10 votes
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How are arbitrary $2\times 2$ matrices decomposed in the Pauli basis?

The Pauli matrices form an orthogonal basis of $\mathcal{M}_2$, this vector space can be endowed with a scalar product called the Hilbert-Schmidt inner product $$ \langle A,B\rangle=\mathrm{Tr}(A^\...
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9 votes
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Inverting the depolarizing channel

The existence of the inverse of a linear map is independent of the way the map affects the trace. Moreover, if an invertible map preserves a property then its inverse necessarily also preserves the ...
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  • 13.5k
8 votes
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Closeness of purifications of states

No dimension-independent bound is possible. Consider states $\rho_A$ and $\sigma_A$ that are close in $p$-norm (for $p>1$) but have relatively low fidelity. Specifically, assume $$ \|\rho_A - \...
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  • 4,538
8 votes
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Why is the transpose of a density matrix positive and trace preserving?

Transposing a matrix is trace preserving since for $\rho = \sum_{a,b} \rho_{a,b} | a \rangle \langle b |$: $$\text{Tr}(\rho)= \sum_c \langle c| \big( \sum_{a,b} \rho_{a,b} | a \rangle \langle b | \...
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7 votes
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Does $\mathcal E^{\otimes n}$ admit a more efficient Stinespring dilation than the one used for $\mathcal E$?

No. The minimal size of the environment is just the rank of the Choi matrix of $\mathcal E$, call it $J(\mathcal E)$. Since $J(\mathcal E^{\otimes n}) = \big(J(\mathcal E)\big)^{\otimes n}$ and $\text{...
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7 votes

Why is the transpose of a density matrix positive and trace preserving?

Trace preservation The trace must be preserved in the transpose of a matrix, because the trace is the sum of the diagonal elements. When transposing a matrix, you do not change the diagonals at all! ...
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7 votes

In Stinespring dilation, can we always use a mixed state as the ancilla?

No, that doesn't work. It's fine to use an arbitrary pure state because the unitary $U$ can always be used to take it to any pure state you want. This argument doesn't work for a mixed state, as ...
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7 votes
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How to express real matrices as linear combinations of unitaries?

You can select a basis of unitary matrices with respect to which you can decompose your matrix. For example, if your matrix $A$ is $2^n\times 2^n$, then you can select the Pauli basis $$ \sigma_y,\...
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  • 47.1k
6 votes
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How to show that Bell states are orthonormal

First, note that $|0\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ and $|1\rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$ and therefore $\langle 0 |1\rangle = \begin{pmatrix} 1 & 0 \end{pmatrix} \...
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  • 12.5k
6 votes
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What is the matrix for a SWAP operation on two qubits?

In the general case I think it's easier to consider the matrix in the form $$ M = \sum_{i_1,\dots,i_n, j_1, \dots j_n} c_{i_1,\dots,j_n} |i_1 \dots i_n\rangle \langle j_1 \dots j_n|, $$ where the $i_1,...
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  • 3,902
6 votes
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Why can any density operator be written this way? (quantum tomography)

From linear algebra, if $v_1, \dots, v_n$ is a basis of the vector space $V$ then every vector $v\in V$ can be written as a linear combination $$ v = a_1 v_1 + \dots + a_n v_n\tag1 $$ where the ...
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  • 13.5k
6 votes
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Does a basis of maximally entangled states exist for two-qubit or two-qutrit system so that the density matrices of the basis states don't commute?

No such (orthonormal) basis can exist. An orthonormal basis $\{|\psi_i\rangle\}$ requires $\langle \psi_i | \psi_j \rangle = 0$ for $i\neq j$, and so clearly \begin{align} [\rho_i, \rho_j] &= |\...
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  • 4,667
6 votes
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Does the controlled Pauli Z gate cause entanglement?

Yes, CZ gate can entangle its inputs. For example $$ CZ|+\rangle|+\rangle = \frac{|0\rangle|+\rangle+|1\rangle|-\rangle}{\sqrt2} $$ which is entangled because the reduced density matrix of either ...
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  • 13.5k
6 votes

Prove that $\rho_{AB} \leq |B|(\rho_A\otimes I_B)$ for any bipartite state $\rho_{AB}$

There's a very quick proof if you can use the properties of the Choi-Jamiołkowski isomorphism. Define a map that acts on subsystem $B$ as $$\Lambda(\rho_B) = \mathrm{Tr}(\rho_B) |B| I_B - \rho_B.$$ ...
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  • 86
6 votes
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What does $ A - \langle A \rangle $ mean?

For any vector $v$ and scalar $\alpha$, we have $\alpha v = \alpha I v$, so multiplication by a scalar $\alpha$ behaves the same way as the linear operator $\alpha I$. Therefore, we interpret $A - \...
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  • 13.5k
5 votes

How to show that Bell states are orthonormal

The most basic but laborious way of checking that Bell states are orthonormal is to carry out the calculations for all sixteen inner products such as $\langle\Phi^+|\Psi^-\rangle$. One way to do this ...
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5 votes
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How do I prove that $P_\pm=\frac12(1\pm U)$ if $U^2=I$?

First, we can start with $U = P_+ - P_-$, since the Hermitian is the sum of the projection operators of the eigenspaces scaled by their eigenvalues. If $U^2 = I$, that means $I = (P_+ - P_-)(P_+ - P_-)...
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5 votes
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Is there an identity for the partial transpose of a product of operators?

Your suspicion is correct, even when $A=B$. Consider the Hilbert space of two qubits and let $^{T_A}$ denote the partial transpose with respect to one of them. Suppose that $$ A=B=\begin{pmatrix} 1 &...
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  • 13.5k
5 votes
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Can we express $\mathrm{tr}_A((A\otimes B)\rho_{AB})$ in terms of $A$, $B$, $\rho_A$ and $\rho_B$?

In general, the knowledge of the marginals $\rho_A$ and $\rho_B$ and the operators $A$ and $B$ is insufficient to compute $\mathrm{tr}_A((A\otimes B)\rho_{AB})$. Indeed, we can find two different ...
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  • 13.5k
5 votes
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Cliffordness of the qutrit Hadamard gate

The answer is no. Define X=[[0,1,0],[0,0,1],[1,0,0]] Z=[[1,0,0],[0,w,0],[0,0,w^2]], w^3=1 Then the Pauli group is generated by X and Z and is of order 27. With H ...
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  • 920
5 votes
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Prove that $|(\langle \psi|_{A} \otimes \langle \phi|_{B})|\theta\rangle_{AB}|^{2}<1$ for entangled $|\theta\rangle_{AB}$

TL;DR: There is no need to use Schmidt decomposition. The non-strict variant of the inequality follows directly from Cauchy-Schwarz inequality and equality is ruled out because $|\theta\rangle_{AB}$ ...
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  • 13.5k
5 votes
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What is the best way to write a tridiagonal matrix as a linear combination of Pauli matrices?

Summary: There is a solution for expressing a tridiagonal matrix of the form you've provided for arbitrary $n$ in terms of Pauli operators using recursion. This procedure is given at the bottom of ...
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  • 4,667
4 votes

Expected value of a Haar random quantum state multiplied by a unitary

I'm writing an alternate proof because it uses some interesting tools, computes the value of these expressions, and gives some insights into how we can interpret the quantities in consideration. The ...
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4 votes
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Expected value of a Haar random quantum state multiplied by a unitary

With the chosen structure of $ U $, i think it's even possible to prove the stronger statement: $$ \langle z| \rho|z \rangle = \langle z| \sigma_\rho|z \rangle, \hspace{0.2em} \text{where} \hspace{0....
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  • 1,326
4 votes
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Depolarization of density operator with zeros in diagonal

Quantum channels are foremost, linear operators. So given a basis for the Hilbert-Schmidt operator space (for example the states $\{|0\rangle\langle 0|,|0\rangle\langle 1|,|1\rangle\langle 0|,|1\...
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4 votes
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Diamond norm distance bound on Stinespring dilations of channels

Yes, in fact there exists Stinespring dilations such that $$\frac{\|N_1-N_2\|_{cb}}{\sqrt{\|N_1\|_{cb}}+\sqrt{\|N_2\|_{cb}}}\leq \|V_1-V_2\|\leq \sqrt{\|N_1-N_2\|_{cb}}$$ where the distance between ...
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  • 1,512
4 votes
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Conjugation of $R_x(\theta)$ with $CNOT$

This is an application of the following identity $$ Be^AB^{-1} = e^{BAB^{-1}}\tag1 $$ where $A$ is any $n\times n$ real or complex matrix and $B$ is any invertible $n\times n$ real or complex matrix. ...
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  • 13.5k
4 votes
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If the eigenvalues of $Z$ are $\pm1$, why are the computational basis states labeled with "$0$" and "$1$"?

The reason is when you make measurement on the state $|\psi \rangle = \alpha |0\rangle + \beta |1\rangle$, the state $|\psi\rangle$ will collapsed onto whether the state $|0\rangle$ or $|1\rangle$. If ...
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  • 12.5k
4 votes

In Stinespring dilation, can we always use a mixed state as the ancilla?

No. If you take $\sigma=\tfrac1{d_A}\mathbb I$, you will have that for $\rho=\tfrac1{d_S}\mathbb I$, $$ \mathrm{tr}_A(U^\dagger(\rho\otimes\sigma)U)=\tfrac1{d_S}\mathbb I\ , $$ and thus, it will not ...
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4 votes

What's the state-of-the-art to calculate $|Ab\rangle$, given a matrix $|A\rangle$ and a vector $|b\rangle$ in QRAM encoding

I do not know what the state-of-the-art for this problem is, but here is my attempt at it. First, I doubt whether the required unitary $U$ can be found for every matrix $A$. The most glaring issue ...
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