# Why are Spin 1/2 particles invariant to $4\pi$ rotation loops while Spin 1 particles are invariant to $2\pi$ loops?

Why do Spin 1/2 particles when turning them by 360 deg get a phase factor of -1 and a loop of 720 deg leads to the identity while for spin 1 particles a loop of 360 deg gives already the identity?

It turns out the it's not the rotational group $$SO(3)$$ that is important in QM but it's universal cover, which happens to be a double cover and in this context is called the spin group, $$Spin(3)$$.

This is also true for relativistic QM when it's not the Lorentz group $$SO(1,3)$$ but it's universal cover, which again is a double cover and is called the spin group, $$Spin(1,3)$$.

Notably, universal covers are simply connected and hence do not have 'holes' in them. This might be one reason why Nature prefers the spin groups to the rotation groups.

Now, the simplest geometric model to see what is going on here is to consider the situation in two dimensions, and here the 2d rotation group, $$SO(2)$$ is just the circle $$S^1$$.

Now over the circle we can build a double cover by taking a Möbius band over the circle and taking the cover as it's single edge and this gives a double cover.

Now if we rotate a vector on the circle it come backs to itself after a single full rotation. Think of this as a zero spin particle.

If we rotate a vector on the Möbius band it will return to itself with a reverse orientation after a single full rotation and hence back to itself after a two full rotations. Think of this as a spin-1/2 particles.

In this model, spin counts the 'twist' in the rotational path: none in spin-0 and one in spin-1/2.

As characteric classes count twists this suggests that the theory of Chern classes may be implicated in the theory of spin.

Higher spins can be constructed by taking a band with more twists. When the band has a whole number of full twists then it is intrinsically similar to a band with no twists; so we can think of all the integral spins as alike on some manner; and when it has an odd number of half-twists then it is intrinsically similar to the Möbius band, and so we can think of all half spin particles as alike, in some manner.

Finally, although we ought not to think of an electron as a particle that is physically spinning it might help heuristically to think of the electron as doing exactly that (this is after all how spin was originally dreamt up by Uhlenbeck & Goudsmit). Then we think of the electron spinning around its axis but as it completes a full rotation we also rotate the axis by half a rotation. This again we require two full rotations to get the electron back to where it was. This is exactly the same as the möbius model above, thought of in a different manner.

• The explanation of the different spins is excellent, but the answer is missing a crucial caveat: The state of a single fermion actually is invariant under a $2\pi$ rotation, too! What changes sign is the representation of the state as a vector in Hilbert space, but you can multiply any state vector by an arbitrary complex number and it still represents the same state (=ray in Hilbert space). You can only detect this "non-invariance under $2\pi$" as a relative phase, cf. my answers at physics.stackexchange.com/a/167470/50583, physics.stackexchange.com/a/322017/50583 – ACuriousMind Jan 20 '19 at 10:16

You ask for "why". Well the reason is that the rotation operator in matrix form does that.

For a spinor $$\left( \begin{array}{} a \\ b \end{array} \right)$$, you find that the rotation operator about an axis defined by the unit vector $$\hat{n}$$ along an angle $$\theta$$ is: $$U(\theta)=cos\left(\frac{\theta}{2}\right)\mathbb{I}-isin\left(\frac{\theta}{2}\right)(\vec{\sigma}\cdot\hat{n})$$

Where $$\sigma_i$$ represent Pauli's matrices. This is equivalent to

$$\left( \begin{array}{cc} \cos\left(\frac{\theta}{2}\right)-i \sin\left(\frac{\theta}{2}\right)n_z & -i\sin\left(\frac{\theta}{2}\right)(n_x-in_y) \\ -i\sin\left(\frac{\theta}{2}\right)(n_x+in_y) & \cos\left(\frac{\theta}{2}\right) + i \sin\left(\frac{\theta}{2}\right) n_z \end{array} \right)$$

For simplicity, let's choose $$\hat{n}=\hat{z}$$, that is, turning about the z axis. The matrix becomes

$$U_z(\theta)=\left( \begin{array}{cc} \cos\left(\frac{\theta}{2}\right)-i \sin\left(\frac{\theta}{2}\right) & 0 \\ 0 & \cos\left(\frac{\theta}{2}\right) + i \sin\left(\frac{\theta}{2}\right) \end{array} \right)=\left( \begin{array}{cc} \exp\left(-i\frac{\theta}{2}\right) & 0 \\ 0 & \exp\left(+i\frac{\theta}{2}\right) \end{array} \right)$$

So, whereas the rottation matrix for objects in real space is the usual one, spins get transformed this way.

This is the "why". Mathematics work like this. But let me point out some ideas on common misconceptions:

1. When the space is rotated by the usul matrices, spins suffer this transformatio, which is not actually a rotation. Real rotations induce these spin transfomations.
2. This matrix, acting on the $$\left( \begin{array}{} a \\ b \end{array} \right)$$ produces a minus sign, that is: $$U_z(2\pi)\left( \begin{array}{} 1 \\ 0 \end{array} \right)=-1\left( \begin{array}{} 1 \\ 0 \end{array} \right)$$

But check that $$-1=\exp(i\pi)$$ is just a global phase that is not relevant. The state keeps being eigenvector of $$S_z$$ with the same eigenvalue of $$\hbar /2$$. This is coherent with the fact that a spin "pointing in $$z$$ is ot affected by a rotation about that axis.

This is a global phase factor. It doesn't affect physics, as it must be. It'd only affect if you rotated only part of the system.

This question involves a few different themes, including how the rotation group is allowed to act on state-vectors, and how the angular momentum operators are related to the operators that implement rotations.

When we use a state-vector $$|\psi\rangle$$ to represent the state of a physical system in quantum theory, the physical meaning of $$|\psi\rangle$$ is not affected when it is multiplied by any non-zero complex number, because physical predictions always ultimately involve some expression of the form $$\frac{\langle\psi|\cdots|\psi\rangle}{\langle\psi|\psi\rangle} \tag{1}$$ where "$$\cdots$$" is a placeholder for an observable. So $$|\psi\rangle$$ and $$-|\psi\rangle$$ represent the same physical state, because the negative sign doesn't affect (1).

In quantum theory, continuous symmetries like rotational symmetry are implemented in terms of unitary operators. A unitary representation of the rotation group is a specific kind of implementation that associates a unitary operator $$U(R)$$ with every rotation $$R$$, subject to these conditions: $$U(R)U(R')=U(RR') \tag{2}$$ where $$RR'$$ denotes the composition of two rotations and where $$U(R)U(R')$$ denotes multiplication of the corresponding unitary operators. With this representation, the effect of a rotation $$R$$ on a state-vector $$|\psi\rangle$$ is $$|\psi\rangle\to U(R)|\psi\rangle. \tag{3}$$ A $$2\pi$$ rotation corresponds to $$R=1$$, and equation (2) implies $$U(1)U(1)=U(1)$$, which in turn implies $$U(1)=1$$. So with a unitary representation of the rotation group, state-vectors are invariant under $$2\pi$$ rotations.

Observables must be invariant (with no sign-change) under any $$2\pi$$ rotation, but as explained above, a state-vector doesn't really need to be invariant under a $$2\pi$$ rotation; it only needs to be invariant up to a non-zero complex factor. So a unitary representation of the rotation group, as described by (2), is not the most general way of implementing rotations in quantum theory.

The group $$SU(2)$$ of $$2\times 2$$ unitary matrices with determinant $$=1$$ is not the same as the $$3$$-d rotation group $$SO(3)$$, but it becomes the same if we agree to treat the two matrices $$M$$ and $$-M$$ as being "equivalent" for each $$M\in SU(2)$$. So instead of considering only unitary representations of the rotation group, we can also consider unitary representations of the group $$SU(2)$$, because $$|\psi\rangle$$ and $$-|\psi\rangle$$ represent the same physical state. An observable $$X$$ transforms as $$X\to U X U^{-1}$$, so observables are still invariant under $$2\pi$$ rotations even if a $$2\pi$$ rotation is represented by $$U=-1$$. So this is an allowable way to implement rotations in quantum theory. It's not a unitary representation of the rotation group, but it's a unitary representation of the double cover $$SU(2)$$ of the rotation group; it's also called a projective representation of the rotation group. A state-vector that transforms according to such a representation changes sign under a $$2\pi$$ rotation, so it is invariant under a $$4\pi$$ rotation.

Finally, to relate this to angular momentum (spin), we need to consider the generators of the given representation. If $$G(\mathbf{n})$$ is the generator of rotations about a particular axis $$\mathbf{n}$$, then the unitary operator $$U(\mathbf{n},\theta)$$ that implements a rotation through angle $$\theta$$ about that axis is $$U(\mathbf{n},\theta)=\exp\Big(iG(\mathbf{n})\theta\Big). \tag{4}$$ By definition, the operator representing angular momentum about the $$\mathbf{n}$$ axis is $$\hbar G(\mathbf{n})$$. In an ordinary unitary representation of the rotation group, we have $$U(\mathbf{n},2\pi)=1$$ for every axis $$\mathbf{n}$$. The smallest representation that satisfies this condition is a representation spanned by just one state-vector $$|\psi\rangle$$ that satisfies $$G(\mathbf{n})|\psi\rangle=0$$ for every axis $$\mathbf{n}$$. This is the spin $$0$$ representation, because this state-vector has zero angular momentum. The next-smallest representation that satisfies this condition is a representation spanned by three state-vectors; this is the spin $$1$$ representation, because we can choose a (non-orthogonal) basis in which each basis vector satisfies $$\hbar G(\mathbf{n})|\psi\rangle=\hbar |\psi\rangle$$ for some $$\mathbf{n}$$, which means that they each have angular momentum $$1\times\hbar=\hbar$$ about some axis.

When we consider unitary representations of $$SU(2)$$ instead (that is, projective representations of the rotation group), then we can get a representation spanned by two state-vectors, each of which has angular momentum $$\hbar/2$$ for some $$\mathbf{n}$$. This is the spin $$1/2$$ representation. According to equation (4), the condition $$\hbar G(\mathbf{n})|\psi\rangle=(\hbar/2)|\psi\rangle$$ implies $$U(\mathbf{n},2\pi\big)|\psi\rangle= \exp\big(iG(\mathbf{n})2\pi\big)|\psi\rangle= \exp\big(i(1/2)2\pi\big)|\psi\rangle= \exp\big(i\pi\big)|\psi\rangle= -|\psi\rangle, \tag{5}$$ so this state-vector changes sign under a $$2\pi$$ rotation.