Wigner symmetry representation theorem tells that if $\mathcal{S}:\mathbb{P}\mathcal{H}\to \mathbb{P}\mathcal{H}$ is a symmetry, then $\mathcal{S}[\Psi]=[\hat{U}\Psi]$ where $\hat{U}:\mathcal{H}\to \mathcal{H}$ is either a linear unitary operator or an antilinear antiunitary operator. But the linear unitary operator $\hat{U}$ is not unique, because for all $\theta \in \mathbb{R}$, $e^{i\theta}\hat{U}$ also does the job. My question is that if there is another linear operator $\hat{V}:\mathcal{H}\to \mathcal{H}$ for which $\mathcal{S}[\Psi]=[\hat{V}\Psi]$, is $\hat{V}$ unitary as well?


1 Answer 1


I suppose that by the brackets you mean the equivalence class of the state, that is, $[\psi]\in\mathbb{PH}$ is the ray to which $\psi\in\mathbb{H}$ belongs. Now, if your question is whether every operator compatible with a symmetry is (anti-)unitary, then the answer is yes, and it is so by definition. In fact, the whole point of Wigner's theorem is to prove the converse, namely that for any symmetry transformation there is an (anti-)unitary operator that is compatible with it.

A bijection $S: \mathbb{PH}\to \mathbb{PH}$ is defined to be a symmetry transformation if it preserves the product of rays, that is, if $S$ is bijective and

$$S[\psi]\cdot S[\phi] = [\psi]\cdot[\phi].\tag{$1$}$$

Consequently, if $S$ is a symmetry and $V: \mathbb{H}\to\mathbb{H}$ is such that

$$S[\psi] = [V\psi]\tag{$2$}$$

then by definition of both symmetry and the ray product

$$[\psi]\cdot [\phi] := \frac{|\langle\psi, \phi\rangle|}{\|\psi\|\|\phi\|}\tag{$3$}$$

we have

$$\frac{|\langle V\psi, V\phi\rangle|}{\|V\psi\|\|V\phi\|} \stackrel{(3)}{=}[V\psi]\cdot [V\phi] \stackrel{(2)}{=} S[\psi]\cdot S[\phi] \stackrel{(1)}{=}[\psi]\cdot [\phi] = \frac{|\langle\psi, \phi\rangle|}{\|\psi\|\|\phi\|}.$$

This only holds for any two states if and only if $$\langle V\psi, V\phi\rangle = \langle\psi, \phi\rangle\quad\text{or}\quad \langle V\psi, V\phi\rangle = \langle\psi, \phi\rangle^*.$$

You might wonder why $\langle V\psi, V\phi\rangle = e^{i\theta}\langle\psi, \phi\rangle$ for arbitrary $\theta \in \mathbb{R}$ is not an option, but note that in that case we would have

$$\frac{|\langle V\psi, V\phi\rangle|}{\|V\psi\|\|V\phi\|} =\frac{|e^{i\theta}\langle \psi, \phi\rangle|}{\sqrt{e^{i2\theta}\langle \psi, \psi\rangle\langle \phi, \phi\rangle}} = e^{-i\theta} \frac{|\langle\psi, \phi\rangle|}{\|\psi\|\|\phi\|}.$$

Actually, this later possibility is absurd in deeper ways, since for any $\theta\neq 0$ it would imply that $\langle V\psi, V\psi\rangle \not\in \mathbb{R}^{\geq0}$, which attempts against the definition of inner product.

  • $\begingroup$ I really appreciate your help and nice explanation. I got it now. Thank you so much. $\endgroup$
    – Mahtab
    Feb 9 at 4:52
  • $\begingroup$ You're welcome! $\endgroup$
    – Albert
    Feb 9 at 8:47
  • $\begingroup$ Sorry to ask another question. Let $S=1$ the indenity function on $PH$ and $U_1$ denotes the corresponding unitary operator (i.e. 1$[\psi]=[U_1 \psi]$). Can we conclude that $U_1\circ U_1=U_1$? The same question for $f\in PH$: Is $U_f \circ U_{f^{-1}}=U_1$? $\endgroup$
    – Mahtab
    Feb 11 at 6:46
  • $\begingroup$ I believe you could not conclude that it is the identity, because what Wigner's theorem guarantees you is only that, for a symmetry $f$ and its associated unitary operator $U_f$, you will have $[\psi] = f^\circ f^{-1}[\psi] = [U_f U_{f^{-1}}\psi]$. Consequently, the most you can say is that $U_f U_{f^{-1}}$ maps $\psi$ to some $e^{i\theta}\psi$, for a possibly non zero $\theta\in\mathbb{R}$, because by definition the elements of a ray $[\psi]$ can differ from each other by a phase factor. $\endgroup$
    – Albert
    Feb 11 at 9:38
  • $\begingroup$ Of course, if we have an operator $U$ compatible with a symmetry $S$, then $e^{i \theta}$ for $\theta \in \mathbb{R}$ is compatible with $S$ too. This apparent subtlety is actually critical, because even though two states differing by a phase factor lead to the same physics, the phases are relevant to linear combinations of states (in the same way the direction of a vector is crucial for vector addition), which are absolutely essential for quantum mechanics. $\endgroup$
    – Albert
    Feb 11 at 9:47

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