2
$\begingroup$

I am trying to understand this paper- https://journals.aps.org/pra/abstract/10.1103/PhysRevA.30.1386 I will try to give my understanding of the paper first. We start with the quantum Langevin equation-

$\frac{da}{dt}=-\frac{i}{\hbar}[a, H_{sys}]-\frac{\gamma}{2}a + \Gamma$

The first term in the RHS comes from the Heisenberg relation. The second term corresponds to the decay of the cavity mode and the last term denotes the noise operator. In order to excite the cavity mode, we need to somehow communicate with the cavity by "partially opening it". This allows both the external drive and noise to couple with the cavity mode. Thus, $\Gamma = \gamma'a_{in}$ indicates the noise due to an external drive.

We can write a time reversed equation similar to the previous one-

$\frac{da}{d(-t)}=\frac{i}{\hbar}[a, H_{sys}]-\frac{\gamma}{2}a + \gamma'a_{out}$

Since the sign of $a_{out}$ is the same as $a_{in}$, I think they assume the following scenario-

enter image description here

Thus, the boundary condition of mode rate allows me to write the following equation- $a=k(a_{in}+a_{out})$.

From this, they take the fourier transform of the original equation and establish a relation that looks like this-

$a_{out}(\omega)=[A+(\frac{\gamma}{2})+ i\omega][-A +(\gamma -i\omega)]^{-1} a_{in}(\omega)$

In general, from what I understand, this procedure is applicable to a general Hamiltonian, where I can have different input and output photon frequencies (For example, I have seen papers where this procedure is followed to convert photon frequency). Therefore, they should not be the frequency of the input and output photons.

My question is: does this $\omega$ mean the frequency at which an input photon will yield an output photon?

$\endgroup$
0
$\begingroup$

To understand what the $\omega$ means, let us take expectation values of the expression in the question

$$\langle a_{out}(\omega) \rangle = \frac{A+(\frac{\gamma}{2})+ i\omega}{-A +(\gamma -i\omega)} \langle a_{in}(\omega)\rangle.$$

This relation can be understood from our intuition from classical optics: $\langle a_{in}(\omega)\rangle$ is the input field amplitude of the Fourier component at frequency $\omega$, $\langle a_{out}(\omega)\rangle$ is the corresponding output. Since the scattering process is elastic in the considered system, the scattering matrix is simply the frequency dependent factor $\frac{A+(\frac{\gamma}{2})+ i\omega}{-A +(\gamma -i\omega)}$.

The above is just for intuition. For non-classical fields, for example even for a Fock state at a certain frequency, one will find that $\langle a_{in}(\omega) \rangle$ can vanish and does not correspond to the classical electromagnetic amplitude. However, instead we can count photons and look at the photon number expectation value $\langle a^\dagger_{in}(\omega) a_{in}(\omega)\rangle$. Due to the linearity of the input-output relation, one simply gets

$$\langle a^\dagger_{out}(\omega) a_{out}(\omega)\rangle = \left|\frac{A+(\frac{\gamma}{2})+ i\omega}{-A +(\gamma -i\omega)}\right|^2\langle a^\dagger_{in}(\omega) a_{in}(\omega)\rangle.$$

This can be expected, since now we are essentially measuring intensities, such that the absolute value of the scattering matrix matters.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.