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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.

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  • $\begingroup$ related: quantumcomputing.stackexchange.com/q/16393/55 $\endgroup$
    – glS
    Mar 8 at 19:18
  • $\begingroup$ Yes, I wrote that one :) @glS $\endgroup$
    – mikanim
    Mar 8 at 19:35
  • $\begingroup$ I'm aware. I added the comment because it creates a two-way link between the questions $\endgroup$
    – glS
    Mar 8 at 20:47
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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}

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  • $\begingroup$ 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? $\endgroup$ Mar 9 at 17:22
  • $\begingroup$ @AdamZalcman indeed, you are right. I fixed the normalisation in the last equations, thanks $\endgroup$
    – glS
    Mar 9 at 21:57
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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.

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

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  • $\begingroup$ Thanks for the answer. Why do we need the phase damping channel? $\endgroup$
    – mikanim
    Mar 8 at 19:00
  • $\begingroup$ As the answer shows, you are asking about the Choi-Jamiołkowski matrix of the phase damping channel. This realization gives us the Kraus representation. $\endgroup$ Mar 8 at 19:09
  • $\begingroup$ 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. $\endgroup$
    – mikanim
    Mar 8 at 19:39
  • $\begingroup$ 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 $(4)$. $\endgroup$ Mar 8 at 19:43
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    $\begingroup$ 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 :-) $\endgroup$ Mar 8 at 19:56

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