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A paper I read talked about local interactions on only a single subsystem, leaving the state of that subsystem invariant, and the observation of this requiring the whole state.

Here is the paper

Now their overall claims of some extension to Quantum Discord not withstanding, or it's similarities to other local computation mixed state work, for example 9, they talk about how to take

$$\frac{1}{4}(|00\rangle \langle00|+|01\rangle \langle01|+|10\rangle \langle10|+|11\rangle \langle11|$$ to $$\frac{1}{2}(|01\rangle \langle01|+|10\rangle \langle10|)$$

via a local operation on subsystem A via the use of the pauli $X$ gate. However, the fully mixed state should be invariant to that interaction along with the first subsystem, as $|00\rangle \langle00| \to |10\rangle \langle10|$, $|10\rangle \langle10| \to |00\rangle \langle00|$, $|01\rangle \langle01| \to |11\rangle \langle11|$ and $|11\rangle \langle11| \to |01\rangle \langle01|$, unless I have made some error here.

They talk about how some third party, Charlie, in this case, could do so for $|00\rangle \langle00|$ and $|11\rangle \langle11|$ by applying $X$ locally, but to avoid doing so for the other two states would require a controlled operation, and as such would no longer be local to subsystem A. Have I made some oversight here, perhaps reading their paper incorrectly or is this part of their paper wrong?

I mean their whole paper seems focused on only local operations that achieve this "quantum house effect", so if they just arbitrarily extend it to non-local ones, that seems to invalidate it for me.

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    $\begingroup$ from a cursory read of the text in example 9 in the paper, it doesn't seem to me like they're applying a local operation. They say Charlie applies $X$ to the first qubit, conditionally to the two qubits being either $|00\rangle$ or $|11\rangle$. In other words, it's a local operation conditioned to the state of both qubits, ie an overall nonlocal operation (or in other words, yes I agree with your assessment). In fact, a local operation cannot create correlations between the two parties, thus it cannot send $I/4$ to that state, which features full (classical) correlations $\endgroup$
    – glS ♦
    Commented May 26, 2022 at 16:52
  • $\begingroup$ Given the mixed state, they'd need a conditional quantum channel, with the channel being conditioned on which of the density operators of the mixed state it's applied to, correct? $\endgroup$
    – GaussStrife
    Commented May 27, 2022 at 9:53
  • $\begingroup$ yes, I'd say so $\endgroup$
    – glS ♦
    Commented May 27, 2022 at 10:45

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Thank you for pointing this out. I've changed in the paper "local operation on subsystem $A$" to "operation on subsystem $A$", and made it clear it merely means that the operation is performed inside Alice's lab. The important thing (the focus of the paper) is that the effect shouldn't be possible according to classical intuition.

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