6
$\begingroup$

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.

$\endgroup$
14
  • 2
    $\begingroup$ What is an anti-unitary gate? $U U^\dagger = -I$? $\endgroup$
    – Rammus
    Aug 17 at 16:15
  • 1
    $\begingroup$ @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 $KK^\dagger=I$ $\endgroup$
    – Mauricio
    Aug 17 at 16:29
  • $\begingroup$ Ahh interesting, maybe it is good to add the definition of antiunitary operators to the question. $\endgroup$
    – Rammus
    Aug 17 at 16:45
  • 1
    $\begingroup$ 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" $\endgroup$
    – glS
    Aug 18 at 12:03
  • 1
    $\begingroup$ @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. $\endgroup$ Aug 18 at 12:09
6
$\begingroup$

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.)

$\endgroup$
6
  • 3
    $\begingroup$ 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. $\endgroup$
    – Mauricio
    Aug 17 at 20:04
  • $\begingroup$ 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). $\endgroup$
    – Mauricio
    Aug 17 at 20:10
  • 3
    $\begingroup$ @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. $\endgroup$ Aug 17 at 21:02
  • $\begingroup$ 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. $\endgroup$
    – Joshua
    Aug 18 at 3:53
  • $\begingroup$ 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?). $\endgroup$
    – M. Stern
    Aug 19 at 6:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.