7 votes

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-...
Greg Kuperberg's user avatar
6 votes
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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 ...
Norbert Schuch's user avatar
6 votes
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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 ...
John Watrous's user avatar
  • 5,887
5 votes
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Clifford group without the phase gate

The real Pauli group is defined as the subgroup of the Pauli matrices with only real entries. In particular, it contains $Y_i Y_j = - X_i Z_i X_j Z_j$. One can show that it is generated by $X_i$ and $...
Markus Heinrich's user avatar
4 votes
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Why can't the eigenvalues of a unitary matrix have the form $e^{i\theta}$?

in case of $e^{2\pi\cdot i\cdot \theta}$ the values of $\theta\in [0,1]$, in the case of $e^{i\cdot \theta}$ the values of $\theta \in [0,2\pi]$. it's just a different convention but they are ...
Sezzart's user avatar
  • 160
4 votes
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Under what conditions are two sets of Pauli operators Clifford-equivalent?

What I would do is to extend S and T into full Clifford tableaus $C_S$ and $C_T$ by discarding linearly dependent products and filling in generators for the unspanned $n$-qubit space. The Clifford $...
Craig Gidney's user avatar
  • 36.7k
4 votes

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

$M(d)^{\otimes k}$ is a linear combination of $U(d)^{\otimes k}$, this is what theorem 35 implies, yes. It's not obvious. In fact, the next theorem 36 gives the description of the span of $U(d)^{\...
Danylo Y's user avatar
  • 7,174
4 votes

Interesting properties of maps whose natural representation is unitary?

It may or may not come at a surprise that the natural representation of channels can only be unitary if the channel itself is unitary. More precisely, we will prove the following statement: Theorem. ...
Frederik vom Ende's user avatar
3 votes

What's the trace distance between $|0\rangle^{\otimes n}$ and $\frac{1}{\sqrt{2}}\big(|0\rangle^{\otimes n} + |1 \rangle^{\otimes n} \big)$?

Writing $|b^n\rangle := |b\rangle^{\otimes n}$ for $b \in \{0,1\}$, we have \begin{equation} \rho_2 - \rho_1 = \frac{1}{2}|0^n\rangle \langle 0^n| - \frac{1}{2}|0^n\rangle \langle 1^n| - \frac{1}{2}|1^...
forky40's user avatar
  • 6,718
3 votes

What can we say about the eigendecomposition of quantum channels?

At the risk of saying something trivial I'd like to add to Norbert's great answer and point out that normal channels are necessarily unital, i.e. to have an orthonormal basis of eigenvectors it is ...
Frederik vom Ende's user avatar
3 votes

How to represent general isometries and unitaries:

$V$ is a linear map from $H_A \rightarrow H_A \otimes H_E$, so we can pick arbitrary orthonormal bases $\{|i_A\rangle\}_{i=1}^{\text{dim}(H_A)}$ and $\{|i_E\rangle\}_{i=1}^{\text{dim}(H_E)}$ for $H_A$ ...
forky40's user avatar
  • 6,718
2 votes

Does the Lie closure of a set of Hamiltonians describe all unitaries you can generate with them?

Theorem. Given $H_1, \ldots, H_k\in\mathbb C^{d\times d}$ Hermitian let $\mathfrak{g} := \langle iH_1, ... iH_k \rangle_{\text{Lie}}$ denote the associated Lie algebra and let $e^{\mathfrak g}$ be the ...
Frederik vom Ende's user avatar
2 votes

The matrix norm $\|A\|=\max_{\langle u|u\rangle=1}|\langle u|A|u\rangle|$ in the proof of Lieb's theorem

The short answer is that there are two problems with your argument: The partial derivative of $u^TAu$ you state is wrong More gravely, the whole approach is flawed because you're treating $u$ as a ...
Frederik vom Ende's user avatar
2 votes

How to take partial trace of a $n - 1$ qubit subsystem from a $n$ qubit system

While the comment of glS contains the mathematical tools needed here, this particular problem features a bit more structure. For simplicity I will treat the case $n=2$ but all that follows does ...
Frederik vom Ende's user avatar
2 votes
Accepted

QuTiP ptrace function results do not recreate original composite system

This would only work if fullsystem is a product state. Let $\rho$ be your full system. For example, say $$\rho = |011\rangle\langle011|\,.\tag{1}$$ Now, $$\rho_1 = \...
FDGod's user avatar
  • 2,391
2 votes

What's the trace distance between $|0\rangle^{\otimes n}$ and $\frac{1}{\sqrt{2}}\big(|0\rangle^{\otimes n} + |1 \rangle^{\otimes n} \big)$?

We can answer this question in the general case. Let $\mathbb{P}_\psi\equiv|\psi\rangle\!\langle\psi|$ and $\mathbb{P}_\phi\equiv|\phi\rangle\!\langle\phi|$ be two arbitrary pure states. We want to ...
glS's user avatar
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1 vote

If $\text{tr}_B \rho \in A$, then $\rho \in A \otimes B$?

It's true in an appropriate formulation. First of all, don't confuse pure states $|\psi\rangle$ from $H$ and density matrices $\rho$ on $H$ (which are $|\psi\rangle\langle\psi|$ for pure states). Let $...
Danylo Y's user avatar
  • 7,174
1 vote

What is the action of $CCZ$ on $X \times I \times I$?

CCZ is defined by $$ CCZ \vert{x_1,x_2,x_3} \rangle = \vert x_1, x_2 \rangle Z^{x_1\cdot x_2} \vert{x_3}\rangle = (-1)^{x_1 x_2 x_3} \vert{x_1,x_2,x_3} \rangle $$ Therefore, \begin{align} (CCZ)(XII)(...
Frederik Ravn Klausen's user avatar
1 vote
Accepted

What is the action of $CCZ$ on $X \times I \times I$?

$CCZ$ will map $X_iI_jI_k$ to $(X_iI_jI_k)(CZ_{j,k})$ by conjugation where the parenthesis indicates the regular product (i.e. gate chaining) and $CZ_{j,k}$ is the $CZ$ operator between qubit $j$ and $...
AG47's user avatar
  • 439
1 vote

Left-canonical matrix product state

Iterative SVDs of the rank-$N$ coefficient tensor $\psi$ (see Eq. (1)) result in the representation of the quantum state as an MPS of the form written in Eq. (2). For a left-canonical MPS, let $A^{[...
jayjay's user avatar
  • 111
1 vote

Minimizing $1 - \text{Tr}(\Phi(\rho,U)^2)$

In your setup, the choice $U={\bf 1}$ of course suffices to minimize your function $$1 - \text{Tr}(\Phi(\rho, U)^2), \tag{1}$$ because we then have $\Phi(\rho, U) = {\rm Tr}_2(\rho) = \overline{\rho}...
Jos Bergervoet's user avatar
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 $f(1)$, ...
Nick Mertes's user avatar
1 vote

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 ...
Frederik vom Ende's user avatar
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 ...
Frederik vom Ende's user avatar
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 ...
Frederik vom Ende's user avatar

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