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

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

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!


4 Answers 4


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 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.

  • $\begingroup$ 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? $\endgroup$
    – fagd
    Commented Jan 29, 2022 at 5:09
  • $\begingroup$ Ah, never mind. I see it, after I expand everything out. $\endgroup$
    – fagd
    Commented Jan 29, 2022 at 5:47
  • $\begingroup$ 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. $\endgroup$ Commented Jan 29, 2022 at 6:16
  • 1
    $\begingroup$ 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. $\endgroup$ Commented Jan 29, 2022 at 6:17
  • $\begingroup$ (For completeness) Re eq $(4)$: We apply eq $(1)$ twice. $\endgroup$ Commented Jan 29, 2022 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 the irrep $\lambda$. 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.

  • $\begingroup$ 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. $\endgroup$
    – fagd
    Commented Jan 29, 2022 at 5:12
  • $\begingroup$ @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). $\endgroup$ Commented Jan 31, 2022 at 11:46
  • $\begingroup$ thanks! another thing: in the second part when you discuss the unitary group, if I understand correctly, $X$ becomes a superoperator, as do the projectors $P_i$. So when you write $\langle P_i,X\rangle$ this is an inner product between superoperators? How do you define it? $\langle P_i,X\rangle=\sum_\alpha \langle P_i(\sigma_\alpha),X(\sigma_\alpha)\rangle\equiv\sum_\alpha{\rm Tr}[P_i(\sigma_\alpha)^\dagger X(\sigma_\alpha)]$ for some orthonormal basis of operators $\{\sigma_\alpha\}$? $\endgroup$
    – glS
    Commented Nov 7, 2022 at 10:21
  • $\begingroup$ @glS indeed it's an inner product on super operators. You have to take one w.r.t. which the representation is unitary. Usually, that's just the trace inner product (remember, superoperators are still just plain old linear operators): $\langle X, Y\rangle = \mathrm{tr}(X^\dagger Y)$. Here, the adjoint is taken w.r.t. to the trace inner product on operators. $\endgroup$ Commented Nov 7, 2022 at 18:53

I will give an extended explanation of Nielsen's proof, i.e. your first ref link. The idea is that, $\rho=\sum_ip_i|i\rangle\langle i|$, we can prove it's depolarizing channel for each $|i\rangle\langle i|$ with same $p$, then we are done.

I start after eq.(10):$$V \mathcal{E}_T(\rho) V^{\dagger}=\mathcal{E}_T\left(V \rho V^{\dagger}\right)\tag{1}$$ For one $|i\rangle \langle i|$, we can choose $V$ to be diagonal block with respect to $|i\rangle \langle i|$ and $I-|i\rangle \langle i|$, you can think it as written $V$ in basis of $|i\rangle$ as $\left( \begin{matrix} a& 0\\ 0& B\\ \end{matrix} \right) $ where $a$ is a number and $B$ is a matrix. Now by eq1 we have $V\mathcal{E} _T\left( |i\rangle \langle i| \right) V^{\dagger}=\mathcal{E} _T\left( |i\rangle \langle i| \right) $, hence we have $\left[ V,\mathcal{E} _T\left( |i\rangle \langle i| \right) \right] =0$. I skip the proof that if $\left[ V,\mathcal{E} _T\left( |i\rangle \langle i| \right) \right] =0$ for all block diagonal unitary of the form mentioned above, we can have $\mathcal{E} _T(|i\rangle \langle i|)=\alpha |i\rangle \langle i|+\beta \left( I-|i\rangle \langle i| \right) $. Notice that $\mathcal{E} _T$ is trace preserving so we can rewrite it as $\mathcal{E} _T(|i\rangle \langle i|)=pI/d+(1-p)|i\rangle \langle i|$ for some $p$. Then we want to show that for different $|i\rangle \langle i|$, the $p$ is the same. To see this, we know that $|\tilde{i}\rangle $ and $|i\rangle $ can be connected with a $U$ such that $|\tilde{i}\rangle \langle \tilde{i}|=U|i\rangle \langle i|U^{\dagger}$. Then we will have $$\mathcal{E} _T(|\tilde{i}\rangle \langle \tilde{i}|)=\mathcal{E} _T(U|i\rangle \langle i|U^{\dagger})=U\mathcal{E} _T(|i\rangle \langle i|)U^{\dagger} \\ =U\left( pI/d+(1-p)|i\rangle \langle i| \right) U^{\dagger} \\ =pI/d+(1-p)|\tilde{i}\rangle \langle \tilde{i}|$$ So for $|\tilde{i}\rangle $ we have the same $p$.

Remark Notice that twirling does not have to be done w.r.t. to unitary group $U(d)$, any group $G$ can have its corresponding twirling operation, but we can see from the end of the reasoning above that if we want to connect any two pure state $|i\rangle$ and $|\tilde i\rangle$, then we must have the twirling w.r.t. $U(d)$.

  • 1
    $\begingroup$ In ". I skip the proof that if ..." you're implicitly using Schur's lemma twice, one for $a|i\rangle i\langle i|$ and the second for $B$ (which each one of them are the irreducible representations of $V$), right? $\endgroup$ Commented Apr 25, 2023 at 15:53
  • 1
    $\begingroup$ @user2820579 I was not thinking so. I was thinking about using eq(3.5) and eq(3.6) in this thesis. By directly using Schur's lemma, I think we cannot get this result. For example, if we have a unitary of form $\mathrm{diag}\left\{ U,U \right\} $ where $U$ is a 2 by 2 unitary matrix. Then the matrices commute with it should also contain $B\otimes I_2$ where $B$ is an arbitrary matrix. $\endgroup$
    – narip
    Commented Apr 26, 2023 at 9:18
  • 1
    $\begingroup$ @user2820579 I will directly answer your question in the comments, and if I have time, I will amend some contents of my posts. $\left( \begin{matrix} a& 0\\ 0& B\\ \end{matrix} \right) $ in eq(3.5) is $\left( A_1\otimes 1 \right) \oplus \left( A_2\otimes 1 \right) =\left( a\otimes 1 \right) \oplus \left( B\otimes 1 \right) $ where $A_1$ is of one-dimension(actually is $a$) and $A_2$ is of 3 dimension(actually is $B$). Therefore, eq(3.6) should be $\left( 1\otimes B_1 \right) \oplus \left( I_3\otimes B_2 \right) =\left( 1\otimes \alpha \right) \oplus \left( I_3\otimes \beta \right) $ $\endgroup$
    – narip
    Commented Apr 26, 2023 at 10:50
  • 1
    $\begingroup$ Hey, thanks a lot for breaking this out into bits! Just a last small question, you write $I_3$ because you are considering that the full space is the Hilbert space of two qubits, right? $\endgroup$ Commented Apr 26, 2023 at 11:40
  • 1
    $\begingroup$ @user2820579 Oh. My bad. Yeah, in this condition I'm considering 2 qubits. I should have said that earlier. I think extending it into general cases should be clear now. $\endgroup$
    – narip
    Commented Apr 26, 2023 at 12:12

I'll present a slightly different wording for the general approach to the proof discussed in this answer, but being more explicit about the application of the general formalism to prove that the twirling operator products depolarizing channels. More specifically, consider the twirling operation $\mathcal T(\mathcal E)\equiv \mathcal E_T$ defined as $$\mathcal T(\mathcal E)\equiv \int dU\, \Phi_{U^\dagger}\circ\mathcal E\circ \Phi_{U},$$ with $\Phi_U(\rho)\equiv U\rho U^\dagger$ the unitary channel associated to $U$. Our goal is to prove that $\mathcal T(\mathcal E)$ is a depolarising channel, for any map $\mathcal E$. More explicitly, this means that $\mathcal T(\mathcal E)(\rho)=p_{\cal E} \operatorname{tr}(\rho) \frac{I}{d} + (1-p_{\cal E}) \rho$, for some $p_{\cal E}$.

Start observing that $\mathcal T^2\equiv \mathcal T\circ\mathcal T=\mathcal T$. This means that $\mathcal T$ is a linear (supersuper)operator, acting on the space of quantum maps. It is not hard to see that $\mathcal T$ is also Hermitian, and thus an orthogonal projection. In summary, $\mathcal T$ projects onto the set of quantum maps that commute with the action of the unitary group. Meaning $\mathcal \Phi_V\circ \mathcal T(\mathcal E)=\mathcal T(\mathcal E)\circ\Phi_V$ for all maps $\mathcal E$ and unitaries $V$. Here $$\mathbf U(d)\ni V\mapsto \Phi_V\in\mathbf{GL}(\mathrm{Lin}(\mathbb{C}^d))$$ is a representation of the unitary group into the space of quantum maps.

These observations allow to use Schur's lemma to characterise $\mathcal T(\mathcal E)$, as described in detail e.g. here. If the representation $U\mapsto \Phi_U$ was irreducible, then we would directly apply the lemma and conclude that $\mathcal T(\mathcal E)$ would have to be a multiple of the identity (here "identity" would mean the identity superoperator). However, this representation is not irreducible, as $\Phi_U(I)=I$ for all $U\in\mathbf U(d)$, which means that $\mathbb{C} I$ is a subspace left invariant by the representation. Similarly you can observe that for any traceless $X\in\operatorname{Lin}(\mathbb{C}^d)$, you have $\operatorname{tr}(\Phi_U(X))=0$, meaning the subspace of traceless operators is also an invariant subspace for the representation. You can then show that these two are both irreducible representations, and thus you get the decomposition $$\mathcal T(\mathcal E)= \alpha\, \Pi_I + \beta\, \Pi_0, \qquad \Pi_0 \equiv \operatorname{Id}-\Pi_I$$ where $\alpha,\beta\in\mathbb{R}$ are to be determined, $\operatorname{Id}$ is the identity superoperator, $\Pi_I:X\mapsto \operatorname{tr}(X)I/d$ is the map projecting onto the subspace generated by $I$, and $\Pi_0$ the map projecting onto the subspace of traceless operators, which can be given the above explicit form.

To find the coefficients $\alpha,\beta$ we now compute the inner product between $\mathcal T(\mathcal E)$ and $\Pi_I,\Pi_0$. For the RHS, we use the relations: $$\langle \Pi_I, \Pi_I \rangle = 1, \qquad\langle \Pi_0,\Pi_0\rangle = \operatorname{tr}(\Pi_0)=d^2-1, \qquad \langle\Pi_I,\Pi_0\rangle=0.$$ For the LHS, we instead use the specific structure of the twirling to observe $$\alpha = \langle \Pi_I, \mathcal T(\mathcal E)\rangle \equiv \frac1{d} \langle \Pi_I(I),\mathcal T(\mathcal E)(I)\rangle = \frac1d\operatorname{tr}[\mathcal T(\mathcal E)(I)] = \frac1d\operatorname{tr}(\mathcal E(I)), \\ (d^2-1)\beta=\langle \Pi_0,\mathcal T(\mathcal E)\rangle =\langle \operatorname{Id},\mathcal T(\mathcal E)\rangle - \langle\Pi_I,\mathcal T(\mathcal E)\rangle = \operatorname{tr}(\mathcal E) - \frac1d\operatorname{tr}(\mathcal E(I)).$$ Note that here we used the inner product between superoperators, defined as $$\langle\Phi,\Psi\rangle\equiv\sum_\sigma \langle\Phi(\sigma),\Psi(\sigma)\rangle\equiv \sum_\sigma \operatorname{tr}[\Phi(\sigma)^\dagger\Psi(\sigma)],$$ for any pair of quantum maps $\Phi,\Psi$ and orthonormal basis of operators $\{\sigma\}\in\operatorname{Lin}(\mathbb{C}^d)$. Similarly, $\operatorname{tr}(\mathcal E)$ is the "superoperator trace", equal to $\operatorname{tr}(\mathcal E)=\sum_\sigma \langle\sigma,\mathcal E(\sigma)\rangle$. We thus concluded that $$\mathcal T(\mathcal E) = \frac{\operatorname{tr}(\mathcal E(I))}{d} \Pi_I + \frac{\operatorname{tr}(\mathcal E)-\operatorname{tr}(\mathcal E(I))/d}{d^2-1} \Pi_0.$$ In the special case where $\mathcal E$ is a quantum channel we furthermore have $\mathcal E(I)=I$, and thus $$\mathcal T(\mathcal E) = \Pi_I + \frac{\operatorname{tr}(\mathcal E)-1}{d^2-1} \Pi_0.$$ Finally we can rewrite this slightly to get the depolarising channel: observe that $\Pi_0=\operatorname{Id} - \Pi_I$, and thus $$\mathcal E_T\equiv \mathcal T(\mathcal E) = \left(\frac{d^2-\operatorname{tr}(\mathcal E)}{d^2-1}\right)\Pi_I + \left(\frac{\operatorname{tr}(\mathcal E)-1}{d^2-1}\right)\operatorname{Id}, \\ \mathcal E_T(\rho) = \left(\frac{d^2-\operatorname{tr}(\mathcal E)}{d^2-1}\right)\operatorname{tr}(\rho) \frac{I}{d} + \left(\frac{\operatorname{tr}(\mathcal E)-1}{d^2-1}\right)\rho.$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.