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A quantum state does not have any classical memories or classical output, period: a classical memory refers eventually in every application to the process of data stored by copying. As it has been proven quantum states cannot be copied. However they can be entangled. A quantum memory may store a quantum state, but not for the purpose of copying in any ...


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What I don't quite understand is how do we relate the parameters of Rabi cycles like Rabi frequency to the intensity of absorption? Why do we in fact see any absorption at all if it's more or less compensated by emission? Rabi oscillation is a concept based on an approximate description of a single atom. This description does not include any ...


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Spontaneous emission rate ($\Gamma$) plays an important role. If $\Gamma=0$ than indeed one would excite atoms and de-excite back in ground state by stimulated emission with emission of photon in same direction. As a result as you pointed out one would not observe absorption. In real life spontaneous emission will lead to emission of photons in random ...


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Strictly speaking a photon cannot be localized and the single particle "wavefunction" (as well as it's corresponding position operator $\hat{r}$) only exists in an approximate sense. The reason for this is quantum electrodynamics (QED), which is the theory that contains photons, is a quantum field theory (QFT) rather than the (non-relativistic) quantum ...


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The formal analogy between a mode of the radiation field and a particle in a harmonic potential stems from the fact that both systems have the Hamiltonian (in appropriate units) $$ H = \frac{1}{2}P^2 + \frac{1}{2}\omega^2 X^2,$$ where the variables $X$ and $P$ obey canonical commutation relations $[X,P] = \mathrm{i}\hbar$. For the radiation field, these ...


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There exists the KISS rule ( keep it simple stupid), and I will. Let us take an excited potassium atom and solve the Schrodinger equation for the transitions available. This will give for this atom a wavefunction, in principle from -infinity to +infinity in space and time. The square of this wavefunction will give the probability for a photon to fall on a ...


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This is a perfect example of the collapse of the wave function. The "potassium" atom makes a transition from higher energy to lower energy, sending out a speherical wavefront into the universe. One hundred years later, that spherical wavefront crosses the photomultiplier tube. Since the energy of that wave is spread over hundreds of square light years, there ...


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Here is a suggestion of how to produce the state in your second formula. Choose to make holes in a screen, NOT at the two points A and B of which I spoke in my first answer, but at two other points, e.g. on top of the upper cone, and on the bottom of the lower cone. Then, pass the upper beam through a beam splitter $BS_1$. Block the transmitted wave, let ...


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Spontaneous Parametric Down Conversion is a process that produces pairs of photons. The process goes like this: a strong beam of ultra-violet (UV) photons, is sent upon a down-conversion crystal (placed inside your black-box). Inside the crystal, the UV photon is SPLIT into two photons, named (I don't know why) SIGNAL and IDLER. In general the two photons ...


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Since this seems a homework exercise, here's a sketch: I'm not sure about the $\mathcal{N}$-part in the formular, but in general: Note that $\sigma_i$ form an orthonormal basis of the Hermitian matrices according to the Hilbert-Schmidt inner-product ($\langle A,B\rangle:=\operatorname{tr} (A^{\dagger}B)\rangle$). This means that you can write $$ ...


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It turns out that this form is just a fancy way of writing a normal Fourier transform. Use the definition of the intensity correlation function $$g^{(2)}(\tau) = \langle n(\tau) n(0) \rangle, $$ where $n(t)$ denotes the (Hermitian) intensity at the position of the detector, evaluated at time $t$ in the Heisenberg picture. The angle brackets $\langle ...



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