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!

  • $\begingroup$ 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 $SU(2)$. $\endgroup$
    – Valac
    Jun 14, 2022 at 4:34

1 Answer 1


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!

  • $\begingroup$ 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 $SU(2)$. $\endgroup$
    – Valac
    Jun 14, 2022 at 4:32
  • $\begingroup$ Yes I know that Wigner's Theorem is important. But my claim is that it is not used in the solution to this problem. $\endgroup$
    – Ruy
    Jun 14, 2022 at 4:36
  • $\begingroup$ Is it perhaps the other way round: that this problem indicates one way to prove Wigner's theorem? $\endgroup$
    – Vincent
    Jun 14, 2022 at 7:10
  • $\begingroup$ 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. $\endgroup$ Jun 15, 2022 at 14:17
  • $\begingroup$ @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. $\endgroup$
    – Ruy
    Jun 17, 2022 at 12:27

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