Show that $\lambda |\phi^+\rangle \langle\phi^+| + (1-\lambda )|\phi^-\rangle \langle\phi^-|$ is the Choi–Jamiołkowski matrix of a quantum channel

I'm curious how to show how this matrix:

$$c = \lambda |\phi^+\rangle \langle\phi^+| + (1-\lambda )|\phi^-\rangle \langle\phi^-|$$

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

• Hint: It's an isomorphism so you need only show that the matrix $c$ satisfies the properties that a Choi matrix of the channel would satisfy. That is, it is positive semidefinite and that one of its partial traces gives the identity matrix on the other subsystem. Alternatively, you could use the explicit inverse map to reconstruct the action of the superoperator and show that this superoperator is a channel. Mar 8 at 10:28
• Thanks for the hint! What exactly do you mean by "gives the identity matrix on the other subsystem"? What other subsystem are you referring to? Mar 8 at 10:32
• Let $X,Y$ be Hilbert spaces, $L(X), L(Y)$ denote linear operators on those hilbert spaces. Then the isomorphism is between maps $\Phi: L(X) \rightarrow L(Y)$ and $L(X\otimes Y)$ ( some also define the isomorphism to $L(Y\otimes X)$). So if you want $c$ to represent a channel you first need $c$ to be a bipartite operator. I'm guessing $|\phi^{(+/-)}\rangle$ are Bell-states so the natural choice would be the two qubit subsystems. The channel it would define would then be a single qubit channel. Then, you will need that if you trace out one of the qubits you are left with the identity matrix. Mar 8 at 10:51
• Thanks @Rammus your comment helped me solve it. Mar 8 at 11:55

I'll use notation inspired from Watrous' book. Let $$\Phi:\mathrm{Lin}(\mathcal X)\to\mathrm{Lin}(\mathcal Y)$$ be a quantum channel (i.e. a CPTP map). Define its Choi representation as the operator $$J(\Phi)\in\mathrm{Lin}(\mathcal Y\otimes\mathcal X)$$ defined by $$J(\Phi) = \sum_{ij} \Phi(E_{ij})\otimes E_{ij},\qquad E_{ij}\equiv |i\rangle\!\langle j|.$$ Then you can verify that $$J(\Phi)\ge0$$ and $$\operatorname{Tr}_{\mathcal Y}(J(\Phi))=I_{\mathcal X}$$.

Vice versa, let $$A\in\mathrm{Lin}(\mathcal Y\otimes\mathcal X)$$ be a positive semidefinite linear operator such that $$\operatorname{Tr}_{\mathcal Y}(A)=I_{\mathcal X}$$, and define the linear map $$\Phi_A\in\mathrm{Lin}(\mathrm{Lin}(\mathcal X),\mathrm{Lin}(\mathcal Y))$$ via $$\Phi_A(E_{j\ell}) \equiv \sum_{ik} \langle i,j|A|k,\ell\rangle E_{ik}, \qquad \Phi_A(X) = \operatorname{Tr}_{\mathcal X}[A(I\otimes X^T)].$$ You can then verify that $$\Phi_A$$ is CPTP. Moreover, the two operations are one the inverse of the other: $$\Phi_{J(\Phi)}=\Phi$$.

In conclusion, a linear operator $$A\in\mathrm{Lin}(\mathcal Y\otimes\mathcal X)$$ is the Choi of a CPTP map iff it is positive semidefinite and satisfies the trace property.

In this case, verifying the positivity is immediate. Moreover, assuming $$|\phi^\pm\rangle$$ here denote maximally entangled states, the partial trace gives the identity for both terms, and thus the other condition is also trivially verified for all $$\lambda\in[0,1]$$.

I should note that in the definition I'm using here of "Choi representation" isn't a state, in that it's not normalised: $$\mathrm{Tr}(J(\Phi))=\operatorname{dim}(\mathcal X)$$. This is however trivially fixed by adding the appropriate normalisation factor in the definition.

• Thanks for the comment! How is the positivity immediate? Mar 8 at 11:45
• @mikanim because the sum of positive semidefinite operators is positive semidefinite, and $|u\rangle\!\langle u|$ is always positive semidefinite
– glS
Mar 8 at 11:45
• right but $|u \rangle$ in this case is a bipartite state if we look at it in the computational basis. I only see that it is positive semidefinite once I calculate out the matrix into the computational basis, which took 2-3 lines. Could you elaborate a bit on how you found it so quick? Mar 8 at 11:54
• adding questions only makes the post too broad and liable to be closed. You can remove them by editing this post and ask the other questions on a separate thread.
– glS
Mar 8 at 12:18
• it doesn't matter what $|u\rangle$ is. The operator $|u\rangle\!\langle u|$ is always positive semidefinite. An easy way to see it is that it has a single nonzero eigenvalue which is equal to $\|u\|$. Equivalently, you can verify that $\langle \alpha|u\rangle\!\langle u|\alpha\ge0$ for all $|\alpha\rangle$, which is again true because you can verify picking $|\alpha\rangle$ from an orthonormal basis containing $|u\rangle$. Equivalently, if $\|u\|=1$, just notice that $|u\rangle\!\langle u|$ is the projection onto $|u\rangle$.
– glS
Mar 8 at 12:21