Can a balanced homodyne detector measure instantaneous change in frequency?

I am reading a paper(https://arxiv.org/abs/1906.03540) where a Balanced Homodyne detector is producing a homodyne signal that is the instantaneous change in the oscillator's frequency.

I know that a homodyne signal is proportional to either quadrature (x or p) of a field mode. Why it is the instantaneous change in oscillator's frequency is not obvious to me. Please help.

if $$\omega_1$$ is the change in angular freq.,

$$x_0(t) = \cos(\omega t),\quad p_0(t) = \sin(\omega t)$$

$$x = cos((\omega+\omega_1)t) = cos(\omega t)cos(\omega_1 t) - sin(\omega t) sin(\omega_1 t) = x_0(t) cos(\omega_1 t) - p_0(t) sin(\omega_1 t)$$

$$p = sin((\omega+\omega_1)t) = sin(\omega t)cos(\omega_1 t) + cos(\omega t) sin(\omega_1 t) = p_0(t) cos(\omega 1 t) + x_0(t) cos(\omega_1 t)$$

Equivalently, $$x = cos((\omega+\omega_1)t) = cos(\omega t)cos(\omega_1 t) - sin(\omega t) sin(\omega_1 t) = x_1(t) cos(\omega t) - p_1(t) sin(\omega t)$$

$$p = sin((\omega+\omega_1)t) = sin(\omega t)cos(\omega_1 t) + cos(\omega t) sin(\omega_1 t) = x_1(t) \sin(\omega t) + p_1(t) cos(\omega t)$$

where $$x_1(t) = cos(\omega_1 t)$$

$$p_1(t) = sin(\omega_1 t)$$

I am not sure how w1 can be the homodyne signal. @jgerber

• Saumya biswas, I've edited your question to give you a start on using mathjax for the equations. Can you take it from here? Commented May 31, 2020 at 3:39
• Thank you for your interest in this article. It is not entirely clear to me what you are asking. I wonder if you are wondering about the text directly above Eq. (8) in the linked article? Commented May 31, 2020 at 4:09
• yes. I know that BHD measures a quadrature like x. S(t) in eq. 8 is the detuning of c as per eq. 6. I do not know how that works out. Commented May 31, 2020 at 4:15
• I think S(t) in eq. 8 is just c^{PM}(t) from (7b), which just happens to be the time dependent detuning. Right? Commented May 31, 2020 at 4:29

OP's comment above is correct. The signal $$S(t)$$ in Eq. (8) comes from $$\hat{c}^{PM}$$ in (7b). The connection with the instantaneous frequency of the cavity is as follows. Because the probe light is on resonance it collects zero phase shift as it passes through the cavity. However, as the mechanical oscillator oscillates, because of the optomechanical interaction, the cavity resonance frequency slightly shifts. Now the probe is detuned by a small amount. This means the cavity light collects a slight phase shift. The light is then phase modulated by the position of the mechanical oscillator. If the phase modulation is small then we may approximate it as Q-quadrature modulation. The homodyne detector then detectors the Q-quadrature of the optical field which can be interpreted as being proportional to the position of the mechanical oscillator.

In some mathematical detail: The cavity transfer function and phase response is is

\begin{align} T(\Delta) =& \frac{\kappa}{\kappa - i \Delta}\\ \phi(\Delta) =& \arg(T(\Delta)) = \arctan\left(\frac{\Delta}{\kappa}\right) \approx \frac{\Delta}{\kappa} \end{align}

For small detunings $$\Delta \ll \kappa$$.

The optomechanical Hamiltonian (before linearization that is expressed in Eq. (5)) is given by

\begin{align} \frac{\hat{H}}{\hbar} =& \omega_a \hat{a}^{\dagger}\hat{a} + \omega_c \hat{c}^{\dagger}\hat{c} + g \hat{X} \hat{c}^{\dagger}\hat{c}\\ =& \omega_a \hat{a}^{\dagger}\hat{a} + \left(\omega_c + g\hat{X}\right)\hat{c}^{\dagger}\hat{c} \end{align}

So we see that the instantaneous cavity frequency is modulated by $$g\hat{X}$$. If the mechanical oscillator is near the ground state then $$\hat{X} \sim 1$$ so for $$g \ll \kappa$$ the cavity modulation is small so we have that the phase shift of the probe beam is roughly

$$\phi(t) = \frac{\Delta(t)}{\kappa} = \frac{g}{\kappa}\hat{X}(t)$$

This shows how the position of the mechanical oscillator 1) modulates the instantaneous cavity frequency thus 2) modulates the probe phase quadrature. Perhaps this answers the question already?

I'll say some more. The phase quadrature can be related to the Q quadrature very simply for small phase modulations. Generally a complex number can be written as

$$S = A e^{i\phi} = I + i Q$$

The real and complex parts can be related to the phase by

$$\phi = \arctan\left(\frac{Q}{I}\right) \approx \frac{Q}{I}$$

In this case the $$I$$ quadrature is roughly constant and the $$Q$$ quadrature is a small modulation. We can then write

$$Q(t) \approx \phi(t)I_0$$

Where $$I_0$$ is the average magnitude of the field amplitude, in this case $$\sqrt{n}$$. The $$Q$$ quadrature of the field is then given by

$$\hat{Q}(t) = \hat{c}^{PM}(t) = \sqrt{n}\frac{g}{\kappa} \hat{X}(t)$$

This much explains how the position of the mechanical oscillator is mapped onto the $$Q$$ quadrature of the optical field. Note that the article does not distinguish between the $$Q$$ quadrature and the phase quadrature, this is standard in the field and always justified for small phase modulations.

Perhaps you are asking how a balanced homodyne detector can measure the phase quadrature of light? This is as follows. Two tones $$S$$ and $$L$$ are split by a beamsplitter and sent onto a photodiode. Suppose $$L=L_0e^{i\theta}$$ and $$S$$ is given by $$S = I + i Q$$. Because of unitarity of the beamsplitter, the two photodiodes see different interferences. We have the measured photo signals

\begin{align} D_{\pm} =& |L \pm S|^2 = |L|^2 + |S|^2 \pm L_0 e^{i\theta}(I-iQ) \pm L_0 e^{-i\theta}(I+iQ)\\ =& |L|^2 + |S|^2 \pm 2I\cos(\theta) \pm 2Q \sin(\theta) \end{align}

The balanced photodetector takes the difference of these two signals. Suppose we set the LO phase $$\theta = \frac{\pi}{2}$$ so that

$$D_{bal} = D_+ - D_- = 4L_0Q$$

In this way, by appropriately tuning the phase of the homodyne local oscillator the balanced homodyne detector can be made to measure the $$Q$$ quadrature of the incident signal field. Above I showed how the $$Q$$ quadrature of the incident is proportional to oscillator position $$\hat{X}(t)$$. This gives the intuition for the homodyne detector measures the instantaneous cavity frequency/the mechanical oscillator position.

• I am asking because $\int dt m(t) S(t) with S(t) being linear in annihilation operator, a can not be a c-number. I would accept a bilinear operator like intensity to be a c-number, but s(t) is linear in a. Now, I know the local oscillator of the BHD does contribute another a or a^{\dagger}, which we replace by a number. I am still wondering if S(t) and q^{u} should have the standing of an quantum mechanical operator. Commented May 31, 2020 at 19:19 • Or has the homodyne signal become an expectation value and is only a c-number? Commented May 31, 2020 at 19:50 • Can you please confirm something about fig 2(a). For each of the 8000 trajectories, the initial state estimates are found from knowing q^{u} at$t_f\$ only? Commented May 31, 2020 at 22:50