# What is the Kraus representation of the quantum channel with Choi $\lambda |\phi^+\rangle \langle\phi^+| + (1-\lambda )|\phi^-\rangle \langle\phi^-|$?

This matrix

$$c_{\lambda} = \lambda |\phi^+\rangle \langle\phi^+| + (1-\lambda )|\phi^-\rangle \langle\phi^-|$$

is the Choi–Jamiołkowski matrix of a quantum channel for any $$\lambda \in [0,1]$$.

The questions I am trying to solve are:

Provide a Kraus (operator-sum) representation of the quantum channel $$T_{\lambda}$$ that is described by $$c_{\lambda}$$ and show that $$T_{1/2}$$ is an entanglement breaking quantum channel and describe its action on the bloch sphere.

• – glS
Mar 8, 2021 at 19:18
• Yes, I wrote that one :) @glS Mar 8, 2021 at 19:35
• I'm aware. I added the comment because it creates a two-way link between the questions
– glS
Mar 8, 2021 at 20:47

Another way is to observe that Choi $$J(\Phi)\in\mathrm{Lin}(\mathcal Y\otimes\mathcal X)$$ and Kraus operators $$\{A_a\}_a\subset\mathrm{Lin}(\mathcal X,\mathcal Y)$$ of a map $$\Phi:\mathrm{Lin}(\mathcal X)\to\mathrm{Lin}(\mathcal Y)$$ are directly related via $$J(\Phi) = \sum_a \operatorname{vec}(A_a)\operatorname{vec}(A_a)^\dagger,$$ where $$\operatorname{vec}(A_a)\in\mathcal Y\otimes\mathcal X$$ is the vector with components $$\operatorname{vec}(A_a)_{ij}=(A_a)_{ij}$$.

Therefore, if a Choi is a rank-one projection of the form $$|u\rangle\!\langle u|$$, then $$\operatorname{vec}(A_a)=|u\rangle$$, i.e. $$A_a=\sum_{ij}\langle i,j|u\rangle |i\rangle\!\langle j|$$, or equivalently $$(A_a)_{ij}=u_{ij}$$.

Finally, $$\sqrt2\langle i,j|\phi^+\rangle = \delta_{ij}$$ and $$\sqrt2\langle i,j|\phi^-\rangle = \delta_{ij}(-1)^i$$, thus \begin{align} 2|\phi^+\rangle\!\langle\phi^+|&\to A_0 = I, \\ 2|\phi^-\rangle\!\langle\phi^-|&\to A_1 = Z. \end{align}

• Am I correct in reading the last two lines as: Choi $|\phi^+\rangle\langle\phi^+|$ corresponds to a single Kraus operator $A_0$ and similarly for the other one? If so, I think normalization is off: I'd expect Kraus operators to satisfy completeness relation $\sum_k A_k^\dagger A_k=I$. Similarly, trace of a Choi should be $d=2$, not $1$. IOW, I think it should say $2|\phi^+\rangle\langle\phi^+|\to A_0 = I, 2|\phi^-\rangle\langle\phi^-|\to A_1 = Z$. Then this answer yields the same Kraus operators as the other two. Or perhaps there is another way of reading the last two lines? Mar 9, 2021 at 17:22
• @AdamZalcman indeed, you are right. I fixed the normalisation in the last equations, thanks
– glS
Mar 9, 2021 at 21:57

There's even a more direct way than the one described by Adam. Note that every pure Choi state corresponds to a superoperator acting by conjugation. Since $$\mathbb{I} \otimes \mathbb{I} |\phi^+\rangle = |\phi^+\rangle, \qquad Z \otimes \mathbb{I} |\phi^+\rangle = |\phi^-\rangle,$$ the inverse of $$c_\lambda$$ under the Choi-Jamiołkowski isomorphism is simply $$T_\lambda = \lambda\, \mathrm{id} + (1-\lambda)\, Z \cdot Z^\dagger.$$ This gives you the Kraus operators $$K_1 := \sqrt{\lambda}\,\mathbb{I}, \qquad K_2:= \sqrt{1-\lambda}\, Z.$$

Remark: In general, you can find the Kraus operators by "reshaping" the pure Choi states, since $$\mathcal{J}(K\cdot K^\dagger) = d^{-1}\, \mathrm{vec}(K)\mathrm{vec}(K)^\dagger,$$ where the vectorisation isomorphism $$\mathrm{vec}:\, L(\mathcal H)\simeq \mathcal H\otimes \mathcal H^* \rightarrow \mathcal H\otimes \mathcal H$$ acts as $$\mathrm{vec}(|x\rangle\langle y|) = |x\rangle \otimes | y\rangle$$ i.e. it applies the Riesz isomorphism to the second factor.

Edit: Sorry, I used two different conventions for the Choi-Jamiołkowski isomorphism, one where the image of a CPTP map has trace one and the other where it has trace $$d$$ (=dimension). Fixed the second part since OP uses first convention.

• ah, sorry, I just noticed you later also included the path via the direct relation between Choi and Kraus
– glS
Mar 9, 2021 at 9:33
• @glS wow, clash of answers :D There was maybe a minute between answering and editing :) Mar 9, 2021 at 14:37
• +1 Very direct and beautiful solution indeed. Mar 9, 2021 at 17:27

Assuming $$|\phi^+\rangle = |00\rangle+|11\rangle$$ and $$|\phi^-\rangle = |00\rangle-|11\rangle$$ we compute

$$c = \begin{pmatrix} 1 & 0 & 0 & 2\lambda-1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 2\lambda-1 & 0 & 0 & 1 \end{pmatrix}\tag1.$$

A useful property of the $$d^2\times d^2$$ Choi-Jamiołkowski matrix is that each of the $$d^2$$ $$d\times d$$ blocks represents the action of the channel on the matrices from the standard basis of the space of $$d\times d$$ matrices. In our case $$d=2$$ and

$$c = \begin{pmatrix} T\begin{pmatrix}1&0\\0&0\end{pmatrix} & T\begin{pmatrix}0&1\\0&0\end{pmatrix}\\ T\begin{pmatrix}0&0\\1&0\end{pmatrix} & T\begin{pmatrix}0&0\\0&1\end{pmatrix} \end{pmatrix}.\tag2$$

Now, recall that the action of the phase damping channel

$$\mathcal{F}(\rho) = p\rho + (1-p)Z\rho Z\tag3$$

on a density matrix $$\rho = \begin{pmatrix}\rho_{00}&\rho_{01}\\\rho_{10}&\rho_{11}\end{pmatrix}$$ is

$$\mathcal{F}(\rho) = \begin{pmatrix} \rho_{00} & (2p-1)\rho_{01} \\ (2p-1)\rho_{10} & \rho_{11} \end{pmatrix}.\tag4$$

Substituting $$\begin{pmatrix}1&0\\0&0\end{pmatrix}$$, $$\begin{pmatrix}0&1\\0&0\end{pmatrix}$$, $$\begin{pmatrix}0&0\\1&0\end{pmatrix}$$ and $$\begin{pmatrix}0&0\\0&1\end{pmatrix}$$ into $$(4)$$ and comparing against $$(1)$$ and $$(2)$$, we see that $$c$$ is the Choi-Jamiołkowski matrix of the phase damping channel with $$p=\lambda$$.

Consequently, $$(3)$$ is the Kraus representation of the channel with Choi-Jamiołkowski matrix $$c$$. The channel shrinks the equator of the Bloch sphere, see e.g. figure 8.9 on page 377 in Nielsen & Chuang. Finally, when $$\lambda=\frac{1}{2}$$ then the output of the channel is separable regardless of the input.

• Thanks for the answer. Why do we need the phase damping channel? Mar 8, 2021 at 19:00
• As the answer shows, you are asking about the Choi-Jamiołkowski matrix of the phase damping channel. This realization gives us the Kraus representation. Mar 8, 2021 at 19:09
• But in my question it isn't obvious that the C-J matrix is for the phase damping channel. I'm not seeing how we should use the phase damping channel. It's just "the quantum channel" in the question. Mar 8, 2021 at 19:39
• It isn't obvious. However, when you calculate the elements of $c$, i.e. $(1)$ then the elements $2\lambda - 1$ are very suggestive since they look like the $2p-1$ factors by which the phase damping channel multiplies the off-diagonal elements of the input density matrix, see $(4)$. Mar 8, 2021 at 19:43
• As usual, there are many paths to solution. IMHO, recognizing and exploiting well-known objects when they appear in a chain of reasoning leads to simpler, faster and more interesting solutions than a run-of-the-mill calculation :-) Mar 8, 2021 at 19:56