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The standard notation of $|X⟩⟩$ for the $4^n$-dimensional vector representing $X$ in the space acted on by superoperators. $E= (|I⟩⟩⟨⟨I|+|X⟩⟩⟨⟨X|)^{⊗2}$.

From this definition, How comes $E|I⟩⟩=(Tr(I) I + Tr(X ⊗ I) X ⊗ I+ Tr(I ⊗ X) I ⊗ X + Tr(X ⊗ X)X ⊗ X)$?

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Let me use the shorter notation $\mathbb{P}(I) \equiv |I\rangle\!\rangle\!\langle\!\langle I|$ and $\mathbb{P}(X) \equiv |X\rangle\!\rangle\!\langle\!\langle X|$ for conciseness.

You have $$ (\mathbb{P}(I)+\mathbb{P}(X))^{\otimes 2}= \mathbb{P}(I)\otimes\mathbb{P}(I) + \mathbb{P}(I)\otimes\mathbb{P}(X) + \mathbb{P}(X)\otimes\mathbb{P}(I) + \mathbb{P}(X)\otimes\mathbb{P}(X), $$ where the operators are by definition such that $\mathbb{P}(X)(Y)=X \langle X,Y\rangle$ for any $Y$. It follows that $$(\mathbb{P}(I)+\mathbb{P}(X))^{\otimes 2} (Y\otimes Z) = (I\otimes I)\langle I,Y\rangle \langle I,Z\rangle + (I\otimes X)\langle I,Y\rangle \langle X,Z\rangle + (X\otimes I) \langle X,Y\rangle \langle I,Z\rangle + (X\otimes X)\langle X,Y\rangle \langle X,Z\rangle.$$ Using $Y=Z=I$ you get your relation.

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  • $\begingroup$ Can you expand $\mathbb{P}(X)(Y)=X⟨X,Y⟩$?, I didn't get it. $\endgroup$
    – karry
    Commented Oct 25, 2023 at 9:19
  • $\begingroup$ that's almost the definition of things like $|X\rangle\!\rangle\!\langle\!\langle X|$. Though technically it should be $\mathbb{P}(X)(Y)=|X\rangle\!\rangle \langle X,Y\rangle$, with the vectorised operator as output, not the operator itself. This is analogous to what a "ketbra" like $|\psi\rangle\!\langle \psi|$ mean: an operator which given any vector $|\phi\rangle$ returns $|\psi\rangle\langle\psi,\phi\rangle$ $\endgroup$
    – glS
    Commented Oct 25, 2023 at 9:34
  • $\begingroup$ But how comes the trace term, $\langle X,Y\rangle=tr(XY^{\dagger})$, it can't produce the $X\otimes Y$ in trace term $tr(X\otimes Y)$ $\endgroup$
    – karry
    Commented Oct 25, 2023 at 13:01
  • $\begingroup$ @Karry $\langle X,Y\rangle=tr(XY^{\dagger})$ is just the definition of inner product in this case, see eg en.wikipedia.org/wiki/Hilbert%E2%80%93Schmidt_operator $\endgroup$
    – glS
    Commented Oct 25, 2023 at 13:03
  • $\begingroup$ thank you, but it seems that I can't get $tr(X\otimes I)$ from your equation $(\mathbb{P}(I)+\mathbb{P}(X))^{\otimes 2} (Y\otimes Z) = (I\otimes I)\langle I,Y\rangle \langle I,Z\rangle + (I\otimes X)\langle I,Y\rangle \langle X,Z\rangle + (X\otimes I) \langle X,Y\rangle \langle I,Z\rangle + (X\otimes X)\langle X,Y\rangle \langle X,Z\rangle$. using $Y=Z=I$ $\endgroup$
    – karry
    Commented Oct 25, 2023 at 13:11

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