Mach-Zehnder interferometer with squeezed vacuum (expectations)
In my previous post here, I discussed the expectation value of the balanced signal \mathbf N_{\lambda_6} - \mathbf N_{\lambda_5}. Now, I turn my attention to the expectation of its square, \left(\mathbf N_{\lambda_6} - \mathbf N_{\lambda_5}\right)^2. Starting from the expression for the balanced signal difference:
\mathbf N_{\lambda_6} - \mathbf N_{\lambda_5} = -\sin(\varepsilon) \left( \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} - \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} \right) + i \cos(\varepsilon) \left( \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2} - \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_1} \right)
To compute the expectation of the squared balanced signal, I need to square this operator expression:
\begin{aligned} \left(\mathbf N_{\lambda_6} - \mathbf N_{\lambda_5}\right)^2 = & \left\{-\sin(\varepsilon) \left[ \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} - \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} \right] + i \cos(\varepsilon) \left[ \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2} - \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_1} \right] \right\}^2 \end{aligned}
As before, input channel (1) is in a squeezed vacuum state, so I replace \mathbf a_{\lambda_1} and \mathbf a_{\lambda_1}^\dag with their expressions in terms of the squeezed vacuum operators \mathbf{A}_{R_{\lambda_1}} and \mathbf{A}_{R_{\lambda_1}}^\dag:
\begin{aligned} \mathbf a_{\lambda_1} &= \mathbf{A}_{R_{\lambda_1}} \cosh(R_{\lambda_1}) - \mathbf{A}_{R_{\lambda_1}}^\dag \sinh(R_{\lambda_1}) \\ \mathbf a_{\lambda_1}^\dag &= \mathbf{A}_{R_{\lambda_1}}^\dag \cosh(R_{\lambda_1}) - \mathbf{A}_{R_{\lambda_1}} \sinh(R_{\lambda_1}) \end{aligned}
Substituting these transformations into the squared balanced signal expression results in a lengthy formula involving products of \mathbf{A}_{R_{\lambda_1}}, \mathbf{A}_{R_{\lambda_1}}^\dag, \mathbf a_{\lambda_2}, and \mathbf a_{\lambda_2}^\dag. For example, the term (\mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1})^2 expands to:
\begin{aligned} \left( \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} \right)^2 = & \left( \cosh^2(R_{\lambda_1}) \mathbf{A}_{R_{\lambda_1}}^\dag \mathbf{A}_{R_{\lambda_1}} - \cosh(R_{\lambda_1}) \sinh(R_{\lambda_1}) \mathbf{A}_{R_{\lambda_1}}^\dag \mathbf{A}_{R_{\lambda_1}}^\dag \right. \\ & \left. - \sinh(R_{\lambda_1}) \cosh(R_{\lambda_1}) \mathbf{A}_{R_{\lambda_1}} \mathbf{A}_{R_{\lambda_1}} + \sinh^2(R_{\lambda_1}) \mathbf{A}_{R_{\lambda_1}} \mathbf{A}_{R_{\lambda_1}}^\dag \right)^2 \end{aligned}
Expanding the full squared expression involves computing terms like (\mathbf a_{\lambda_1}^\dag)^2, (\mathbf a_{\lambda_1})^2, (\mathbf a_{\lambda_1}^\dag)^2 \mathbf a_{\lambda_1}, and \mathbf a_{\lambda_1}^\dag (\mathbf a_{\lambda_1})^2 in terms of \mathbf{A}_{R_{\lambda_1}} and \mathbf{A}_{R_{\lambda_1}}^\dag. These calculations, while algebraically intensive, are necessary to fully characterize the quantum noise properties of the balanced signal in this interferometer setup.
The resulting expression for \left(\mathbf N_{\lambda_6} - \mathbf N_{\lambda_5}\right)^2 in terms of squeezed vacuum operators is quite complex. In a subsequent step, I will evaluate the expectation value of this expression with respect to the input state | \boldsymbol \Psi_{\text{in}} \rangle = | \mathbf 0, R_\lambda \rangle_1 \otimes | \boldsymbol \alpha_{\lambda} \rangle_2. This will allow me to understand how the quantum fluctuations of the squeezed vacuum and the coherent state contribute to the variance of the balanced detector signal.
For more insights into this topic, you can find the details here.