# Tag Info

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I will give an utter matrix method, to be more clear, the notation $\cdot$ will stand for matrix multiplication. Mind that $$Tr\left( \left( M\otimes I \right) \rho ^{AB} \right) =\sum_{ij}{\langle ij|\left( M\otimes I \right) \rho ^{AB}|ij\rangle} \\ =\sum_{ij}{\left( \langle i| \right) \cdot \left( I\otimes \langle j| \right) \cdot \left( M\otimes I \right)... 1 If you run the following code you will see the output statevector plotted in Bloch sphere. qc = QuantumCircuit(1) qc.ry(3 * np.pi/4, 0) sim = Aer.get_backend('statevector_simulator') result = execute(qc, sim).result() output_state = result.get_statevector(qc) plot_bloch_multivector(output_state) Now from the picture you can see that the state lies near ... 2 This is basically up to you: which elements are you transposing? If you're talking about transposing just the third system, then you'd be talking about$$ |abc\rangle\langle xyz|\mapsto |abz\rangle\langle xyc| $$but you could do this on any of the three individual subsystems, or any of the three pairs of subsystems. Of course, if you're talking about doing ... 4 The equality follows by hitting both sides of$$ X_A\mathrm{tr}_B(Y_{AB})=\mathrm{tr}_B((X_A\otimes I_B)Y_{AB})\tag1 $$with the trace and setting X_A=M and Y_{AB}=\rho^{AB}. We can establish (1) for Y_{AB} of the form Y_{AB}=Y_A\otimes Y_B using$$ \mathrm{tr}_B(A\otimes B)=A\mathrm{tr}(B)\tag2 $$as follows$$ \begin{align} X_A\mathrm{tr}_B(Y_{...

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If you're tracing over two systems, $A$ and $B$, you can split this into two steps $$\text{Tr}(Q_{AB})=\text{Tr}\left(\text{Tr}_B(Q_{AB})\right)$$ (To see this, let the basis you use for taking the first trace be the standard basis, $|ij\rangle$. All I'm doing here is separating out the sums over $i$ and $j$.) So, if you let $Q_{AB}=\tilde M\rho^{AB}$, ...

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It's easy to see that it's true for $\rho^{AB} = A \otimes B$ for any matrices $A,B$ (even when $\rho^{AB}$ is not a state, but just a matrix). Any matrix (not only states) is a linear combination of such products, that is $\rho^{AB} = \sum_i A_i \otimes B_i$, where $A_i,B_i$ are some matrices. Thus $tr(M\rho^A)=tr((M\otimes I_B)\rho^{AB})$ since both sides ...

0

After the first Hadamard gate you have the state $\frac{1}{\sqrt{2}}(|000\rangle+|100\rangle)$ (qubit zero being the leftmost). After the controlled $O_1$ gate, you get $\frac{1}{\sqrt{2}}(|000\rangle+|111\rangle)$. And the next Hadamard leaves you with: \begin{align} |\psi\rangle &= \frac{1}{\sqrt{2}}\left(\frac{1}{\sqrt{2}}(|000\rangle+|100\rangle)+\... 2 Write the eigendecomposition of the state as \rho=\sum_k p_k u_k u_k^\dagger, where \{u_k\}_k is a family of orthonormal vectors in the underlying space. Suppose there is some u_k\notin \operatorname{supp}(\Pi), that is, some u_k such that \Pi u_k\neq u_k. Then\operatorname{Tr}(\Pi u_k u_k^\dagger)=\| \Pi u_k\|^2<\|u_k\|^2=1,$$and thus$$\...

6

Yes its true. Define another orthogonal projector $\Pi_\perp$ such that $\Pi + \Pi_\perp = I$ and write $\rho$ in terms of a spectral decomposition $$\rho = \sum_k \lambda_k(\rho) |\lambda_k\rangle \langle \lambda_k| \tag{1}$$ where we're fine to just consider components such that $\lambda_k(\rho) >0$. Then we have $$1 = \text{... 2 The claim does not specify what protocols for distinguishing quantum states are acceptable. In particular, it does not state whether we are allowed to err or reserve judgment. Below, we note success probability for protocols allowed to err and compute success probability for an error-free protocol. The success probability of the former is not bounded by 4\... 1 Yes, this is exactly what happens. Instead of saying "I want to measure this specific value," you specify a set of POVM elements asking "which of these values will I get?". When you are measuring the value of the first qubit alone to see if it is in state |0\rangle or |1\rangle, the two POVM elements can be specified by \Pi^{(0)}=|0\... 2 You don't state this explicitly, but I'm guessing this is the crucial part: How do |A_i\rangle relate to A? I assume that |A_i\rangle correspond to normalized rows, i of matrix A, while \|A_i\| is the weight of the row i such that \|A_i\||A_i\rangle would have all the elements corresponding to the i^{th} row of matrix A. In other words,$$...

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Summary: The expression you're looking for is: $$\frac{1}{4} \left[ (III + IZZ + ZIZ + ZZI) + (XXX - XYY - YXY - YYX)\right]$$ where Pauli string notation like $XYX$ denotes $\sigma_1 \otimes \sigma_2 \otimes \sigma_1$, for example. To start, we need to write the $n$-qubit GHZ state as an operator, namely |\psi^{(n)}\rangle\langle\psi^{(n)...

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A direct while brute-force way is by writing state $|\psi\rangle$ as density matrix $\rho$. Then by noticing that $\rho=\frac{1}{2^3}\sum_{ijk}t_{ijk}\sigma_{i}\otimes\sigma_{j}\otimes\sigma_{k}$ for $i,j,k=1,2,3,4$ stands for Pauli matrices and identity matrix, find all the coefficients $t_{ijk}$ by $Tr(\rho\sigma_{i}\otimes\sigma_{j}\otimes\sigma_{k})$. If ...

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We have, \begin{align}\begin{aligned}\newcommand{\th}{\frac{\theta}{2}}\\\begin{split}Ry(\theta) = \exp\Big(-i \th Y\Big) = \begin{pmatrix} \cos{\th} & -\sin{\th} \\ \sin{\th} & \cos{\th} \end{pmatrix}\end{split}\end{aligned}\end{align} If the input state is $|0\rangle$, then the probability of getting $|0\rangle$ as a ...

1

To build on the other answer, we can in fact characterise the set of coefficients that can fit into a convex decomposition of a density matrix. Given $\rho$, we can write $$\rho = \sum_k a_k |u_k\rangle\!\langle u_k|$$ for some set of (not necessarily orthogonal) states $\{|u_k\rangle\}$ if and only if $\mathbf a\preceq \boldsymbol{\lambda}(\rho)$, meaning ...

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Given a multipratite state $\rho$, its reduced density matrix is the partial trace with respect to some of its degrees of freedom. If $\rho$ is bipartite, then the partial trace with respect to its first degree of freedom is the matrix with elements $$[\operatorname{Tr}_1(\rho)]_{ij} = \sum_k \rho_{ik,jk}.$$ If you have a multipartite state (e.g. a three-...

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No. In the ensemble interpretation of a density matrix $$\rho=\sum_k p_k|\psi_k\rangle\langle \psi_k|$$ the states $|\psi_k\rangle$ are not necessarily orthogonal. Also, the probabilities $p_k$ are not necessarily eigenvalues. Ensembles vs eigendecomposition A density matrix $\rho$ encodes multiple ensembles $\{p_k, |\psi\rangle_k\}$, see theorem $2.3$ on ...

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That's an appropriate example, yes. The underlying reason this happens is that the set of entangled states is not convex, which roughly speaking means you sum entangled states and obtain (upon renormalisation) separable ones. As a slight generalisation of the example you already provide, you can consider any pair of entangled states with Schmidt ...

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Yes, that's correct, and your example is exactly what I had in mind after reading the question title. More generally, the four Bell states form a basis in the space of all 2-qubit states, so you can express any state using a linear combination of them, including all unentangled states.

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The relation $U\otimes I=I\otimes U$ is wrong, unless $U=\alpha I$ for some $\alpha\in\mathbb C$. You have $$(U\otimes I)|\Phi\rangle =\sum_i (U\otimes I)|i,i\rangle = \sum_i (U|i\rangle)\otimes |i\rangle = \sum_{ij} U_{ji} (|j\rangle\otimes|i\rangle) \\= \sum_{j} |j\rangle\otimes \left(\sum_i U_{ji}|i\rangle\right) = \sum_j |j\rangle\otimes(U^T |j\rangle) =... 1 Yes, both these answers should work. My guess is, the quiz author was looking for alternative answers that looked similar to the correct one but were incorrect, and accidentally got a second correct one instead. If there is a way to provide feedback on that page, you can try letting the authors know - I know I'm always happy when someone catches a bug that I ... 1 Explicit indices The difficulty here arises from making indices implicit in tensor product expressions. For example, the unitary U corresponding to controlled-NOT gate on qubits 1 and 2 and identity on qubit 3 is often written down as$$ U=\text{CNOT}\otimes I\tag1  but a similar unitary $U'$ corresponding to controlled-NOT on qubits $1$ and $3$ ...

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