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In Nielsen and Chuang, 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

$\frac{\mathbb{1}}{2} = \frac{\rho + X\rho X + Y\rho Y +Z\rho Z}{4}$.

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

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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 $\sigma$.

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  • $\begingroup$ 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. $\endgroup$ – user1936752 Dec 16 '18 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? $\endgroup$ – Norbert Schuch Dec 16 '18 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). $\endgroup$ – user1936752 Dec 16 '18 at 22:26
  • $\begingroup$ @user1936752 Have edited. $\endgroup$ – Norbert Schuch Dec 17 '18 at 0:17
  • $\begingroup$ Thank you very much $\endgroup$ – user1936752 Dec 17 '18 at 11:41

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