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Currently doing a PhD in Controlled Quantum Dynamics at Imperial College and the University of Oxford.


1d
comment What is the Quantum Transition Time for Photon Emission?
@CuriousOne To be clear, I was agreeing with you; time dependence is not the explanation. I was just pointing out that the Jaynes-Cummings model is a good place to start along the lines you were suggesting. And indeed that the answer I linked already contains a lot of the necessary elements.
1d
comment What is the Quantum Transition Time for Photon Emission?
@CuriousOne BTW, the purpose of my comment was simply to point out that the Jaynes-Cummings Hamiltonian is easy to understand but also gives extremely good agreement with experiments at optical frequencies, so no need for a contrived model Hamiltonian. I think this answer gives a nice exposition of the relevant details.
1d
comment What is the Quantum Transition Time for Photon Emission?
@CuriousOne The standard description of this process in quantum optics goes via the Jaynes-Cummings model, which is time-independent. Talking about the time-dependent field only makes sense when the light field is in a semi-classical (e.g. coherent) state. In the full quantum treatment the role of the oscillating electric field is replaced (formally speaking) by the oscillatory time dependence $e^{-iEt/\hbar}$ of the photon states. Nevetheless, the Hamiltonian is time-independent; the point is that the interaction mixes the $n$-photon and atomic eigenstates so transitions occur between them.
Oct
17
awarded  Good Answer
Oct
17
comment Entanglement Hamiltonian for two 1/2 spin system
@Antonio_phy Setting $k_B T=1$ just means that you use units of energy equal to $k_B T$, it doesn't necessarily mean that the temperature cannot change. However, if you like you can think of $T$ being constant (but finite, importantly), and imagine varying the external magnetic field instead of the temperature. The results are the same, because the only thing that matters is the dimensionless ratio $\mu/k_BT$.
Oct
17
revised Entanglement Hamiltonian for two 1/2 spin system
added 22 characters in body
Oct
17
revised Entanglement Hamiltonian for two 1/2 spin system
added 400 characters in body
Oct
17
answered Entanglement Hamiltonian for two 1/2 spin system
Oct
16
comment Why does an external laser drive only couples certain levels?
@garyp I think the OP refers to an atomic level scheme similar to the first figure on this page.
Oct
16
comment Why does an external laser drive only couples certain levels?
@KobyYavilberg Often the two ground states will have non-zero spin projections in different directions. You can selectively couple to just one transition by using polarised light, which changes the angular momentum projection by a definite amount according to the direction of polarisation and the selection rules Emilio has described. So even if the ground sub-levels are completely degenerate, and the two lasers are at the same, resonant frequency, each laser can be made to couple to only one transition by choosing different polarisation states.
Oct
16
awarded  Yearling
Sep
30
awarded  Explainer
Sep
24
awarded  Autobiographer
Aug
23
awarded  Revival
Aug
22
comment Stabilization of von Neumann equation
@XingdongZuo You're welcome. Yes, a probability distribution that does not evolve in time can be called a stationary state or steady state in physics.
Aug
22
comment Stabilization of von Neumann equation
@XingdongZuo This expansion holds whether or not $[H,\rho(0)] = 0$. However, $[H,\rho(0)] = 0$ implies that $\rho_{mn} = 0$ for $E_m\neq E_n$, and therefore there are no oscillations. That is, if $[H,\rho(0)]=0$, it follows that $\rho_{mn}(t) = \rho_{mn}(0)e^{i(E_m-E_n)t} = \rho_{mn}(0)$, since either $e^{i(E_m-E_n)t}=1$ or $\rho_{mn}(0) = 0$.
Aug
22
comment Stabilization of von Neumann equation
@XingdongZuo But the state of a system does not reach thermal equilibrium as $t\to 0$ under unitary evolution. If $[\rho,H]\neq 0$ you will just see coherent oscillations forever. There is no damping of fluctuations without some external influence.
Aug
22
comment Stabilization of von Neumann equation
Hi @XingdongZuo, I'm not sure I understand what you mean by "stabilised". Do you mean reaching thermal equilibrium?
Aug
16
comment Unknown letter ℑ used in an equation
This question is about typesetting and so would be appropriate for our sister site LaTeX StackExchange.
Aug
15
comment What is the physical meaning of commutators in quantum mechanics?
@AlexeyBobrick Cool, I think we both agree, just wanted to be clear. Thanks :)