Which group acts on spin $1/2$ states?

Question

I am trying to get a better undersdanding of spins and of the role of the group $SU(2)$. I read many times something that looks like this: to fully rotate a spin, we should make a rotation of $2\times 360$ degrees, and that the group really acting on spins is the double cover of $SO(3)$, that is to say $SU(2)$. Here is what bothers me.

The possible states of the spin can be represented by elements of the Hilbert space $\mathbb{C}^2$. From what I understood, $\psi\in H$ and $\alpha \psi$ with $\alpha \in \mathbb{C}$ represent the same state. Hence Given $M\in SU(2)$, $M\psi$ and $-M\psi$ represent the same state: when $SU(2)$ acts on the set of states, $M$ and $-M$ cannot be distinguished. Hence the group acting on spin states is really $SU(2)/{\pm I}$, which is in fact isomorphic to $SO(3)$. Hence a rotation of $360$ degrees send spin states on themselves.

I hope that I make sense. If I do I would appreciate some help to clarify this.

Answer 1

You need to distinguish between group actions on the space of quantum states, and (linear) group representations on the Hilbert space.

As you say, the space of states for a spin-1/2 particle is not the Hilbert space $\mathbb C^2$, but rather the projective Hilbert space obtained by identifying two elements $\psi,\phi\in \mathbb C^2$ as equivalent if $\psi=\lambda \phi$ for some nonzero $\lambda\in \mathbb C$. However, the dynamics and manipulations which occur in quantum mechanics happen at the level of the Hilbert space, not the state space. Once we understand what happens to the vectors, we can translate that into a statement about what happens to the states.

It's easy to see that a linear group representation on the vector space gives rise to a unique group action on the state space, but the reverse is not true. Given a group action on the state space, we may not be able to find a corresponding linear representation on the vector space - but of course, we don't necessarily need one. The group action at the level of the vector space need only satisfy the group composition rules up to a phase (since the phase is lost when we descend to the level of the state space), which is why we are interested in projective linear representations rather than only true representations.

As it turns out, there is no faithful linear representation of $SO(3)$ which acts on $\mathbb C^2$, but there is a projective representation of $SO(3)$ (or equivalently, a true representation of $SO(3)$'s universal cover, $SU(2)$).

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The point here is the operator on $\mathbb C^2$ which implements a $2\pi$ rotation about the $\hat z$-axis is $\exp[2\pi i \sigma_z/2] = -\mathbf 1$, which maps a vector $\psi \mapsto -\psi$. It maps the state of the system to itself, but the vector "avatar" which we use to compute the dynamics of the state gets a sign change.

All by itself, such a global change of phase is unobservable. However, if we have a composite system and rotate only one of its parts, we can obtain interference effects which make these changes relevant. See this nice explanation of neutron interferometry by ACuriousMind.

Answer 2

Which group acts on spin 1/2 states?

It is a given irreducible representation of the group that rotates states of a given spin. The dimension of the irreducible representation is $2s+1$, where $s$ is the spin. So, for spin 1/2, the rotation matrices are two-dimensional and have the form: $$ R(\theta \hat n) = \cos(\theta/2) - i\sin(\theta/2)\vec{\sigma}\cdot \hat n \;, $$ where $\vec \sigma$ are the Pauli matrices, $\theta$ is the angle of rotation in radians, and $\hat n$ points along the axis of rotation such that the rotation direction is given by the right hand rule.

Hence a rotation of $360$ degrees send spin states on themselves.

Yes, but there can be an overall phase factor. For example, for spin 1/2: $$ R(2\pi)|+z> = -|+z>\;, $$ which is the same state up to an overall phase ($e^{i\pi} = -1$).