In Nielsen and Chuang (page:379), it is shown that the operator sum representation of a depolarizing channel $\mathcal{E}(\rho) = \frac{pI}{2} + (1-p)\rho$ is easily seen by substituting the identity matrix with

$$\frac{\mathbb{I}}{2} = \frac{\rho + X\rho X + Y\rho Y +Z\rho Z}{4}.$$

What is the more systematic way to see this result? Particularly, for the higher dimensional analogue, I cannot see how to proceed.


2 Answers 2


This really depends where you want to start from. For instance, you can construct the Choi state of $\mathcal E$, i.e., $$ \sigma = (\mathcal E \otimes \mathbb I)(|\Omega\rangle\langle\Omega|)\ , $$ with $\Omega = \tfrac{1}{\sqrt{D}}\sum_{i=1}^D |i,i\rangle$, and then extract the Kraus operators of $\mathcal E(\rho)=\sum M_i\rho M_i^\dagger$ by taking any decomposition $$ \sigma = \sum |\psi_i\rangle\langle\psi_i|\ ,\tag{*} $$ and writing $|\psi_i\rangle = (M_i\otimes\mathbb I)|\Omega\rangle$ (which is always possible).

Note that the decomposition $(*)$ is highly non-unique (any $|\phi_j\rangle = \sum V_{ij} |\psi_i\rangle$, with $V$ an isometry, is also a valid decomposition), which relates to the fact that the Kraus decomposition is equally non-unique. Obviously, the eigenvalue decomposition is a simple choice (which, moreover, minimizes the number of Kraus operators).

Let's look at your example in a bit more detail. Here, $D=2$. You have that $$ \mathcal E(X)=p\mathrm{tr}(X)\,\frac{\mathbb I}{2}+(1-p)X $$ for any $X$ (due to linearity) -- the $\mathrm{tr}(X)$ is required to make this trace-preserving for general $X$.

We now have that \begin{align} \sigma &= (\mathcal E \otimes \mathbb I)(|\Omega\rangle\langle \Omega|) \\ & = \tfrac1D \sum_{ij} \mathcal E(|i\rangle\langle j|)\otimes |i\rangle\langle j|\ \end{align} inserting the definition of $|\Omega\rangle$ and using linearity.

This yields $$ \sigma = \frac{p}{2D}\mathbb I\otimes \sum_{i}|i\rangle\langle i| + (1-p)\frac1D \sum_{ij}|i\rangle\langle j|\otimes |i\rangle\langle j|\ . $$ The second term is just $(1-p)|\Omega\rangle\langle\Omega|$, and the first term is $\frac{p}{2D}\mathbb I\otimes\mathbb I$.

You can now see that one possible eigenvalue decomposition of $\sigma$ is given by the four Bell states (I leave it to you to work out the weights), and it is well known and easy to check that that the four Bell states can be written as $$ (\sigma_k\otimes \mathbb I)|\Omega\rangle\ , $$ where $\sigma_k$ are the three Pauli matrices or the identity.

Thus, you get that the $M_i$ in the Kraus representation are the Paulis and the identity, with the weight given by the eigenvalue decomposition of $\sigma$.

  • $\begingroup$ Could you add to your answer exactly how this works for the specific example of the two qubit state and depolarizing channel (such that one obtains the Pauli matrices)? I am not sure how to express $\mathcal{E}$ in the first equation you have written? I assume $\Omega$ is the Bell state for two qubits but I'm not sure what exactly $(\mathcal{E}\otimes\mathbb{1})(\vert\Omega\rangle\langle\Omega\vert)$ looks like. $\endgroup$ Commented Dec 16, 2018 at 22:06
  • $\begingroup$ @user1936752 Well, you have to know how you are given the channel. But in whichever form you are given the channel, you should have a way to apply it to an input state. -- Maybe could you first explain what you tried to apply this to your example? $\endgroup$ Commented Dec 16, 2018 at 22:08
  • $\begingroup$ I see. I'm only given the effect of the channel i.e. $\mathcal{E}(\rho) = \frac{pI}{2} + (1-p)\rho$. Is this what you mean? The presentation of this in Nielsen and Chuang is that one can see that the identity operator can be expressed (as in the question) as a summation of Pauli operators. This gives immediately the Kraus operators. However, I cannot see how to get the same result through your suggestion (and also I don't know how to generalize this to higher dimensions). $\endgroup$ Commented Dec 16, 2018 at 22:26
  • $\begingroup$ @user1936752 Have edited. $\endgroup$ Commented Dec 17, 2018 at 0:17

While the procedure in the existing answer, based on channel-state duality, applies to general channels, there's a more direct way to obtain Kraus operators for this particular case of the depolarising channel, $\mathcal{E}(\rho) = (1-p)\rho + p \frac{I}{d}$, where $d$ is the Hilbert space dimension. Since $\mathrm{tr}(\rho) = 1$ for any density operator, we can write $\frac{I}{d} = \frac{I}{d}\mathrm{tr}(\rho)$. For any orthonormal basis $\{|1\rangle, \dots, |d\rangle\}$, $I = \sum_i |i\rangle \langle i|$ and $\mathrm{tr}(\rho) = \sum_i \langle i|\rho |i\rangle$, so $$ \frac{I}{d} = \frac{1}{d}\sum_i|i\rangle\langle i| \sum_j \langle j|\rho|j\rangle = \frac{1}{d} \sum_{i,j} |i\rangle \langle j|\rho|j \rangle\langle i|,$$ from which we see that a possible set of Kraus operators for $\mathcal{E}$ are $\{|i\rangle \langle j|/\sqrt{d}\}$, with $i,j \in \{1,\dots, d\}$. (You can check that these satisfy the completeness relation.)

For the one-qubit case, note that $\{I/2,X/2,Y/2,Z/2\}$ are related to $\{|0\rangle\langle 0|/\sqrt{2}, |0\rangle\langle 1|/\sqrt{2}, |1\rangle\langle 0|/\sqrt{2},|1\rangle\langle 1|/\sqrt{2}\}$ by a unitary, so by using the unitary freedom in the Kraus representation, we can get from the above representation to the standard $\frac{I}{2} = \frac{\rho + X\rho X + Y\rho Y + Z\rho Z}{4}$.

  • $\begingroup$ “note that $\{I,X,Y,Z\}$ are related to $\{|0\rangle\langle 0|/\sqrt{2}, |0\rangle\langle 1|/\sqrt{2}, |1\rangle\langle 0|/\sqrt{2},|1\rangle\langle 1|/\sqrt{2}\}$ by a unitary” how come? $\endgroup$ Commented Jan 12, 2023 at 6:32
  • $\begingroup$ it should have said $\{I/2,X/2,Y/2,Z/2\}$ (just edited the answer); does it make sense to you now? $\endgroup$
    – kianna
    Commented Feb 6, 2023 at 13:54
  • $\begingroup$ I'm not sure what you mean by "related to". Are you saying there is some unitary that maps $\{|0\rangle\langle 0|/\sqrt{2}, |0\rangle\langle 1|/\sqrt{2}, |1\rangle\langle 0|/\sqrt{2},|1\rangle\langle 1|/\sqrt{2}\}$ to $\{I/2,X/2,Y/2,Z/2\}$? (i.e. a change of basis matrix that is unitary). If this is what you are stating, how do you prove it? Can you also extend your proof to higher dimensions? (example, show it for a n-qubit depolarizing channel) $\endgroup$ Commented Feb 23, 2023 at 2:41
  • 1
    $\begingroup$ See Theorem 8.2: (Unitary freedom in the operator-sum representation) in Nielsen & Chuang, which says that $\{E_i\}_i$ and $\{F_i\}_i$ are sets of Kraus operators for the same quantum channel iff $E_i = \sum_j u_{ij} F_j$ for some unitary matrix $u$. $\endgroup$
    – kianna
    Commented Feb 23, 2023 at 13:27
  • 2
    $\begingroup$ Here, $$\begin{pmatrix}I/2 \\ X/2 \\ Y/2 \\ Z/2 \end{pmatrix} = \frac{1}{\sqrt{2}}\begin{pmatrix}1 &0 &0 &1 \\ 0 &1 & 1 & 0 \\ 0 &i &-i &0 \\ 1 &0 &0 &-1 \end{pmatrix} \begin{pmatrix} |0\rangle\langle 0|/\sqrt{2} \\ |0\rangle\langle 1|/\sqrt{2} \\ |1\rangle\langle 0|/\sqrt{2}\\ |1\rangle\langle 1|/\sqrt{2} \end{pmatrix}.$$ The $4\times 4$ matrix (with the factor of $1/\sqrt{2}$ is clearly unitary. You can probably also find a unitary relating Paulis to $|i\rangle\langle j|$ for higher dimensions that are powers of 2, but I haven't thought about it. $\endgroup$
    – kianna
    Commented Feb 23, 2023 at 13:33

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