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6

What would I be missing in Cavity QED if I had never learnt about quantum electrodynamcis? Excellent question, especially the way it is posed! The bottom line is that what the OP calls "standard quantum optics" is essentially the non-relativistic sector of QED. Many textbooks indeed do not come from the QED perspective and instead introduce effective ...


8

The terminology "cavity QED" is used to signal that one is treating the electromagnetic field as itself quantised and there is a cavity with mirrors in the problem. The alternative would be to treat the electric field of the cavity mode as a classical variable. Instead one uses raising and lowering operators for both the field mode and the internal state of ...


4

Well if you are representing light as a quantum field and not as a classical field, then I suppose cavity QED is QED. The only difference is that cQED is usually done with nonrelativistic particles, so maybe cQED is best understood as nonrelativistic QED in a strongly coupled environment. What this also means is that things like pair-production, ...


1

After reviewing the comments I believe KF Gauss is correct in their statement that the atom picks ups angular momentum with respect to the light propagation axis. See Eq. 5.448 in Quantum and Atom Optics by Steck regarding the mechanical force on an atom by an optical field. $$ \langle \boldsymbol{F}\rangle = \frac{i\hbar|\Omega(\boldsymbol{r})|^2}{4\left(\...


1

The key to understanding how orbital angular momentum (OAM) can be conserved when a photon that carries OAM is absorbed by a small particle is to understand OAM itself a little better. Note that OAM in a paraxial light beam is always defined with respect to the propagation axis of that beam. If one would pick some random axis pointing in some random ...


-1

where does the OAM go when an atom happens to absorb a photon with non-zero OAM It is simple : the photon can raise an electron to a higher energy level only if all the quantum numbers are conserved,including angular momentum. It is not enough to have the energy difference ( within the width of the energy level). If the photon is head on with the atom, ...


1

This is going to be a bit of a half baked answer but here it is. The usual picture for light with orbital angular momentum is light in a Laguerre-Gaussian Mode mode with non-zero azimuthal index. This sort of field has phase wrapping azimuthally around the propagation axis. We suppose without loss of generality that the atom is on the propagation axis of ...


0

The difference is that multi-photon physics incorporates quantum phenomena whereas classical nonlinear optics does not. For example, multi-photon phenomena can yield entangled pairs of photons which have no classical analog.


0

I think your confusion relies on a misunderstanding: a homodyne detection does not directly measures $\langle q \rangle$, but a value of the operator $\hat q$, by projecting the state on one of its eigenstates, an infinitely squeezed state. This is clear when you use a time resolved homodyne detection, as needed for a quantum communication protocol like ...


1

Edit: 2nd § significantly changed A squeezed state, squeezed in $p$ or $q$ has a diagonal covariance matrix $\Gamma$ (se below) and is not thermal. $$\begin{align}\Gamma&=\begin{bmatrix} S & 0 \\ 0 & \frac{1}{S}\end{bmatrix} & \text{for $S>0$ and $s\neq1$} \end{align}$$. For a single-mode thermal sate, we have $\Gamma_{\text{th}}=\...


1

Actually, I figured out the answer on my own. As it turns out, the two operators have different eigenstates (which is perhaps obvious, given the observed behaviour). To understand what the difference is between the two approaches, we'll consider a simplified scenario. We'll fix the time at a specific value $t$, and divide the space into discrete points so ...


0

You are correct, reflection at the QM level is not absorption and re-emission. Absorption and re-emission would change the energy level of the photons, their relative phase, relative angle (thus creating a diffuse reflection). https://en.wikipedia.org/wiki/Diffuse_reflection Mirror reflection is elastic scattering. That is the only way to keep the energy ...


0

Here is how I understand it. The only reason why a coherent state can remain the same after an annihilation operator acted on it is because when it is expressed as an expansion over number states the summation runs all the way to infinity. If I have a state with an infinity number of photons, then the removal of one photon would not have any effect. The ...


0

Which means that although I entered a single photon with a |+⟩ polarization, I could measure 2 photons with |−⟩ polarization. This is of no concern to the no-cloning theorem. As you correctly point out, if $|0\rangle\mapsto|00\rangle$ and $|1\rangle\mapsto|11\rangle$ then you cannot have $|+\rangle\mapsto|++\rangle$. However, there is no problem with ...


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