Timeline for Matrix representation of continuous-variable gates
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Aug 2, 2018 at 21:47 | comment | added | Norbert Schuch | @Mithrandir24601 The underlying process for squezzing is nonlinear (which involves a pump and so on). But the effective Hamiltonian you get in "linear optics". This is pretty much standard. "Gaussian" refers to the fact that exponentials of quadratic "things" (like Hamiltonians) are Gaussians (in the sense of the Gaussian distribution), which describes thermal states, time evolution, and so on, of these systems. | |
| Aug 2, 2018 at 21:44 | comment | added | Mithrandir24601♦ | @NorbertSchuch There seems to be different ways to use the words 'linear' and 'nonlinear' in quantum optics - personally, I've only ever heard of squeezing referred to as a nonlinear process but I wouldn't be surprised if there are people that disagree with that. I have heard it (and any quadratic term) referred to as a 'Gaussian' process but where the use of the word 'Gaussian' comes from here, I have no idea | |
| Aug 2, 2018 at 21:36 | comment | added | Norbert Schuch | @Mithrandir24601 You are right - in some way. However, in quantum optics, what is called linear optics is everything which can be described by linear + quadratic Hamiltonians in the creation/annihilation operators (pretty much for the reason you give in your comment). Thus, $H=a^\dagger a^\dagger + a a$ describes a linear optical process, only cubic and higher terms are non-linear optics. | |
| Aug 2, 2018 at 21:16 | comment | added | Mithrandir24601♦ | @NorbertSchuch Having a unitary that's an exponential of a term such as $aa - a^\dagger a^\dagger$ (squeezing) is nonlinear but expanding $S^\dagger aS$ gives linear terms (e.g. $\left[a^\dagger a^\dagger, a\right] = -2a^\dagger$, which is linear. However, generally if the term in the exponential is cubic (or higher order), the commutation relations that arise from expanding the exponential and multiplying everything together give nonlinear results, so I'm referring to the term being exponentiated as 'too nonlinear' (it's hard for me to describe in a comment, so this might deserve an edit) | |
| Aug 2, 2018 at 20:44 | comment | added | Norbert Schuch | What do you mean by "too nonlinear"? It is simply nonlinear, that's why it doesn't fit the framework. | |
| Aug 2, 2018 at 19:19 | comment | added | user1271772 No more free time | I'm now confused even more. I have no idea what's going on here. @meowzz, since you un-accepted my answer and accepted this one, I presume you understand what's going on? What is the matrix representation for D using this method? In my method you plug the matrices for $a$ and $a^\dagger$ that I gave, into the formula you gave for D, and voila. Contrarily in this answer, I don't see what the matrix is for D. Is it the 3x3 matrix? That gives bj = aj + alpha. I'm sorry I don't understand. Mithrandir2: I'm not saying your answer is wrong, I'm just trying to understand it. | |
| Aug 2, 2018 at 10:48 | comment | added | Mithrandir24601♦ | @user1271772 for that example, it's $b_j = D^\dagger a_j D$ (and the others follow in the same way) | |
| Aug 2, 2018 at 9:20 | comment | added | user1271772 No more free time | @Mithrandir2, how do these even relate to the operators in the original question? For example for D, you have bj = aj + alpha, how does this even relate to the form given in the original question? | |
| Aug 1, 2018 at 22:49 | vote | accept | user820789 | ||
| Aug 1, 2018 at 22:27 | history | answered | Mithrandir24601♦ | CC BY-SA 4.0 |