# How can electrons spin if they have no volume? [duplicate]

My understanding of electrons is that they have no volume, eg they are point particles. If this is true, how can a point spin?

## marked as duplicate by StephenG, Chris♦, David Z♦May 8 '18 at 4:19

The electron spin corresponds to a classical angular momentum, but it is not related to any "rotation" of the electron. The spin of the electron is a quantum mechanical phenomenon and can only have the value of + or - $\frac {\hbar}{2}$ in a given direction. The electron is a point-like particle, it has no finite spatial dimensions that could rotate.

• it does have finite volume however: en.wikipedia.org/wiki/Point_particle – JohnS May 8 '18 at 1:36
• @JohnS - You are not right. The wave function (orbital) of an electron in an atom, giving its probability of finding it at a position in space, can have a non-zero spatial volume. But the electron itself is a point-like particle and has no volume. – freecharly May 8 '18 at 1:47
• Quoting: "In quantum mechanics, the concept of a point particle is complicated by the Heisenberg uncertainty principle, because even an elementary particle, with no internal structure, occupies a nonzero volume. " I think you have a different definition of particle volume than Wikipedia. And me. – JohnS May 8 '18 at 2:05
• @JohnS - Heisenberg's uncertainty principle only relates to finding the particle at a location on space, it doesn't relate to the proper particle extensions. Even regarding the QM uncertainty principle the point particle can have an exact point location in space when the conjugate momenta are completely undetermined. – freecharly May 8 '18 at 2:22
• I think your argument is more ontological than physical. Whether the electron has internal structure or not is an open question. But there is no need to believe that particles with no internal structure occupy zero volume. Can we know more about the electron than its wavefunction? Just as the mass of the photon may be non-zero (the Proca Lagrangian). The goal of physics should be to put an increasingly refined upper bound on this. In fact there is no reason to believe that QM needs to be formulated in terms of continuua anyway. See Chris Isham's recent work. – JohnS May 8 '18 at 2:41

If you ignore the location/velocity of an electron, each electron carries one bit (or, more precisely, q-bit) of information: whether it is spin up or spin down with respect to a chosen axis. This information is encoded in a vector with two components and complex components. That is, the "spin state" of the electron can be written as

$$(\alpha, \beta) \in \mathbb{C}^2$$

where $\alpha$ and $\beta$ are complex numbers, and $|\alpha|^2 + |\beta|^2 = 1$. There is nothing "spinning," nothing is "dynamical." If you rotate the electron (or even just rotate your reference frame) the spin state of the electron will have to "rotate" accordingly as well. (How do you rotate an electron? You can manipulate it with a magnetic field, or you can just give up and mathematically pretend you are rotating it.)

Any appropriate "rotation" of a vector in $\mathbb{C}^2$ will be an $SU(2)$ matrix. The 3 dimensional rotation group, $SO(3)$, must somehow be mapped into $SU(2)$. (The map can only be accomplished up to a physically unmeasurable sign ambiguity, but whatever. This is called the "spin 1/2" (projective) representation of $SO(3)$.)

If that went over your head, consider this: the spin state is, in an extremely crude sense, pointing in some direction, and this little arrow, pointing in space, must be rotated somehow. In classical mechanics, aside from velocity, the only arrow you can really associate with an object is its angular momentum, which will will only be non-zero if the object is "spinning." Unlike a classical object, which can be made to stop spinning, a spin 1/2 particle like an electron, always has this little arrow pointing somewhere that we call spin.