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I was trying to solve an exercise from Vazirani's course "Qubits, Quantum Mechanics and Computers":

A mathematically nice, but unphysical, way to detect entanglement is to use the state inversion operator, $T$, which, for all states $n$ acts as $T\vert n\rangle = \vert -n\rangle$. Why can’t this operator be realized physically?

Can anyone help me with this problem? Many thanks! The problem is from https://inst.eecs.berkeley.edu/~cs191/fa10/homework/hw7.pdf

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    $\begingroup$ I'm a bit confused about what exactly the state $|n\rangle$ (and conversely, $|-n\rangle$) exactly is. I assume it's not $|-n\rangle = |(-1)n\rangle = (-1)|n\rangle$? Because then it is just an (irrelevant) global phase... $\endgroup$
    – JSdJ
    Nov 11 '20 at 18:59
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    $\begingroup$ Also, for HW problems, I'd recommend you show some work / attempt to talk about your underlying logic $\endgroup$
    – C. Kang
    Nov 11 '20 at 19:18
  • $\begingroup$ Suppose there are N superpositions. Then I guess |-n> denotes |N - n> for n = 0, 1, ..., N-1. $\endgroup$ Nov 11 '20 at 20:06
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    $\begingroup$ @DavidLeiden As far as I can tell, that's not what the operator $T$ does (or if your definition is correct, at least refer to a document that contains this definition -- the one you attach above doesn't). It seems that the action of $T$ is to invert the Bloch vector, namely, take a state $|\vec{n}\rangle \overset{T}{\mapsto} |-\vec{n}\rangle$; as an example, it should flip the $\sigma_z$ eigenstates $|\vec{z+}\rangle \overset{T}{\mapsto} |\vec{z-}\rangle$. For example, $\sigma_x$ would be a state-inversion operator for these states. $\endgroup$ Nov 12 '20 at 2:11
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    $\begingroup$ @keisuke.akira after some thought I also think this is the case. Your argument together with imposing linearity on $T$ could already proof the statement though. $\endgroup$
    – JSdJ
    Nov 12 '20 at 8:50
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If you're talking about a single-qubit transformation where the Bloch vector is changed from $\vec{n}$ to $-\vec{n}$, then I would way think about how the Bloch vector is changing. You want it to go from $$ (n_x,n_y,n_z)\rightarrow(-n_x,-n_y,-n_z). $$ However, if we do any unitaries we want on the initial state, this doesn't change the feasibility of the operation, so let me start by applying pauli-$Y$. Hence we have achieved the transformation $$ (n_x,n_y,n_z)\rightarrow(-n_x,n_y,-n_z), $$ and the only step left is an effective $$ (n_x,n_y,n_z)\rightarrow(n_x,-n_y,n_z). $$ However, this is the transpose operation which, given the question, I'm assuming you already know to be impossible, and hopefully know how to prove is impossible (probably based on completely positive maps).

This is a reduction to a known problem. You could probably prove it more directly now. For example, take a suitable two-qubit state and show that the action of $T$ on one qubit does not leave a proper quantum state.

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  • $\begingroup$ Good answer. Thus, the map $T$ on Hilbert space is anti-unitary, $T=\mathcal{C}Y$ with $\mathcal{C}$ the complex conjugation. $\endgroup$ Nov 12 '20 at 15:35
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Well for starters it violates the no-communication theorem:

circuit

This is because under the hood the UniversalNot operation uses a conjugation. This breaks the symmetry between the imaginary and real parts of the wavefunction, so you can detect the UniversalNot happening anywhere else by preparing a local state like $|0\rangle + i|1\rangle$ and noticing the $|1\rangle$ component changing w.r.t. the $|0\rangle$ component.

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