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8 votes
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How are two photons entangled in the Xanadu computer?

QML researcher at Xanadu here. Our X-series chip produce entangled states by squeezing light and then combining it at beam splitters: those 'cables' are waveguides in a chip, which when they are close ...
Ziofil's user avatar
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6 votes
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What are spin-coherent states?

These are also known as SU(2)-coherent states; one original reference is https://doi.org/10.1103/PhysRevA.6.2211. In a spin system, with states labeled by the eigenvalue of the total angular momentum ...
Quantum Mechanic's user avatar
5 votes
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Why does joint ground state not change under action of beam splitting unitary operator?

Calculate $$ \begin{align} \hat{U}|00\rangle &= \exp\left(-igt(\hat{a}^\dagger_2\hat{a}_1+\hat{a}^\dagger_1\hat{a}_2)\right)|00\rangle \\ &= \sum_{k=0}^\infty \frac{(-igt)^k}{k!}(\hat{a}^\...
Adam Zalcman's user avatar
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4 votes

How is this expression for a GHZ state obtained in the nature paper by Pan et al. (2000)?

First, in order to express $|H'\rangle$, $|V'\rangle$, $|R\rangle$ and $|L\rangle$ in terms of $|H\rangle$ and $|V\rangle$, add and subtract the pair of equations $(2)$ and add and subtract the pair $(...
Adam Zalcman's user avatar
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4 votes
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Why is a state Gaussian if and only if its covariance matrix satisfies $\boldsymbol\sigma+i\boldsymbol\Omega\ge0$?

From the definition of the covariance matrix, $${\sigma}_{ij}=\left\langle \frac{x_i x_j+ x_j x_i}{2}\right\rangle -\langle x_i\rangle\langle x_j\rangle,$$ where we define $$\boldsymbol{x}=(q_1,p_1,\...
Quantum Mechanic's user avatar
4 votes

Why does joint ground state not change under action of beam splitting unitary operator?

There's more than one way, and I'll suggest two of them here: Expand $\hat{U}$ using the formula for the Taylor series of an exponential ($e^\hat{A}$) centered around $\hat{A}=\hat{0}$, and then you ...
user1271772 No more free time's user avatar
3 votes
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Trace distance between thermal states

Since the multimode Fock states $|\mathbf{n}\rangle=|n_1\rangle\otimes |n_2\rangle\otimes\cdots\otimes |n_M\rangle$ form an orthonormal basis, one can treat each state by their probability ...
Quantum Mechanic's user avatar
3 votes

Comparing technical limitations of major quantum computing approaches

For a general overview about hardware and the difficulties they present, I recommend section three of Quantum Computing: An Overview Across the System Stack. Another good introduction would be Quantum ...
epelaez's user avatar
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3 votes
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How is measurement performed on a stream of polarized photons?

One idea is to do polarimetry. By using a polarizing beam splitter, the polarization qubit can have each of its polarization components directed to a different detector for photon counting (ideally a ...
Quantum Mechanic's user avatar
3 votes
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Relation between Wigner quasi-probability distribution and statistical second-moment

You mean something like $$W_{G}(\mathbf{r}) =\frac{2^{n}}{\pi^{n} \sqrt{\operatorname{Det} \sigma}} \mathrm{e}^{-(\mathbf{r}-\overline{\mathbf{r}})^{\top} \boldsymbol{\sigma}^{-1}(\mathbf{r}-\overline{...
keisuke.akira's user avatar
2 votes
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How do we restrict to a limited number of dimensions, say 3 for qutrits, while using OAM states of light?

As for any platform, one has to choose a suitable $d$-dimensional "computational" subspace. Suitability depends on your application, but generally it means that one should be able to perform ...
Markus Heinrich's user avatar
2 votes

Shor's Code: Understanding how it satisfies Knill Laflamme Theorem

Strictly, what you have to calculate is that for all $i$ and $j$ $$ \langle 0_L|U_iU_j|1_L\rangle=0 $$ and $$ \langle 0_L|U_iU_j|0_L\rangle=\langle 1_L|U_iU_j|1_L\rangle. $$ (I've ignored the ...
DaftWullie's user avatar
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2 votes

Why does joint ground state not change under action of beam splitting unitary operator?

Let $|\psi\rangle$ be an eigenstate of an operator $A$, $A|\psi\rangle=\lambda|\psi\rangle$. Then $$e^A |\psi\rangle = \sum_{k=0}^\infty \frac{A^k}{k!}|\psi\rangle = \sum_{k=0}^\infty \frac{\lambda^k}{...
glS's user avatar
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2 votes
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How to write the covariance matrix of a quantum gaussian state as a function of photon numbers?

Recall that $a=(q+ip)/\sqrt{2}$ in some dimensionless units (Weedbrook might change the units because I think they like $\hbar=2$; I'm using $[q,p]=i$ and $[a,a^\dagger]=1$). We can thus find the ...
Quantum Mechanic's user avatar
2 votes
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What are the antisymmetric terms in $\sigma_{mn}$ in the expression for the Fisher information?

Antisymmetric means that $\sigma_{nm}=-\sigma_{mn}$. Since the sum ranges over all values of $m$ and $n$, adding an antisymmetric term adds something proportional to $$|\langle \psi_m^\lambda|\...
Quantum Mechanic's user avatar
2 votes
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Does the covariance matrix of $\rho_1-\rho_2$ have a simple expression in terms of the individual covariance matrices?

There are many things being overlapped here that likely have no business together. A Gaussian state is uniquely determined by its covariance matrix and its displacement vector, any linear combination ...
Quantum Mechanic's user avatar
2 votes
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How do the Type I and Type II fusions work?

NOTE: I've seen this line of reasoning and those calculations somewhere, but I don't remember where. Here is my reconstruction of them. Type I: Let's have in mind the sketch of fusion type I from the ...
David Dentelski's user avatar
2 votes
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Are there resources to learn about quantum optics that focus on math and/or are addressed to those familiar with the qubit formalism?

There are a number of rather well regarded review papers, such as: https://doi.org/10.48550/arXiv.quant-ph/0503237, https://doi.org/10.1016/j.physrep.2007.04.005, https://doi.org/10.1103/RevModPhys.84....
Quantum Mechanic's user avatar
1 vote

Hong Mandel Effect - question on math of indistinguishability of photons

Your state is perfectly normalized if you use that $(a^\dagger)^2|0\rangle = \sqrt{2}|2\rangle$ (where $|0\rangle$ and $|2\rangle$ are normalized states): Its norm is $$ (t^2-r^2)^2 + (\sqrt{2}tr)^2 + ...
Norbert Schuch's user avatar
1 vote

How to compute a partial trace of the form ${\rm Tr}_S [s^\dagger b- s b^\dagger, \vert \psi \rangle \langle \psi \vert \otimes \rho_B]$?

This can be computed directly by using the partial trace $Tr_S(\rho)=\sum_j\langle \omega_j|\rho|\omega_j\rangle$. I will drop the subscripts S and I and use the labels $\omega$ and $v$ to specify the ...
Quantum Mechanic's user avatar
1 vote
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Two-photon N00N state through Mach-Zehnder interferometer

There are some corrections to the calculations, but it is more important to focus on the motivation: why are you sending a NOON state through a Mach-Zehnder interferometer? The idea of the NOON state ...
Quantum Mechanic's user avatar
1 vote

Nielsen and Chuang: Solving equation of motion for amplitude damping

Think about how the density matrix is represented with the Bloch vector: $$\tilde{\rho} = \frac{1}{2}(I + \vec{n} \cdot \vec{\sigma})$$ Then $$\frac{\mathrm{d}\tilde{\rho}}{\mathrm{d}t} = \dot{\vec{n}}...
physicsprune's user avatar
1 vote

Diagonal elements of the transition dipole moment

The diagonal elements of the Hamiltonian give you information about the energy of a certain configuration. Not about the transitions. For example, let's consider the Pairing Hamiltonian used in the ...
Andrés Ruiz's user avatar
1 vote
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Does the entanglement have the conventional meaning in optical interferometers?

Entanglement is entanglement, there's just one definition of it (well, there's more complex variations like genuine multipartite entanglement etc, but that's not relevant here). However, (bipartite) ...
glS's user avatar
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1 vote

What are the measurement operators $F_k$ corresponding to a homodyne measurement?

One thing that I feel is worth noting is that there are two ways to understand what "measurement" means in this context. On the one hand, you have "measurements" in the sense of ...
glS's user avatar
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1 vote
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Why do we use 'modes' in quantum optics?

Modes are governed by eigenfunctions, I agree. In quantum optics, we need more than just eigenfunctions to describe a state of light: we need to know how many photons have properties corresponding to ...
Quantum Mechanic's user avatar

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