Balanced Homodyne Detection
In the formalism of the light quantization (here) the electric field can be written as:
\begin{aligned} \mathbf{E}(\mathbf{r}) & = \sum_\lambda i\mathbf{e}_\lambda \mathscr E^{(1)}_\lambda \left(\mathbf a_\lambda e^{i\mathbf{k}_\lambda \cdot \mathbf{r}} - \mathbf a_\lambda^\dag e^{-i\mathbf{k}_\lambda \cdot \mathbf{r}} \right) \\ & = \mathbf{E}^{(+)}(\mathbf{r},t) + \mathbf{E}^{(-)}(\mathbf{r},t) \end{aligned}
Considering a single mode in the quasi-classical state:
| \boldsymbol{\Psi}_{qc} (t) \rangle = | \alpha_\lambda e^{-i \omega_\lambda t} \rangle
It average value oscillate at a very high frequency:
\begin{aligned} \langle \mathbf{E} \rangle (\mathbf{r},t) & = \langle \boldsymbol{\Psi}_{qc}(t) | \mathbf{E}(\mathbf{r}) | \boldsymbol{\Psi}_{qc}(t) \rangle \\ & = i \mathscr{E}_\lambda^{(1)} \alpha_\lambda \mathbf{e}_\lambda e^{i(\mathbf{k}_\lambda \cdot \mathbf{r} - \omega_\lambda t)} + \text{c.c.} \end{aligned}
We can’t directly measure the electric and magnetic fields of visible light (like orange light at 5 \times 10^{14} \text{ Hz}) because they oscillate too rapidly for existing detectors.
Even the fastest detectors average this signal over a picosecond, effectively smoothing out the rapid oscillations and giving a zero reading. Therefore, quantum optics experiments rely on measuring photons (photoelectric measurements) instead.
Given a single-mode quasi-classical state and using the Heisenberg picture (where observables evolve in time), we need to show that the probabilities of single and joint photon detection per unit time are constant.
First we compute the electric field as function of time, so it is possible to use the Heisenberg formalism:
\mathbf{E}(\mathbf{r}, t) = \sum_\lambda i\mathbf{e}_\lambda \mathscr E^{(1)}_\lambda \left(\mathbf a_\lambda(0) e^{i(\mathbf{k}_\lambda \cdot \mathbf{r} - \omega_\lambda t)} + \mathbf a_\lambda^\dag(0) e^{-(i\mathbf{k}_\lambda \cdot \mathbf{r} - \omega_\lambda t)} \right)
The only state to consider is the initial state |\boldsymbol{\Psi} (0) \rangle:
| \boldsymbol{\Psi}_{qc} (0) \rangle = | \boldsymbol {\alpha}_\lambda \rangle
The property of coherent states is \mathbf a_\lambda | \boldsymbol {\alpha}_\lambda \rangle = \alpha_\lambda | \boldsymbol {\alpha}_\lambda \rangle, where \alpha_\lambda is a complex number.
Using the Heisenberg formalism, the probabilities per unit time of single photodetection is:
\begin{aligned} w^{(1)}(\mathbf{r},t) & = s \left\| \mathbf{E}^{(+)}(\mathbf{r}, t) | \boldsymbol{\Psi}(0) \rangle \right\|^2 \\ & = s \left[\mathscr E^{(1)}_\lambda\right]^2 \left\|\mathbf a_\lambda e^{i(\mathbf{k}_\lambda \mathbf r - \omega_\lambda t)}| \boldsymbol {\alpha}_\lambda \rangle \right\|^2 \\ & = s \left[\mathscr E^{(1)}_\lambda\right]^2 \left\| e^{i(\mathbf{k}_\lambda \mathbf r - \omega_\lambda t)} \left(\mathbf a_\lambda | \boldsymbol {\alpha}_\lambda \rangle \right)\right\|^2 \\ & = s \left[\mathscr E^{(1)}_\lambda\right]^2 \left\| e^{i(\mathbf{k}_\lambda \mathbf r - \omega_\lambda t)} \alpha_\lambda | \boldsymbol {\alpha}_\lambda \rangle \right\|^2 \\ &= s \left[\mathscr E^{(1)}_\lambda\right]^2 \left| e^{i(\mathbf{k}_\lambda \mathbf r - \omega_\lambda t)} \alpha_\lambda \right|^2 \| | \boldsymbol {\alpha}_\lambda \rangle \|^2 \\ & = s \left[\mathscr E^{(1)}_\lambda\right]^2 \left| e^{i(\mathbf{k}_\lambda \mathbf r - \omega_\lambda t} \right|^2 |\alpha_\lambda|^2 \times 1 \\ & = s \left| \alpha_\lambda \right|^2 \left[\mathscr E^{(1)}_\lambda\right]^2 \end{aligned}
The probability for joint photodetections is:
\begin{aligned} w^{(2)}(\mathbf r_1,t_1,\mathbf r_2,t_2)& = s^2\left\| \mathbf{E}^{(+)}(\mathbf{r_2},t_2) \mathbf{E}^{(+)}(\mathbf{r_1}, t_1)| \boldsymbol{\Psi}(0) \rangle\right\|^2 \\ &= s^2 \left[\mathscr E^{(1)}_\lambda\right]^4 \left\| e^{i(\mathbf{k}_\lambda \mathbf r_1 - \omega_\lambda t_1)} e^{i(\mathbf{k}_\lambda \mathbf r_2 - \omega_\lambda t_2)} \mathbf a_\lambda \mathbf a_\lambda | \boldsymbol {\alpha}_\lambda \rangle \right\|^2 \\ &= s^2 \left[\mathscr E^{(1)}_\lambda\right]^4 \left\| e^{i(\mathbf{k}_\lambda \mathbf r_1 - \omega_\lambda t_1)} e^{i(\mathbf{k}_\lambda \mathbf r_2 - \omega_\lambda t_2)} \mathbf a_\lambda (\mathbf a_\lambda | \boldsymbol {\alpha}_\lambda \rangle) \right\|^2 \\ &= s^2 \left[\mathscr E^{(1)}_\lambda\right]^4 \left\| e^{i(\mathbf{k}_\lambda \mathbf r_1 - \omega_\lambda t_1)} e^{i(\mathbf{k}_\lambda \mathbf r_2 - \omega_\lambda t_2)} \mathbf a_\lambda (\alpha_\lambda | \boldsymbol {\alpha}_\lambda \rangle) \right\|^2 \\ &= s^2 \left[\mathscr E^{(1)}_\lambda\right]^4 \left\| e^{i(\mathbf{k}_\lambda \mathbf r_1 - \omega_\lambda t_1)} e^{i(\mathbf{k}_\lambda \mathbf r_2 - \omega_\lambda t_2)} \alpha_\lambda (\mathbf a_\lambda | \boldsymbol {\alpha}_\lambda \rangle) \right\|^2 \\ &= s^2 \left[\mathscr E^{(1)}_\lambda\right]^4 \left\| e^{i(\mathbf{k}_\lambda \mathbf r_1 - \omega_\lambda t_1)} e^{i(\mathbf{k}_\lambda \mathbf r_2 - \omega_\lambda t_2)} \alpha_\lambda (\alpha_\lambda | \boldsymbol {\alpha}_\lambda \rangle) \right\|^2 \\ &= s^2 \left[\mathscr E^{(1)}_\lambda\right]^4 \left\| e^{i(\mathbf{k}_\lambda \mathbf r_1 - \omega_\lambda t_1)} e^{i(\mathbf{k}_\lambda \mathbf r_2 - \omega_\lambda t_2)} \alpha_\lambda^2 | \boldsymbol {\alpha}_\lambda \rangle \right\|^2 \\ &=s^2 \left[\mathscr E^{(1)}_\lambda\right]^4 \left| e^{i(\mathbf{k}_\lambda \mathbf r_1 - \omega_\lambda t_1)} e^{i(\mathbf{k}_\lambda \mathbf r_2 - \omega_\lambda t_2)} \alpha_\lambda^2 \right|^2 \| | \boldsymbol {\alpha}_\lambda \rangle \|^2 \\ &=s^2 \left[\mathscr E^{(1)}_\lambda\right]^4 \left| e^{i(\mathbf{k}_\lambda \mathbf r_1 - \omega_\lambda t_1)} \right|^2 \left| e^{i(\mathbf{k}_\lambda \mathbf r_2 - \omega_\lambda t_2)} \right|^2 |\alpha_\lambda^2|^2 \times 1 \\ &=s^2 \left[\mathscr E^{(1)}_\lambda\right]^4 \cdot 1 \cdot 1 \cdot |\alpha_\lambda|^4 \\ & = s^2 \left[\mathscr E^{(1)}_\lambda\right]^4 |\alpha_\lambda|^4 \end{aligned}
So both quantities are constant.
There are observables directly related to the electric field which can be measured even in the case of visible light, which are the quadrature observables of a mode of the field.
To determine them, we can use a similar apparatus to the one used for the laser beat note here.
It consists of entering the field to analyze in the input one of a beam-splitter | \boldsymbol{\Psi}_1 \rangle and a large quasi-classical field in the second input of the beam splitter | \boldsymbol {\alpha}_2 \rangle. This second field is called the local oscillator, and the measured quantities is the difference between the photocurrents. For an balanced beam splitter with equal coefficient of transmission and reflection, this is called a balanced detection.
If the frequency of the local oscillator has a frequency different from the one in input one, that is an heterodyne detection scheme. In this analysis, we rather consider the case where the frequency is the same as the mode in input one (\omega_2 = \omega_1), and this is a homodyne detection scheme.
It is possible to compute the photoelectric signals of the output space by using quantities in the input space, in a similar fashion as done here.
Considering a lossless semi-reflecting mirror with:
- reflectivity R is the fraction of incident power that is reflected, and for a semi-reflecting mirror R = 0.5 = \frac{1}{2},
- transmissivity T is the fraction of incident power that is transmitted, and for a semi-reflecting mirror T = 0.5 = \frac{1}{2},
- coefficient of reflection |r| = \sqrt{R} = \frac{1}{\sqrt{2}},
- coefficient of transmission |t| = \sqrt{T} = \frac{1}{\sqrt{2}}.
Then the output fields can be written as:
\begin{aligned} \mathbf E_3^{(+)} & = \frac{1}{\sqrt 2}\left(\mathbf E_1^{(+)} + \mathbf E_2^{(+)} \right) \\ \mathbf E_4^{(+)} & = \frac{1}{\sqrt 2}\left(\mathbf E_1^{(+)} - \mathbf E_2^{(+)}\right) \end{aligned}
The rate of photodetection is:
\begin{aligned} w^{(1)}(\mathbf r_4,t) & = s \left\| \mathbf E_4^{(+)} | \boldsymbol{\Psi}_{34} \rangle \right\|^2 \\ & = s \left\| \frac{1}{\sqrt 2}\left(\mathbf E_1^{(+)} - \mathbf E_2^{(+)}\right) | \boldsymbol{\Psi}_{12} \rangle \right\|^2 \end{aligned}
With the specific conditions and assumptions made about the input modes and the homodyne detection setup:
- same Frequency Modes: The input modes 1 and 2 having the same frequency eliminates any time-dependent factors in the photoelectric signal. This is because the interference between the two modes, which is responsible for the signal, becomes stationary when the frequencies are identical;
- mirror Image k-vectors: The k-vectors of the two modes being mirror images of each other implies that they have the same magnitude (and therefore the same wavelength). This, along with the matched polarizations , leads to a uniform spatial distribution of the interference pattern (and thus the probability density) in the detection region. In simpler terms, the light intensity is the same everywhere within the detector;
- matched Polarizations: The polarizations of the two modes also being mirror images ensures that they contribute equally and consistently to the interference. If the polarizations were different, the interference pattern (and hence the signal) would vary depending on the relative polarization directions. Since they are matched, the polarization dependence disappears from the signal.
As a consequence of the above points, the probability of photodetection per unit time becomes constant. This means that w^{(1)}(\mathbf r_4,t), which is related to this probability density, is also constant.
Then the total detection (multiplying for the volume given by the area S and the time T):
ST w^{(1)}(\mathbf r_4,t) = \eta \langle \boldsymbol {\Psi} | \mathbf N_4 | \boldsymbol {\Psi} \rangle
The average current is proportional to the mean rate of photodetection:
\langle \mathbf i_4 \rangle(t) = \frac{q_eS w^{(1)}(\mathbf r_4,t)}{T} = q_e \eta \langle \boldsymbol {\Psi} | \mathbf N_4 | \boldsymbol {\Psi} \rangle = q_e \eta \langle \boldsymbol {\Psi} | \mathbf a_4^\dag \mathbf a_4 | \boldsymbol {\Psi} \rangle
This result is independent of T since the mode volume is proportional to T. Since the operator \mathbf a_4 transform like the electric field:
\mathbf a_4 = \frac{1}{\sqrt 2}\left(\mathbf a_1 - \mathbf a_2\right)
The average current can be expressed as:
\langle \mathbf i_4 \rangle(t) = \frac{1}{2} q_e \eta \langle \boldsymbol {\Psi} | \left(\mathbf a_1^\dag - \mathbf a_2^\dag\right)\left(\mathbf a_1 - \mathbf a_2\right) | \boldsymbol {\Psi} \rangle = \frac{1}{2} q_e \eta \langle \boldsymbol {\Psi} | \left(\mathbf a_1^\dag\mathbf a_1 + \mathbf a_2^\dag \mathbf a_2 - \mathbf a_1^\dag\mathbf a_2 - \mathbf a_2^\dag\mathbf a_1\right)| \boldsymbol {\Psi} \rangle
A similar calculation for the input (3) gives a rate of photodetection is:
\begin{aligned} w^{(1)}(\mathbf r_3,t) & = s \left\| \mathbf E_3^{(+)} | \boldsymbol{\Psi}_{34} \rangle \right\|^2 \\ & = s \left\| \frac{1}{\sqrt 2}\left(\mathbf E_1^{(+)} + \mathbf E_2^{(+)}\right) | \boldsymbol{\Psi}_{12} \rangle \right\|^2 \end{aligned}
which is also constant, with a total detection:
ST w^{(1)}(\mathbf r_3,t) = \eta \langle \boldsymbol {\Psi} | \mathbf N_3 | \boldsymbol {\Psi} \rangle
The average current is proportional to the mean rate of photodetection:
\langle \mathbf i_3 \rangle(t) = \frac{q_eS w^{(1)}(\mathbf r_3,t)}{T}= q_3 \eta \langle \boldsymbol {\Psi} | \mathbf N_3 | \boldsymbol {\Psi} \rangle = q_e \eta \langle \boldsymbol {\Psi} | \mathbf a_3^\dag \mathbf a_3 | \boldsymbol {\Psi} \rangle
Since the operator \mathbf a_3 transform like the electric field:
\mathbf a_3 = \frac{1}{\sqrt 2}\left(\mathbf a_1 + \mathbf a_2\right)
The average current can be expressed as:
\langle \mathbf i_3 \rangle(t) = \frac{1}{2} q_e \eta \langle \boldsymbol {\Psi} | \left(\mathbf a_1^\dag + \mathbf a_2^\dag\right)\left(\mathbf a_1 + \mathbf a_2\right) | \boldsymbol {\Psi} \rangle = \frac{1}{2} q_e \eta \langle \boldsymbol {\Psi} | \left(\mathbf a_1^\dag\mathbf a_1 + \mathbf a_2^\dag \mathbf a_2 + \mathbf a_1^\dag\mathbf a_2 + \mathbf a_2^\dag\mathbf a_1\right)| \boldsymbol {\Psi} \rangle
The difference in the current measured is:
d = \langle \mathbf i_3 - \mathbf i_4 \rangle(t) = q_e \eta \langle \boldsymbol {\Psi} | \left(\mathbf N_3 - \mathbf N_4 \right)| \boldsymbol {\Psi} \rangle = q_e \eta \langle \boldsymbol {\Psi} | \left(\mathbf a_1^\dag\mathbf a_2 + \mathbf a_2^\dag\mathbf a_1\right)| \boldsymbol {\Psi} \rangle
The input state is a tensor product:
| \boldsymbol {\Psi} \rangle = | \boldsymbol {\Psi}_{12} \rangle = | \boldsymbol {\Psi}_1 \rangle \otimes | \boldsymbol {\alpha}_\lambda \rangle
The difference becomes:
d = q_e \eta \left( \langle \boldsymbol \Psi_1 | \mathbf a_1^\dag |\boldsymbol \Psi_1 \rangle \langle \boldsymbol \alpha_\lambda | \mathbf a_2 |\boldsymbol \alpha_\lambda \rangle + \langle \boldsymbol \Psi_1 | \mathbf a_1 |\boldsymbol \Psi_1 \rangle \langle \boldsymbol \alpha_\lambda | \mathbf a_2^\dag |\boldsymbol \alpha_\lambda \rangle \right)
Using the property of coherent states \mathbf a_2 |\boldsymbol \alpha_\lambda \rangle = \alpha_\lambda |\boldsymbol \alpha_\lambda \rangle and \langle \boldsymbol \alpha | \mathbf a_2^\dag = \langle \boldsymbol \alpha_\lambda | \overline{\alpha}_\lambda:
\begin{aligned} & \langle \boldsymbol \alpha_\lambda | \mathbf a_2 |\boldsymbol \alpha_\lambda \rangle = \langle \boldsymbol \alpha_\lambda | \left(\alpha_\lambda |\boldsymbol \alpha_\lambda \rangle\right) = \alpha_\lambda \langle \boldsymbol \alpha_\lambda |\boldsymbol \alpha_\lambda \rangle = \alpha_\lambda \\ & \langle \boldsymbol \alpha_\lambda | \mathbf a_2^\dag |\boldsymbol \alpha_\lambda \rangle = \left(\langle \boldsymbol \alpha_\lambda | \overline {\alpha}_\lambda \right)|\boldsymbol \alpha_\lambda \rangle = \overline {\alpha}_\lambda \langle \boldsymbol \alpha_\lambda |\boldsymbol \alpha_\lambda \rangle = \overline {\alpha}_\lambda \end{aligned}
Substituting:
\langle \mathbf i_3 - \mathbf i_4 \rangle(t) = q_e \eta \left( \alpha_\lambda \langle \boldsymbol \Psi_1 | \mathbf a_1^\dag |\boldsymbol \Psi_1 \rangle + \overline{\alpha}_\lambda\langle \boldsymbol \Psi_1 | \mathbf a_1 |\boldsymbol \Psi_1 \rangle \right)
This result is only depending on the average of the creation and annihilation operators in mode one, taken in the state |\boldsymbol \Psi_1 \rangle. This result is independent of time and therefore it is not oscillating at high frequencies and can be measured.
The sum of these two quantities in parenthesis is an Hermitian operator and can be measured. Writing \alpha_\lambda as:
\alpha_\lambda = \left|\alpha_\lambda\right|e^{i\varphi_\lambda}
and summing it can be written as:
\alpha_\lambda \langle \boldsymbol \Psi_1 | \mathbf a_1^\dag |\boldsymbol \Psi_1 \rangle + \overline{\alpha}_\lambda\langle \boldsymbol \Psi_1 | \mathbf a_1 |\boldsymbol \Psi_1 \rangle = \left|\alpha_\lambda\right| \langle \boldsymbol \Psi_1 | \left(e^{-i\varphi_\lambda}\mathbf a_1 + e^{i\varphi_\lambda} \mathbf a_1^\dag \right) |\boldsymbol \Psi_1 \rangle
which is Hermitian and therefore an observable:
d = q_e \eta \left|\alpha_\lambda\right| \langle \boldsymbol \Psi_1 | \left(e^{-i\varphi_\lambda}\mathbf a_1 + e^{i\varphi_\lambda} \mathbf a_1^\dag \right) |\boldsymbol \Psi_1 \rangle
This can be rewritten using complex formulas as:
d = q_e \eta \left|\alpha_\lambda\right| \left[\cos\left(\varphi_\lambda\right) \langle \boldsymbol \Psi_1 | \left( \mathbf a_1 + \mathbf a_1^\dag \right) |\boldsymbol \Psi_1 \rangle + \sin\left(\varphi_\lambda\right) \langle \boldsymbol \Psi_1 | \left( \frac{\mathbf a_1 - \mathbf a_1^\dag}{i} \right) |\boldsymbol \Psi_1 \rangle \right]
and taking the phase \varphi_\lambda equal to 0 or \frac{\pi}{2} these two quadrature operators can be measured.
To compute the standard deviation of the measurement, which are the fluctuations of the photocurrent difference, and, at the end, the noise in the balanced homodyne detection, we need to compute the variance:
\Delta^2\left(\mathbf i_3 - \mathbf i_4\right) = \langle \left(\mathbf i_3 - \mathbf i_4\right)^2 \rangle - \left(\langle \mathbf i_3 - \mathbf i_4 \rangle \right)^2
As often in quantum mechanics, it is proportional to the difference between the average of the square minus the square of the average:
\begin{aligned} & \langle \left(i_3 - i_4\right)^2 \rangle = q_e^2 \eta^2 \langle \boldsymbol {\Psi} | \left(\mathbf N_3 - \mathbf N_4 \right)^2| \boldsymbol {\Psi} \rangle\\ & \left(\langle i_3 - i_4 \rangle \right)^2 = d^2 = q_e^2 \eta^2 \left(\langle \boldsymbol {\Psi} | \left(\mathbf N_3 - \mathbf N_4 \right)| \boldsymbol {\Psi} \rangle \right)^2 \end{aligned}
Substituting:
\begin{aligned} \Delta^2\left(\mathbf i_3 - \mathbf i_4\right) & = \langle \left(\mathbf i_3 - \mathbf i_4\right)^2 \rangle - \left(\langle \mathbf i_3 - \mathbf i_4 \rangle \right)^2 \\ & = q_e^2 \eta^2 \langle \boldsymbol {\Psi} | \left(\mathbf N_3 - \mathbf N_4 \right)^2| \boldsymbol {\Psi} \rangle - q_e^2 \eta^2 \left(\langle \boldsymbol {\Psi} | \left(\mathbf N_3 - \mathbf N_4 \right)| \boldsymbol {\Psi} \rangle \right)^2\\ & = q_e^2 \eta^2 \left[\langle \boldsymbol {\Psi} | \left(\mathbf N_3 - \mathbf N_4 \right)^2| \boldsymbol {\Psi} \rangle - \left(\langle \boldsymbol {\Psi} | \left(\mathbf N_3 - \mathbf N_4 \right)| \boldsymbol {\Psi} \rangle \right)^2\right]\\ & = q_e^2 \eta^2 \Delta^2\left(\mathbf N_3 - \mathbf N_4\right) \end{aligned}
The variance is proportional to \Delta^2\left(\mathbf N_3 - \mathbf N_4\right).
Computing the first term:
\begin{aligned} (\mathbf{N}_3 - \mathbf{N}_4)^2 & = (\mathbf{a}_1^\dagger \mathbf{a}_2 + \mathbf{a}_2^\dagger \mathbf{a}_1)^2 \\ & = (\mathbf{a}_1^\dagger)^2 (\mathbf{a}_2)^2 + (\mathbf{a}_2^\dagger)^2 (\mathbf{a}_1)^2 + \mathbf{a}_1^\dagger \mathbf{a}_1 \mathbf{a}_2^\dagger \mathbf{a}_2 + \mathbf{a}_1^\dagger \mathbf{a}_1 \mathbf{a}_2^\dagger \mathbf{a}_2 \\ & = (\mathbf{a}_1^\dagger)^2 (\mathbf{a}_2)^2 + (\mathbf{a}_2^\dagger)^2 (\mathbf{a}_1)^2 + \mathbf{a}_1^\dagger \mathbf{a}_1 \mathbf{a}_2^\dagger \mathbf{a}_2 + \mathbf{a}_1^\dagger \mathbf{a}_1 (1 + \mathbf{a}_2^\dagger \mathbf{a}_2) \\ & = (\mathbf{a}_1^\dagger)^2 (\mathbf{a}_2)^2 + (\mathbf{a}_2^\dagger)^2 (\mathbf{a}_1)^2 + \mathbf{a}_1^\dagger \mathbf{a}_1 \mathbf{a}_2^\dagger \mathbf{a}_2 + \mathbf{a}_1^\dagger \mathbf{a}_1 + \mathbf{a}_1^\dagger \mathbf{a}_1 \mathbf{a}_2^\dagger \mathbf{a}_2) \end{aligned}
Using:
\boldsymbol{\Psi}_{12} = | \boldsymbol{\Psi}_1 \rangle \otimes | \boldsymbol \alpha_{\lambda} \rangle
It becomes:
\begin{aligned} \langle \boldsymbol{\Psi}_{\text{in}} | (\mathbf{N}_3 - \mathbf{N}_4)^2 | \boldsymbol{\Psi}_{\text{in}} \rangle & = (\alpha_\lambda)^2 \langle \left( \mathbf{a}_1^\dagger\right)^2 \rangle + (\overline \alpha_\lambda)^2 \langle \mathbf{a}_1^2 \rangle + |\alpha_\lambda |^2 \langle \mathbf{a}_1^\dagger \mathbf{a}_1 + \mathbf{a}_1^\dagger \mathbf{a}_1 \rangle + \langle \mathbf{a}_1^\dagger \mathbf{a}_1 \rangle \\ & = |\alpha_\lambda|^2 \langle \boldsymbol{\Psi}_1 | \left(e^{i \varphi_2} \mathbf{a}_1^\dagger + e^{-i \varphi_2} \mathbf{a}_1 \right)^2 | \boldsymbol{\Psi}_1 \rangle + \langle \boldsymbol{\Psi}_1 | \mathbf{a}_1^\dagger \mathbf{a}_1 | \boldsymbol{\Psi}_1\rangle \end{aligned}
Since |\alpha_\lambda|^2 \gg 1, the last term can be ignored, and the value of the noise is:
\langle \boldsymbol{\Psi}_{\text{in}} | (\mathbf{N}_3 - \mathbf{N}_4)^2 | \boldsymbol{\Psi}_{\text{in}} \rangle = |\alpha_\lambda|^2 \langle \boldsymbol{\Psi}_1 | \left(e^{i \varphi_2} \mathbf{a}_1^\dagger + e^{-i \varphi_2} \mathbf{a}_1 \right)^2 | \boldsymbol{\Psi}_1 \rangle Using this expression and the expression previously computed we can arrive at the variance:
\begin{aligned} \Delta^2\left(\mathbf N_3 - \mathbf N_4\right) & = \langle \boldsymbol {\Psi} | \left(\mathbf N_3 - \mathbf N_4 \right)^2| \boldsymbol {\Psi} \rangle - \left(\langle \boldsymbol {\Psi} | \left(\mathbf N_3 - \mathbf N_4 \right)| \boldsymbol {\Psi} \rangle \right)^2 \\ & = |\alpha_\lambda|^2 \langle \boldsymbol{\Psi}_1 | \left(e^{i \varphi_2} \mathbf{a}_1^\dagger + e^{-i \varphi_2} \mathbf{a}_1 \right)^2 | \boldsymbol{\Psi}_1 \rangle - \left(\left|\alpha_\lambda\right| \langle \boldsymbol \Psi_1 | \left(e^{-i\varphi_\lambda}\mathbf a_1 + e^{i\varphi_\lambda} \mathbf a_1^\dag \right) |\boldsymbol \Psi_1 \rangle \right)^2 \\ & = \Delta^2 \left( e^{-i\varphi_\lambda}\mathbf a_1 + e^{i\varphi_\lambda} \mathbf a_1^\dag \right) \end{aligned}
and it is a value that can also be measured. Therefore, it provides a way of measuring the field quadrature observables, and also their fluctuation, bypassing the technical limitation of the response time of the detector. If the local oscillator is intense sufficiently, it is possible to have a much stronger signal than the noise of the detectors. and to gain access of the fluctuations of the radiation.