The generators of the unitary representation of $SU(2)$ on the internal spin Hilbert space of (say) a spin-$1/2$-particle are typically said to represent components of spin along various spatial axes. For example, if $|+\rangle$ is the eigenvector of $S_z$ with eigenvalue $1/2$, then (according to the standard story) if the system is in the state represented by $|+\rangle$, the component of the system's spin along the $z$-axis is $1/2$.

In general, textbook wisdom has it that the component of spin along the spatial axis $\vec{n}$ is given by $$\vec{n}\cdot \vec{S} = n_xS_x + n_yS_y + n_zS_z,$$ where $\vec{n}$ is a unit vector.

I find this puzzling, for at least two reasons.

  1. The generators of $SU(2)$ are self-adjoint operators on an internal spin Hilbert space that has no spatial degrees of freedom at all. (In the case of spin-$1/2$-particles, this is the two-dimensional Hilbert space of Pauli spinors.) It is hard to see what connection these generators have with axes in physical space.

  2. One may think that the coincidence between the number of spatial dimensions and the number of basis elements of the $SU(2)$ Lie algebra is suggestive of interpreting the latter as representing components of spin along spatial axes. However, it is my understanding that the fact that there are three $SU(2)$-generators is a result of the structure of $SU(2)$, and not of the fact that the relevant physical space is three-dimensional. Indeed, the "components" of spin of a spin-$1/2$-particle in four spatial dimensions would also be represented by triples of linearly independent $SU(2)$-generators.

My question is therefore: what is the basis for the identification of the generators of the unitary $SU(2)$-representation on the internal spin Hilbert space with the components of spin along spatial axes?

(Note that I am aware of the Bloch sphere illustration of the Hilbert space of a spin-$1/2$ system. I have also read this thread: Do spins have spatial directions?. Neither contains an answer to my question.)

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    $\begingroup$ Spin-orbit coupling pegs the orientation of S to that of L which is linked to spatial orientation, no? $\endgroup$ May 20, 2018 at 22:22
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    $\begingroup$ Recall S· L is a rotational invariant. Any similarity transform of the 3 Pauli matrices will do! $\endgroup$ May 21, 2018 at 0:39

2 Answers 2


An $SU(2)$ symmetry, a priori, has absolutely nothing to do with spatial rotations. For instance, the symmetry between protons and neutrons is described by isospin, an $SU(2)$ symmetry that treats the proton and neutron just like the spin up and spin down states of a spin $1/2$ particle. Similarly there is weak isospin, also described by $SU(2)$, which relates the electron with the electron neutrino. Neither of these are related to rotations, except for the math being the same; you cannot do a physical rotation to turn a proton into a neutron. (However, the analogy with spin is so useful for conceptualizing what's going on that they all have the word 'spin' in their names.)

If a textbook says "we've identified an $SU(2)$ symmetry, so it must physically correspond to rotational symmetry", then that book is being sloppy. The argument should be phrased in reverse. Below I'll show how that goes very explicitly.

We start with experimental data that we want to understand. For example, suppose we want to model the precession of the magnetic dipole moment of a nucleus in a time-varying magnetic field. (It turns out this dipole moment is proportional to the spin, so this is exactly what you're asking about.) To make a quantum mechanical model we must define a Hilbert space and a Hamiltonian, as well as operators $\mu_i$ corresponding to the components of the magnetic dipole moment.

Of course, there is no unique mathematical prescription for this. For example, you may choose the Hilbert space to be zero-dimensional, but then it clearly wouldn't fit the data. It turns out that for some nuclei the model works if we choose the Hilbert space to be two-dimensional, with the states $|\uparrow\rangle$ and $|\downarrow\rangle$ corresponding to the dipole moment pointing vertically up and down. In other words, this defines $\mu_z$ as $$\mu_z |\uparrow \rangle = \mu_0 |\uparrow\rangle, \quad \mu_z |\downarrow\rangle = - \mu_0 |\downarrow \rangle.$$ For convenience we'll ignore spatial degrees of freedom. You can think of the nucleus as nailed down at the origin, if you want.

Next, we define rotation operators that physically rotate the system around. We know that classically, the rotations of space form a group $SO(3)$. Because of issues with quantum phases, this means the rotation operators must necessarily be a representation of the group $SU(2)$. This is a bit weird, but something you get used to after a while -- there is generally a turn-the-crank mathematical procedure for figuring out which group to use in the quantum case.

However, we still don't know which representation of $SU(2)$ it is. For example, the rotational operators could conceivably all do nothing -- that is exactly the right choice for isospin symmetry (ignoring the spins of the proton and neutron) because you can't rotate a proton into a neutron. But it's not the right choice for the magnetic moments, because we can observe that tilting the magnet and running the experiment again gives a different result.

We also know that a rotation about the $z$ axis should fix $|\uparrow \rangle$ and $|\downarrow \rangle$, again by observation, while $180^\circ$ rotations about the $x$ and $y$ axes should interchange these states. (Everything here is just up to phases.) We can't proceed further and get explicit expressions because they will vary depending on the phases of the states. But in the standard phase conventions, one can continue with this logic to show that the $\mu_i$ operators must all be proportional to $\sigma_i$, and the rotation operators are exponentials of $\sigma_i$. This is described in detail here.

This is a pretty long and very explicit argument to make one simple point: in physics, we do not blindly do mathematics and put the physical interpretation in at the end. We start with a physical system we want to describe and define mathematical objects accordingly. We know that rotations must be described by $SU(2)$ by general principles, so we define a Hilbert space with a representation of $SU(2)$. Which one? Whatever works.

  • $\begingroup$ One more comment that I don't want to cram into this already long answer. If we include the spatial degrees of freedom in, then our physical rotation operators will act on both the spatial and spin degrees of freedom. Of course there are also formally operators that act only on the spatial and only on the spin degrees of freedom, and also form $SU(2)$'s. These are not associated with any physical rotations. $\endgroup$
    – knzhou
    May 20, 2018 at 22:28
  • $\begingroup$ If you hand me a Hilbert space with separate "spatial rotational operators" and "spin rotation operators" and absolutely nothing else, then I indeed cannot ascribe physical meaning to the spin orientations, i.e. for all I know the $S_z$ really measures spin in the physical $x$ direction. You need the operators that actually represent physical rotations to make the correspondence. $\endgroup$
    – knzhou
    May 20, 2018 at 22:29
  • $\begingroup$ Thank you, knzhou, this is very helpful! Just to be sure, let me briefly recap: the physical rotations of a spin system about, say, the $z$-axis are generated by $J_z = L_z + S_z$, where $L_z$ is the generator of the rotation group acting on the spatial degrees of freedom and $S_z$ is the generator of the rotation group acting on the spin degrees of freedom. We cannot make sense of what it would be to rotate only the spin degrees of freedom, or to only rotate the spatial degrees of freedom. Is this correct? $\endgroup$
    – Al_Gebra
    May 21, 2018 at 2:41
  • $\begingroup$ ... and so essentially: our antecedent empirical grasp is that of $\vec{J}$, the spatial vector quantity of total angular momentum, which (as empirical testing reveals) is the sum of orbital and spin angular momentun. So the latter better also be a spatial vector quantity. $\endgroup$
    – Al_Gebra
    May 21, 2018 at 2:50
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    $\begingroup$ @ACuriousMind Yup, that is exactly the experiment I had in mind. Also some spin precession experiments, I guess. Later I might edit and put in proper names for things. $\endgroup$
    – knzhou
    May 21, 2018 at 10:15

The logic is as follows:

  1. $SU(2)\cong SPIN(3)$.

  2. More generally, $G:=SPIN(n)$ is the is the double cover of the rotation group $SO(n)$.

  3. $G$ acts via the adjoint representation on its Lie algebra $\mathfrak{g}:=\mathfrak{so}(n)$.

  4. Each element of the Lie algebra $\mathfrak{g}:=\mathfrak{so}(n)$ generates an infinitesimal rotation, i.e. it is an angular momentum.

  5. The Lie algebra $\mathfrak{g}:=\mathfrak{so}(n)\cong \bigwedge^2 V$ can be identified with the set of 2-planes (i.e. rotation planes) in $n$-space $V\cong \mathbb{\mathbb{R}^n}$. See also this related Phys.SE post.

  6. For $n=3$, we have $\bigwedge^2 V\cong V$, i.e. the elements of $\mathfrak{su}(2)\cong\mathfrak{so}(3)$ can be identified with 3-space $V$ itself.


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