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I recently asked this question about the meaning of a quantum classical channel in a paper I read.

The answer I accepted provided an explanation for the 1-LOCC norm (which I asked about) which is defined to be:

$$ || \: | \psi \rangle \langle \psi | \: ||_{1-LOCC} = max_{\Lambda_{B \rightarrow M}} || (I_A \otimes \Lambda_{B \rightarrow M}) (| \psi \rangle \langle \psi | )||_1 $$

And the explanation was

Basically, the norm represents the largest trace norm of X that can be obtained after performing measurements solely on subsystem B.

But I'm not sure if my original intuition of this is correct.

What I'm thinking right now is this:

Suppose we have some quantum state, $\psi = \psi_1 \otimes \psi_2$. And say we measure $\psi_2$ to be, for example, $1$ after we're given the state.

For a concrete example, let's say we are given the state $\psi = | + + \rangle$ (the hadamard transform on $| 00 \rangle$). After measurement, $\psi = | + 1 \rangle$. If we calculate the singular values (and their sum), both turn out to be $1$. But I don't see how measuring a qubit can actually change the trace norm of a state.

But the paper states

$|| X ||_{1-LOCC} \leq ||X||_1$

So in what case can the $1$-LOCC norm be less than the trace norm?

Improve this question
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  • $\begingroup$ if $\rho$ is a state (normalised density matrix) then $\|\rho\|_1=1$. And if $\Lambda$ is a channel and $X$ a state, then $(I\otimes\Lambda)(X)$ is a state. So maybe here $X$ are not states? The definition is later used (eg 3.2) with things like $X=\phi-\sigma$, so that seems to be the case $\endgroup$
    – glS ♦
    Commented Apr 5, 2023 at 8:55
  • $\begingroup$ @glS The authors define the 1-LOCC norm on all matrices. I guess the inequality is coming from a result like contractivity of the trace-norm under quantum channels. $\endgroup$
    – Rammus
    Commented Apr 5, 2023 at 11:44
  • $\begingroup$ @glS Oh, that looks true. But how would we use a quantum-classical channel on a matrix that isn't a density matrix? $\endgroup$
    – Loic Stoic
    Commented Apr 5, 2023 at 18:42
  • $\begingroup$ @LoicStoic the same way you would otherwise. Any quantum map or channel can be applied to an arbitrary linear operator $\endgroup$
    – glS ♦
    Commented Apr 5, 2023 at 18:52
  • $\begingroup$ Oh, I guess I have to study more about this. $\endgroup$
    – Loic Stoic
    Commented Apr 5, 2023 at 18:53

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