Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

We know a vector field is a $(\frac{1}{2},\frac{1}{2})$ representation of Lorentz group, which should describe both spin-1 and spin-0 particles. However many of the articles(mostly lecture notes) I've read, when they talk about the relation between types of fields and spins of particles, they'll always say something like

[...] scalar fields describe spin-0 particles, vector fields describe spin-1 particles. [...]

Is there a good reason to omit the spin-0 part of vector fields? If one really wants to talk about spin-1 only, why not talk about tensor fields in the $(1,0)$ or $(0,1)$ representation?

share|improve this question
add comment

2 Answers

up vote 11 down vote accepted

It's because 4-vector fields in the $(1/2,1/2)$ representation don't produce any spin-0 excitations, at least not in consistent theories.

Electromagnetism is the canonical example. The vector field $A_\mu$ would create both positive-norm ($A_i$) and negative-norm ($A_0$) polarizations. The latter is time-like. However, probabilities can't be negative, so for this electromagnetic potential as well as any other 4-vector field, there must exist a gauge symmetry that decouples the spin-0, timelike component. In fact, it decouples two components – the time-like one and the longitudinal one. The longitudinal one may get restored by the Higgs mechanism in the case of massive vectors such as the W-bosons.

At any rate, whether one talks about massless particles created such fields – e.g. photons or gluons – or about the massive ones such as W- and Z-bosons, there is no physical spin-0 polarizations which is why the massive ones in particular are known as vector bosons.

Incidentally, for electromagnetism and other gauge theories, the $(1,0)$ or $(0,1)$ representations appear, too. $F_{\mu\nu}$, the gauge-invariant field strength of the $(1/2,1/2)$ potential, transforms as $(1,0)\oplus (0,1)$. In the Minkowski signature, one may impose a real projection on this representation to get 6 real components (electric and magnetic field strength); in other signatures, one may separate $(0,1)$ and $(1,0)$.

share|improve this answer
I need a clarification for one more point. Is it that a vector field can describe either spin-0 or spin-1 but not both, or is it that it can only describe spin-1? –  Jia Yiyang Dec 25 '12 at 15:58
add comment

I believe that you are confusing SO(4) irreps with SO(3) irreps. The (1/2,1/2) representation of SO(4) is irreducible, so it corresponds to a single spin, spin 1. You get a vector and a scalar in the Clebsch-Gordan decomposition of two spinors of SO(3).

Of course, these two notions are related because the (1/2,1/2) irrep of SO(4) is formed from two spinors of SO(3), which I imagine is the source of your confusion. This means that under an SO(3) subgroup of SO(4) the vector will decompose as a product of an SO(3) vector and a scalar. For instance, under the subgroup of spatial rotations the space-like components of a 4-vector become a 3-vector and the time-like component becomes a scalar. However, Lubos has explained that in a good theory even the SO(3) scalar corresponding to the timelike component does not contribute any physical excitations. I'm not sure what answer you were looking for but I thought I should add this answer in case it was a simple case of mixing up SO(3) and SO(4) irreps.

The moral of the story is that you must always remember which group you are dealing. When people say that the (1/2,1/2) irrep corresponds to a spin 1 field they are talking about spin 1 of SO(4). When people talk about addition of angular momentum in quantum mechanics the group is SO(3).

share|improve this answer
I agree that it's that when restricting to subgroup $SO(3)$ we get a C-G decomposition of spin-1 and spin-0, but I don't see what's wrong with calling these two components as spin-1 and spin-0 particles. After all spin is defined via the representation of little group and in our case(say massive) it is $SO(3)$. It seems to me that irreducibility in the full Lorentz group and reducibility in $SO(3)$ mean a boost may mix the states of spin-1 and spin-0 and a rotation cannot, which is not quite physical, but still it has nothing to do with the fact we have two types of spins. –  Jia Yiyang Dec 26 '12 at 2:03
Ok, I see what is going on here. You have to differentiate between representations of the Lorentz group and representations of the Poincare group. Of course you are right that the spin is defined by the representation of the little group. However, to determine the spin of a field you do not decompose the representation of the Lorentz group into representations of the little group. The little group is used to construct representations of the Poincare group in which single particle states transform. –  ald5657 Dec 26 '12 at 15:21
So, you can see that a massive vector field (Lorentz group) is spin 1 because it creates single particle states that transform in the spin 1 representation of the little group SO(3), not because it decomposes into the spin 1 representation under the SO(3) subgroup. –  ald5657 Dec 26 '12 at 15:22
When restricting to rotation subgroup, there's a close relation between of representation of Lorentz group using field and using single particle states, i.e. the field representation must contain single particle state representation, c.f. Weinberg chap 5.1 page 196. Anyway this is not quite the point, the point is in the first place how do you know $(\frac{1}{2},\frac{1}{2})$ can describe spin-1?It is because $\frac{1}{2}+\frac{1}{2}=1$, then what makes you throw away $\frac{1}{2}-\frac{1}{2}=0$? –  Jia Yiyang Dec 27 '12 at 6:49
For our purposes you should completely disregard what the (1/2,1/2) irrep decomposes into under SO(3). Yes, it is true that this decomposition must always contain the correct spin representation, but that is not necessary to understand the spin. The point is that if you take a (1/2,1/2) field like the photon and calculate the creation and annihilation operators you will find that these operators always create spin 1 single particle states and never spin 0 states. That's why we say that the (1/2,1/2) irrep is spin 1. –  ald5657 Dec 27 '12 at 15:02
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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