I often hear about subatomic particles having a property called "spin" but also that it doesn't actually relate to spinning about an axis like you would think. Which particles have spin? What does spin mean if not an actual spinning motion?

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    $\begingroup$ Was there something in particular you didn't understand in the wikipedia article? en.wikipedia.org/wiki/Spin_%28physics%29 $\endgroup$
    – j.c.
    Commented Nov 2, 2010 at 19:11
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    $\begingroup$ The Wikipedia article doesn't really explain how spin manifests itself experimentally or how to measures it. That's something I would like to see discussed more in the answers. $\endgroup$
    – Michael
    Commented Nov 9, 2010 at 10:31
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    $\begingroup$ abstrusegoose.com/342 $\endgroup$
    – Qmechanic
    Commented Feb 16, 2015 at 20:08

6 Answers 6


Spin is a technical term specifically referring to intrinsic angular momentum of particles. It means a very specific thing in quantum/particle physics. (Physicists often borrow loosely related everyday words and give them a very precise physical/mathematical definition.)

Since truly fundamental particles (e.g. electrons) are point entities, i.e. have no true size in space, it does not make sense to consider them 'spinning' in the common sense, yet they still possess their own angular momenta. Note however, that like many quantum states (fundamental variables of systems in quantum mechanics,) spin is quantised; i.e. it can only take one of a set of discrete values. Specifically, the allowed values of the spin quantum number $s$ are non-negative multiples of 1/2. The actual spin momentum (denoted $S$) is a multiple of Planck's constant, and is given by $S = \hbar \sqrt{s (s + 1)}$.

When it comes to composite particles (e.g. nuclei, atoms), spin is actually fairly easy to deal with. Like normal (orbital) angular momentum, it adds up linearly. Hence a proton, made of three constituent quarks, has overall spin 1/2.

If you're curious as to how this (initially rather strange) concept of spin was discovered, I suggest reading about the Stern-Gerlach experiment of the 1920s. It was later put into the theoretical framework of quantum mechanics by Schrodinger and Pauli.


Imagine going to the rest frame of a massive particle. In this frame, there is rotational symmetry, which means that the Lie algebra of rotations acts on the wave function. So the wave function is a vector in a representation of Lie(SO(3)) = Lie(SU(2)). "Spin" is the label of precisely which representation this is. Note that while SO(3) and SU(2) share a Lie algebra, they are different as groups, and it is a fact of life ("the connection between spin and statistics") that some particles -- fermions, with half-integral spin -- transform under representations of SU(2) while others -- bosons, with integral spin -- transform under SO(3).

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    $\begingroup$ An accurate answer, but if the poster doesn't understand the actual concept of spin (not to mention group theory), this is all but useless. $\endgroup$
    – Noldorin
    Commented Nov 2, 2010 at 19:32
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    $\begingroup$ It would be more accurate to say that both fermion and boson wavefunctions transform under projective unitary representations of SO(3). Projective means “up to a phase factor”, which we know are not too important in QM. The group involved is that if 3D rotations, SO(3). It just so happens that the projective unitary reps of SO(3) are the unitary reps of its universal covering group, SU(2). $\endgroup$
    – Andrea
    Commented Jul 21, 2021 at 7:48

I'm trying to give a less technical answer. It's not rigorous but should give you the idea how spin and the regular rotation related.

Maxwell's equations say in order to have magnetic field, you need a ring current.

This can be achieved by giving angular momentum to charged particles. This can be orbital or simply because the particle is spinning. This was the original thought hence the name 'spin'.

So in the classical picture, if you spin a tiny charged ball you'll have a spinning magnet. The axis of spinning and the north pole of the magnet pointing to the same direction.

If you put this spinning magnet into a magnetic field. The field will apply torque on it to turn it into the direction of the field (this is how compasses work).

But since the our magnet is spinning this torque cause the axis of spinning precess around the magnetic field. This means the component of the rotation axis that is parallel to the magnetic field (typically referred as the Z component) won't change while the other two components (X,Y) will circle around this axis.

On the other hand if the magnetic field inhomogeneous there will be a net force on the particle that will move it (that's why magnets can snap and repel each other). This force is proportional to the Z component. So the axis perpendicular to the magnetic field there will be no force, if it's parallel there will be maximum force (basically a dot product). This allows us measuring the Z component of the rotation axis.

That's the point of the Stern–Gerlach experiment. We would normally expect that particles will spin in a whole variety of random axes. So we would expect to measure random values for the Z component.

But in reality they have measured only two possible values corresponding to the Z angular momentum component: $ħ/2$ and $-ħ/2$ (for electrons). And not any other random values. Here the classical picture breaks down, angular momentum is also quantized. You can see spin is not the classical rotation vector. It's something you can dot multiply a vector to and you can only get two possible values. The positive component typically referred as the 'up' spin component while the negative is the 'down' spin.

Precession renders all axes other than the one being measured uncertain. This is how uncertainty principle plays role here: if you measure the Z component first, then measure the X component, then the Z again, you get random up/down results again, because the measurement of the X components precessed the Y and Z component. Also, you cannot cheat here: you may want to use weaker magnetic field to reduce the precession, the displacement will be too weak to distinguish between the up and down spins. If you try to use timing; you cannot cheat again because if you measure the time accurately, then the energy so the precession rate becomes uncertain.


Spin is angular momentum of particles. The lowest possible spin is 1/2 h-bar. It's impossible for any particle with angular momentum to have a lower angular momentum than this, and whatever angular momentum a particle does have must be an integer multiple of this. Consider it the angular momentum building block. It's value is 340 dB below a kilogram meter squared radian per second.

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    $\begingroup$ Yes, but one must keep in mind that while spin is indeed angular momentum, this momentum does not come from a spinning particle and does not have a classical analogy as such. $\endgroup$ Commented Oct 7, 2012 at 12:38
  • $\begingroup$ "The lowest possible spin is 1/2 h-bar" - this is what has been observed experimentally. But, why this value is the lowest? $\endgroup$
    – phenomenon
    Commented Nov 3, 2017 at 8:28
  • $\begingroup$ You're assuming that all particles have semi-integer spin, which isn't the case. $\endgroup$
    – gented
    Commented May 14, 2019 at 13:53

All particles have spin. Though it may be zero.

At the most basic level, spin tells you how a particle transforms under rotations. For a spin $S$ particle there are $2S+1$ states, which transform into one another when it's rotated (or when the system is rotated around it). So a spin 0 particle like the Higgs is just one state, a spin $1 \over 2$ particle like an electron has two ('up' and 'down'), a spin one particle like the $Z$ has 3, and so on.


More than ten years later, time is ripe for a more up-to-date answer.

Let's consider the case of the electron. Other particles have non-zero spin, and the following considerations will be valid for them, too, with minor changes.

From a formal point of view, the spin $\frac12$ of the electron tells us that we need more than one wave function to describe its properties. Indeed, in a classical regime (Pauli's equation), we need a two-component wavefunction, while in a relativistic regime (Dirac's equation), we need four components. This fact can be nicely re-expressed in terms of the dimensionality of the irreducible representation of the Poincaré group, but it doesn't help to build intuition about the spin.

I find it more helpful to start from the relation between the angular momentum of an electron in an atom and the magnetic moment. We know from the solution of Schrödnger's equation for the Hydrogen atom that the magnetic dipole moment of the eigenstates is proportional to the eigenvalue $m$ of the $L_z$ component of angular momentum. We can explain the presence of a magnetic dipole with a non-zero probability density current (and then a corresponding electric density current) associated with the non-zero-$m$ wavefunctions. Therefore, the normal angular momentum is associated with wavefunctions carrying a non-zero probability density current.

The argument can be extended to the intrinsic spin relating it to a space-independent part of the angular momentum originating from the probability current in the two-component Pauli wavefunction. Details can be found in a paper by Mita ( American Journal of Physics 68, 259 (2000); DOI: 10.1119/1.19421 ) and also in some of its references, in particular, the paper by Ohanian.

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    $\begingroup$ Is it not too ridiculous to consider the spin as a sort of .. "sex" ? When two Hydrogen atoms are colliding, they can make a molecule H2 if one electron is spin up, and the other is spin down. Exactly like two people can make a couple if a boy meets a girl (no homo on the atomic level). If two hydrogen atoms are colliding with their electrons having the same spin, they will not make a molecule. $\endgroup$
    – Maurice
    Commented Sep 18, 2022 at 19:06

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