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I've been curious recently about considering quantum channels whose Choi matrices are strictly real in the computational basis. Given the Choi matrix of a quantum channel $$C=\sum_{i,j=1}^d \mathcal E(|i\rangle\langle j|) \otimes |i\rangle\langle j|$$ we may calculate the action of the channel on a pure state as $$\mathcal E(|\psi\rangle\langle \psi|)= \text{Tr}_2[C(I \otimes |\psi \rangle\langle \psi|^T)].$$ This is interesting as these channels will take states which have real density matrices to other states which have real density matrices. My question is, given two pure qubit states $|\psi \rangle = xe^{i\theta}|0\rangle +y |1\rangle$ and $|\phi \rangle = ze^{i\alpha}|0\rangle +t |1\rangle$ where $x,y,z,t,\theta,\alpha \in \mathbb R$ and $x^2+y^2=z^2+t^2=1$, when does there exist a real Choi matrix $C$ such that $\mathcal E(|\psi\rangle\langle \psi|)=|\phi\rangle\langle \phi|$?

I've tried doing this explicitely by writing out the general form of a real Choi matrix: $$C = \begin{bmatrix}c_1 & c_2 & c_3 & c_4 \\ c_2 & c_5 & c_6 & c_7 \\ c_3 & c_6 & c_8 & c_9 \\ c_4 & c_7 & c_9 & c_{10} \\\end{bmatrix}$$ where $c_k \in \mathbb R$ for each $k=1,..,10$. The channel must be trace preserving so $$\text{tr}_1[C] =\begin{bmatrix} c_1+c_8 & c_2+c_9 \\ c_2 + c_9 & c_5 + c_{10}\end{bmatrix} = \begin{bmatrix}1 &0 \\ 0 & 1\end{bmatrix}$$ and thus $c_8 = 1-c_1, c_9 = -c_2$ and $c_{10}=1-c_5$. Therefore $$C = \begin{bmatrix}c_1 & c_2 & c_3 & c_4 \\ c_2 & c_5 & c_6 & c_7 \\ c_3 & c_6 & 1-c_1 & -c_2 \\ c_4 & c_7 & -c_2 & 1-c_5 \\\end{bmatrix}.$$ Now, $$\text{Tr}_2[C(I \otimes |\psi \rangle\langle \psi|^T)] = \begin{bmatrix}x^2 c_1 +2xy c_2 \cos(\theta)+y^2 c_5 & x^2 c_3 +c_4xye^{i\theta}+c_6 xy e^{-i\theta} +y^2 c_7 \\ x^2 c_3 +c_4xye^{-i\theta}+c_6 xy e^{i\theta} +y^2 c_7 & x^2 (1-c_1) -2xy c_2 \cos(\theta)+y^2 (1-c_5) \end{bmatrix} = \begin{bmatrix}z^2 & zte^{i \alpha} \\ zte^{-i\alpha} & t^2 \end{bmatrix}$$ which yields the equations $$c_1 x^2 + 2c_2 xy \cos(\theta) = c_5 y^2 = z^2$$ $$c_3 x^2 +(c_4+c_6)xy\cos(\theta) + c_7 y^2 = zt \cos(\alpha)$$ $$(c_4-c_6)xy \sin(\theta) = zt \sin(\alpha).$$ We therefore have 7 unknown variables and only 3 equations to determine the Choi matrix which does the job. Is there a general way to find conditions on when a real Choi matrix exists which does the transformation in terms of the given states ($x,y,\theta,z,t,\alpha$)? How would we impose the positivity of $C$ on the variables $c_1,..,c_7$?

Given Norbert Schuch's comment, we can expand the input state in terms of the Pauli basis $$|\psi\rangle\langle \psi| = \frac 12(I + u\sigma_1+v\sigma_2+w\sigma_3)$$ where $u^2+v^2+w^2=1$. We can then think of a completely positive map as a $\mathbb T$ matrix acting on the Bloch vector $\vec b = (1,u,v,w)^T$ where $$\mathbb T = \begin{bmatrix} 1 & 0 & 0 & 0\\ t_1 &k_1& k_2& k_3 \\ t_2 &k_4&k_5& k_6\\ t_3 &k_7&k_8&k_9 \\\end{bmatrix}.$$ The action of the qubit channel is then defined by the action of $\mathbb T$ on the Bloch vector, i.e. $\cE(|\psi\rangle\langle \psi|)$ has bloch vector $\mathbb T \vec b$. The associated Choi matrix to this operation is $$C = \frac 12\begin{bmatrix} 1+t_3+k_9 & k_7+ik_8 & t_1 + k_3-i(t_2+k_6) & k_1+k_5+i(k_2-k_4) \\ k_7-ik_8 & 1+ t_3-k_9 & k_1-k_5-i(k_2+k_4) & t_1-k_3-i(t_2-k_6) \\ t_1+k_3+i(t_2-k_6) & k_1-k_5+i(k_2+k_4) & 1-t_3-k_9 & -k_7-ik_8 \\ k_1+k_5-i(k_2-k_4) & t_1-k_3+i(t_2-k_6) & -k_7+ik_8 & 1-t_3 +k_9\end{bmatrix}.$$ We therefore restrict ourselves to operations with $k_2=t_2=k_4=k_6=k_8=0$. The T-matrix then takes the form $$\mathbb T = \begin{bmatrix} 1 & 0 & 0 & 0\\ t_1 &k_1& 0& k_3 \\ 0 &0&k_5& 0\\ t_3 &k_7&0&k_9 \\\end{bmatrix}$$ and the Choi matrix is $$C = \frac 12\begin{bmatrix} 1+t_3+k_9 & k_7 & t_1 + k_3 & k_1+k_5 \\ k_7 & 1+ t_3-k_9 & k_1-k_5 & t_1-k_3 \\ t_1+k_3 & k_1-k_5 & 1-t_3-k_9 & -k_7 \\ k_1+k_5 & t_1-k_3 & -k_7 & 1-t_3 +k_9\end{bmatrix}.$$ How then do we know if there is a Choi matrix of this form such that we can transform $|\psi\rangle$ to $|\phi\rangle$? It seems I am in the same place as before as I must impose positivity constraints on $C$.

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  • $\begingroup$ Why would you think such a Choi matrix exists in the first place? $\endgroup$ – Norbert Schuch Aug 21 '17 at 10:53
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    $\begingroup$ There are some succinct characterizations of qubit quantum channels (in particular positivity of the Choi matrix) using their representation in the Pauli basis. Might be helpful here. $\endgroup$ – Norbert Schuch Aug 21 '17 at 23:25
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    $\begingroup$ Why should T have this diagonal form? I think all you can say is that the first row is (1,0,0,0) (=trace preserving). $\endgroup$ – Norbert Schuch Aug 22 '17 at 11:11
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    $\begingroup$ T is in this form only up to a unitary transformation on the channel (see here ). So finding the case where $t_2=0$ does not guarantee that the true channel (after applying the appropriate unitary) will be real. $\endgroup$ – Joel Klassen Aug 22 '17 at 11:41
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    $\begingroup$ The same, but with less zeros. $\endgroup$ – Norbert Schuch Aug 24 '17 at 14:36

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