# Tag Info

5

$\mathrm{X}$ is not equivalent to a $\mathrm{CNOT}$ gate. The former is a 1-qubit gate whereas the 2nd is a 2-qubit gate (in essence, a controlled-$\mathrm{X}$). The $\mathrm{X}$ basically flips the state of qubit B i.e., $|0\rangle_B\to|1\rangle_B$ and $|1\rangle\to|0\rangle_B$, and does not depend on the state of qubit A.

4

The Hadamard gate is: $$\frac{1}{\sqrt 2} \left(|0\rangle \langle 0 | + |0\rangle\langle 1| + |1\rangle \langle 0| - |1\rangle \langle 1|\right)$$ And since $|+\rangle$ is $\frac{1}{\sqrt 2}\left(|0\rangle + |1\rangle \right)$, you can work out that $H(|+\rangle) = |0\rangle$ So, $$CNOT(H|+\rangle \otimes |+\rangle)$$ $$= CNOT(|0\rangle \otimes |+\... 3 First suggestion: don't try it on a Bell state! This is already in the Schmidt basis, and lots of things that you do to the state will keep in Schmidt decomposed as well! Instead, why not try an example such as$$ |\psi\rangle=\frac{3}{5\sqrt{2}}|00\rangle+\frac{3}{5\sqrt{2}}|01\rangle+\frac{4}{5\sqrt{2}}|10\rangle-\frac{4}{5\sqrt{2}}|11\rangle $$So, as ... 3 As you wrote at point 3, CNOT is the sum of two tensor products, each one involving two matrices. Consider the first tensor product and apply the two matrices respectively to |0\rangle and to |0\rangle (the |00\rangle pair in your state obtained at 2.) and perform the tensor product, so you get the first quarter of the final expansion. Apply the same ... 3 In the teleportation protocol, the two parties share the entangled Bell State and it is implemented via a CNOT gate between the state to be sent (suppose Alice is sending the state) and the part of entangled Bell state Alice has. The CNOT gate creates another entangled state whose measurement Alice will send to the second party, say it's Bob, to perform the ... 3 Perhaps the following diagram helps to show what acts where. I think you've got a little muddled. 3 No, you wouldn't find 0.9 again. To make the partial trace calculation simpler you can note that the state |\psi'\rangle is separable under the bipartition a_1b_1 | a_2 b_2, i.e. |\psi'\rangle = |00\rangle \otimes (\sqrt{a} |00\rangle + \sqrt{1-a} |11\rangle). So irrespective of the value of a we have \operatorname{Tr}_{a_2b_2}[|\psi'\rangle\... 3 If we measure states |\Psi^-\rangle and |\Psi^+\rangle in computational basis, both look identical: if one qubit is measured |0\rangle, the other is measured |1\rangle; if one qubit is measured |1\rangle, the other is measured |0\rangle. Similarly, if we measure states |\Phi^-\rangle and |\Phi^+\rangle in computational basis, both look ... 3 Simply start by writing out everything$$ |B_{00}\rangle_{13}|B_{00}\rangle_{24}=\frac12\left(|00\rangle_{13}|00\rangle_{24}+|00\rangle|11\rangle+|11\rangle|00\rangle+|11\rangle|11\rangle\right) $$Let me rearrange each of these terms$$ \frac12\left(|00\rangle_{12}|00\rangle_{34}+|01\rangle|01\rangle+|10\rangle|10\rangle+|11\rangle|11\rangle\right). $$Now ... 3 The most direct way to do this using normal notation is to simply write the Bell projections using the same convention for subscripts:\def\ket#1{\lvert#1\rangle}\def\bra#1{\langle#1\rvert}\def\idop{\mathbf 1}$$\begin{aligned} \bra{\Phi^+}_{1,5} \;&=\; \tfrac{1}{\sqrt 2}\Bigl(\,\bra{0}_1\bra{0}_5 \,+\, \bra{1}_1 \bra{1}_5\,\Bigr), \\ \bra{\Phi^-}_{1,...

3

Q1) The qubits in the $|\Phi^+\rangle$ state are entangled - this means that (by definition) you can not represent the state of one of them individually without talking about the second one (mathematically this would be represented as tensor product of two single-qubit states). The best description of the individual qubits received by Alice and Bob is that ...

3

The four Bell states are $$|\Phi_{\pm}\rangle=(|00\rangle\pm|11\rangle)/\sqrt{2}\qquad |\Psi_{\pm}\rangle=(|01\rangle\pm|10\rangle)/\sqrt{2}.$$ So, let's consider what happens then we try and measure in the Bell basis, i.e. project onto one of these four states. If we started with the state $|00\rangle$, then we can write it as $$|00\rangle=\frac{1}{\sqrt{... 3 There is a more direct characteristic that makes the state of an entangled pair of qubits distinct from a non-entangled pair (which is also known as a separable state). When two qubits are not entangled, the state of the one qubit can be described without any knowledge of the state of the other qubit. Mathematically we can write: \begin{equation} |\psi\... 3 Suppose you have 2-qubit state |\psi\rangle_{AB}. Entangled or not, you can always write it as$$|\psi\rangle_{AB}=a_0|0\rangle_A|\psi_0\rangle_{B}+a_1|1\rangle_A|\psi_1\rangle_{B}$$If you measure qubit A in state |0\rangle, then the qubit B is in state |\psi_0\rangle, and you can compute the probability of qubit B being in state |0\rangle as ... 3 I find it a little tough to understand your calculations directly. I am especially confused by the circuit diagrams in your question; why they are there and what you are using them for. If you are performing calculations on theoretical data (without noise), then I feel you can make do with an easier approach for quantum state tomography. As per my answer on ... 3 In case you are interested in an alternative proof... For a state |\psi\rangle as given, the reduced density matrices of each of the three qubits must all be of the form \sum_i\lambda_i|i_X\rangle\langle i_X|, where the basis |i_X\rangle is orthonormal. That means that all three density matrices have the same spectrum (eigenvalues \lambda_i). Now, ... 2 Any scheme of back-in-time information transfer based on entangled particles runs up against the no-signalling theorem. Initially it's not clear how you conclude that "they do not need to compare their results with Victor, in order to distinguish between the cases when Victor entangles his photons or measures them independently." For example, in your ... 2 From my understanding of what you are asking, you may take the product of two depolarization operations, using the reduced density matrix of each qubit in the Bell state in the expression. Let's denote our two qubits as \mathrm{A} and \mathrm{B}. The Bell state of these two qubits is then:$$ |\beta_{00} \rangle =\frac{|0 \rangle_\mathrm{A} \otimes |0 \...

2

I think your reasoning is just fine and I checked that the Bell states are indeed eigenvectors of $$M=U^\dagger(Z\otimes Z)U,$$ as \begin{align} M|\phi^+\rangle =& \phantom{{}-{}}|\phi^+\rangle,\\ M|\phi^-\rangle =& -|\phi^-\rangle,\\ M|\psi^+\rangle =& -|\psi^+\rangle,\\ \text{and } M|\psi^-\rangle =& \phantom{{}-{}}|\psi^-\rangle \end{... 2 I guess the way that I'd start (aside from just getting a computer to do it!) is to remember that the Bell states form an orthonormal basis. So, you can ask, for example, about what the |\Phi^+\rangle^{AD} component is: \langle\Phi^+|^{AD}|\Phi\rangle^{ABCD}=-\frac12|\Phi^+\rangle^{BC}. $$You do this for each of the four states, and you can use that to ... 1 I will answer the question in a different way. Let's assume your two qubits are represented as below. Consider your |00\rangle, represented by |q_1q_0\rangle, is the state of above circuit before applying any gates, this is same as your original state |0\rangle \otimes |0\rangle. Now applying hadamard in your step 2 is same as applying hadamard on ... 1 I would recommend to use a direct matrix representation. An input state |00\rangle can be writen as vector$$x= \begin{pmatrix} 1\\0\\0\\0 \end{pmatrix} $$First step, i.e. Hadamard gate on first qubit and "nothing" on second qubit is described by operation$$ H \otimes I = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}...

1

As also discussed here you can write your channel as $$\mathcal E = p\mathcal E_{dp} + (1-p) \operatorname{Id}, \quad \mathcal E_{dp}(\rho)\equiv\operatorname{Tr}(\rho)I/d.$$ where $p\in[0,1]$ and, in your case, $d=2$. You thus have \mathcal E\otimes\mathcal E=p^2 \mathcal E_{dp}\otimes \mathcal E_{dp} + p(1-p) [\mathcal E_{dp} \otimes \operatorname{Id} + ...

1

So the Bell states are states of entanglement this means that the state of qubit one is now correlated to the state of qubit two, if you measure one you know information about the other. If you have a two qubit system, for which both qubits are in indepent superpostion, i.e $|q_1\rangle = |q_2\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle$, so both ...

1

There is no measurement in the picture; the picture shows how to construct 4 Bell states $|\Phi^+\rangle$, $|\Phi^-\rangle$, $|\Psi^+\rangle$ and $|\Psi^-\rangle$ using Hadamard-CNOT circuit.

1

Generally, if you want to inverse a quantum algorithm you simply read it from other side and replace all quantum gates with their inverses, i.e. conjugate transposed operators. So, if your original algorithm is described by series of gates $A_{n}A_{n-1}\dots A_2A_1$, the inverse algorithm is $A_{1}^\dagger A_{2}^\dagger \dots A_{n-1} ^\dagger A_{n} ^\... 1 I have found my mistake, and hopefully it may help more people. It lies on the fact that$(AB)^\dagger=B^\dagger A^\dagger$. Having done so, my result is as expected. But any more enlightening answer is more than welcome. 1 It is not true that$ |00 \rangle + | 11 \rangle = |\phi_0 \phi_0 \rangle + | \phi_1 \phi_1 \rangle$for an arbitrary change of basis. For example, let$|\phi_0 \rangle = |0\rangle $and$ |\phi_1 \rangle = i|1 \rangle $. You can, however, pick$ \phi_1 $so that this equality is true, given$ \phi_0 $. If$ |\phi_0 \rangle = \kappa | 0 \rangle + \mu | 1 \...

Only top voted, non community-wiki answers of a minimum length are eligible