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According to Wigner’s theorem, every symmetry operation must be represented in quantum mechanics by a 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

$$|\langle\psi|\psi\rangle'|^2=|\langle O\psi|O\psi \rangle'|^2$$

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

$$\langle A\psi| A\psi'\rangle= \langle\psi|\psi'\rangle^* ,$$ 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.

Could one build a version of quantum mechanics using only anti-unitary operations?

Any anti-unitary operation is of the form $A=UK$, where $U$ is a unitary operator and $K$ is the complex conjugation operator (which is itself anti-unitary).

For a state $|\psi\rangle=\sum_\lambda c_\lambda |\lambda\rangle$ in some base $\{|\lambda\rangle\}$, then $K|\psi\rangle = \sum_\lambda c_\lambda^* |\lambda\rangle$.

A common example of an anti-unitary operator is the time-reversal operator.

Assumption

It seems to me that I can plug $K$ all over quantum mechanics to make any unitary transformation anti unitary and get back the same results, as it seems to preserve the actual probabilities that can be measured.

However, part of this question came from an earlier question in Quantum Computing Stack Exchange question of mine, where the answer showed that if anti-unitary operations existed, you could build faster-than-light messaging devices.

So what is wrong with my assumption? Is anti-unitary quantum mechanics problematic?

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    $\begingroup$ There is no theory with only anti-unitaries. The product of two anti-unitaries is always unitary. $\endgroup$ Aug 18, 2021 at 10:29
  • $\begingroup$ @AccidentalFourierTransform Good point. $\endgroup$
    – Mauricio
    Aug 18, 2021 at 10:33
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    $\begingroup$ You can't make the time evolution operator $U(t)$ anti-unitary for all $t$ since $U(0)$ needs to be the identity. $\endgroup$
    – jacob1729
    Aug 18, 2021 at 10:39
  • $\begingroup$ I agree with the other comments: it is not possible to use only anti unitary operators. However, this identity $A|\psi \rangle = |A\psi \rangle$ is obscure, I cannot interpret it. $\endgroup$ Aug 18, 2021 at 11:00
  • $\begingroup$ @ValterMoretti Is more of a definition, the problem is that $A$ is not your usual matrix because it is antilinear. $\endgroup$
    – Mauricio
    Aug 18, 2021 at 11:02

1 Answer 1

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You can't make continuous transformations anti-unitary because the product of two anti-unitary transformations must be unitary. As a special case, you cannot make transformations that are continuously connected to unity anti-unitary because an identity operator is not anti-unitary.

Thus, the only transformations that can be represented by anti-unitary operators are discrete. And indeed, this does show up in quantum mechanics. For example, the time-reversal operation is represented by an anti-unitary matrix.

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    $\begingroup$ @NDewolf By a discrete operator, I mean one that belongs to a discrete group. Unlike continuous transformations, there aren't any continuous parameters that you can use to label the elements of such a group. $\endgroup$
    – ACat
    Aug 18, 2021 at 17:39
  • $\begingroup$ Note that the "parity" element of O(2) can be anti-unitary. So the copy of SO(2) disconnected from the identity (which is a continuous set, but not a group) can be anti-unitary. This group of symmetries is important in practice (e.g. cond-mat systems with symmetry $T\times U(1)$). So "discrete operator" might not be the best choice of words. Anti-unitary operators can and often do form continuous sets. Given any theory with one anti-unitary operator $T$ and some continuous group $U$, you can consider symmetries of the form $Tu$ with $u\in U$. These are all anti-unitary yet continuous in $u$. $\endgroup$ Aug 23, 2021 at 15:27
  • $\begingroup$ @AccidentalFourierTransform Yeah, I was thinking about it but I think it's still fair to say "discrete" (or so it seems to me) because, for example, in your example, the continuous parameter is really labeling $U$, not $T$, so the continuity seems "fake" in some sense? I don't know what's the best language to use to convey what I mean but I hope the point is getting across. $\endgroup$
    – ACat
    Aug 23, 2021 at 15:35

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