# Tag Info

2

We take the Wigner function $$W(\alpha)=\frac{1}{\pi^2}\int \text{e}^{\alpha \beta^*-\alpha^*\beta}\text{Tr}\left(\hat \rho \hat D(\beta) \right) \text{d}^2\beta,$$ and write the displacement operator as $\hat D(\beta)=e^{\beta\hat a^\dagger-\beta^*\hat a}=e^{-\beta^*\hat a}e^{\beta\hat a^\dagger}e^{\frac 1 2|\beta|^2}$ using the BCH formula such that $$... 2 @Sunyam has mapped out your homework for you, but here are the two explicit steps that should allow you to unfold it,$$ \frac{1}{\pi^2}\int {e}^{\alpha \beta^*-\alpha^*\beta}\text{Tr}\left(\hat \rho e^{ -\beta^*\hat a} e^{\beta \hat a^\dagger } e^{|\beta|^2/2}\right) \text{d}^2\beta= \\ \frac{1}{\pi^2}\int {e}^{\alpha \beta^*-\alpha^*\beta} e^{|\...

0

No, two single photons coming from two different sources cannot be combined to form a superposition that represents one photon with one spatial mode. In other words, $$\text{two photon state} \neq |a\rangle + |b\rangle .$$ Even though the two photons come from different sources, they still give you a two-photon state. Therefore, what you have is $$\text{... 2 There is a way because a set of n qubits allows universal computation on Hilbert spaces of up to 2^n dimensions. But the word "universal" has hidden in it some issues surrounding computational complexity. The question becomes, can qubits efficiently simulate continuous variables. The answer to that is that if a continuous variable calculation can be ... 0 OP's comment above is correct. The signal S(t) in Eq. (8) comes from \hat{c}^{PM} in (7b). The connection with the instantaneous frequency of the cavity is as follows. Because the probe light is on resonance it collects zero phase shift as it passes through the cavity. However, as the mechanical oscillator oscillates, because of the optomechanical ... 0 @Vadim has given a good answer that certainly goes in the right direction. However, the formalism they refer to is still for closed quantum systems. Even though it may be applicable or extendible to open quantum systems, this constitutes a non-trivial task. I would like to fill this gap by the following review: L. M. Sieberer, M. Buchhold, S. Diehl, ... 1 Well, let \hat{n}_j = \hat{a}^{\dagger}_j\hat{a}_j be the number operator, and remember the commutation relations, the only non vanishing being:$$ \left[ \hat{a}_i , \hat{a}_j^{\dagger}\right] = \delta_{ij}, $$Now let's use this to work the first term to write it in terms of the number operators. First of all, swap \hat{a}_j^{\dagger} and bring it in ... 0 There's a useful relation for commutators of products. It's quite easy to derive if you write out the product and add 0 (i.e add and take away the same term). It's [AB,C]=A[B,C]+[A,C]B there's also [A,BC]=B[A,C]+[A,B]C or you can just flip the first one and take the negative. There's a similar identity relating anti-commutators and commutators too 1 I necro because i think this part was not answered properly: A related question: Can we say that light transmitted and light absorbed by a linear polarizer have the same phase? Or is this question meaningless because we don't have information about the absorbed part? I assume you actually mean: "transmitted light when the polarizer is set to full ... 1 If the two inputs are not in phase, the answer is "no". If they are in phase, they can enter in orthogonal linear polarizations, and be combined to form a single mode at 45 degrees polarization. But if they are not in phase, the resultant polarization will vary randomly so the resultant will be mixed mode. 2 I think the question will be much more clear if you specify some of the remaining basis vectors, for instance the \vert{\uparrow 0}\rangle. I recommend to write the state as follows.$$\vert{i}\rangle=\dfrac{1}{\sqrt{2}}(\vert{\uparrow 0}\rangle_A\vert{\downarrow 0}\rangle_B+\vert{\downarrow 0}\rangle_A\vert{\uparrow 0}\rangle_B)$$Note that it lives in ... 1 If you send two optical pulses, in phase, into the two inputs of a 2:1 coupler, they will be combined into a single mode on the other side. If, however the two signals are out of phase, the two signals will not be combined. The best way to think about it is this. Lets say all outputs/inputs of your 2:1 coupler are identical single-mode waveguides, and let ... 1 I try to unravel the question as far as I know about the topic. I try to answer the question of when evolution is markovian or not. We recall that in general a quantum evolution is described by a one-parameter family of dynamical maps \Phi_t which are CPT (completely positive and trace-preserving) maps from the set of states. At this point, to define ... 1 Start from the definitions D(\alpha)\equiv\exp(\alpha a^\dagger - \alpha^* a) and S(\xi)\equiv\exp[\frac12(\xi a^{\dagger 2}-\xi^* a^2)]. Remember the identity$$e^A B e^{-A}=e^{\operatorname{ad}_A}B\equiv \sum_{k=0}^\infty\frac{1}{k!}[\underbrace{A\cdots A}_k,B],$$with [\underbrace{A\cdots A}_k,B]\equiv \operatorname{ad}_A^k B denoting the iterated ... 1 You are probably having problems with bases and representations. For example, take your spectacularly unnormalized formal "state"$$ \rho = a^\dagger a = \hat N = a a^\dagger -1\!\! 1= \rho_A, $$so the subscript A means we have rewritten the operator in the anti-normal-ordered representation, with the creators on the right. The two forms are strictly ... 1 There is rich literature on using Green's functions for non-equilibrium systems - this typically implies resorting to the Keldysh formalism, see here for the list of sources. In this list it is worth paying special attention to the series of papers by Meir, Wingreen and Jauho - these focus on excluding the external degrees of freedom from a Green's function ... 0 Certain open system processes can be described within the Greens function formalism by adding imaginary terms to the Hamiltonian, namely all processes where something only exits the open system and nothin comes back from the environment. So, basically decay processes. Other non-unitary processes such as dephasing cannot be directly accounted for by a Green ... 2 I would guess this has to do with the fact that a^{\dagger} a = N is the photon number operator which is such that N|n> = n|n>. Assuming that a = (a^{\dagger})^{\dagger}, and that a^{\dagger}|n> = \alpha_n |n+1>, you can find the prefactor using that relation. Note that you could also define another operator a' such that a'^{\dagger}|... 0 The drive is just a coherent monochromatic light source with angular frequency  \omega_{d} . The cavity can be thought of selective amplifier which only allows discrete spectrum of frequencies to sustain, given by  \lambda_{n} = \frac{2L}{n}  which is nothing but the allowed modes of standing waves in cavity of length  L . Waves having any other ... 3 The coherent drive represents an electromagnetic field interacting with the light/atom in the cavity. Say you have a drive (electromagnetic field):$$E\propto \epsilon b +\epsilon^*b^\dagger$$and the light in the cavity has creation/annihilation operators a, a^\dagger. Then, the interaction between the drive and the light in the cavity would be ... 0 Interaction terms H_\mathrm{int} in the Hamiltonian induce state-to-state transitions at some random moments of time. Considering the time evolution of the system, for example, by means of S-matrix$$ \hat{T}\exp\left\{-\frac{i}\hbar\int\limits_0^t H_\mathrm{int}(t')dt'\right\}=\sum_{n=0}^\infty\left(-\frac{i}\hbar\right)^n\int dt_1\ldots dt_n\:\hat{T}H_\...

1

This will not qualify as a full answer, but it is too long for comments. The approximation of frequency-independent coupling is widely used in tunneling problems, where it is called broad band limit. Typically one deals with a tunneling rate like $$\Gamma = 2\pi|V(E)|^2\rho(E),$$ where the matrix element $V(E)$ and the density-of-states $\rho(E)$ are energy ...

2

Manufacturing. The laws of quantum physics don't exclude a perfect unitary transformation. Of course, any real device will have finite losses. (On a different note, I would not call this "measurement" - there is no information obtained. It is rather decoherence.)

0

There is already a nice answer but I feel that some important aspects deserve additional attention. My answer is simply a list of observations: Master equations involve approximations: It is intuitive that the tracing out procedure that kicks out the bath to give you a Master equation comes at a loss of generality. Typical approximations include the bath ...

Top 50 recent answers are included