculate the correlator.

I hope this helps. Let me know if you have any further questions or need clarification. Good luck with your calculations!