$\DeclareMathOperator\tr{tr} $Is the following statement true?

Conjecture: Let $\cal E_1,\cal E_2$ be completely positive trace-preserving maps (quantum superoperators). Assume that for any positive Hermitean operator $\rho$ with $\tr\rho=1$ (density operator), we have $\cal E_1(\rho)=\cal E_2(\rho)$. Then $\cal E_1=\cal E_2$.

The intuitive meaning of this is: if two operations on a quantum system have exactly the same effect ($\forall\rho.\cal E_1(\rho)=\cal E_2(\rho)$), then they also have the same effect when acting on a subsystem of a larger composite system ($\cal E_1=\cal E_2$).

At the first glance one would expect this to hold, but I cannot prove it.

What I found out myself:

  • Every Hermitean operator operator can be decomposed into a linear combination of density operators, hence we have $\cal E_1(\rho)=\cal E_2(\rho)$ for all Hermitean $\rho$. It remains only to show that this implies $\cal E_1(\rho)=\cal E_2(\rho)$ for anti-Hermitean $\rho$ as well.

Given that the $\mathcal E_i$ are linear operators, i.e., $\mathcal E_i(\alpha A + B)=\alpha \mathcal E_i(A) + \mathcal E_i(B)$ for all $A$ and $B$, it is true that $\mathcal E_1(\rho)=\mathcal E_2(\rho)$ for all $\rho\ge0$ implies that $\mathcal E_1=\mathcal E_2$.

The argument goes as follows: Every operator $A$ can be written as $A=X+iY$, with $X=\tfrac12(A+A^\dagger)$ and $Y=\tfrac{-i}{2}(A-A^\dagger)$ hermitian. As you already noticed, every hermitian operator can be written as difference of two positive operators. Using linearity, the claim follows.

Note that your "counterexample" $\mathcal E_2(A) = A^\dagger$ is not linear.

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  • $\begingroup$ Thanks. I agree that my counterexample is incorrect, only the transpose is linear, not the conjugate transpose. I removed my counterexample. $\endgroup$ – Dominique Unruh Dec 17 '14 at 13:08

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