Quantum Optics - Measuring Squeezed States Of Light

Measuring Squeezed States of Light

Mach-Zehnder interferometer

Gravitational wave detection using optical interferometers is limited by laser shot noise. Squeezed light was theoretically proposed in the early 1980s to enhance interferometer sensitivity beyond this limit, potentially enabling gravitational wave detection with feasible instruments.

While gravitational wave detectors use more complex interferometers, the principle of sensitivity improvement with squeezed light can be understood using a Mach-Zehnder interferometer as a simpler example.

I have already explained how the interferometers works here, I will change the vacuum port from (2) to (1) to align with the description of the homodyne detection here.

Vacuum in channel one

In a Mach-Zehnder interferometer, a laser beam (quasi-classical state | \boldsymbol \alpha_\lambda \rangle) enters input channel (2), and vacuum enters input channel (1):

| \boldsymbol \Psi_{\text{in}} \rangle = | \mathbf 0 \rangle_1 \otimes | \boldsymbol \alpha_{\lambda} \rangle_2

This setup, resembling balanced homodyne detection, allows for calculating photon counts \mathbf N_{\lambda_5} and \mathbf N_{\lambda_6}. The average photon number in channel (2) is |\alpha_\lambda|^2:

\mathbf N_{\lambda_2} = \langle \boldsymbol \alpha_\lambda | \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} | \boldsymbol \alpha_\lambda \rangle = \left| \alpha_\lambda \right|^2

It is possible to derive the operators \mathbf a_{\lambda_6} and \mathbf a_{\lambda_5}. We assume a balanced beam splitter (R = T = 0.5) and a minus sign for the amplitude reflection coefficient from mode (2) to (4) and from mode (3) to (5).

Let’s use the beam splitter matrix for the first beam splitter BS_1:

{BS}_1 = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}

and for the second beam splitter BS_2:

{BS}_2 = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 1 \\ -1 & 1 \end{bmatrix}

The input modes are \mathbf a_{\lambda_1} and \mathbf a_{\lambda_2}. After the first beam splitter, the modes \mathbf b_1 and \mathbf b_2 are:

\begin{bmatrix} \mathbf b_1 \\ \mathbf b_2 \end{bmatrix} = {BS}_1 \begin{bmatrix} \mathbf a_{\lambda_1} \\ \mathbf a_{\lambda_2} \end{bmatrix} = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} \mathbf a_{\lambda_1} \\ \mathbf a_{\lambda_2} \end{bmatrix} = \frac{1}{\sqrt{2}} \begin{bmatrix} \mathbf a_{\lambda_1} + \mathbf a_{\lambda_2} \\ \mathbf a_{\lambda_1} - \mathbf a_{\lambda_2} \end{bmatrix}

So,

\begin{aligned} \mathbf b_1 & = \frac{1}{\sqrt{2}} (\mathbf a_{\lambda_1} + \mathbf a_{\lambda_2}) \\ \mathbf b_2 & = \frac{1}{\sqrt{2}} (\mathbf a_{\lambda_1} - \mathbf a_{\lambda_2}) \end{aligned}

Let’s introduce phase shifts in paths 3 and 4. As the beam propagating through arm 3 of length L_3, it has an accumulated phase factor e^{ikL_3}. This phase factor arises from the wave nature of light. As a wave with wave number k propagates a distance L_3, its phase advances by kL_3, then the amplitude of the wave is multiplied by e^{ikL_3}. Similarly, e^{ikL_4} is the phase factor for arm 4 of length L_4.

\begin{aligned} \mathbf c_1 = e^{ikL_3} \mathbf b_1 & = \frac{1}{\sqrt{2}} e^{ikL_3} (\mathbf a_{\lambda_1} + \mathbf a_{\lambda_2}) \\ \mathbf c_2 = e^{ikL_4} \mathbf b_2 & = \frac{1}{\sqrt{2}} e^{ikL_4} (\mathbf a_{\lambda_1} - \mathbf a_{\lambda_2}) \end{aligned}

After the second beam splitter, the output modes \mathbf a_{\lambda_6} and \mathbf a_{\lambda_5} are:

\begin{bmatrix} \mathbf a_{\lambda_6} \\ \mathbf a_{\lambda_5} \end{bmatrix} = {BS}_2 \begin{bmatrix} \mathbf c_1 \\ \mathbf c_2 \end{bmatrix} = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 1 \\ -1 & 1 \end{bmatrix} \begin{bmatrix} \mathbf c_1 \\ \mathbf c_2 \end{bmatrix} = \frac{1}{\sqrt{2}} \begin{bmatrix} \mathbf c_1 + \mathbf c_2 \\ -\mathbf c_1 + \mathbf c_2 \end{bmatrix}

Therefore:

\begin{aligned} \mathbf a_{\lambda_6} & = \frac{1}{\sqrt{2}} (\mathbf c_1 + \mathbf c_2) = \frac{1}{2} \left[ e^{ikL_3} (\mathbf a_{\lambda_1} + \mathbf a_{\lambda_2}) + e^{ikL_4} (\mathbf a_{\lambda_1} - \mathbf a_{\lambda_2}) \right] \\ \mathbf a_{\lambda_5} & = \frac{1}{\sqrt{2}} (-\mathbf c_1 + \mathbf c_2) = \frac{1}{2} \left[ -e^{ikL_3} (\mathbf a_{\lambda_1} + \mathbf a_{\lambda_2}) + e^{ikL_4} (\mathbf a_{\lambda_1} - \mathbf a_{\lambda_2}) \right] \end{aligned}

Rearranging the terms:

\begin{aligned} \mathbf a_{\lambda_6} & = \frac{1}{2} \left[ (e^{ikL_3} + e^{ikL_4}) \mathbf a_{\lambda_1} + (e^{ikL_3} - e^{ikL_4}) \mathbf a_{\lambda_2} \right] \\ \mathbf a_{\lambda_5} & = \frac{1}{2} \left[ (-e^{ikL_3} + e^{ikL_4}) \mathbf a_{\lambda_1} + (-e^{ikL_3} - e^{ikL_4}) \mathbf a_{\lambda_2} \right] \end{aligned}

Factor out e^{ik(L_3+L_4)/2}:

\begin{aligned} \mathbf a_{\lambda_6} = & \frac{1}{2} e^{ik(L_3+L_4)/2} \left[ (e^{ik(L_3-L_4)/2} + e^{-ik(L_3-L_4)/2}) \mathbf a_{\lambda_1} \right. \\ & \left. + (e^{ik(L_3-L_4)/2} - e^{-ik(L_3-L_4)/2}) \mathbf a_{\lambda_2} \right] \\ \mathbf a_{\lambda_5} = & \frac{1}{2} e^{ik(L_3+L_4)/2} \left[ (-e^{ik(L_3-L_4)/2} + e^{-ik(L_3-L_4)/2}) \mathbf a_{\lambda_1} \right. \\ & \left. + (-e^{ik(L_3-L_4)/2} - e^{-ik(L_3-L_4)/2}) \mathbf a_{\lambda_2} \right] \end{aligned}

Using \cos(x) = \frac{e^{ix} + e^{-ix}}{2} and \sin(x) = \frac{e^{ix} - e^{-ix}}{2i}:

\begin{aligned} \mathbf a_{\lambda_6} & = e^{ik(L_3+L_4)/2} \left[ \cos\left(\frac{k(L_3 - L_4)}{2}\right) \mathbf a_{\lambda_1} + i \sin\left(\frac{k(L_3 - L_4)}{2}\right) \mathbf a_{\lambda_2} \right] \\ \mathbf a_{\lambda_5} & = e^{ik(L_3+L_4)/2} \left[ -i \sin\left(\frac{k(L_3 - L_4)}{2}\right) \mathbf a_{\lambda_1} - \cos\left(\frac{k(L_3 - L_4)}{2}\right) \mathbf a_{\lambda_2} \right] \end{aligned}

Defining \delta:

\delta \equiv k\left(L_3 - L_4 \right) = 2\pi \frac{L_3 - L_4}{\lambda}

The formulas simplify to:

\begin{aligned} \mathbf a_{\lambda_6} & = e^{ik \frac{(L_3+L_4)}{2}} \left[ \mathbf a_{\lambda_1} \cos\left(\frac{\delta}{2}\right) + i\mathbf a_{\lambda_2} \sin\left(\frac{\delta}{2}\right) \right] \\ \mathbf a_{\lambda_5} & = e^{ik \frac{(L_3+L_4)}{2}} \left[ -i\mathbf a_{\lambda_1} \sin\left(\frac{\delta}{2}\right) - \mathbf a_{\lambda_2} \cos\left(\frac{\delta}{2}\right) \right] \end{aligned}

We can compute the expectation value of the operators \mathbf N_{\lambda_6} and \mathbf N_{\lambda_5} with respect to the input state |\boldsymbol \Psi_{\text{in}} \rangle = | \mathbf 0 \rangle_1 \otimes | \boldsymbol \alpha_{\lambda} \rangle_2.

Since input channel (1) is the vacuum (\langle \mathbf 0 | \mathbf a_{\lambda}^\dag = \mathbf a_{\lambda} | \mathbf 0 \rangle=0) and the input channel (2) is quasi-classical (\langle \boldsymbol \alpha_{\lambda} | \mathbf a_{\lambda}^\dag = \bar \alpha_{\lambda} \langle \boldsymbol \alpha_{\lambda} | and \mathbf a_\lambda | \boldsymbol \alpha_\lambda \rangle = \alpha_{\lambda} | \boldsymbol \alpha_\lambda \rangle), the following holds:

\begin{aligned} \langle \boldsymbol \Psi_{\text{in}} | \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} | \boldsymbol \Psi_{\text{in}} \rangle & = {}_1 \langle \mathbf 0 | {}_2 \langle \boldsymbol \alpha_{\lambda} | \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} | \mathbf 0 \rangle_1 | \boldsymbol \alpha_{\lambda} \rangle_2 = \langle \mathbf 0 | \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} | \mathbf 0 \rangle \langle \boldsymbol \alpha_{\lambda} | \boldsymbol \alpha_{\lambda} \rangle = 0 \\ \langle \boldsymbol \Psi_{\text{in}} | \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} | \boldsymbol \Psi_{\text{in}} \rangle & = {}_1 \langle \mathbf 0 | {}_2 \langle \boldsymbol \alpha_{\lambda} | \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} | \mathbf 0 \rangle_1 | \boldsymbol \alpha_{\lambda} \rangle_2 = \langle \mathbf 0 | \mathbf 0 \rangle \langle \boldsymbol \alpha_{\lambda} | \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} | \boldsymbol \alpha_{\lambda} \rangle = |\alpha_{\lambda}|^2 \\ \langle \boldsymbol \Psi_{\text{in}} | \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2} | \boldsymbol \Psi_{\text{in}} \rangle & = {}_1 \langle \mathbf 0 | {}_2 \langle \boldsymbol \alpha_{\lambda} | \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2} | \mathbf 0 \rangle_1 | \boldsymbol \alpha_{\lambda} \rangle_2 = \langle \mathbf 0 | \mathbf a_{\lambda_1}^\dag | \mathbf 0 \rangle \langle \boldsymbol \alpha_{\lambda} | \mathbf a_{\lambda_2} | \boldsymbol \alpha_{\lambda} \rangle = 0 \times \alpha_{\lambda} = 0 \\ \langle \boldsymbol \Psi_{\text{in}} | \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_1} | \boldsymbol \Psi_{\text{in}} \rangle & = {}_1 \langle \mathbf 0 | {}_2 \langle \boldsymbol \alpha_{\lambda} | \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_1} | \mathbf 0 \rangle_1 | \boldsymbol \alpha_{\lambda} \rangle_2 = \langle \boldsymbol \alpha_{\lambda} | \mathbf a_{\lambda_2}^\dag | \boldsymbol \alpha_{\lambda} \rangle \langle \mathbf 0 | \mathbf a_{\lambda_1} | \mathbf 0 \rangle = \bar \alpha_{\lambda} \times 0 = 0 \end{aligned}

For \mathbf a_{\lambda_6}:

\begin{aligned} \langle \boldsymbol \Psi_{\text{in}} | \mathbf N_{\lambda_6} | \boldsymbol \Psi_{\text{in}} \rangle = & \langle \boldsymbol \Psi_{\text{in}} | \mathbf a_{\lambda_6}^\dag \mathbf a_{\lambda_6} | \boldsymbol \Psi_{\text{in}} \rangle \\ = & \langle \boldsymbol \Psi_{\text{in}} | \left\{ e^{-ik \frac{(L_3+L_4)}{2}} \left[ \mathbf a_{\lambda_1}^\dag \cos\left(\frac{\delta}{2}\right) - i\mathbf a_{\lambda_2}^\dag \sin\left(\frac{\delta}{2}\right) \right] \right\} \\ & \cdot \left\{ e^{ik \frac{(L_3+L_4)}{2}} \left[ \mathbf a_{\lambda_1} \cos\left(\frac{\delta}{2}\right) + i\mathbf a_{\lambda_2} \sin\left(\frac{\delta}{2}\right) \right] \right\} | \boldsymbol \Psi_{\text{in}} \rangle \\ = & \langle \boldsymbol \Psi_{\text{in}} | \left[ \mathbf a_{\lambda_1}^\dag \cos\left(\frac{\delta}{2}\right) - i\mathbf a_{\lambda_2}^\dag \sin\left(\frac{\delta}{2}\right) \right] \\ & \cdot \left[ \mathbf a_{\lambda_1} \cos\left(\frac{\delta}{2}\right) + i\mathbf a_{\lambda_2} \sin\left(\frac{\delta}{2}\right) \right] | \boldsymbol \Psi_{\text{in}} \rangle \\ = & \langle \boldsymbol \Psi_{\text{in}} | \left[ \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} \cos^2\left(\frac{\delta}{2}\right) + i \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2} \cos\left(\frac{\delta}{2}\right) \sin\left(\frac{\delta}{2}\right) \right. \\ & \left. - i \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_1} \sin\left(\frac{\delta}{2}\right) \cos\left(\frac{\delta}{2}\right) + \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} \sin^2\left(\frac{\delta}{2}\right) \right] | \boldsymbol \Psi_{\text{in}} \rangle \\ = & \cos^2\left(\frac{\delta}{2}\right) \langle \boldsymbol \Psi_{\text{in}} | \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} | \boldsymbol \Psi_{\text{in}} \rangle + \sin^2\left(\frac{\delta}{2}\right) \langle \boldsymbol \Psi_{\text{in}} | \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} | \boldsymbol \Psi_{\text{in}} \rangle \\ & + i \cos\left(\frac{\delta}{2}\right) \sin\left(\frac{\delta}{2}\right) \langle \boldsymbol \Psi_{\text{in}} | \left( \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2} - \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_1} \right) | \boldsymbol \Psi_{\text{in}} \rangle \\ = & \langle \boldsymbol \Psi_{\text{in}} | \mathbf N_{\lambda_6} | \boldsymbol \Psi_{\text{in}} \rangle = \left| \alpha_{\lambda}\right|^2 \sin^2\left(\frac{\delta}{2}\right) \end{aligned}

Similarly for \mathbf N_{\lambda_5}:

\begin{aligned} \langle \boldsymbol \Psi_{\text{in}} | \mathbf N_{\lambda_5} | \boldsymbol \Psi_{\text{in}} \rangle = & \langle \boldsymbol \Psi_{\text{in}} | \mathbf a_{\lambda_5}^\dag \mathbf a_{\lambda_5} | \boldsymbol \Psi_{\text{in}} \rangle \\ = & \langle \boldsymbol \Psi_{\text{in}} | \left\{ e^{-ik \frac{(L_3+L_4)}{2}} \left[ i\mathbf a_{\lambda_1}^\dag \sin\left(\frac{\delta}{2}\right) - \mathbf a_{\lambda_2}^\dag \cos\left(\frac{\delta}{2}\right) \right] \right\} \\ & \cdot \left\{ e^{ik \frac{(L_3+L_4)}{2}} \left[ -i\mathbf a_{\lambda_1} \sin\left(\frac{\delta}{2}\right) - \mathbf a_{\lambda_2} \cos\left(\frac{\delta}{2}\right) \right] \right\} | \boldsymbol \Psi_{\text{in}} \rangle \\ = & \langle \boldsymbol \Psi_{\text{in}} | \left[ i\mathbf a_{\lambda_1}^\dag \sin\left(\frac{\delta}{2}\right) - \mathbf a_{\lambda_2}^\dag \cos\left(\frac{\delta}{2}\right) \right] \\ & \cdot \left[ -i\mathbf a_{\lambda_1} \sin\left(\frac{\delta}{2}\right) - \mathbf a_{\lambda_2} \cos\left(\frac{\delta}{2}\right) \right] | \boldsymbol \Psi_{\text{in}} \rangle \\ = & \langle \boldsymbol \Psi_{\text{in}} | \left[ \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} \sin^2\left(\frac{\delta}{2}\right) + i \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2} \sin\left(\frac{\delta}{2}\right) \cos\left(\frac{\delta}{2}\right) \right. \\ & \left. - i \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_1} \cos\left(\frac{\delta}{2}\right) \sin\left(\frac{\delta}{2}\right) + \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} \cos^2\left(\frac{\delta}{2}\right) \right] | \boldsymbol \Psi_{\text{in}} \rangle \\ = & \sin^2\left(\frac{\delta}{2}\right) \langle \boldsymbol \Psi_{\text{in}} | \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} | \boldsymbol \Psi_{\text{in}} \rangle + \cos^2\left(\frac{\delta}{2}\right) \langle \boldsymbol \Psi_{\text{in}} | \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} | \boldsymbol \Psi_{\text{in}} \rangle \\ & + i \cos\left(\frac{\delta}{2}\right) \sin\left(\frac{\delta}{2}\right) \langle \boldsymbol \Psi_{\text{in}} | \left( \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2} - \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_1} \right) | \boldsymbol \Psi_{\text{in}} \rangle \\ = & \left|\alpha_{\lambda}\right|^2 \cos^2\left(\frac{\delta}{2}\right) \end{aligned}

Therefore the number of photons are shows a sinusoidal variation as a function of the phase difference \delta, and it can be adjusted with a piezoelectric transducer action on mirror \mathbf M_3.

To measure small arm length variations, we can subtract the two output signals \mathbf N_6 - \mathbf N_5:

\begin{aligned} \langle \boldsymbol \Psi_{\text{in}} | \mathbf N_{\lambda_6} - \mathbf N_{\lambda_5} | \boldsymbol \Psi_{\text{in}} \rangle = & \langle \boldsymbol \Psi_{\text{in}} | \mathbf N_{\lambda_6} | \boldsymbol \Psi_{\text{in}} \rangle - \langle \boldsymbol \Psi_{\text{in}} | \mathbf N_{\lambda_5} | \boldsymbol \Psi_{\text{in}} \rangle \\ = & \left| \alpha_{\lambda}\right|^2 \sin^2\left(\frac{\delta}{2}\right) - \left| \alpha_{\lambda}\right|^2 \cos^2\left(\frac{\delta}{2}\right) \\ = & - \left| \alpha_{\lambda}\right|^2 \cos\left(\delta\right) \end{aligned}

Operating around a phase difference \delta = \pi/2 where the derivative is the maximum then gives the highest sensitivity:

\delta = k\left(L_3 - L_4 \right) = \frac{\pi}{2} + \varepsilon

Using the trigonometric identity \cos(a+b) = \cos(a)\cos(b) - \sin(a)\sin(b), we get:

\begin{aligned} \cos(\pi/2 + \varepsilon) & = \cos(\pi/2)\cos(\varepsilon) - \sin(\pi/2)\sin(\varepsilon) \\ & = 0 \cdot \cos(\varepsilon) - 1 \cdot \sin(\varepsilon) = -\sin(\varepsilon) \end{aligned}

Using the trigonometric identity \sin(a+b) = \sin(a)\cos(b) + \cos(a)\sin(b), we get:

\begin{aligned} \sin(\pi/2 + \varepsilon) & = \sin(\pi/2)\cos(\varepsilon) + \cos(\pi/2)\sin(\varepsilon) \\ & = 1 \cdot \cos(\varepsilon) + 0 \cdot \sin(\varepsilon) = \cos(\varepsilon) \end{aligned}

The equation for the photon number difference becomes:

\langle \boldsymbol \Psi_{\text{in}} | \mathbf N_{\lambda_6} - \mathbf N_{\lambda_5} | \boldsymbol \Psi_{\text{in}} \rangle = - \left| \alpha_{\lambda}\right|^2 (-\sin(\varepsilon)) = \left| \alpha_{\lambda}\right|^2 \sin(\varepsilon)

For small variations \varepsilon, we can approximate \sin(\varepsilon) \approx \varepsilon. So,

\langle \boldsymbol \Psi_{\text{in}} | \mathbf N_{\lambda_6} - \mathbf N_{\lambda_5} | \boldsymbol \Psi_{\text{in}} \rangle \approx \left| \alpha_{\lambda}\right|^2 \varepsilon = \mathbf N_{\lambda_2} \varepsilon

where \mathbf N_{\lambda_2} = |\alpha_\lambda|^2 is the average number of photons entering input channel (2).

While a large \mathbf N_{\lambda_2} appears to increase sensitivity, noise limits the measurement. Shot noise, arising from the photoelectric effect, causes fluctuations in \mathbf N_{\lambda_5} and N\mathbf N_{\lambda_6}.

In a semi-classical approach, since they are independent Poisson variables, the variance of their difference \mathbf N_{\lambda_6} - \mathbf N_{\lambda_5} is the sum of their variances, resulting in a noise level of \sqrt{\mathbf N_{\lambda_5} + \mathbf N_{\lambda_6}} = \sqrt{\mathbf N_{\lambda_2}}:

\Delta_{\mathbf N_{\lambda_6}-\mathbf N_{\lambda_5}} = \sqrt{\mathbf N_{\lambda_5} + \mathbf N_{\lambda_6}} = \sqrt{\mathbf N_{\lambda_2}} = \left| \alpha_\lambda \right|

and the signal to noise ratio:

\text{SNR} = \frac{\langle \mathbf N_{\lambda_5} - \mathbf N_{\lambda_6} \rangle}{\Delta_{\mathbf N_{\lambda_6}-\mathbf N_{\lambda_5}}} = \varepsilon \sqrt{\mathbf N_{\lambda_2}}

The signal-to-noise ratio (\text{SNR}) is \varepsilon \sqrt{\mathbf N_{\lambda_2}}, allowing detection of a dephasing \epsilon = 1/\varepsilon \sqrt{\mathbf N_{\lambda_2}} for \text{SNR} =1. Increasing photon number \mathbf N_{\lambda_2} improves sensitivity, but laser power cannot be increased indefinitely, and longer measurement times reduce detection bandwidth, hindering broadband signal detection, as the signal of interest could happen to a broad range of frequencies.

To overcome limitations of increasing laser power or measurement time for enhanced sensitivity, a quantum description of noise is necessary to understand how squeezed light improves performance.

Using the previously expression for \mathbf a_{\lambda_6} we can compute the expectation number of photon \mathbf N_{\lambda_6}:

\begin{aligned} \mathbf N_{\lambda_6} = & \mathbf a_{\lambda_6}^\dag\mathbf a_{\lambda_6} \\ = & \left(\mathbf a_{\lambda_1}^\dag \cos\left(\frac{\delta}{2}\right) - i\mathbf a_{\lambda_2}^\dag \sin\left(\frac{\delta}{2}\right) \right) \left(\mathbf a_{\lambda_1} \cos\left(\frac{\delta}{2}\right) + i\mathbf a_{\lambda_2} \sin\left(\frac{\delta}{2}\right) \right) \\ = & \cos^2\left(\frac{\delta}{2}\right) \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} + i \cos\left(\frac{\delta}{2}\right) \sin\left(\frac{\delta}{2}\right) \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2} \\ & - i \sin\left(\frac{\delta}{2}\right) \cos\left(\frac{\delta}{2}\right) \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_1} + \sin^2\left(\frac{\delta}{2}\right) \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} \\ = & \cos^2\left(\frac{\delta}{2}\right) \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} + \sin^2\left(\frac{\delta}{2}\right) \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} + i \cos\left(\frac{\delta}{2}\right) \sin\left(\frac{\delta}{2}\right) \left( \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2} - \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_1} \right) \end{aligned}

Similarly, using the previously expression for \mathbf a_{\lambda_5} we can compute the expectation number of photon \mathbf N_{\lambda_5}:

\begin{aligned} \mathbf N_{\lambda_5} = & \mathbf a_{\lambda_5}^\dag\mathbf a_{\lambda_5} \\ = & \left( i\mathbf a_{\lambda_1}^\dag \sin\left(\frac{\delta}{2}\right) - \mathbf a_{\lambda_2}^\dag \cos\left(\frac{\delta}{2}\right) \right) \left( -i\mathbf a_{\lambda_1} \sin\left(\frac{\delta}{2}\right) - \mathbf a_{\lambda_2} \cos\left(\frac{\delta}{2}\right) \right) \\ = & \sin^2\left(\frac{\delta}{2}\right) \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} - i \sin\left(\frac{\delta}{2}\right) \cos\left(\frac{\delta}{2}\right) \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2} + i \cos\left(\frac{\delta}{2}\right) \sin\left(\frac{\delta}{2}\right) \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_1} + \cos^2\left(\frac{\delta}{2}\right) \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} \\ = & \sin^2\left(\frac{\delta}{2}\right) \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} + \cos^2\left(\frac{\delta}{2}\right) \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} + i \cos\left(\frac{\delta}{2}\right) \sin\left(\frac{\delta}{2}\right) \left( \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_1} - \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2} \right) \end{aligned}

Then the difference \mathbf N_{\lambda_6} - \mathbf N_{\lambda_5} can be expressed as:

\begin{aligned} \mathbf N_{\lambda_6} - \mathbf N_{\lambda_5} = & \left( \cos^2\left(\frac{\delta}{2}\right) - \sin^2\left(\frac{\delta}{2}\right) \right) \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} + \left( \sin^2\left(\frac{\delta}{2}\right) - \cos^2\left(\frac{\delta}{2}\right) \right) \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} \\ & + i \cos\left(\frac{\delta}{2}\right) \sin\left(\frac{\delta}{2}\right) \left( \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2} - \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_1} \right) - i \cos\left(\frac{\delta}{2}\right) \sin\left(\frac{\delta}{2}\right) \left( \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_1} - \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2} \right) \\ = & \cos(\delta) \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} - \cos(\delta) \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} + 2 i \cos\left(\frac{\delta}{2}\right) \sin\left(\frac{\delta}{2}\right) \left( \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2} - \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_1} \right) \\ = & \cos(\delta) \left( \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} - \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} \right) + i \sin(\delta) \left( \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2} - \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_1} \right) \end{aligned}

While input channel (1) is in a vacuum state, its inclusion (\mathbf a_{\lambda_1}) is important and highlights the non-trivial role of vacuum in quantum optical systems.

We can express this difference now as function of the dephasing \varepsilon with \delta = \pi/2 + \varepsilon:

\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)

We can take the take its expectation when input channel (1) is empty and a quasi-classical state is injected in input channel (2):

\begin{aligned} \langle \boldsymbol \Psi_{\text{in}} | \mathbf N_{\lambda_6} - \mathbf N_{\lambda_5} | \boldsymbol \Psi_{\text{in}} \rangle = & \langle \boldsymbol \Psi_{\text{in}} | -\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) | \boldsymbol \Psi_{\text{in}} \rangle \\ = & -\sin(\varepsilon) \langle \boldsymbol \Psi_{\text{in}} | \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} | \boldsymbol \Psi_{\text{in}} \rangle + \sin(\varepsilon) \langle \boldsymbol \Psi_{\text{in}} | \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} | \boldsymbol \Psi_{\text{in}} \rangle \\ & + i \cos(\varepsilon) \langle \boldsymbol \Psi_{\text{in}} | \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2} | \boldsymbol \Psi_{\text{in}} \rangle - i \cos(\varepsilon) \langle \boldsymbol \Psi_{\text{in}} | \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_1} | \boldsymbol \Psi_{\text{in}} \rangle \end{aligned}

For the first term:

-\sin(\varepsilon) \langle \boldsymbol \Psi_{\text{in}} | \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} | \boldsymbol \Psi_{\text{in}} \rangle = -\sin(\varepsilon) \langle \mathbf 0 |_1 \langle \boldsymbol \alpha_{\lambda} |_2 \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} | \mathbf 0 \rangle_1 | \boldsymbol \alpha_{\lambda} \rangle_2 = -\sin(\varepsilon) \langle \mathbf 0 |_1 \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} | \mathbf 0 \rangle_1 \langle \boldsymbol \alpha_{\lambda} |_2 | \boldsymbol \alpha_{\lambda} \rangle_2 = 0

because \mathbf a_{\lambda_1} | \mathbf 0 \rangle_1 = 0.

For the second term:

\sin(\varepsilon) \langle \boldsymbol \Psi_{\text{in}} | \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} | \boldsymbol \Psi_{\text{in}} \rangle = \sin(\varepsilon) \langle \mathbf 0 |_1 \langle \boldsymbol \alpha_{\lambda} |_2 \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} | \mathbf 0 \rangle_1 | \boldsymbol \alpha_{\lambda} \rangle_2 = \sin(\varepsilon) \langle \mathbf 0 |_1 | \mathbf 0 \rangle_1 \langle \boldsymbol \alpha_{\lambda} |_2 \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} | \boldsymbol \alpha_{\lambda} \rangle_2 = \sin(\varepsilon) |\alpha_{\lambda}|^2

because \langle \boldsymbol \alpha_{\lambda} |_2 \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} | \boldsymbol \alpha_{\lambda} \rangle_2 = \langle \boldsymbol \alpha_{\lambda} |_2 \mathbf a_{\lambda_2}^\dag | \boldsymbol \alpha_{\lambda} \rangle_2 \langle \boldsymbol \alpha_{\lambda} |_2 \mathbf a_{\lambda_2} | \boldsymbol \alpha_{\lambda} \rangle_2 = \bar \alpha_{\lambda} \alpha_{\lambda} = |\alpha_{\lambda}|^2.

For the third term:

i \cos(\varepsilon) \langle \boldsymbol \Psi_{\text{in}} | \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2} | \boldsymbol \Psi_{\text{in}} \rangle = i \cos(\varepsilon) \langle \mathbf 0 |_1 \langle \boldsymbol \alpha_{\lambda} |_2 \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2} | \mathbf 0 \rangle_1 | \boldsymbol \alpha_{\lambda} \rangle_2 = i \cos(\varepsilon) \langle \mathbf 0 |_1 \mathbf a_{\lambda_1}^\dag | \mathbf 0 \rangle_1 \langle \boldsymbol \alpha_{\lambda} |_2 \mathbf a_{\lambda_2} | \boldsymbol \alpha_{\lambda} \rangle_2 = 0

because \langle \mathbf 0 |_1 \mathbf a_{\lambda_1}^\dag = 0.

For the fourth term:

- i \cos(\varepsilon) \langle \boldsymbol \Psi_{\text{in}} | \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_1} | \boldsymbol \Psi_{\text{in}} \rangle = - i \cos(\varepsilon) \langle \mathbf 0 |_1 \langle \boldsymbol \alpha_{\lambda} |_2 \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_1} | \mathbf 0 \rangle_1 | \boldsymbol \alpha_{\lambda} \rangle_2 = - i \cos(\varepsilon) \langle \boldsymbol \alpha_{\lambda} |_2 \mathbf a_{\lambda_2}^\dag | \boldsymbol \alpha_{\lambda} \rangle_2 \langle \mathbf 0 |_1 \mathbf a_{\lambda_1} | \mathbf 0 \rangle_1 = 0

because \mathbf a_{\lambda_1} | \mathbf 0 \rangle_1 = 0.

Overall only one term is non-zero:

\langle \boldsymbol \Psi_{\text{in}} | \mathbf N_{\lambda_6} - \mathbf N_{\lambda_5} | \boldsymbol \Psi_{\text{in}} \rangle = \sin(\varepsilon) |\alpha_{\lambda}|^2 = \sin(\varepsilon) \langle \mathbf N_{\lambda_2} \rangle

Which coincide with the result previously derived with a semi-classical approach. A quantum treatment reveals that noise originates from the interference of the laser input in input channel (2) with vacuum fluctuations in input channel (1). Unlike the classical view of noise as detector randomness, quantum noise is attributed to vacuum fluctuations entering the interferometer’s empty port. This understanding suggests that injecting a tailored quantum state into input channel (1) could lead to noise reduction and improved sensitivity.

In order to compute the standard deviation of the observable \mathbf N_{\lambda_6} - \mathbf N_{\lambda_5} we need to square it:

\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 \\ = & \sin^2(\varepsilon) \left( \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} - \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} \right)^2 - \cos^2(\varepsilon) \left( \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2} - \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_1} \right)^2 \\ & - i \sin(\varepsilon) \cos(\varepsilon) \left[ \left( \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} - \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} \right) \left( \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2} - \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_1} \right) + \left( \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2} - \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_1} \right) \left( \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} - \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} \right) \right] \\ = & \sin^2(\varepsilon) \left[ (\mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1})^2 - 2 (\mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1}) (\mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2}) + (\mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2})^2 \right] \\ & - \cos^2(\varepsilon) \left[ (\mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2})^2 - (\mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2}) (\mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_1}) - (\mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_1}) (\mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2}) + (\mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_1})^2 \right] \\ & - i \sin(\varepsilon) \cos(\varepsilon) \left[ (\mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1}) (\mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2}) - (\mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1}) (\mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_1}) - (\mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2}) (\mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2}) + (\mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2}) (\mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_1}) \right. \\ & \left. + (\mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2}) (\mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1}) - (\mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2}) (\mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2}) - (\mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_1}) (\mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1}) + (\mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_1}) (\mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2}) \right] \\ = & \sin^2(\varepsilon) \left[ \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} - 2 \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} + \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} \right] \\ & - \cos^2(\varepsilon) \left[ \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2} \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2} - \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2} \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_1} - \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_1} \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2} + \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_1} \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_1} \right] \\ & - i \sin(\varepsilon) \cos(\varepsilon) \left[ \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2} - \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_1} - \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2} + \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_1} \right. \\ & \left. + \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2} \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} - \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2} \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} - \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_1} \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} + \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_1} \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} \right] \end{aligned}

In this formula many terms are in normal order with respect to \mathbf a_{\lambda_1} and will gives zero when applied to the vacuum and can remove them. It remains:

\begin{aligned} \left(\mathbf N_{\lambda_6} - \mathbf N_{\lambda_5}\right)^2 = & \sin^2(\varepsilon) \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} \\ & - \cos^2(\varepsilon) \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2} \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2} \\ & + \cos^2(\varepsilon) \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_1} \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2} \\ & + i \sin(\varepsilon) \cos(\varepsilon) \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2} \\ & + i \sin(\varepsilon) \cos(\varepsilon) \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2} \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} \end{aligned}

We can now compute the expectation:

\begin{aligned} \langle \boldsymbol \Psi_{\text{in}} | \left(\mathbf N_{\lambda_6} - \mathbf N_{\lambda_5}\right)^2 | \boldsymbol \Psi_{\text{in}} \rangle = & \langle \boldsymbol \Psi_{\text{in}} | \sin^2(\varepsilon) \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} | \boldsymbol \Psi_{\text{in}} \rangle \\ & - \langle \boldsymbol \Psi_{\text{in}} | \cos^2(\varepsilon) \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2} \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2} | \boldsymbol \Psi_{\text{in}} \rangle \\ & + \langle \boldsymbol \Psi_{\text{in}} | \cos^2(\varepsilon) \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_1} \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2} | \boldsymbol \Psi_{\text{in}} \rangle \\ & + \langle \boldsymbol \Psi_{\text{in}} | i \sin(\varepsilon) \cos(\varepsilon) \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2} | \boldsymbol \Psi_{\text{in}} \rangle \\ & + \langle \boldsymbol \Psi_{\text{in}} | i \sin(\varepsilon) \cos(\varepsilon) \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2} \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} | \boldsymbol \Psi_{\text{in}} \rangle \end{aligned}

Let us consider each term separately:

\begin{aligned} \langle \boldsymbol \Psi_{\text{in}} | \sin^2(\varepsilon) \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} | \boldsymbol \Psi_{\text{in}} \rangle = & \sin^2(\varepsilon) \langle \boldsymbol \alpha_{\lambda_2} |_2 \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} | \boldsymbol \alpha_{\lambda_2} \rangle_2 \langle \mathbf 0 |_1 | \mathbf 0 \rangle_1 \\ = & \sin^2(\varepsilon) \langle \boldsymbol \alpha_{\lambda_2} | (\mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2}) (\mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2}) | \boldsymbol \alpha_{\lambda_2} \rangle \\ = & \sin^2(\varepsilon) \langle \boldsymbol \alpha_{\lambda_2} | \mathbf N_{\lambda_2}^2 | \boldsymbol \alpha_{\lambda_2} \rangle \\ = & \sin^2(\varepsilon) |\alpha_{\lambda_2}|^4 \\ - \langle \boldsymbol \Psi_{\text{in}} | \cos^2(\varepsilon) \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2} \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2} | \boldsymbol \Psi_{\text{in}} \rangle = & - \cos^2(\varepsilon) \langle \boldsymbol \alpha_{\lambda_2} |_2 \langle \mathbf 0 |_1 \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2} \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2} | \mathbf 0 \rangle_1 | \boldsymbol \alpha_{\lambda_2} \rangle_2 \\ = & - \cos^2(\varepsilon) \langle \boldsymbol \alpha_{\lambda_2} |_2 \mathbf a_{\lambda_2} \mathbf a_{\lambda_2} | \boldsymbol \alpha_{\lambda_2} \rangle_2 \langle \mathbf 0 |_1 \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1}^\dag | \mathbf 0 \rangle_1 \\ = & 0 \\ + \langle \boldsymbol \Psi_{\text{in}} | \cos^2(\varepsilon) \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_1} \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2} | \boldsymbol \Psi_{\text{in}} \rangle = & \cos^2(\varepsilon) \langle \boldsymbol \alpha_{\lambda_2} |_2 \langle \mathbf 0 |_1 \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_1} \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2} | \mathbf 0 \rangle_1 | \boldsymbol \alpha_{\lambda_2} \rangle_2 \\ = & \cos^2(\varepsilon) \langle \boldsymbol \alpha_{\lambda_2} |_2 \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} | \boldsymbol \alpha_{\lambda_2} \rangle_2 \langle \mathbf 0 |_1 \mathbf a_{\lambda_1} \mathbf a_{\lambda_1}^\dag | \mathbf 0 \rangle_1 \\ = & \cos^2(\varepsilon) \langle \boldsymbol \alpha_{\lambda_2} | \mathbf N_{\lambda_2} | \boldsymbol \alpha_{\lambda_2} \rangle \langle \mathbf 0 | (\mathbf I + \mathbf N_{\lambda_1}) | \mathbf 0 \rangle \\ = & \cos^2(\varepsilon) |\alpha_{\lambda_2}|^2 (1 + 0) \\ = & \cos^2(\varepsilon) |\alpha_{\lambda_2}|^2\\ + \langle \boldsymbol \Psi_{\text{in}} | i \sin(\varepsilon) \cos(\varepsilon) \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2} | \boldsymbol \Psi_{\text{in}} \rangle = & i \sin(\varepsilon) \cos(\varepsilon) \langle \boldsymbol \alpha_{\lambda_2} |_2 \langle \mathbf 0 |_1 \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2} | \mathbf 0 \rangle_1 | \boldsymbol \alpha_{\lambda_2} \rangle_2 \\ = & i \sin(\varepsilon) \cos(\varepsilon) \langle \boldsymbol \alpha_{\lambda_2} |_2 \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} \mathbf a_{\lambda_2} | \boldsymbol \alpha_{\lambda_2} \rangle_2 \langle \mathbf 0 |_1 \mathbf a_{\lambda_1}^\dag | \mathbf 0 \rangle_1 \\ = & 0\\ + \langle \boldsymbol \Psi_{\text{in}} | i \sin(\varepsilon) \cos(\varepsilon) \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2} \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} | \boldsymbol \Psi_{\text{in}} \rangle = & i \sin(\varepsilon) \cos(\varepsilon) \langle \boldsymbol \alpha_{\lambda_2} |_2 \langle \mathbf 0 |_1 \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2} \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} | \mathbf 0 \rangle_1 | \boldsymbol \alpha_{\lambda_2} \rangle_2 \\ = & i \sin(\varepsilon) \cos(\varepsilon) \langle \boldsymbol \alpha_{\lambda_2} |_2 \mathbf a_{\lambda_2} \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} | \boldsymbol \alpha_{\lambda_2} \rangle_2 \langle \mathbf 0 |_1 \mathbf a_{\lambda_1}^\dag | \mathbf 0 \rangle_1 \\ = & 0 \end{aligned}

Summing all terms:

\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) |\alpha_{\lambda_2}|^4 + \cos^2(\varepsilon) |\alpha_{\lambda_2}|^2 \end{aligned}

Calculating the variance of \mathbf N_{\lambda_6} - \mathbf N_{\lambda_5} for \varepsilon = 0, it is equivalent to previous calculations for balanced homodyne detection, as a Mach-Zehnder interferometer at \varepsilon = 0 behaves like a balanced beam splitter:

\langle \boldsymbol \Psi_{\text{in}} | \left(\mathbf N_{\lambda_6} - \mathbf N_{\lambda_5}\right)^2 | \boldsymbol \Psi_{\text{in}} \rangle = \sin^2(0) |\alpha_{\lambda_2}|^4 + \cos^2(0) |\alpha_{\lambda_2}|^2 = |\alpha_{\lambda_2}|^2 = \mathbf N_{\lambda_2}

Which is the same as the classical result. This a quantum calculation reveals that signal fluctuations originate from the interference of the laser input (channel (2)) with vacuum fluctuations entering input channel (1). This contrasts with a semi-classical view of noise as detector randomness, instead identifying vacuum fluctuations as the noise source. This quantum understanding suggests that modifying the input channel (1) could potentially improve the signal-to-noise ratio.

Generic state in channel one

We consider now the case where the state in input channel (1) could be any state | \boldsymbol \psi \rangle_1 but we have a quasi-classical state | \boldsymbol \alpha_\lambda \rangle_2 in input channel (2), with \alpha_\lambda \in \mathbb R (^{}\langle \boldsymbol \alpha_{\lambda} | \mathbf a_{\lambda}^\dag = \alpha_{\lambda} \langle \boldsymbol \alpha_{\lambda} | and \mathbf a_\lambda | \boldsymbol \alpha_\lambda \rangle = \alpha_{\lambda} | \boldsymbol \alpha_\lambda \rangle), so that the computation will be slightly simpler.

We first compute the expectation for the state in input channel (2):

\begin{aligned} {}_2 \langle \boldsymbol \alpha_\lambda | \mathbf N_{\lambda_6} - \mathbf N_{\lambda_5}| \boldsymbol \alpha_\lambda \rangle_2 = & {}_2 \langle \boldsymbol \alpha_\lambda | \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]| \boldsymbol \alpha_\lambda \rangle_2 \\ = & -\sin(\varepsilon) \left( {}_2 \langle \boldsymbol \alpha_\lambda | \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} | \boldsymbol \alpha_\lambda \rangle_2 - {}_2 \langle \boldsymbol \alpha_\lambda | \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} | \boldsymbol \alpha_\lambda \rangle_2 \right) \\ & + i \cos(\varepsilon) \left( {}_2 \langle \boldsymbol \alpha_\lambda | \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2} | \boldsymbol \alpha_\lambda \rangle_2 - {}_2 \langle \boldsymbol \alpha_\lambda | \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_1} | \boldsymbol \alpha_\lambda \rangle_2 \right) \\ = & -\sin(\varepsilon) \left( \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} - \alpha_{\lambda_2}^2 \right) + i \cos(\varepsilon) \left( \mathbf a_{\lambda_1}^\dag \alpha_{\lambda_2} - \alpha_{\lambda_2} \mathbf a_{\lambda_1} \right) \end{aligned}

For the particular case \varepsilon = 0, we obtain:

{}_2 \langle \boldsymbol \alpha_\lambda | \mathbf N_{\lambda_6} - \mathbf N_{\lambda_5}| \boldsymbol \alpha_\lambda \rangle_2 = i \alpha_{\lambda_2} \left( \mathbf a_{\lambda_1}^\dag - \mathbf a_{\lambda_1} \right) =\alpha_{\lambda_2} \sqrt{\frac{2}{\hbar}} \mathbf P_{\lambda_1}

using:

\mathbf P_{\lambda_1} = \sqrt{\frac{\hbar}{2}}i (\mathbf a_{\lambda_1}^\dag - \mathbf a_{\lambda_1})

It is the result that corresponds to measuring the \mathbf P quadrature of the input state in input channel (1), a result that is consistent with what is expected from a balanced beam splitter measurement:

\langle \boldsymbol \Psi_{\text{in}} | \mathbf N_{\lambda_6} - \mathbf N_{\lambda_5}| \boldsymbol \Psi_{\text{in}} \rangle = \alpha_{\lambda_2} \sqrt{\frac{2}{\hbar}} {}_1 \langle \boldsymbol \psi | \mathbf P_{\lambda_1}| \boldsymbol \psi \rangle_1

We want now to compute the expectation of the squared balanced signal:

We then again first compute the expectation for the state in input (2):

\begin{aligned} {}_2 \langle \boldsymbol \alpha_\lambda | \left(\mathbf N_{\lambda_6} - \mathbf N_{\lambda_5} \right)^2 | \boldsymbol \alpha_\lambda \rangle_2 = & {}_2 \langle \boldsymbol \alpha_\lambda | \left[ \right.\sin^2(\varepsilon) \left[ \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} - 2 \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} + \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} \right] \\ & - \cos^2(\varepsilon) \left[ \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2} \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2} - \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2} \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_1} - \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_1} \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2} + \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_1} \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_1} \right] \\ & - i \sin(\varepsilon) \cos(\varepsilon) \left[ \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2} - \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_1} - \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2} + \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_1} \right. \\ & \left. + \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2} \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} - \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2} \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} - \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_1} \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} + \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_1} \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} \right] \left. \right] | \boldsymbol \alpha_\lambda \rangle_2 \end{aligned}

We compute the terms separately. The first term is:

\begin{aligned} & {}_2 \langle \boldsymbol \alpha_\lambda | \left[ \right.\sin^2(\varepsilon) \left[ \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} - 2 \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} + \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} \right] \left. \right] | \boldsymbol \alpha_\lambda \rangle_2 \\ = & \sin^2(\varepsilon) \left[ {}_2 \langle \boldsymbol \alpha_\lambda | \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} | \boldsymbol \alpha_\lambda \rangle_2 - 2 {}_2 \langle \boldsymbol \alpha_\lambda | \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} | \boldsymbol \alpha_\lambda \rangle_2 + {}_2 \langle \boldsymbol \alpha_\lambda | \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} | \boldsymbol \alpha_\lambda \rangle_2 \right] \\ = & \sin^2(\varepsilon) \left[ \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} {}_2 \langle \boldsymbol \alpha_\lambda | 1 | \boldsymbol \alpha_\lambda \rangle_2 - 2 \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} {}_2 \langle \boldsymbol \alpha_\lambda | \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} | \boldsymbol \alpha_\lambda \rangle_2 + {}_2 \langle \boldsymbol \alpha_\lambda | \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} | \boldsymbol \alpha_\lambda \rangle_2 \right] \\ = & \sin^2(\varepsilon) \left[ \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} - 2 \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} \alpha_{\lambda_2}^2 + {}_2 \langle \boldsymbol \alpha_\lambda | (\mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2})^2 | \boldsymbol \alpha_\lambda \rangle_2 \right] \\ = & \sin^2(\varepsilon) \left[ \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} - 2 \alpha_{\lambda_2}^2 \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} + (\alpha_{\lambda_2}^4 + \alpha_{\lambda_2}^2) \right] \end{aligned}

The second term is:

\begin{aligned} & - \cos^2(\varepsilon) {}_2 \langle \boldsymbol \alpha_\lambda | \left[ \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2} \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2} - \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2} \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_1} - \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_1} \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2} + \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_1} \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_1} \right] | \boldsymbol \alpha_\lambda \rangle_2 \\ = & - \cos^2(\varepsilon) \left[ {}_2 \langle \boldsymbol \alpha_\lambda | \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2} \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2} | \boldsymbol \alpha_\lambda \rangle_2 - {}_2 \langle \boldsymbol \alpha_\lambda | \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2} \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_1} | \boldsymbol \alpha_\lambda \rangle_2 \right. \\ & \left. - {}_2 \langle \boldsymbol \alpha_\lambda | \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_1} \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2} | \boldsymbol \alpha_\lambda \rangle_2 + {}_2 \langle \boldsymbol \alpha_\lambda | \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_1} \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_1} | \boldsymbol \alpha_\lambda \rangle_2 \right] \\ = & - \cos^2(\varepsilon) \left[ \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1}^\dag {}_2 \langle \boldsymbol \alpha_\lambda | \mathbf a_{\lambda_2} \mathbf a_{\lambda_2} | \boldsymbol \alpha_\lambda \rangle_2 - \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} {}_2 \langle \boldsymbol \alpha_\lambda | \mathbf a_{\lambda_2} \mathbf a_{\lambda_2}^\dag | \boldsymbol \alpha_\lambda \rangle_2 \right. \\ & \left. - \mathbf a_{\lambda_1} \mathbf a_{\lambda_1}^\dag {}_2 \langle \boldsymbol \alpha_\lambda | \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} | \boldsymbol \alpha_\lambda \rangle_2 + \mathbf a_{\lambda_1} \mathbf a_{\lambda_1} {}_2 \langle \boldsymbol \alpha_\lambda | \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2}^\dag | \boldsymbol \alpha_\lambda \rangle_2 \right] \\ = & - \cos^2(\varepsilon) \left[ \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1}^\dag \alpha_{\lambda_2}^2 - \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} {}_2 \langle \boldsymbol \alpha_\lambda | \mathbf a_{\lambda_2} \mathbf a_{\lambda_2}^\dag | \boldsymbol \alpha_\lambda \rangle_2 - \mathbf a_{\lambda_1} \mathbf a_{\lambda_1}^\dag \alpha_{\lambda_2}^2 + \mathbf a_{\lambda_1} \mathbf a_{\lambda_1} \alpha_{\lambda_2}^2 \right] \\ = & - \cos^2(\varepsilon) \left[ \alpha_{\lambda_2}^2 (\mathbf a_{\lambda_1}^\dag)^2 - \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} {}_2 \langle \boldsymbol \alpha_\lambda | (\mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} + 1) | \boldsymbol \alpha_\lambda \rangle_2 - \alpha_{\lambda_2}^2 (\mathbf a_{\lambda_1} \mathbf a_{\lambda_1}^\dag) + \alpha_{\lambda_2}^2 (\mathbf a_{\lambda_1})^2 \right] \\ = & - \cos^2(\varepsilon) \left[ \alpha_{\lambda_2}^2 (\mathbf a_{\lambda_1}^\dag)^2 - \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} (\alpha_{\lambda_2}^2 + 1) - \alpha_{\lambda_2}^2 (\mathbf a_{\lambda_1} \mathbf a_{\lambda_1}^\dag) + \alpha_{\lambda_2}^2 (\mathbf a_{\lambda_1})^2 \right] \end{aligned}

The third term is:

\begin{aligned} & - 2 i \sin(\varepsilon) \cos(\varepsilon) {}_2 \langle \boldsymbol \alpha_\lambda | \left( \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} - \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} \right) \left( \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2} - \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_1} \right) | \boldsymbol \alpha_\lambda \rangle_2 \\ = & - 2 i \sin(\varepsilon) \cos(\varepsilon) {}_2 \langle \boldsymbol \alpha_\lambda | \left[ \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2} - \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_1} - \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2} + \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_1} \right] | \boldsymbol \alpha_\lambda \rangle_2 \\ = & - 2 i \sin(\varepsilon) \cos(\varepsilon) \left[ {}_2 \langle \boldsymbol \alpha_\lambda | \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2} | \boldsymbol \alpha_\lambda \rangle_2 - {}_2 \langle \boldsymbol \alpha_\lambda | \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_1} | \boldsymbol \alpha_\lambda \rangle_2 \right. \\ & \left. - {}_2 \langle \boldsymbol \alpha_\lambda | \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_2} | \boldsymbol \alpha_\lambda \rangle_2 + {}_2 \langle \boldsymbol \alpha_\lambda | \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_1} | \boldsymbol \alpha_\lambda \rangle_2 \right] \\ = & - 2 i \sin(\varepsilon) \cos(\varepsilon) \left[ \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} \mathbf a_{\lambda_1}^\dag {}_2 \langle \boldsymbol \alpha_\lambda | \mathbf a_{\lambda_2} | \boldsymbol \alpha_\lambda \rangle_2 - \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} \mathbf a_{\lambda_1} {}_2 \langle \boldsymbol \alpha_\lambda | \mathbf a_{\lambda_2}^\dag | \boldsymbol \alpha_\lambda \rangle_2 \right. \\ & \left. - \mathbf a_{\lambda_1}^\dag {}_2 \langle \boldsymbol \alpha_\lambda | \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} \mathbf a_{\lambda_2} | \boldsymbol \alpha_\lambda \rangle_2 + \mathbf a_{\lambda_1} {}_2 \langle \boldsymbol \alpha_\lambda | \mathbf a_{\lambda_2}^\dag \mathbf a_{\lambda_2} \mathbf a_{\lambda_2}^\dag | \boldsymbol \alpha_\lambda \rangle_2 \right] \\ = & - 2 i \sin(\varepsilon) \cos(\varepsilon) \left[ \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} \mathbf a_{\lambda_1}^\dag \alpha_{\lambda_2} - \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} \mathbf a_{\lambda_1} \alpha_{\lambda_2} - \mathbf a_{\lambda_1}^\dag \alpha_{\lambda_2}^3 + \mathbf a_{\lambda_1} \alpha_{\lambda_2}^3 \right] \\ = & - 2 i \sin(\varepsilon) \cos(\varepsilon) \alpha_{\lambda_2} \left[ \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} \mathbf a_{\lambda_1}^\dag - \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} \mathbf a_{\lambda_1} - \alpha_{\lambda_2}^2 \mathbf a_{\lambda_1}^\dag + \alpha_{\lambda_2}^2 \mathbf a_{\lambda_1} \right] \\ = & - 2 i \sin(\varepsilon) \cos(\varepsilon) \alpha_{\lambda_2} \left[ (\mathbf a_{\lambda_1}^\dag)^2 \mathbf a_{\lambda_1} - \mathbf a_{\lambda_1}^\dag (\mathbf a_{\lambda_1})^2 - \alpha_{\lambda_2}^2 (\mathbf a_{\lambda_1}^\dag - \mathbf a_{\lambda_1}) \right] \end{aligned}

Overall then:

\begin{aligned} {}_2 \langle \boldsymbol \alpha_\lambda | \left(\mathbf N_{\lambda_6} - \mathbf N_{\lambda_5} \right)^2 | \boldsymbol \alpha_\lambda \rangle_2 = & \sin^2(\varepsilon) \left[ \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} - 2 \alpha_{\lambda_2}^2 \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} + (\alpha_{\lambda_2}^4 + \alpha_{\lambda_2}^2) \right] \\ & - \cos^2(\varepsilon) \left[ \alpha_{\lambda_2}^2 (\mathbf a_{\lambda_1}^\dag)^2 - \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} (\alpha_{\lambda_2}^2 + 1) - \alpha_{\lambda_2}^2 (\mathbf a_{\lambda_1}\mathbf a_{\lambda_1}^\dag) + \alpha_{\lambda_2}^2 (\mathbf a_{\lambda_1})^2 \right] \\ & - 2 i \sin(\varepsilon) \cos(\varepsilon) \alpha_{\lambda_2} \left[ (\mathbf a_{\lambda_1}^\dag)^2 \mathbf a_{\lambda_1} - \mathbf a_{\lambda_1}^\dag (\mathbf a_{\lambda_1})^2 - \alpha_{\lambda_2}^2 (\mathbf a_{\lambda_1}^\dag - \mathbf a_{\lambda_1}) \right] \end{aligned}

The square of the expectation value is:

\begin{aligned} \left({}_2 \langle \boldsymbol \alpha_\lambda | \mathbf N_{\lambda_6} - \mathbf N_{\lambda_5}| \boldsymbol \alpha_\lambda \rangle_2\right)^2 = &\left[ -\sin(\varepsilon) \left( \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} - \alpha_{\lambda_2}^2 \right) + i \cos(\varepsilon) \left( \mathbf a_{\lambda_1}^\dag \alpha_{\lambda_2} - \alpha_{\lambda_2} \mathbf a_{\lambda_1} \right)\right]^2 \\ = & \sin^2(\varepsilon) \left( \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} - \alpha_{\lambda_2}^2 \right)^2 - \cos^2(\varepsilon) \left( \mathbf a_{\lambda_1}^\dag \alpha_{\lambda_2} - \alpha_{\lambda_2} \mathbf a_{\lambda_1} \right)^2 \\ & - 2i \sin(\varepsilon) \cos(\varepsilon) \left( \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} - \alpha_{\lambda_2}^2 \right) \left( \mathbf a_{\lambda_1}^\dag \alpha_{\lambda_2} - \alpha_{\lambda_2} \mathbf a_{\lambda_1} \right) \\ = & \sin^2(\varepsilon) \left[ (\mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1})^2 - 2 \alpha_{\lambda_2}^2 \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} + \alpha_{\lambda_2}^4 \right] \\ & - \cos^2(\varepsilon) \left[ (\mathbf a_{\lambda_1}^\dag \alpha_{\lambda_2})^2 - 2 (\mathbf a_{\lambda_1}^\dag \alpha_{\lambda_2}) (\alpha_{\lambda_2} \mathbf a_{\lambda_1}) + (\alpha_{\lambda_2} \mathbf a_{\lambda_1})^2 \right] \\ & - 2i \sin(\varepsilon) \cos(\varepsilon) \left[ \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} (\mathbf a_{\lambda_1}^\dag \alpha_{\lambda_2} - \alpha_{\lambda_2} \mathbf a_{\lambda_1}) - \alpha_{\lambda_2}^2 (\mathbf a_{\lambda_1}^\dag \alpha_{\lambda_2} - \alpha_{\lambda_2} \mathbf a_{\lambda_1}) \right] \\ = & \sin^2(\varepsilon) \left[ (\mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1})^2 - 2 \alpha_{\lambda_2}^2 \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} + \alpha_{\lambda_2}^4 \right] \\ & - \cos^2(\varepsilon) \left[ \alpha_{\lambda_2}^2 (\mathbf a_{\lambda_1}^\dag)^2 - 2 \alpha_{\lambda_2}^2 \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} + \alpha_{\lambda_2}^2 (\mathbf a_{\lambda_1})^2 \right] \\ & - 2i \sin(\varepsilon) \cos(\varepsilon) \left[ \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} \mathbf a_{\lambda_1}^\dag \alpha_{\lambda_2} - \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} \alpha_{\lambda_2} \mathbf a_{\lambda_1} - \alpha_{\lambda_2}^2 \mathbf a_{\lambda_1}^\dag \alpha_{\lambda_2} + \alpha_{\lambda_2}^3 \mathbf a_{\lambda_1} \right] \end{aligned}

Subtracting:

\begin{aligned} \Delta^2_{\mathbf N_{\lambda_6} - \mathbf N_{\lambda_5}} & = \sin^2(\varepsilon) \left[ \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} - 2 \alpha_{\lambda_2}^2 \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} + (\alpha_{\lambda_2}^4 + \alpha_{\lambda_2}^2) \right] \\ & - \cos^2(\varepsilon) \left[ \alpha_{\lambda_2}^2 (\mathbf a_{\lambda_1}^\dag)^2 - \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} (\alpha_{\lambda_2}^2 + 1) - \alpha_{\lambda_2}^2 (\mathbf a_{\lambda_1} \mathbf a_{\lambda_1}^\dag) + \alpha_{\lambda_2}^2 (\mathbf a_{\lambda_1})^2 \right] \\ & - 2 i \sin(\varepsilon) \cos(\varepsilon) \alpha_{\lambda_2} \left[ (\mathbf a_{\lambda_1}^\dag)^2 \mathbf a_{\lambda_1} - \mathbf a_{\lambda_1}^\dag (\mathbf a_{\lambda_1})^2 - \alpha_{\lambda_2}^2 (\mathbf a_{\lambda_1}^\dag - \mathbf a_{\lambda_1}) \right] \\ & - \sin^2(\varepsilon) \left[ (\mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1})^2 - 2 \alpha_{\lambda_2}^2 \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} + \alpha_{\lambda_2}^4 \right] \\ & + \cos^2(\varepsilon) \left[ \alpha_{\lambda_2}^2 (\mathbf a_{\lambda_1}^\dag)^2 - 2 \alpha_{\lambda_2}^2 \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} + \alpha_{\lambda_2}^2 (\mathbf a_{\lambda_1})^2 \right] \\ & + 2i \sin(\varepsilon) \cos(\varepsilon) \left[ \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} \mathbf a_{\lambda_1}^\dag \alpha_{\lambda_2} - \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} \alpha_{\lambda_2} \mathbf a_{\lambda_1} - \alpha_{\lambda_2}^2 \mathbf a_{\lambda_1}^\dag \alpha_{\lambda_2} + \alpha_{\lambda_2}^3 \mathbf a_{\lambda_1} \right] \\ & = \sin^2(\varepsilon) \left[ \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} - (\mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1})^2 + \alpha_{\lambda_2}^2 \right] \\ & \cos^2(\varepsilon) \left[ \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} + \alpha_{\lambda_2}^2 (\mathbf a_{\lambda_1} \mathbf a_{\lambda_1}^\dag) - \alpha_{\lambda_2}^2 \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} \right] \\ & + 2i \sin(\varepsilon) \cos(\varepsilon) \left[ \alpha_{\lambda_2} \left[ \mathbf a_{\lambda_1}^\dag (\mathbf a_{\lambda_1})^2 - (\mathbf a_{\lambda_1}^\dag)^2 \mathbf a_{\lambda_1} + \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} \mathbf a_{\lambda_1}^\dag - \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} \mathbf a_{\lambda_1} \right] \right. \\ & + \left. \alpha_{\lambda_2}^3 (\mathbf a_{\lambda_1}^\dag - \mathbf a_{\lambda_1}) - \alpha_{\lambda_2}^2 \mathbf a_{\lambda_1}^\dag \alpha_{\lambda_2} + \alpha_{\lambda_2}^3 \mathbf a_{\lambda_1} \right] \end{aligned}

Let’s consider again the case \varepsilon = 0 for a generic |\boldsymbol \Psi_{\text{in}} \rangle:

We already evaluated the second term to be \alpha_{\lambda_2} \sqrt{\frac{2}{\hbar}} \mathbf P_{\lambda_1}:

If the laser beam in in input channel (2) is intense, then \alpha_{\lambda_2}^2 \gg 1 and therefore (\alpha_{\lambda_2}^2 + 1) \approx \alpha_{\lambda_2}^2:

\begin{aligned} \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 \langle {}_1\langle \mathbf 0 | -\left[ (\mathbf a_{\lambda_1}^\dag)^2 - \mathbf a_{\lambda_1}^\dag \mathbf a_{\lambda_1} - \mathbf a_{\lambda_1} \mathbf a_{\lambda_1}^\dag + (\mathbf a_{\lambda_1})^2 \right] |\mathbf 0 \rangle_1\\ & - \alpha_{\lambda_2}^2 \frac{2}{\hbar} \left(\langle \mathbf P_{\lambda_1} \rangle\right)^2 \\ = & \alpha_{\lambda_2}^2 {}_1\langle \mathbf 0 | \left( i (\mathbf a_{\lambda_1}^\dag - \mathbf a_{\lambda_1})\right)^2 |\mathbf 0 \rangle_1 - \alpha_{\lambda_2}^2 \frac{2}{\hbar} \left(\langle \mathbf P_{\lambda_1} \rangle\right)^2 \\ = & \alpha_{\lambda_2}^2 {}_1\langle \mathbf 0 | \left( \sqrt{\frac{2}{\hbar}} \mathbf P_{\lambda_1} \right)^2 |\mathbf 0 \rangle_1 - \frac{2}{\hbar} \alpha_{\lambda_2}^2 \left(\langle \mathbf P_{\lambda_1} \rangle\right)^2 \\ = & \alpha_{\lambda_2}^2 \frac{2}{\hbar} \langle \mathbf P_{\lambda_1}^2 \rangle - \alpha_{\lambda_2}^2 \frac{2}{\hbar} \left(\langle \mathbf P_{\lambda_1} \rangle\right)^2 \\ = & \alpha_{\lambda_2}^2 \frac{2}{\hbar} \left[ \langle \mathbf P_{\lambda_1}^2 \rangle - \left(\langle \mathbf P_{\lambda_1} \rangle\right)^2\rangle\right] \end{aligned}

Returning to the case a balanced Mach-Zehnder interferometer with a quasi-classical state in input channel (2) (with \alpha_\lambda \in \mathbb R), measuring the difference signal effectively measures the \mathbf P_{\lambda_1} quadrature of input channel (1) field.

Repeating this measurement allows determining the average and dispersion of that quadrature for the input state in channel one. Consequently, the measurement of \mathbf N_{\lambda_6} - \mathbf N_{\lambda_5} can be understood as measuring the \mathbf P_{\lambda_1}quadrature of vacuum, scaled by a factor, when the input channel (1) is | \mathbf 0 \rangle:

\begin{aligned} & \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 \frac{2}{\hbar} \langle \mathbf 0 | \mathbf P_{\lambda_1}^2 | \mathbf 0 \rangle \\ & \langle \boldsymbol \Psi_{\text{in}} | \Delta_{\mathbf N_{\lambda_6} - \mathbf N_{\lambda_5}} | \boldsymbol \Psi_{\text{in}} \rangle = \alpha_{\lambda_2}^2 \sqrt{\frac{2}{\hbar}} \langle \mathbf 0 | \mathbf P_{\lambda_1} | \mathbf 0 \rangle = 0 \end{aligned}

We can show it in the complex plane representation of the quadratures of the vacuum.

Phasor plane representation of vacuum quadratures