Mach-Zehnder interferometer with squeezed vacuum (variance)
Building upon my previous posts (here and here) where I derived the expectation value of the balanced signal and set up the framework for squared signal analysis, I now focus on the variance of the output signal difference \mathbf N_{\lambda_6} - \mathbf N_{\lambda_5}. After extensive calculations, the expectation value of the squared balanced signal \langle \boldsymbol \Psi_{\text{in}} | \left(\mathbf N_{\lambda_6} - \mathbf N_{\lambda_5} \right)^2 | \boldsymbol \Psi_{\text{in}} \rangle is found to be:
\begin{aligned} \langle \boldsymbol \Psi_{\text{in}} | \left(\mathbf N_{\lambda_6} - \mathbf N_{\lambda_5} \right)^2 | \boldsymbol \Psi_{\text{in}} \rangle = & \sin^2(\varepsilon) \left[ \sinh^4(R_{\lambda_1}) - 2 \alpha_{\lambda_2}^2 \sinh^2(R_{\lambda_1}) + (\alpha_{\lambda_2}^4 + \alpha_{\lambda_2}^2) \right] \\ & + \cos^2(\varepsilon) \left[ \sinh^2(R_{\lambda_1}) + \alpha_{\lambda_2}^2 e^{2R_{\lambda_1}} \right] \end{aligned}
From this, I can compute the variance \langle \boldsymbol \Psi_{\text{in}} |\left(\Delta_{\mathbf N_{\lambda_6} - \mathbf N_{\lambda_5}}\right)^2 | \boldsymbol \Psi_{\text{in}} \rangle = \langle \boldsymbol \Psi_{\text{in}} | \left(\mathbf N_{\lambda_6} - \mathbf N_{\lambda_5} \right)^2 | \boldsymbol \Psi_{\text{in}} \rangle - \left(\langle \boldsymbol \Psi_{\text{in}} |\mathbf N_{\lambda_6} - \mathbf N_{\lambda_5}| \boldsymbol \Psi_{\text{in}} \rangle\right)^2, which simplifies to:
\langle \boldsymbol \Psi_{\text{in}} |\left(\Delta_{\mathbf N_{\lambda_6} - \mathbf N_{\lambda_5}}\right)^2 | \boldsymbol \Psi_{\text{in}} \rangle = \alpha_{\lambda_2}^2 \left( \sin^2(\varepsilon) + e^{2R_{\lambda_1}} \cos^2(\varepsilon) \right) + \sinh^2(R_{\lambda_1}) \cos^2(\varepsilon)
To assess the performance enhancement, I examine the signal-to-noise ratio (SNR), defined as the ratio of the expected signal to the standard deviation of the noise:
\text{SNR} = \frac{\langle \mathbf N_{\lambda_5} - \mathbf N_{\lambda_6} \rangle}{\Delta_{\mathbf N_{\lambda_6}-\mathbf N_{\lambda_5}}}
Substituting the expressions for the expectation value and the standard deviation (square root of the variance), I find:
\text{SNR} = \frac{\sin(\varepsilon) \left[ \alpha_\lambda^2 - \sinh^2(R_{\lambda_1}) \right]}{\sqrt{\alpha_{\lambda_2}^2 \left( \sin^2(\varepsilon) + e^{2R_{\lambda_1}} \cos^2(\varepsilon) \right) + \cos^2(\varepsilon) \sinh^2(R_{\lambda_1})}} \approx e^{-R_{\lambda_1}} \varepsilon \sqrt{\mathbf N_{\lambda_2}}
For small dephasings \varepsilon and a strong coherent signal in input channel (2), the SNR is approximately increased by a factor e^{-R_{\lambda_1}} compared to a standard interferometer without squeezed vacuum. Since for squeezed vacuum R_{\lambda_1} < 0, this factor is greater than unity, indicating an enhanced sensitivity. The minimum detectable dephasing is reduced, demonstrating that the sensitivity with a squeezed vacuum input surpasses the standard quantum limit.
As a consistency check, consider the case where \varepsilon = 0. In this scenario, the variance becomes:
\langle \boldsymbol \Psi_{\text{in}} |\left(\Delta_{\mathbf N_{\lambda_6} - \mathbf N_{\lambda_5}}\right)^2 | \boldsymbol \Psi_{\text{in}} \rangle|_{\varepsilon=0} = \alpha_{\lambda_2}^2 e^{2R_{\lambda_1}} = e^{2R_{\lambda_1}} \langle \mathbf N_{\lambda_2} \rangle
This result aligns with the expected variance of the squeezed momentum quadrature \mathbf P_{\lambda_1}, confirming the validity of my calculations.
In summary, my detailed quantum analysis reveals that employing a squeezed vacuum state at the input of a Mach-Zehnder interferometer leads to a significant improvement in sensitivity. The variance of the balanced signal is modified by the squeezing parameter, resulting in a higher signal-to-noise ratio and the ability to detect smaller phase shifts. This demonstrates the practical advantage of using squeezed light for precision quantum measurements.
For more insights into this topic, you can find the details here.