# Why does the twirl of a quantum channel give a depolarizing channel?

I would like to understand in detail why the twirl of a quantum channel gives depolarizing channel, which is the starting point of randomized benchmarking. To be self-contained, let me set up the notation.

Let $$\hat{U}$$ denote a superoperator that acts on the density matrix as $$\hat{U}(\rho)\equiv U\rho U^\dagger$$ where $$U$$ (without the hat) is the corresponding unitary. Let $$\hat{\Lambda}$$ be a quantum channel such that $$\hat{\Lambda}(\rho)=\sum_kA_k\rho A_k^\dagger$$ where $$A_k$$ is the Kraus operator. We use $$\circ$$ to denote the composition of superoperators: $$\hat{U}_1\circ\hat{U}_2(\rho)=U_1U_2\rho U_2^\dagger U_1^\dagger$$. The twirl of a quantum channel is defined as $$\hat{\Lambda}_t\equiv\int dU\hat{U}\circ\hat{\Lambda}\circ\hat{U}^\dagger$$ which is equal to a depolarizing channel in the sense that

$$$$\hat{\Lambda}_t(X) = (1-p_d)X + \frac{p_d}{D}\text{tr}(X)I$$$$

for any operator $$X$$. I would like to understand the derivation of this fact.

What I know is that the twirl commutes with arbitrary unitary superoperator $$\hat{U}\circ\hat{\Lambda}_t = \hat{\Lambda}_t\circ\hat{U}$$ and it hints that I should use some sort of Schur's lemma in the natural representation of $$\hat{\Lambda}$$, but I am not sure how to proceed...

Resources that I have found:

1. Nielsen's paper, but I don't understand his argument below Eq. (10).
2. The original RB paper, I don't understand their Eq. (46), so I guess I am missing some group theory here.
3. Meier's thesis, essentially following 2 but with a slight different representation, which I could not follow as well.

Any help to fill in the gap is really appreciated!

• Jan 28 at 12:13

Nielsen's paper cited in the question simplifies the arguments originally laid out in two papers by Horodecki family. This answer sketches the original arguments and is meant to complement the nice explanation based on representation theory written by @Markus Heinrich by requiring less background knowledge and hopefully providing some additional insight into the relationship between depolarizing channels and twirling. It also demonstrates the use of state-channel duality.

## High level summary

The argument uses state-channel duality to translate twirling of channels to $$U\otimes U^*$$ twirling of states. By unitary invariance of the Haar measure, twirling is idempotent, so the Choi matrix of a twirled channel is invariant under $$U\otimes U^*$$ twirling of states. However, it turns out that the only states invariant under $$U\otimes U^*$$ twirling of states are the so-called noisy singlets which under state-channel duality correspond to depolarizing channels.

## Noisy singlet

Consider two systems with the Hilbert spaces of the same finite dimension $$N$$. Let $$|\psi\rangle:=\frac{1}{\sqrt{N}}\sum_{i=1}^N|i\rangle|i\rangle$$. It is easy to check that for any linear operator $$A$$

$$(A\otimes I)|\psi\rangle = (I\otimes A^T)|\psi\rangle.\tag1$$

Now, for $$p\in[0,1]$$, we define the noisy singlet $$\rho_p$$ to be the bipartite state

$$\rho_p:=p|\psi\rangle\langle\psi|+(1-p)\frac{I\otimes I}{N^2}.\tag2$$

## Twirling

Twirling of states sends a bipartite state $$\rho$$ to

$$\rho_t := \int dU (U\otimes U^*)\rho(U^\dagger\otimes U^T)\tag3$$

where $$U^*$$ denotes the complex conjugate of $$U$$. Using $$(1)$$, we can show that the Choi matrix $$J(\hat\Lambda_t)$$ of a twirled channel $$\hat\Lambda_t$$ is the result of twirling of states applied to the Choi matrix $$J(\hat\Lambda)$$ of the original channel $$\hat\Lambda$$

\begin{align} J(\hat\Lambda_t)&=\hat\Lambda_t\otimes\hat{I}(N|\psi\rangle\langle\psi|)\\ &=\left(\int dU\hat{U}\circ\hat{\Lambda}\circ\hat{U}^\dagger\right)\otimes\hat{I}(N|\psi\rangle\langle\psi|)\\ &=\left(\int dU(\hat{U}\otimes\hat{I})\circ(\hat{\Lambda}\otimes\hat{I})\circ(\hat{U}^\dagger\otimes\hat{I})\right)(N|\psi\rangle\langle\psi|)\\ &=\int dU(\hat{U}\otimes\hat{I})\circ(\hat{\Lambda}\otimes\hat{I})\left((U^\dagger\otimes I)N|\psi\rangle\langle\psi|(U\otimes I)\right)\\ &=\int dU(U\otimes I)\left[\hat{\Lambda}\otimes\hat{I}\left((U^\dagger\otimes I)N|\psi\rangle\langle\psi|(U\otimes I)\right)\right](U^\dagger\otimes I)\\ &=\int dU(U\otimes I)\left[\hat{\Lambda}\otimes\hat{I}\left((I\otimes U^*)N|\psi\rangle\langle\psi|(I\otimes U^T)\right)\right](U^\dagger\otimes I)\\ &=\int dU(U\otimes U^*)\left[\hat{\Lambda}\otimes\hat{I}\left(N|\psi\rangle\langle\psi|\right)\right](U^\dagger\otimes U^T)\\ &=\int dU(U\otimes U^*)J(\hat\Lambda)(U^\dagger\otimes U^T)\\ &=J(\hat\Lambda)_t. \end{align}\tag4

Another fact we can easily prove using $$(1)$$ is that every noisy singlet $$(2)$$ is invariant under twirling of states $$\rho_{p,t}=\rho_p$$. In fact, it turns out that noisy singlets are the only states with this property. See section $$V$$ in this paper for a proof of this fact.

## Depolarizing channel

Depolarizing channel is a CPTP map defined by

$$\hat\Delta_p(\rho) = p\rho + (1-p)\frac{I}{N}\mathrm{tr}\rho.\tag5$$

A short calculation shows that the Choi matrix of $$\hat\Delta_p$$ is

$$J(\hat\Delta_p)=(\hat\Delta_p\otimes\hat I)(N|\psi\rangle\langle\psi|)=N\rho_p\tag6$$

where $$\rho_p$$ is a noisy singlet.

## Putting it all together

Finally, unitary invariance of the Haar measure implies that twirling a channel twice yields the same result as twirling it once

$$(\hat\Lambda_t)_t=\hat\Lambda_t.\tag7$$

Therefore, by $$(4)$$

$$J(\hat\Lambda_t)=J((\hat\Lambda_t)_t)=J(\hat\Lambda_t)_t\tag8$$

i.e. the Choi matrix of $$\hat\Lambda_t$$ is invariant under twirling of states. But noisy singlets are the only states with this property. Therefore, $$J(\hat\Lambda_t)$$ is (a scalar multiple of) a noisy singlet

$$J(\hat\Lambda_t)=N\rho_p\tag9$$

for some $$p\in[0,1]$$. However, $$N\rho_p=J(\hat\Delta_p)$$, so by injectivity of $$J$$, we have

$$\hat\Lambda_t=\hat\Delta_p\tag{10}$$

which was to be proven.

• Thanks for the great answer! That is really helpful. From what you illustrated, the only gaps for me now are a) the noisy singlet is the only state that is invariant under twirling b) the choi matrix is injective. I can look up these two myself. Maybe the following is easier for you to elaborate a bit more: in Eq. 4, how did you get from line 6-7?
– fagd
Jan 29 at 5:09
• Ah, never mind. I see it, after I expand everything out.
– fagd
Jan 29 at 5:47
• You're welcome! Thank you for an interesting question! Re a) I left this step out since it is rather technical. Essentially, we plug in a few interesting types of unitary under $U$ in $(U\otimes U^*)\rho(U^\dagger\otimes U^T)=\rho$ until we manage to narrow the matrix elements of $\rho$ down to the form $(2)$. See section $V$ starting on page $6$ in the cited paper for details. Jan 29 at 6:16
• Re b) It is easier to see that $J$ is injective under the other convention $$J(\Phi)=\sum_{ij}|i\rangle\langle j|\otimes\Phi(|i\rangle\langle j|)=\begin{bmatrix}\Phi(|0\rangle\langle0|)&\dots&\Phi(|0\rangle\langle N-1|)\\&\dots&\\\Phi(|N-1\rangle\langle0|)&\dots&\Phi(|N-1\rangle\langle N-1|)\end{bmatrix}.$$ Thus, the matrix $J(\Phi)$ consists of blocks which explicitly specify the action of the channel $\Phi$ on the standard basis. The change of convention back to $J(\Phi)=\sum_{ij}\Phi(|i\rangle\langle j|)\otimes|i\rangle\langle j|$ changes the clarity of the picture, but not the conclusion. Jan 29 at 6:17
• (For completeness) Re eq $(4)$: We apply eq $(1)$ twice. Jan 29 at 6:19

I hope you do not mind if I zoom out a bit and talk about representation theory. I think a more general approach helps understanding the essential bits and will be helpful if you encounter similar expressions in the future.

Let $$G$$ be a (compact) group equipped with its Haar measure and $$\rho$$ a (finite-dimensional unitary) representation on a Hilbert space $$H$$. The operator $$\Pi_G(X) := \int_G \rho(g)X\rho(g)^\dagger\,\mathrm{d}g,$$ is the orthogonal projection onto the commutant $$\rho'$$ of $$\rho$$, i.e. on all operators $$X\in L(H)$$ which commute with $$\rho$$. We can write $$\Pi_G$$ using an orthogonal basis of this subspace.

Clearly, we want to use Schur's lemma but $$\rho$$ is generally a reducible representation. Hence, let's decompose $$\rho$$ into irreps $$\rho = \bigoplus_\lambda \rho_\lambda \otimes \mathrm{id}_{n_\lambda},$$ where $$n_\lambda$$ is the multiplicity of irrep $$n_\lambda$$. We can now apply Schur's lemma irrep-wise: any $$X\in L(H)$$ can be written into matrix blocks as follows $$X = \bigoplus_{\lambda,\lambda'} X_{\lambda,\lambda'}.$$ If $$X$$ is in the commutant $$\rho'$$, $$X_{\lambda,\lambda'} = 0$$ if $$\lambda\neq\lambda'$$ because of Schur's lemma. Moreover, if $$\lambda=\lambda'$$, then $$X$$ can still be non-identity on the multiplicity space, hence $$X = \bigoplus_\lambda \mathrm{id}_\lambda \otimes X_{\lambda}, \qquad X_{\lambda} \in \mathbb{C}^{n_\lambda\times n_\lambda}.$$ Special case: if an irrep is multiplicity-free, $$n_\lambda=1$$ and thus $$X_{\lambda}\in\mathbb C$$ is just a number.

Remark: From the above formula it is easy to see that the dimension of the commutant $$\rho'$$ is $$\dim\rho'=\sum_{\lambda}n_\lambda^2$$. It is also a one-line proof using a character formula.

Next, writing out the projection onto the commutant is simpler in the multiplicity-free case, so let's do this first. Then, any $$X\in\rho'$$ is of the form $$X = \bigoplus_\lambda x_\lambda \mathrm{id}_\lambda, \qquad x_\lambda\in\mathbb C.$$ However, note that the projectors $$P_\lambda\in L(H)$$ onto the irreps of $$\rho$$ in $$H$$ have the form $$P_\lambda = 0 \oplus \dots \oplus 0 \oplus \mathrm{id}_\lambda \oplus 0 \oplus \dots \oplus 0.$$ Hence, $$X = \sum_\lambda x_\lambda P_\lambda.$$ Moreover, the $$P_\lambda$$ are orthogonal and $$\|P_\lambda\|_2 = \sqrt{d_\lambda}$$, such that $$X = \sum_\lambda \frac{1}{d_\lambda}\mathrm{tr}(P_\lambda X) P_\lambda.$$ Hence, our formula for the projection onto $$\rho'$$ is $$\Pi_G(X) = \sum_\lambda \frac{1}{d_\lambda}\mathrm{tr}(P_\lambda X) P_\lambda.$$

Now, let's apply this to the unitary group. So we have $$G=U(d)$$ and $$\rho(U) = U(\cdot)U^\dagger$$ acting on $$H=L(\mathbb C^d)$$. The irreps of $$\rho$$ are the trivial one, spanned by $$I$$ and the the traceless subspace of dimension $$d^2-1$$. Both are multiplicity-free and have the projectors $$P_1(X) = \frac{1}{d}\mathrm{tr}(X) I, \qquad P_0(X) = X - P_1(X) = X - \frac{1}{d}\mathrm{tr}(X) I.$$ To evaluate the channel twirl of a channel $$\Lambda$$, we note that $$\mathrm{tr}(P_1\Lambda) = \frac{1}{d}\mathrm{tr}(\Lambda(I)) = 1, \qquad \mathrm{tr}(P_0\Lambda) = \mathrm{tr}(\Lambda) - 1,$$ and write $$p_\Lambda := \mathrm{tr}(P_0\Lambda)/(d^2-1)$$. Then: $$\Pi_{U(d)}(\Lambda)(X) = \mathrm{tr}(P_1\Lambda) P_1(X) + \frac{1}{d^2-1}\mathrm{tr}(P_0\Lambda) P_0(X) \\ = \mathrm{tr}(X)\frac{I}{d} + p_\Lambda\left(X - \frac{1}{d}\mathrm{tr}(X) I\right)\\ = \left(1 - p_\Lambda\right)\mathrm{tr}(X) \frac{I}{d} + p_\Lambda X.$$ This is a depolarizing channel with parameter $$p_\Lambda$$ (or $$1-p_\Lambda$$ whichever you prefer).

PS: If we have multiplicities, then there is only a canonical decomposition into $$\lambda$$-isotypes. We have to make a choice how to decompose these further into copies of $$\rho_\lambda$$ and define and orthogonal basis for the multiplicity space. The rest stays the same.

• Thanks for the wonderful answer! I think the second expression for $\Pi_G(X)$ is a good starting point for me. Could you please elaborate a bit more on why the irreps of $\rho$ for the unitary group are the ones you mention? In particular, I don't quite get the second irrep with dimension $d^2-1$. It seems to be the main point that I missed from the original RB paper.
– fagd
Jan 29 at 5:12
• @fagd Unitaries preserve the trace, $\mathrm{tr}(UXU^\dagger) = \mathrm{tr}(X)$, in particular the subspace of traceless matrices is invariant under the representation. The action on this subspace is essentially the adjoint representation of $U(d)$ which turns out the be irreducible. (Note that we have the orthogonal decomposition $X = X_0 + \mathrm{tr}(X) I/d$ where $X_0$ is traceless). Jan 31 at 11:46