In Quantum Mechanics spin appears as one type of angular momentum. Indeed, in Quantum Mechanics one angular momentum on the state space $\mathcal{E}$ is a triplet of observables $\mathbf{J}=(J_1,J_2,J_3)$ such that

$$[J_i,J_j]=i\hbar \epsilon_{ijk}J_k.$$

From this general definition one deduces that $J^2 = \sum_i J_i^2$ and $J_3$ commute, and then we consider their common eigenvectors $|k,j,m\rangle$ defined by the equations

$$J^2|k,j,m\rangle = j(j+1)\hbar^2 |k,j,m\rangle, $$ $$J_3|k,j,m\rangle = m\hbar|k,j,m\rangle.$$

We then show that $j\geq 0$ and $j$ is one integer or half integer, and for a given $j$, the only possible values for $m$ are $-j,-j+1,\dots,j-1,j$.

Now, from a physical standpoint, spin is one intrinsic property of particles which was experimentally observed in experiments like the Stern-Gerlach experiment, and which should be accounted for theoretically.

The observations then lead to the usual "spin-half", being theoretically defined as the special case of angular momentum $\mathbf{S}$ whose only value for $s$ is $s = 1/2$ and hence we have $m=\pm 1/2$. In that case we usually consider the "spin state space" as the state space $\mathcal{E}_S$ where the set $\{S^2,S_z\}$ is a Complete Set of Commuting Observables and hence has dimension $2$ with basis composed of $|+\rangle = |1/2,1/2\rangle$ and $|-\rangle = |1/2,-1/2\rangle$.

On the other hand we have the so called spin group, denoted $\operatorname{Spin}(n)$ defined as the "double cover of the special orthogonal group $\operatorname{SO}(n)$" such that there is a short exact sequence of Lie groups:

$$1\to \mathbb{Z}_2\to\operatorname{Spin}(n)\to \operatorname{SO}(n)\to1.$$

Now, this definition of the spin group is overly abstract, but I believe there is a connection between it and the spin from Quantum Mechanics. This is suggest firstly by the name of the group and secondly because both spin and the spin group are somehow related to rotations.

As one angular momentum, spin is a generator of rotations, while $\operatorname{Spin}(n)$ is being defined in terms of the group of rotations.

So, is there a relation between the spin from Quantum Mechanics and the spin group? How can we understand this relation intuitively and how this relation is connected to this overly abstract definition of $\operatorname{Spin}(n)$?


2 Answers 2


In quantum mechanics, the relevant representations of symmetry groups on the space of states are not our usual linear representation, but projective representations on the Hilbert space. The projective representations of a semi-simple Lie group - such as the rotation group $\mathrm{SO}(n)$ - are in bijection to linear representations of its universal cover. For a detailed discussion and a derivation of these facts, see this Q&A of mine. The "intuition" for the appearance fo projective representation is that states really are not vectors but rays in Hilbert space and hence "phases don't matter".

Now, the rotation group in $n > 2$ dimensions has fundamental group $\mathbb{Z}/2\mathbb{Z}$, meaning its universal cover is just a double cover. Therefore, in $n>2$ dimensions, the spin group is by definition its double cover and hence the group we need to linearly represent on the Hilbert space to have a projective representation of the rotation group. Our beloved half-integer "spin" $s$ is now nothing but the number uniquely labeling an irreducible linear representation of $\mathrm{Spin}(3)$ by $L_x^2 + L_y^2 + L_z^2 = s(s+1)$.

  • 2
    $\begingroup$ Hi, sorry for being naive, but is Spin(3) then just $SU(2)$? $\endgroup$ Commented Dec 15, 2019 at 14:23
  • 2
    $\begingroup$ @user2723984 Yes. $\endgroup$
    – ACuriousMind
    Commented Dec 15, 2019 at 14:26

The spin group is related to spin-half objects, called spinors. If you rotate a spinor by 360 degrees, you get back the negative of the spinor you started with. Now it would be nice if you could represent the action of this rotation by saying that an element of $SO(n)$ is acting on the spinor. However, this cannot be done because a rotation by 360 degrees is the same as the identity element of $SO(n)$, and so the action of this rotation must be to leave the spinor invariant, contrary to what we know happens. Thus there is no way to accurately represent the action of $SO(n)$ rotations on a spinor.

However, if you had a bigger group, where a rotation of 360 degrees didn't take you back to the identity element, then you might be able to make a one to one correspondence between elements of this bigger group and the linear transformation it causes on spinors.

The spin group is this bigger group. Since the spin group is a double cover of $SO(n)$, a rotation of 360 degrees only takes you half way around the spin group, and so the group element corresponding to a 360 degree rotation is not constrained to act as the identity on the spinor, but instead can multiply the spinor by $-1$, as it should.

So to summarize, you can make a group out of all the finite rotations that can be applied to a spinor. This group is not $SO(n)$, since the identity rotation and a 360 degree rotation act differently on the spinor but are the same in $SO(n)$. So the group of finite rotations that can be applied to a spinor must be bigger than $SO(n)$. In fact, this group turns out to be the spin group, which is a double cover of $SO(n)$

  • $\begingroup$ So the punchline is, if you want to study spin in $n$ spatial dimensions, study the representations of $\text{Spin}(n)$? $\endgroup$
    – knzhou
    Commented Jun 18, 2016 at 3:10
  • 1
    $\begingroup$ It appears to me that you just shifted the question to why the hell we should consider spinors. $\endgroup$
    – ACuriousMind
    Commented Jun 18, 2016 at 9:22
  • $\begingroup$ @knzhou Yes, I think that is right. $\endgroup$ Commented Jun 18, 2016 at 11:23
  • $\begingroup$ @ACuriousMind I interpreted his question to by what is the relationship between the spin group and spinors, and I assumed he already understood that spinors were necessary to describe spin 1/2 particles. So I didn't go into the question of why we need spinors in the first place. $\endgroup$ Commented Jun 18, 2016 at 11:27

Not the answer you're looking for? Browse other questions tagged or ask your own question.