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I would like to compute the diamond norm of the difference of two single-qubit channels $\Phi_1$ and $\Phi_2$. This difference is equal to, for any $2\times2$ complex matrix $\rho$: $$\DeclareMathOperator{tr}{tr}\Phi_1(\rho)-\Phi_2(\rho)=\beta\rho+\gamma Z\rho Z+\delta\tr(\rho)|0\rangle\!\langle0|+\varepsilon\tr(\rho)|1\rangle\!\langle1|$$ with $\beta+\gamma+\delta+\varepsilon=0$.

Is it possible to obtain a closed-form expression of $\left\|\Phi_1-\Phi_2\right\|_{\diamond}$?

Since this difference is Hermitian preserving, I intended to use the following formula: $$\left\|\Phi_1-\Phi_2\right\|_{\diamond}=\max_{|\psi\rangle\in\mathbb{C}^4}\left\|\left(\Phi_1-\Phi_2\right)(|\psi\rangle\!\langle\psi|)\right\|_1$$ However, the computations get quite overwhelming, not to mention that I would have to optimize over the coefficients of $|\psi\rangle$ thereafter. Furthermore, since I'm interested in a closed form expression, I can't use the semidefinite problem the diamond norm is associated with.

Is there a way for me to compute this diamond norm? If there's not, is it possible if I set some coefficients to $0$?

Tristan Nemoz
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    doesn't answer the question, but I think you can remove at least one of the parameters in the general expression: $\alpha$ seems redundant with $\delta$ and $\varepsilon$. – glS Feb 23 '23 at 17:34
  • @glS Definitely! I thought about putting $\delta$ and $\varepsilon$ inside $\alpha$, which doesn't work, and for some reason didn't even think about doing the converse. Thanks! – Tristan Nemoz Feb 23 '23 at 17:51

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