Linked Questions

5 votes
3 answers
966 views

Show that, averaging over uniformly random unitaries, $\mathbb{E}[UXU^\dagger]=\operatorname{Tr}(X)\frac{I}{d}$

As mentioned e.g. in this answer, if we compute the average $$\Phi(X)\equiv \mathbb{E}_U[UXU^\dagger] \equiv \int_{\mathbf U(d)} d\mu(U) UXU^\dagger,$$ where $d\mu(U)$ is the Haar measure over the ...
glS's user avatar
  • 26.3k
4 votes
2 answers
235 views

Are continuous probability distributions over quantum channels possible?

I am not an expert in the subject and apologize in advance for a strange question and (possible) abuse of the terminology. I have learned that any convex combination of quantum channels (CPTP maps, ...
trurl's user avatar
  • 141
5 votes
1 answer
371 views

Quantum channels that commute with any unitary channel

Consider a quantum channel $\Phi$ that maps from density operators $\mathcal{S}(\mathcal{H}_A)$ to itself, that commutes with any unitary channel $\mathcal{U}$ on $\mathcal{S}(\mathcal{H}_A)$, i.e. $\...
Shadumu's user avatar
  • 343
2 votes
2 answers
724 views

How to find the operator-sum representation of the depolarizing channel?

Consider a circuit built as follows: take two ancilla states and an operator $U$ made of a series of controlled gates which act on a pure state $\rho$ as follows: $X$ if the ancilla is in $|00\rangle$...
Oilobobolus's user avatar
3 votes
1 answer
507 views

Density matrices of multiples copies of a single Haar-Random state

In Pseudorandom States, Non-Cloning Theorems and Quantum Money, the authors state that: Let $\rho_\mu^m$ be the density matrix of a random $|\psi\rangle^{\otimes m}$ for $|\psi\rangle$ chosen from ...
Tristan Nemoz's user avatar
  • 7,302
3 votes
1 answer
491 views

Averaging over a single Haar-random unitary applied $t$ times

I'm trying to compute the following integral: $$\int U^{\otimes t}\left|x_1,\cdots,x_t\middle\rangle\middle\langle x'_1,\cdots,x_n'\right|\left(U^\dagger\right)^{\otimes t}\,\mathrm{d}\mu(U)$$ Where $\...
Tristan Nemoz's user avatar
  • 7,302
4 votes
1 answer
352 views

Prove that uniformly random states have moments ${\bf E}_\psi|\langle x|\psi\rangle|^{2t}\sim1/\binom d t$

Im looking for the moments of Haar random states. Is it true that $\textbf{E}_{\psi\sim \text{Haar}}|\langle x| \psi\rangle|^{2t}\sim \frac{1}{\binom{d}{t}}?$ How does one prove this?
postasguest's user avatar
1 vote
2 answers
440 views

Does applying a random Pauli matrix to a density matrix result in the identity?

Nielsen and Chuang's textbook, Equation 8.101 (section 8.3.4 'Depolarizing Channel') shows that applying a random Pauli to a density matrix representing one qubit equals the identity (times one half): ...
Quantum Guy 123's user avatar
3 votes
1 answer
223 views

Why does Schur's lemma imply that $\int \sigma^{\otimes n}_{HK} d(\sigma)$ must be a multiple of the identity on the symmetric subspace?

I am trying to understand Lemma 2 in this paper. Consider a state $\tau_{H^n}=\int \sigma^{\otimes n}_{H} \mu(\sigma)$ where $\mu(\sigma)$ is the measure on the space of density operators on a single ...
user1936752's user avatar
  • 3,137
4 votes
1 answer
143 views

Is the "unitary twirling operation" physically realizable?

In this neat answer by Markus Heinrich, it is shown that twirling an arbitrary quantum channel $\Lambda$ over the unitary group $U(d)$ yields a depolarizing channel $\tilde{\Lambda}$ given by $$ \...
Eric Kubischta's user avatar
1 vote
1 answer
143 views

What is the definition of twirled superoperator?

I am trying find the definition of twirled (super)operator. One such is Definition 2.3.16 on p. 33 of Christoph Dankert, Efficient Simulation of Random Quantum States and Operators. However, the ...
Hans's user avatar
  • 231
1 vote
0 answers
124 views

Prove that the twirling operation on a channel gives a decomposition $\int dU\, U^\dagger\mathcal E(U\rho U^\dagger)U=\alpha P+\beta Q$

The twirled operation of a quantum channel $\mathcal E$ is defined as \begin{align} \mathcal E_T(\rho) &= \int dU U^\dagger \mathcal E(U \rho U^\dagger)U, \end{align} where the integral is over ...
Michael.Andy's user avatar