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In the introduction to continuous-variable quantum computing by Strawberry Fields (Xanadu), it lists the primary CV gates (rotation, displacement, squeezing, beamsplitter, cubic phase) along with their unitary:

cv gates w/ unitary

What are the matrix representations of these gates?

user820789
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    You usually don't want to use matrix representations, since working with transformations in terms of creation/annihiliation operators, and acting with them on the latter, is usually the far more powerful and easy to deal with formalism. – Norbert Schuch Jul 13 '18 at 20:52
  • @NorbertSchuch Thanks for the comment. New to CV & still (obviously) have much to learn! Creation/annihilation operators definitely interest me. Do you have any good suggestions on learning more? – user820789 Jul 13 '18 at 20:55
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    Start learning basic quantum mechanics, then the harmonic oscillator, then the 2nd quantization description. Pretty much any introductory quantum mechanics book is fine. In principle, one can also learn this purely algebraically (without knowing "normal" QM) but I'm not sure if there is any good reference, or whether that is even a good idea from a physics education point of view. (Clearly, knowing more about your background would help.) – Norbert Schuch Jul 13 '18 at 20:58

2 Answers2

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Background

Often, in quantum optics, the Heisenberg picture is used, where instead of considering equations of motion of states, equations of motions of operators are looked at instead. When considering creation/annihilation operators, this is often considerably easier as the matrices that determine the evolution (assuming it can be written in terms of matrices) are, for one, finite.

The Heisenberg equations of motions are calculated using $$\frac{dA}{dt} = \frac i\hbar \left[H, A\right] + \frac{\partial A}{\partial t},$$ for an operator $A$ evolving under a Hamiltonian $H$.

Here, the operator $a_j \left(a_j^\dagger\right)$ is the annihilation (creation) operator for spatial mode $j$. For a single mode, this allows for an effective Hamiltonian (acting on the operators) to be written as $$i\frac{d}{dt}\begin{pmatrix}a \\ a^\dagger\end{pmatrix} = H_{\text{eff}}\begin{pmatrix}a \\ a^\dagger\end{pmatrix}.$$ This naturally extends to writing their transformation as $$\begin{pmatrix}b \\ b^\dagger\end{pmatrix} = M\begin{pmatrix}a \\ a^\dagger\end{pmatrix}$$ for input modes $a \left(a^\dagger\right)$ and output modes $b \left(b^\dagger\right)$.

When it exists, this transformation matrix $M$ can be calculated using $U^\dagger A_jU = \sum_{k}M_{jk}A_k$ for unitary evolution $U$. For the Unitaries in the question, this gives:

Transformations

Displacement $$D_j\left(\alpha\right):\begin{pmatrix}b_j\\b_j^\dagger\\I\end{pmatrix} = \begin{pmatrix}1 && 0 && \alpha\\0&&1&&\alpha^*\\0&&0&&1\end{pmatrix} \begin{pmatrix}a_j\\a_j^\dagger\\I\end{pmatrix}$$

Rotation $$R_j\left(\phi\right):\begin{pmatrix}b_j\\b_j^\dagger\end{pmatrix} = \begin{pmatrix}e^{-i\phi} && 0\\0&&e^{i\phi}\end{pmatrix} \begin{pmatrix}a_j\\a_j^\dagger\end{pmatrix}$$

Squeezing $$S_j\left(\xi=re^{i\theta}\right):\begin{pmatrix}b_j\\b_j^\dagger\end{pmatrix} = \begin{pmatrix}\cosh r && -e^{i\theta}\sinh r\\-e^{-i\theta}\sinh r&&\cosh r\end{pmatrix} \begin{pmatrix}a_j\\a_j^\dagger\end{pmatrix}$$

Beamsplitter $$B_{jk}\left(\zeta = te^{i\varphi}\right):\begin{pmatrix}b_j\\b_k\\b_j^\dagger\\b_k^\dagger\end{pmatrix} = \begin{pmatrix}t && re^{-i\varphi}&&0&& 0\\re^{i\varphi}&&t&&0&&0\\0&&0&&t&&re^{i\varphi}\\0&&0&&re^{-i\varphi}&&t\end{pmatrix} \begin{pmatrix}a_j\\a_k\\a_j^\dagger\\a_k^\dagger\end{pmatrix},$$ where $r=\cos\left|\zeta\right|$.

Cubic Phase

Unfortunately, this is too nonlinear to write in the above way in matrix form. As $x = \frac{1}{\sqrt 2}\left(a+a^\dagger\right)$, even to first order, $V^\dagger a^\dagger V$ will include terms such as $\left[a^3, a^\dagger\right] = 3a^2$, which cannot be written in terms of $\alpha a^\dagger+\beta a+\gamma$.

Mithrandir24601
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  • @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? – user1271772 No more free time Aug 02 '18 at 09:20
  • @user1271772 for that example, it's $b_j = D^\dagger a_j D$ (and the others follow in the same way) – Mithrandir24601 Aug 02 '18 at 10:48
  • 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. – user1271772 No more free time Aug 02 '18 at 19:19
  • What do you mean by "too nonlinear"? It is simply nonlinear, that's why it doesn't fit the framework. – Norbert Schuch Aug 02 '18 at 20:44
  • @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) – Mithrandir24601 Aug 02 '18 at 21:16
  • @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. – Norbert Schuch Aug 02 '18 at 21:36
  • @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 – Mithrandir24601 Aug 02 '18 at 21:44
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    @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. – Norbert Schuch Aug 02 '18 at 21:47
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The link you gave says:

The CV model is a natural fit for simulating bosonic systems (electromagnetic fields, harmonic oscillators, phonons, Bose-Einstein condensates, or optomechanical resonators) and for settings where continuous quantum operators – such as position & momentum – are present.

Which means you can have many different different matrix representations for the CV gates. They then point out:

The most elementary CV system is the bosonic harmonic oscillator.

This means that for any values of the scalar (non-matrix) parameters $\alpha, \gamma, \phi, z, \theta, \gamma$, you can just calculate the formula they gave you, using the following matrix representations for the creation and annihilation operators for a bosonic harmonic oscillator:

enter image description here

The number operator $\hat{n}$ is just $a^\dagger a$.

Keep in mind that any matrix representation is basis-dependent, meaning that you can take these matrix representations and (for example) diagonalize them, and they would be a perfectly valid matrix representation in a new basis. However the matrices I gave you here are quite "standard" for quantum harmonic oscillators.

  • Would it be possible to generalize the "many many, different different" representations somehow (primarily interested in squeezed states)? – user820789 Jul 13 '18 at 00:37
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    What you have here is the matrix representation for the CV gates of a bosonic harmonic oscillator. What else do you want the matrix representation for? A particular Bose-Einstein condensates? What do the creation and annihilation operators look like? – user1271772 No more free time Jul 13 '18 at 11:46
  • "What do the creation and annihilation operators look like?" I may make this a new question. I would like the matrix representation for the squeeze gate. I can also make that it's own question. – user820789 Jul 13 '18 at 14:41
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    But @meowzz, there will be many matrix representations depending on what type of CV system you're considering. For example above we have the answer for the case where the CV system is a harmonic oscillator. If it's an anharmonic oscillator, it will be a different matrix entirely! CV quantum computing is not the same as "standard QC" where gates like CNOT have a simple matrix representation. – user1271772 No more free time Jul 13 '18 at 15:53
  • Mods: please don't delete above comments, since they are part of the answer – user1271772 No more free time Jul 13 '18 at 17:34
  • The statement that this is different for different CV systems is plain wrong. $a$ and $a^\dagger$ always have the same meaning & properties, and thus always the same matrix representation (up to a basis choice), and thus their exponentials in the question above again have the same representation. – Norbert Schuch Jul 14 '18 at 08:10
  • @Norbert: is "a" the same for a Morse oscillator as for a Harmonic oscillator? – user1271772 No more free time Jul 14 '18 at 13:51
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    What do you mean by "the same"? All what matters are the algebraic properties of $a$, namely $[a,a^\dagger]=1$. Everything can be derived from there, including the representation theory on Fock spaces. So yes, everything touched upon by the question, and all the math, is the same. That's the power of the algebraic approach, and one should use it rather than makeing things look more complicated than they are. – Norbert Schuch Jul 14 '18 at 14:15
  • @NorbertSchuch: The question asks for the matrix representation. the creation operator for a Morse oscillator does not have the same matrix representation as for the harmonic oscillator. The matrices are different. the OP wants a matrix. I wrote that they have to be careful because the matrix depends on the basis. But the matrix also depends on whether we have a harmonic oscillator or a Morse oscillator or some other system. I gave the matrix representation for a harmonic oscillator because it's easy and it's very common. – user1271772 No more free time Jul 14 '18 at 18:18
  • Well, I guess we can agree we disagree. – Norbert Schuch Jul 14 '18 at 20:59
  • It looks like you are thinking of a as the result of writing the Hamiltonian as something like $Q^\dagger Q$ and calling that $a$. That is not the same as $a$ defined purely from two canonically conjugate variables (without even knowing the Hamiltonian). Of course $Q$ has different commutation relations. – AHusain Jul 16 '18 at 16:29
  • @AHusain I fully concede that for a non-harmonic oscillator, the Hamitonian is not $a^\dagger a +1/2$. But the question asks about operators expressed in terms of $a$ and $a^\dagger$, which are always the same and thus have identical representations. – Norbert Schuch Jul 16 '18 at 21:13
  • I was referring to @user1271772, I was guessing the misunderstanding they might have in terms of that Morse oscillator comment. I should have tagged. – AHusain Jul 16 '18 at 22:04
  • @AHusain: I don't think I was thinking of $Q^\dagger Q$ as $a$. I don't think I know what your Q's are but if they are position I was thinking of $Q = a^\dagger + a$ for the QHO (quantum harmonic oscillator). I think the matrix representation of these gates (which is what the question asks), can be obtained for a QHO by simply substituting the matrices I gave into the formulas for each gate. However the system doesn't have to be a QHO, it could be a Morse oscillator for example, and then the matrix representation would be different. – user1271772 No more free time Jul 16 '18 at 23:45
  • @user1271772 Oh, you were implicitly using the basis that comes from the eigenstates of the Hamiltonian you were thinking of. Yes, the matrix representation changes, only because the basis changed. Even if you have a Morse oscillator, you can write everything in terms of the harmonic oscillator basis, but now I see that's not what you were saying. Okay. – AHusain Jul 17 '18 at 00:09
  • @AHusain Isn't the Fock basis just the natural basis for any pair of creation/annihilation operators satisfying the CCR, and thus any system described by such operators would have the same representation in the (canonical) Fock basis? – Norbert Schuch Jul 20 '18 at 15:26