# Tag Info

### In Schur-Weyl's duality, why is the commutant of $\pi_k(S_k)$ spanned by $U(d)^{\otimes k}$ matrices?

While I agree with Markus Heinrich that the argument is non-trivial, actually it can be presented in familiar quantum computing terms. The matrix algebra $M(d)^{\otimes n} \cong M(d^k)$ has a self-...
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### What can we say about the eigendecomposition of quantum channels?

As you observe correctly, $\mathbb N$ is a linear map. Thus, the same holds as for any eigendecomposition of linear maps. In particular, there need not be a complete basis of eigenvectors (there can ...
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### In Schur-Weyl's duality, why is the commutant of $\pi_k(S_k)$ spanned by $U(d)^{\otimes k}$ matrices?

The goal is to prove that the subspace spanned by $U^{\otimes n}$ for all $d\times d$ unitary matrices $U$ includes $M^{\otimes n}$ for every $d\times d$ matrix $M$. Here's my attempt to make this as ...
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1 vote

### How is this step performed in Deutsch's algorithm?

Your misunderstanding is that $|x\rangle$ is only meant to denote either $|0\rangle$ or $|1\rangle$ -- not the superposition $H|0\rangle$. So $f(x)$ respectively denotes either $f(0)$ or $f(1)$, ...
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### How do I show that a reduced density matrix of $1$ is $\rho_{12}^{1} = \text{Tr}_{2}[\rho_{12}] = \sum_{i}\langle i_{2} | \rho | i_{2} \rangle$?

Complementing FDGod's comment, this question would make sense if you had first defined the partial trace via the duality ${\rm tr}({\rm tr}_1(\rho)\omega)={\rm tr}(\rho({\bf1}\otimes\omega))$ as is ...
1 vote

### What are examples of quantum maps with complex eigenvalues?

Strictly speaking this is not an answer to what you asked, but in your question you claim that "for any channel, $\Phi$ has real eigenvalues iff it has a Kraus decomposition in terms of Hermitian ...
1 vote

### Show that while calculating partial traces the probability is independent of the basis of one of the measurements

Independence of the slightly more general expression $\sum_n\langle x\otimes \beta_n|\rho_{AB}|y\otimes \beta_n\rangle$ from the orthonormal basis $\{\beta_n\}_n$ boils down to the fact that the same ...

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