# Tag Info

1

There's a few things that you should be careful not to confuse here : If the universe obeys laws, this isn't the same thing as the universe processing those laws in a computer-like fashion to produce itself. The laws that we ascribe to the universe aren't the same thing as what is actually happening or what actually exists, if such concepts even make sense ...

1

To answer this, I first go through via an example why normal entanglement does not allow faster-than-light communication, look at how the entanglement swapping scenario is similar to the normal case, see how witnesses work, and why a witness unfortunately cannot help in faster-than-light communication. I confess I write it this way to ensure it is clear to ...

1

Assuming spin basis, prepare an electron with spin up in the z direction using Stern-Gerlach setup. It should be in the desired state along x or y.

0

Writing the state in the form $|A\rangle=\sum_{ij}c_{ij}|ij\rangle$, you can associate to each set of coefficients a matrix $C\equiv (c_{ij})_{ij}$. In your case, this is $$C=\frac{1}{2\sqrt2}\begin{pmatrix}1 & \sqrt3 \\ \sqrt3 & 1\end{pmatrix}.$$ You want to find the SVD of this matrix. In this case this amounts to diagonalising it, because the ...

3

You are looking for the SWAP test. The basic idea is the following: First, the overlap $\omega=\mathrm{tr}[\rho\sigma]$ of two quantum states (e.g. the pure states in your question) is equal to $$\omega=\mathrm{tr}[(\rho\otimes\sigma)\mathbb F]\ ,$$ with $\mathbb F$ the SWAP gate. On the other hand, the expectation value of any unitary gate $U$ in a ...

0

You don't define $\hat H_{\Xi}$, but from the equations you write it seems that it's a Hamiltonian that only acts nontrivially on the second system, that is, $\hat H_\Xi\equiv I\otimes \hat H_\Xi$. The main step is the following: $$\exp\{-it[\hat{O}_{\Gamma} \otimes \hat{O}_{\Xi} + g \hat{H}_\Xi]\} \left( |\gamma \rangle \otimes |\Xi_D\rangle \right) = ... 5 The (nominal) advantage of Deutsch’s algorithm isn’t that you only need to measure one qubit. It’s that you only need to evaluate the function f (which might be slow to do) once, whereas in the classical version you need to evaluate it twice before plugging the outputs into the XOR. 1 It's the beamsplitter unitary (a.k.a. the QFT in two dimensions, a.k.a. the Hadamard gate), represented via its action on two-mode Fermion states. With two modes, there are four possible fermionic states: |11\rangle\equiv c_1^\dagger c_2^\dagger |\text{vac}\rangle (one fermion per mode), |01\rangle\equiv c_2^\dagger |\text{vac}\rangle (a fermion in the ... 2 The notation \vert0\rangle and \vert1\rangle denotes a mode with zero or one fermions, respectively. Then, \vert ij\rangle denotes two modes, where i denotes the number of fermions in the first mode and j the number of fermions in the second mode. So if you apply the fSWAP, what you swap is whatever is in the first and the second mode. But this ... 1 It is possible, if the non-diagonal element corresponds to a measurable quantity. Indeed, the expectation value of any operator is $$\langle A\rangle = \textrm{tr} (\hat{\rho}A).$$ Let us look at a single spin with density matrix \hat{\rho} = \begin{pmatrix} \rho_{11} & \rho_{12} \\ \rho_{21} & \rho_{22} \... 0 I can't connect to the site with the paper, but I don't think there's much to it. If you have two numbers \alpha,\beta > 0 such that \alpha^2+\beta^2=1, you can always find x\in\mathbb{R} such that$$ \alpha = \sqrt{\frac{1-\tanh(x)}{2}}, \qquad \beta = \sqrt{\frac{1+\tanh(x)}{2}} $$The formula for this x is$$ x= {\rm artanh}(1-2\alpha^2) = {\...

-1

Evaluate the partial trace over the environment to get the krauss operators

1

The OP pointed out some similarities in the Hamiltonian of the two cases. There is, however, a clear physical distinction between the two cases. Vacuum Rabi oscillations are induced by strong coupling of the transition the the light field/cavity mode. They are thus a property of the Hamiltonian (see this question for details and related regimes), where the ...

1

I am with this question in my mind too. I am not sure that my interpretation is right, but I will share it with you. If $M_{j}$ it's an observable from the basis$\left\{ M_{i}\right\}$, so: $$tr\left(M_{j}\rho^{A}\right)=tr\left(M_{j}f\left(\rho^{AB}\right)\right)$$ Then, using the expansion from $eq \left(2.183\right)$: tr\left(M_{j}\rho^{A}\right)=...

0

There are many proposals for using Majorana zero modes to encode a qubit; one of the most popular is called the hexon, discussed in this paper: https://arxiv.org/pdf/1610.05289.pdf. The main idea here is that a one-dimensional topological superconductor hosts Majorana zero modes at its ends. Two Majorana modes combine into a Dirac fermion, so the ground ...

1

The $G^{(n)}$ are the expectation values of the products of $2n$ of the $E(x_i)$ variables, since $\mathrm{Tr}(\rho X)$ is how one writes the expectation value for $X$ in a mixed state with density matrix $\rho$. In particular $G^{(1)}$ is the expectation value of a product of two random variables. Since we can usually set the expectation value of the fields ...

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