# Show That Wigner’s Theorem Defines a Surjective Homomorphism $\operatorname{U}(2) \rightarrow \operatorname{SO}(3)$

## Preliminary Knowledge

We are working on the finite dimensional Hilbert space $$\mathbb{C}^2$$. The projective Hilbert space is given by

$$\mathbb{P}(\mathbb{C}^2)=\big(\mathbb{C}^2 \backslash \{0\}\big)/\mathbb{C}^*, \tag{1}$$

where the equivalence relation $$\psi \sim \phi$$ is satisfied when $$\psi=\lambda \phi$$, for some $$\lambda \in \mathbb{C}^*$$ and $$\psi ,\phi \in \mathbb{C}^2 \backslash \{0\}$$.

An important structure is quantum mechanics is the transition probability, which is the map $$p:\mathbb{P}(\mathbb{C}^2) \times \mathbb{P}(\mathbb{C}^2) \longrightarrow \mathbb{R}$$ defined by

$$p([\psi],[\phi])=\frac{|\langle \psi , \phi \rangle |^2}{\langle \psi , \psi \rangle \langle\phi , \phi \rangle}. \tag{2}$$

Furthermore, a quantum mechanical symmetry is defined as an invertible map $$f:\mathbb{P}(\mathbb{C}^2) \longrightarrow \mathbb{P}(\mathbb{C}^2)$$ which preserves the transition probability:

$$p(f([\psi]),f([\phi]))=p([\psi],[\phi]), \quad \text{for} \space [\psi],[\phi] \in \mathbb{P}(\mathbb{C}^2).\tag{3}$$

And finally, Wigner's theorem states that for any quantum mechanical symmetry $$f:\mathbb{P}(\mathbb{C}^2) \longrightarrow \mathbb{P}(\mathbb{C}^2)$$ there exists a unitary or anti-unitary operator on $$\mathbb{C}^2$$ that induces $$f$$ on $$\mathbb{P}(\mathbb{C}^2).$$

## What we know

Consider now the set of all one-dimensional projectors in $$\mathbb{C}^2$$:

$$\mathbb{E} := \{e \in \operatorname{End}(\mathbb{C}^2) : e^2=e=e^*, 0 \ne e \ne I, \operatorname{dim} \operatorname{im} (e)=1\}, \tag{4}$$

where $$I$$ is the identity. By identifying $$\mathbb{E}$$ with the space of all $$2 \times 2$$ complex matrices $$\operatorname{M}_{2 \times 2}(\mathbb{C})$$ with trace $$1$$ (while excluding the obviously trivial zero and identity matrix), it is straightforward to show that any $$e \in \mathbb{E}$$ can be written as

$$e(\vec{x})=\frac{1}{2}I + \frac{1}{2}\sum_{i=1}^{3}x_i \sigma_i, \quad||\vec{x}||=1, \tag{5}$$

where $$\sigma_i$$ are the $$3$$ Pauli matrices. Equation $$(5)$$ turns out to define an isomorphism $$S^2 \longrightarrow \mathbb{E}$$, and thus $$S^2 \cong \mathbb{E}$$. One can also show that $$\mathbb{E} \cong \mathbb{P}(\mathbb{C}^2)$$, and so we have $$\mathbb{P}(\mathbb{C}^2) \cong S^2$$. Then, by using the formula $$p(e,f)=\operatorname{Tr}(ef)$$ for any $$e,f \in \mathbb{E}$$, we can show that transition probability is given by

$$p(\vec{x},\vec{y})=\frac{1}{2}(1+ \langle\vec{x},\vec{y}\rangle), \quad \text{for} \space\vec{x},\vec{y} \in S^2. \tag{6}$$

Next, consider the group of all quantum mechanical symmetries:

$$G:= \{f:S^2 \longrightarrow S^2 \space | \space f \space \text{invertible and} \space p(f(\vec{x}),f(\vec{y}))=p(\vec{x},\vec{y}) \space \forall \vec{x},\vec{y} \in S^2\}, \tag{7}$$

as well as the orthogonal group $$\operatorname{O}(3)$$, given by

$$\operatorname{O}(3; \mathbb{R}) = \{\rho \in \operatorname{GL}(3; \mathbb{R}) \space | \space \langle \rho a, \rho b \rangle = \langle a,b \rangle \space \forall a,b \in \mathbb{R}^3 \}. \tag{8}$$

One can then show that $$G \cong \operatorname{O}(3)$$, and hence identify $$\operatorname{O}(3)$$ with $$G$$.

## The Problem

We now restrict to $$\operatorname{SO}(3) \subset \operatorname{O}(3)$$. Show that Wigner's theorem defines a surjective homomorphism $$\operatorname{U}(2) \longrightarrow \operatorname{SO}(3)$$ sending $$u \mapsto R$$ defined by

$$ue(\vec{x})u^*=e(R(\vec{x})). \tag{9}$$

## My Question

I do not understand the meaning of equation $$(9)$$ and I have no idea how to use it; I expected to be given the explicit formula of the homomorphism! I tried inverting it by using $$e^{-1}$$ to isolate $$R$$, but to no avail. How then can I show that this map is a homomorphism using equation $$(9)$$? More importantly, I do not see how Wigner's theorem relates to the problem; I'm trying to find a surjective group homomorphism, how can the unitary/anti-unitary operator that induces $$f:S^2 \longrightarrow S^2$$ help me with that? The theorem doesn't seem to state anything useful in particular.

Any hints/answers will be well appreciated. Thank you!

• The Wigner's theorem actually says that the symmetry group in QM is the central extension of $SO(3)$ by $U(1)$, which is precisely $SU(2)$. Commented Jun 14, 2022 at 4:34

Given $$u$$ in $$U(2)$$, we have for every $$\vec x$$ in $$S^2$$ that $$e(\vec x)$$ is a $$2\times 2$$ matrix that is a projection. Therefore the product of matrices $$ue(\vec x)u^*$$ is also a projection and hence must be of the form $$e(\vec y)$$ for a unique vector $$\vec y\in S^2$$.

This defines a function $$R=R_u:\vec x\in S^2\mapsto \vec y\in S^2.$$

So the problem consists in showing that

• $$R$$ is the restriction of a unique linear transformation (also denoted by $$R$$ by abuse of language), lying is $$SO(3)$$.

• We then get a map $$u\in U(2)\mapsto R\in SO(3),$$ which should be proven to be a group homomorphism.

Finally I don't see either what is the relevance of Wigner's Theorem here!

• It is relevant because Wigner's theorem actually says that in QM is symmstry group actually becomes the central extension of $SO(3)$ by $U(1)$, which is exactly $SU(2)$. Commented Jun 14, 2022 at 4:32
• Yes I know that Wigner's Theorem is important. But my claim is that it is not used in the solution to this problem.
– Ruy
Commented Jun 14, 2022 at 4:36
• Is it perhaps the other way round: that this problem indicates one way to prove Wigner's theorem? Commented Jun 14, 2022 at 7:10
• Hi! Thank you for the answer. I don't see how $u \mapsto R$ can be proven to be a homomorphism if I don't have an explicit formula for the mapping. I suppose equation $(9)$ has something to do with this, but I don't understand what. Commented Jun 15, 2022 at 14:17
• @ShikiRyougi. The purpose of my answer was to clarify how does (9) define $R(\vec x)$ unambiguously in terms of $\vec x$. Not all definitions in Math can be given through explicit formulas but I believe the present definition is enough to allow for proving what you need.
– Ruy
Commented Jun 17, 2022 at 12:27