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According to Wigner’s theorem, every symmetry operation must be represented in quantum mechanics by an unitary or an anti-unitary operator. To see this, we can see that given any two states $|\psi\rangle$ and $|\psi'\rangle$, you would like to preserve

$$|\langle\psi|\psi\rangle'|^2=|\langle O\psi|O\psi \rangle'|^2 \tag{1}$$

under some transformation $O$. If $O$ is unitary, that works. Yet, we can verify that an anti-unitary $A$ operator such that

$$\langle A\psi| A\psi'\rangle= \langle\psi|\psi'\rangle^* ,\tag{2}$$ works too; where ${}^*$ is the complex conjugate. Note that I cannot write it as $\color{red}{\langle \psi| A^\dagger A|\psi\rangle'}$ as $A$ does not behave as usual unitary operators, it is only defined on kets $|A\psi\rangle=A|\psi\rangle$ not bras.

I was wondering if one could make an anti-unitary gate? Would that have any influence on what can be achieved with a quantum computer?

Example

I was thinking on some kind of time reversal gate $T$. The time reversal operator is the standard example of anti-unitarity.

As a qubit is a spin-1/2 like system, for a single qubit we need to perform $$T=iYK\tag{3}$$ where $K$ is the complex conjugate. If you have an state $$|\psi\rangle=a|0\rangle+b|1\rangle\tag{4}$$ then $$K|\psi\rangle=a^*|0\rangle+b^*|1\rangle.\tag{5}$$ I do not think you can build the operator $K$ from the usual quantum gates without knowing the state of the qubit beforehand. The gate has to be tailored to the state of the qubit.

Experiments

I know now of one preprint Arrow of Time and its Reversal on IBM Quantum Computer (2018) where the authors claim to have achieved this on a IBM backend. But I do not know if their time reversal algorithm is the same as making an anti-unitary gate.

Mauricio
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    What is an anti-unitary gate? $U U^\dagger = -I$? – Rammus Aug 17 '21 at 16:15
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    @Rammus I think it is more than that. You need some anti-linearity. An example is the complex conjugate operation $K$. If you have a statevector $[a,b]$ then $K[a,b]=[a^,b^]$ . It can be antilinear and still $KK^\dagger=I$ – Mauricio Aug 17 '21 at 16:29
  • Ahh interesting, maybe it is good to add the definition of antiunitary operators to the question. – Rammus Aug 17 '21 at 16:45
  • @Rammus maybe what I said is not entirely correct. Now I am not sure of what $KK^\dagger$ does. – Mauricio Aug 17 '21 at 16:57
  • Anti-unitary gates do not commute with complex conjugation, eg https://physics.stackexchange.com/a/187980/291677 – Quantum Mechanic Aug 17 '21 at 18:39
  • @Rammus also https://physics.stackexchange.com/q/16193/291677 – Quantum Mechanic Aug 17 '21 at 18:45
  • Anti-unitaries are not completely positive, since they require complex conjugation, or equivalently, transposition. The common consensus is that as such they do not correspond to physical operations. – Markus Heinrich Aug 18 '21 at 08:05
  • would such a thing even be a linear operator? At least in finite dimensions, it doesn't seem possible. You'd have e.g. $e^{i\phi}=\langle Ae_1,A e^{i\phi}e_1\rangle=\langle e_1,A^\dagger A (e^{i\phi}e_1\rangle= e^{-i\phi}$. EDIT: sorry, posted comment at the same time. I agree with you, but then, if it's not even a linear operator, how does it make sense to "apply it to states"? – glS Aug 18 '21 at 09:20
  • @glS no, anti unitary operators are anti linear. Edit: sorry, we are talking past each other. Check Wigners theorem to know its utility. My example is that you could try to build a time reversal operation. The solution below also links to a paper to build an orthogonal operator. – Mauricio Aug 18 '21 at 09:21
  • @glS well, the map $\phi(X) = V X V^\dagger$ is linear for anti-unitary $V$. In fact, you can always rewrite it as $\phi(X)=UX^\top U^\dagger$ for a suitable unitary $U$. The transposition there renders it non-CP though (see above comment). – Markus Heinrich Aug 18 '21 at 11:56
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    Mauricio: ah, I think I understand now. An antilinear map still defines a proper action on the projective space, thus it's still meaningful to ask how it acts on quantum states. I never thought about it this way, it's really interesting. Also, an antilinear $f:V\to V$ is still linear when thought of as $f:\bar V\to V$ with $\bar V$ complex conjugate space, so it might make sense to describe it as a matrix maybe? Granted, care would likely be needed when performing things such as basis changes on such a "matrix representation" – glS Aug 18 '21 at 12:03
  • @MarkusHeinrich interesting! I suppose this matches with what I wrote above: because you still have a well-defined action of antilinear maps on rays, you'll get a well-defined action (which now oughts to be linear) of antilinear on density matrices, which is, if I understand correctly, what you are pointing out. It not being CP I suppose shows that it is well defined only as long as you don't think it applied to entangled states, which shows that the overall mapping is indeed non-physical – glS Aug 18 '21 at 12:05
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    @glS indeed, the action on rays is equivalent to action on projectors ... And yes, the action on density matrices is well-defined since the mapping is positive. Finally, you're completely right, applying it to one half of an entangled state will bring you out of state space. – Markus Heinrich Aug 18 '21 at 12:09
  • In non-relativistic quantum mechanics, a closed system evolves unitarily, so unless the anti-unitary gate is unitary, it doesn't seem appropriate to use as a "gate"! – user1271772 No more free time Aug 18 '21 at 13:00

1 Answers1

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No, you can't perform anti-unitary gates. Or rather: if they are possible, then faster than light communication is possible.

For example, here is a circuit that remotely detects whether or not a conjugation gate was performed:

enter image description here

This works because conjugating sqrt(X) is equivalent to inverting it. The conjugation between the X^(1/2) and the X^(-1/2) makes them add up to an X instead of cancelling to an I. So they act as conjugation detectors.

(It might be a bit silly to associate a conjugation with a specific qubit, as it is a global effect. Perhaps it is coming along for the ride as part of another operation, such as the "Universal Not".)

Of course, this diagram assumes a particular notion of simultaneity (indicated by the dashed vertical line). But that means antiunitary operations would reveal a preferred rest frame to the universe! In fact it's even worse than that. Antiunitary operations also break rotational symmetry! The X rotations act as conjugation detectors, as do Z rotations, but Y rotations don't, so you could find the preferred Y axis of the universe. (Well... okay, technically the exact thing you expect to break, and exactly how you expect it to break, will depend on what your favorite interpretation of quantum mechanics is.)

Craig Gidney
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    Sorry, could you explain a bit more the details of the circuit ? I am not sure I am getting what it’s supposed to do. – Mauricio Aug 17 '21 at 20:04
  • Also thanks for the paper! I would also argue that their universal NOT (a gate that finds an orthogonal state to any state) is exactly the time reversal $T$ I defined above (as long as we talk of single qubits). – Mauricio Aug 17 '21 at 20:10
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    @Mauricio There's not much to say about the circuit. The sqrt(X) gate has the property that its conjugation is equal to its inverse, so when you put a conjugation between it and its inverse you un-inverse the inverse so you get a full X gate instead of cancelling to an I gate. So the two sqrt X pieces act as a conjugation detector that works no matter where the conjugation occurs. – Craig Gidney Aug 17 '21 at 21:02
  • Fun fact: QM threw a monkey wrench into what was thought to be a proof that FTL communication was impossible, and it's still undecided. We believe FTL is simply not possible but can't prove it. – Joshua Aug 18 '21 at 03:53
  • Maybe it would help to clean up the circuit. You could remove all the unneccesary registers (all but the first and second last). Also you can remove the "off" and "on" (what do they mean anyway?). – M. Stern Aug 19 '21 at 06:16
  • @M.Stern The intermediate qubits are for showing distance is irrelevant. The Off and On are state displays showing that it actually works. This is just a screenshot of what's shown in the linked simulator. – Craig Gidney Aug 19 '21 at 14:13