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In p.289 of Sakurai's book on modern quantum mechanics (the latest edition):

In order for (4.410) to be satisfied, it is of physical interests to consider just two types of transformations---unitary or anti-unitary. Other possibilities are related to either of the preceding via trivial phase changes. The proof of this assertion is actually very difficult and will not be discussed here. See Quantum Mechanics: Fundamentals Kurt Gottfried, Tung-Mow Yan (2003).

I believe that he meant to consider the transformation acts on the states in the Hilbert space, say: $$|a\rangle \to |\tilde{a}\rangle $$ $$|b\rangle \to |\tilde{b}\rangle $$

We demand the transformation is indeed a symmetry of the system if it obeys (the (4.410) of Sakurai book): $$|\langle b|a\rangle | = |\langle \tilde{b}|\tilde{a}\rangle|.$$

My question is that how to give such a proof (sorry, I do not have Gottfried-Yan (2003))? And does this imply that we prove that any quantum theory only has symmetry transformations of unitary and antiunitary types? Only unitary symmetries or antiunitary symmetries?

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    $\begingroup$ For the love of god, please, use \ langle and \rangle if you use bras and kets. Another reference perhaps handy for Bargmann's proof of Wigner's theorem is here: physics.stackexchange.com/q/342251 $\endgroup$
    – DanielC
    May 13, 2021 at 23:04

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This is the statement of Wigner's theorem. You can find (apparently two) proofs of Wigner's theorem here.

The broad summary is as follows. First, (pure) states are not elements of the Hilbert space $\mathcal H$. The set of physical states $\Sigma$ can be defined as the equivalence classes of $\mathcal H$ under the equivalence relation $\sim$, where $\psi \sim \phi \iff$ there exists some nonzero $\lambda \in \mathbb C$ such that $\phi=\lambda \psi$.

Let $\Phi,\Psi$ be elements of $\Sigma$. We define the ray product $(\Phi,\Psi)$ by choosing any two representatives $\phi$ and $\psi$ of $\Phi$ and $\Psi$, and defining $$(\Phi,\Psi) := \frac{|\langle \phi,\psi\rangle|}{\Vert \phi\Vert \Vert \psi \Vert}$$

Having made this definition, it turns out that the mathematical predictions of the theory - decay rates, the probabilities of various measurement outcomes, etc - can be expressed in terms of ray products of physical states. Therefore, a symmetry transformation $T:\Sigma\rightarrow\Sigma$ is a bijective transformation on the space of physical states which leaves all ray products invariant, i.e. $(T\Phi,T\Psi)=(\Phi,\Psi)$.

Given some operator $\hat A$ on the Hilbert space, we can induce a corresponding transformation $\underline A$ on $\Sigma$ in the natural way. Furthermore, if $\hat U$ is a unitary or antiunitary operator on $\mathcal H$, one can see right away that $\underline U$ is a symmetry transformation on $\Sigma$, because

$$(\underline U\Phi,\underline U\Psi) := \frac{|\langle \hat U\phi,\hat U\psi\rangle|}{\Vert \hat U\phi\Vert \Vert \hat U\psi\Vert} = \frac{|\langle \phi,\psi\rangle|}{\Vert \phi\Vert \Vert \psi \Vert} = (\Phi,\Psi)$$

Note also that this association between operators on $\mathcal H$ and maps on $\Sigma$ is not one-to-one. Multiple different operators may induce precisely the same map. For example, the identity operator on $\mathcal H$ and the operator $\psi \mapsto e^{i\theta} \psi$ both induce the identity transformation on $\Sigma$ because they map all of the equivalence classes to themselves.


With these preliminaries out of the way, Wigner's theorem is a pseudo-converse of the above. It turns out that any symmetry transformation on $\Sigma$ can be induced by some unitary or antiunitary operator on $\mathcal H$. That doesn't mean that only (anti)unitary operators induce symmetry transformations, but rather that if some operator $\hat A$ induces a symmetry transformation, then we can find a unitary or antiunitary operator $\hat U$ which induces the same transformation. As a result, it is sufficient to restrict your attention to (anti)unitary operators.

Incidentally, this is also why we are interested in projective representations of groups, c.f. my related answer here.

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  • $\begingroup$ Thanks I voted up $\endgroup$ May 13, 2021 at 23:52
  • $\begingroup$ @anniemarieheart If I answered your question, you can accept the answer by ticking the box next to it. If I haven't, feel free to clarify your question. $\endgroup$
    – J. Murray
    May 14, 2021 at 16:12
  • $\begingroup$ I will be happy to accept in a few days if no better answers! (p.s. I asked a new question maybe you also know.) $\endgroup$ May 14, 2021 at 19:31

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