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3 votes
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How to prove that $\max_{T\in{\bf U}(X)}|\langle T,A\rangle|\le \|A\|_1$?

The converse also follows from the singular value decomposition: for all $T\in\mathbf U(X)$ \begin{align*} |\langle T,A\rangle|&=|{\rm tr}(T^*U\Sigma V^*)|=|{\rm tr}(V^*T^*U\Sigma)|\\ &=\Big|\...
Frederik vom Ende's user avatar
3 votes

If eigenvalues of two matrices are equal then the matrices are equal?

A simple counterexample are the matrices $$ \rho=\begin{pmatrix} 1&0\\0&0 \end{pmatrix}\qquad\omega=\begin{pmatrix} 0&0\\0&1 \end{pmatrix} $$ which have the same eigenvalues ($0,1$) ...
Frederik vom Ende's user avatar
3 votes
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If eigenvalues of two matrices are equal then the matrices are equal?

By a density matrix, I will assume you mean a Hermitian matrix with unit trace. Such a matrix has the property that all eigenvalues are real, and there is guaranteed to be a spectral decomposition: $$...
Graham Van Goffrier's user avatar
3 votes

unitary that transform $\sigma^x \pm \alpha \sigma^z$ into $\sigma^x$ and $\sigma^z$

Let $A$ and $B$ be your two initial matrices. You have $$ \text{Tr}(AB)=2(1-\alpha^2). $$ After the unitary, you have $$ \text{Tr}(UAU^\dagger UBU^\dagger)=\text{Tr}(AB)=2(1-\alpha^2). $$ Your aim is ...
DaftWullie's user avatar
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3 votes
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unitary that transform $\sigma^x \pm \alpha \sigma^z$ into $\sigma^x$ and $\sigma^z$

Short answer: Such a unitary $U$ exists if and only if $\alpha=1$ or $\alpha=-1$ Long answer: Because $\sigma_x-\alpha\sigma_z$ are traceless matrices with determinant $-\alpha^2-1$ their eigenvalues ...
Frederik vom Ende's user avatar
2 votes

partial trace of a unitary matrix over the zero state of ancilla qubit

TL;DR I am sharing your confusion. IMO, the nomenclature "partial trace" is misleading in this case. $A$ is not the result of what is usually called a partial trace over $U$. What a partial ...
qubitzer's user avatar
  • 747
2 votes

How to prove that $\max_{T\in{\bf U}(X)}|\langle T,A\rangle|\le \|A\|_1$?

Write the SVD of $A$ as $$A = \sum_k s_k |u_k\rangle\!\langle v_k|,$$ for some pair of orthonormal bases $\{|u_k\rangle\}$ and $\{|v_k\rangle\}$. If $T$ is unitary, then it can always be written as $T=...
glS's user avatar
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1 vote

What does it mean for a gate to "commute with a measurement"?

I think that mathematically you have an observable and this observable is described by an operator. The gate commute with measurement if the commutator between the gate and the operator associated to ...
Simona99's user avatar
  • 200
1 vote
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i have problem in using deltakronker

You don't just substitute $m=n-1$. Instead, you perform the sum over $m$. The only non-zero term that you get is if $m=n-1$, so that means the second sum vanishes, and you have $$ \sum_{n=1}^{\infty}\...
DaftWullie's user avatar
  • 59.8k

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