# What does spin 0 mean exactly?

I heard two definitions:

(1) Spin 0 means that the particle has spherical symmetry, without any preferred axis.

(2) The spin value tells after which angle of rotation the wave function returns to itself: 2 pi / spin = angle. Therefore, spin 1/2 returns to itself after 4 pi, spin 1 after 2 pi, and spin 0 after an infinite rotation angle.

These seem to contradict each other: a sphere returns to itself even after infinitesimal rotations. Can somebody clarify?

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The formula for the return angle is also valid for spin 2: such systems return to their original state after a rotation by pi. –  Clara Jul 2 '12 at 4:59
Hi Clara. You can edit the question to include the comment and then delete the comment (if not, you'll be able later). The tags are also slightly off. Also notice that for a quantitative understanding, explaining Spin in a purely mathematical context is probably conceptually easier first. You mention the term "particle" and thereby introduce potential spatial dependencies and then the term "spherical symmetry" gets a little more confusing. Keywords: The spins label the irreducible representations of the rotation group. I'd say the angle dependency is merely inherited. (don't know if helpful) –  Nick Kidman Jul 2 '12 at 8:23

There is no contradiction, though your statement (2) for spin 0 is incorrect. The wavefunction of a particle of spin $j$, when rotated by $2\pi$ radians, is multiplied by a phase $(-1)^{2j}$, so that integer spins return to themselves after $2\pi$ rotations but half-integer spins need $4\pi$. Thus for spin $j=0$, the wavefunction remains invariant under $2\pi$ rotations. Spin $j=0$ wavefunctions are additionally invariant under all rotations, which is consistent with the special $2\pi$ case.

The full story is that for a rotation of angle $\theta$ about an axis with unit vector $\hat{\mathbf{n}}$, the wavefunction $|\psi\rangle$ of a spin-$j$ particle transforms to $$e^{i\frac{1}{2}\theta\mathbf{J}\cdot\hat{\mathbf{n}}}|\psi\rangle,$$ where $\mathbf{J}$ is the angular momentum vector operator, which obeys $\mathbf{J}^2=\hbar^2j(j+1)$. For $2\pi$ rotations, this reduces to the rule above. For spin $j=0$, you get $\mathbf{J}=0$ and thus the exponential equals unity for all angles.

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The first definition is the most correct one, although the more general transformation group considered is the Lorentz group.

As Nick Kidman said, the idea is that we want particles to be in an irreducible representation of that group, so that the equations of motion can behave nicely under Lorentz transformations, which should leave the physics invariant in the context of relativistic particle physics.

The lowest dimensional representation of that group is the trivial group that contains only the identity. This is what we call a scalar, or a spin 0. In more basic terms, the spin 0 is the one that does not change under any Lorentz transformation (boosts or rotations). It is indeed invariant under rotations.

The second definition you give is more a property of nonzero spins than a definition, I guess. It does indeed not work for spin 0, but I do not think there is a deep explanation for that (if I'm wrong, I'd actually love to hear it !!)

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The two definations tell the same thing. consider a rotation by a angle $\phi$ about z-axis, the spin-s system described by $\mid \alpha >$ becomes $$\exp(iS_z \phi) \mid \alpha >$$. $\phi$ is unchanged by any rotation Since $S_z=0$ which means the state has spherical symmetry. –  Craig Thone Jul 2 '12 at 14:41
As finite compact Lie-groups are very well understood, I'd say you can give a deep explaination for everything in this subject, hehe. The raising of the eigenvalues and the angle dependence is "visible" in the explicit construction of all the according representations, see e.g. Wigners matrices. They fulfill a nice eigenvalue equation and so higher spins "turn faster". But from a conceptual pov I'd rather want to start with the labeling and then "discover" the angle-thing. That's why I said "I'd say the angle dependency is merely inherited." –  Nick Kidman Jul 2 '12 at 15:15
@CraigThone, your comment is approximately what I was about to write up as an answer. I think that's the easiest way to explain this issue. I think you should expand it into an answer. –  Colin McFaul Jul 2 '12 at 16:23
Emilio Pisanty has done some of that. If you have much to say, I suggest you write a full answer. I am waiting for it. –  Craig Thone Jul 3 '12 at 1:46

I think the more exact answer is that a spin $n$ quantum particle is a quantum of a quantized mode of a rank $n$ tensor field.

Let's break this down into manageable pieces. The rank of a tensor is, loosely speaking, the number of times you must apply a coordinate transformation to transform the components.

So, for a familiar example, a vector is a rank 1 tensor and so, to transform the components of a vector, you apply the coordinate transformation once. This means that the vector rotates ("spins") at the same "rate" as the coordinates do under a rotation (keep that in mind when you think about spin 1).

The spacetime metric of GR is a rank 2 tensor and so, to transform it's components, you apply the coordinate transformation twice. This means that the metric rotates ("spins") at twice the "rate" as the coordinates do under a rotation (keep that in mind when you think about spin 2).

Finally, a scalar is a rank 0 tensor and so, to "transform" a scalar, you apply the coordinate transformation zero times, i.e., it doesn't rotate ("spin") at all under a coordinate transformation (keep that in mind when you think about spin 0).

If you take a classical rank $n$ tensor field and quantize it, you get spin $n$ "particles" (quanta).

In the case of fermions, we can loosely say that a spinor field is a rank 1/2 "tensor" or, in some sense, the square root of a vector. A spinor rotates ("spins") at half the "rate" as the coordinates do under a coordinate transformation.

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The only comment is that the rank n tensor has to be symmetric, as antisymmetric tensors transform at lower spin than their rank. –  Ron Maimon Jul 2 '12 at 16:26
Thanks for the clarification Ron! –  Alfred Centauri Jul 2 '12 at 16:44