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1

Thanks to Ján Lalinský for getting me to double check my faulty memory. My assumption of $\langle\chi_n|[H_0,z]|\chi_m\rangle = 0$ was wrong. Here is the solution after double checking my work. $$\langle\chi_n|[H_0,z]|\chi_m\rangle=\langle\chi_n|H_0\,\,z|\chi_m\rangle-\langle\chi_n|z\,\,H_0|\chi_m\rangle$$ $$\langle\chi_n|[H_0,z]|\chi_m\rangle=\langle\...


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I will try to answer the integral question. A continuous function on a bounded interval is bounded. Throughout this answer I will use Riemann's definition for integrability. Using Cauchy-Schwarz Inequality- \begin{equation} \int_{z\in\mathbb{C}}|F_{\phi}(z)|^2 e^{-|z|^2}\leq\Bigg|\int_{z\in\mathbb{C}}|F_{\phi}(z)|^2 e^{-|z|^2}\Bigg|\leq\sqrt{\int_{z\in\...


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Quadrature has a very clear and precise meaning in Quantum Mechanis. Some quantities don't commute, i.e., you cannot measure both of them with unlimeted precision. This is related to Heisenbeg unsertainty. Let's put this two variables in the Cartesian plan and call this observables as X and Y.. if X is position then Y is momentum. If X is electric field ...


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I wanted this post to be a comment, but made it an answer instead. I cant delete it yet, so i'll extend it a bit. You might want to check Quantum mechanics for mathematicians by L. Takhtajan; the result is also in this book, but I don't remember if he proves it. In any case, you are right in that the uniqueness only holds for antiholomorphic functions. ...


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If you have a Poisson distribution with a probability $$P(n)=\frac{\lambda^n}{n!}e^{-\lambda}$$ that there will be $n$ events per bin, then $\lambda$ is the mean number of events per bin. You can get this via a direct calculation, $$ ⟨n⟩ =\sum_{n=0}^\infty nP(n) =\sum_{n=0}^\infty n \frac{\lambda^n}{n!}e^{-\lambda} =\lambda e^{-\lambda}\sum_{n=1}^\infty \...


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The light-matter coupling Hamiltonian you have written is correct in general for a running-wave laser within the dipole approximation (and two-level approximation); it contains no assumption about the trap potential. In fact, it's not true that it connects only internal states. It also connects the motional states due to the $e^{i k \hat{Z}}$ term. This term ...


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I found how to find it again [this is not a proof]. Here $\xi=re^{i\theta}$. We have: $a^\dagger D(\alpha) = D(\alpha) a^\dagger + \alpha^*D(\alpha)$ $a D(\alpha) = D(\alpha) a + \alpha D(\alpha)$ $a^\dagger S(\xi) = S(\xi) a^\dagger ch(r) - S(\xi)ae^{-i\theta}sh(r)$ $a S(\xi) = S(\xi) a ch(r) - S(\xi)a^\dagger e^{i\theta}sh(r)$ Then, applying $a$ to "$...


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The phrase "$r\pi$ pulse" (for $r\in\mathbb R$, usually a multiple of $1/2$) is a specific convention in quantum optics, and particularly it is more specific than the definition you quote, from an initial state $\left|{\uparrow}\right\rangle$ we can obtain the superposition state $\frac{1}{\sqrt{2}}\left(\left|{\downarrow}\right\rangle + \left|{\uparrow}\...


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The extra one simply reflects the commutation relationship $[a,\,a^\dagger] = \mathrm{id}$ of the quantum mechanical harmonic oscillator. If each annihilation operator in the expression for $E^+$ acts only on its corresponding mode, and likewise for the creation operators in $E^-$, then the action of $E^-\,E^+ -E^+\,E^-$ on a state involving only excitations ...


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It seems like you have established that your photons go through both paths! Not only do you not know which path a photon went through, you know it went through both because of the changes in interference pattern you see. You might want to do this in a controlled way!


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Someone please correct me if I'm wrong, but I suspect it might just be a bit ambiguous, like "beam splitter". The $\pi/2$ part of the name refers to the amount of rotation in the Bloch sphere, but single-qubit quantum operations have an extra degree of freedom on top of the amount of rotation and the axis of rotation: the global phase factor. For example, ...


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Coherent states are classical in a precise way which hasn't been stated explicitly yet, although Rod suggests at it. Suppose you want to time-evolve the interaction between a coherent EM state and matter. This amounts to solving the Schroedinger equation for: $$ i\hbar\frac{d}{dt} |\psi \rangle= H |\psi \rangle$$ with $$\hat{H}= \hat{H}_{0A}+\sum_k\omega_k ...


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In general, a 3D lattice allows tunnelling along all three directions. One then makes a 2D lattice (or really a series of 2D lattices) just by turning up the power in one of the lattice lasers to isolate the sheets. There may be specific configurations in which gravity prevents tunnelling between vertical layers (presumably by creating Bloch oscillations ...


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Let me actually collect most of my comments in an answer attempting to be more coherent than they, or your labile question. In fact, you are piling up three different questions, logically distinct, but with strong and natural connections, so it might be worth splitting them apart, before bringing them back together in the final coda. First, there is plain ...


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The Wigner function of the thermal state is given as $W(\alpha)=\frac{1}{\pi(\langle n\rangle+ \frac{1}{2})} \exp \left( -\frac{|\alpha|^2}{\langle n\rangle+\frac{1}{2}} \right)$ The source is Quantum optics, the author is Girish S. Agarwal


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Energy loss by photons occurs in a dielectric medium when the photons interact with the atoms of which the dielectric is made. This interaction does not normally result in a red-shift of the photons, but rather an attenuation of a beam of photons due to photons being absorbed. Depending on the nature of the material and the energy of the photons, the ...



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