# Wigner's Theorem and discrete Symmetries

According to my skript:

A pure state is a ray: $\quad$ $\{λψ\}$, where $ψ ∈ \mathcal H$, $||ψ|| =1$ fixed and $λ ∈ \mathbb C$, $|λ| = 1$.
Pure states are uniquely given by 1-dimensional orthogonal projectors $Π$ on $\mathcal H$: $\quad$ $Πφ = |ψ\rangle\langleψ|φ\rangle$.
Let $Π(\mathcal H)$ be the set of all such projectors.
Definition. A symmetry is a map $S: Π(\mathcal H) \rightarrow Π(\mathcal H'), Π \rightarrow Π'$, such that $$tr(Π_1Π_2) = tr(Π_1'Π_2').$$
Furthermore one can express the expectation value of an observable $A$ through $Π$: $$\langle A\rangle_Π = tr(AΠ) = \langleψ|A|ψ\rangle.$$

Theorem. (Wigner) Every symmetry is represented as $$Π' = S(Π) = UΠU^*$$
(i.e. $ψ' = Uψ$) with $U$ either being a linear an antilinear isometry and $U$ is unique apart form multiplication with a factor $c ∈ \mathbb C ,|c|=1$.

Discrete Symmetries. Considering first a classical particle where $\vec{p} = m\vec{x}$.The discrete symmetries space-inversion $P$ and time-inversion $T$ are
$$P : (\vec{x},\vec{p}) \rightarrow (-\vec{x},-\vec{p})$$ $$T: (\vec{x},\vec{p}) \rightarrow (-\vec{x},-\vec{p})$$

Now their expecation values $\langle x_i\rangle_Π$, $\langle p_i\rangle_Π$ have to transform according to the Theorem: $$U_P^*x_iU_P = -x_i \quad …$$

Question:
How does one get the last line $$U_P^*x_iU_P = -x_i?$$

Attempt:
I am a bit confused by the usage of operators and the projectors, moreover in general by the notation. I mean the symmetry $P$ acts on operators (here $x_i , p_i$), right?
So one symmetry would then be $S = S_P = P$? But I thought that the symmetry above generally acts on projectors, i.e. $S(Π) = Π'$.

Does the last line now follow by: $$tr (-x_i) = tr(P(x_i)) ≡ tr(U_Px_iU_P^*)?$$

Maybe one could help me with the notation and how the bracketing works.

Wigner's theorem says that a symmetry in the sense of a 1-to-1 correspondence of rays in the projective Hilbert space lifts to a (anti)unitary operator. Now the action on states dualises to observables and so the action of $P$ on $x^i$ turns out to be $P(x^i) = -x^i$. But by Wigner's theorem, $P$ is unitarily implemented, and the way this unitary acts is by the covariance relation $U^*x^iU = P(x^i) = - x^i$.
• I'm not sure then what your question is about. The expectation value $\langle x^i\rangle_\Pi$ transforms into $\langle x^i\rangle_{\Pi'} = Tr(\Pi'x^i\Pi')=Tr(U\Pi U^*x^iU\Pi U^*)$. Using the properties of the trace you then have $Tr(\Pi U^*x^iU\Pi)=-Tr(\Pi x^i\Pi) = -\langle x^i\rangle_\Pi$. Dec 28, 2014 at 23:40