A Why Are the Peaks at Δ = 0 and Δ = ω_m Equally High?

LionCereals
Messages
1
Reaction score
0
TL;DR Summary
How can I explain the average occupation numbers of a coupled cavity-mirror system?
I am considering the following Hamiltonian:
$$H = -\Delta a^{\dagger}a + \omega_m b^{\dagger}b + g_0 * a^{\dagger}a (b + b^{\dagger})$$
which is the interaction picture optomechanical Hamiltonian for a cavity with movable end mirror. The mirror vibrations are quantized, leading to phonons that are annihilated / created using ##b, b^{\dagger}##. Photons in one cavity mode are created with ##a^{\dagger}##.

I have plotted the steady-state solution for the average photon number in the cavity, ##n_C##, and the average phonon number in the mirror, ##n_M## as a function of detuning. For this, I used QUTIP and obtained, with the y-axis on log scale (see attachment).This is the typical Lorentzian resonance curve. Now, we clearly see that there are peaks corresponding to ##\Delta = \omega_m$## and ##\Delta = 0##.
I understand the peak at ##\Delta = 0## as then there are lots of photons inside the cavity, so we have a high average photon number and thus a high effective coupling to the mirror, as ##g_0 * n_a## is large.

What I don't understand, however, is why the second peak at ##+\omega_m## is as high as the first peak. Clearly there, we have ##\omega_{in} = \omega_C + \omega_m##, such that the photon - phonon scattering process is resonant. But why has it the same height?

From what I see, if we have (on average) as many phonons in the mirror at ##\Delta = \omega_m## as at ##\Delta = 0##, then the effective coupling strength at both detunings should be equal. But clearly, in the former case there are less photons in the cavity, as they are not resonant with the cavity frequency?
 

Attachments

  • HW2_3_3b_final.png
    HW2_3_3b_final.png
    13.1 KB · Views: 148
Physics news on Phys.org
LionCereals said:
I am considering the following Hamiltonian:
$$H = -\Delta a^{\dagger}a + \omega_m b^{\dagger}b + g_0 * a^{\dagger}a (b + b^{\dagger})$$
which is the interaction picture optomechanical Hamiltonian for a cavity with movable end mirror.
May I know where did you find this Hamiltonian? Would you cite the paper/book?
 
Not an expert in QM. AFAIK, Schrödinger's equation is quite different from the classical wave equation. The former is an equation for the dynamics of the state of a (quantum?) system, the latter is an equation for the dynamics of a (classical) degree of freedom. As a matter of fact, Schrödinger's equation is first order in time derivatives, while the classical wave equation is second order. But, AFAIK, Schrödinger's equation is a wave equation; only its interpretation makes it non-classical...
I asked a question related to a table levitating but I am going to try to be specific about my question after one of the forum mentors stated I should make my question more specific (although I'm still not sure why one couldn't have asked if a table levitating is possible according to physics). Specifically, I am interested in knowing how much justification we have for an extreme low probability thermal fluctuation that results in a "miraculous" event compared to, say, a dice roll. Does a...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. Towards the end of the first lecture for the Qiskit Global Summer School 2025, Foundations of Quantum Mechanics, Olivia Lanes (Global Lead, Content and Education IBM) stated... Source: https://www.physicsforums.com/insights/quantum-entanglement-is-a-kinematic-fact-not-a-dynamical-effect/ by @RUTA
Back
Top