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The Klein-Gordan equation describing a spinless scalar field is one of the first things one studies in a QFT course, but there are no elementary spin-0 fields in nature.

Is the scalar field to QFT like the frictionless plane in high school physics, an aid to get us started?

Or is it deeper than that?

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I recall that the Higgs is a spinless scalar field, so the Klein-Gordon equation describes it. –  namehere Dec 2 '12 at 16:06
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Even without Higgs there are plenty of particles described by scalar fields. They are not fundamental particles, but nevertheless. –  Kostya Dec 3 '12 at 14:30
    
The Higgs field exists, as far as I know . P.S. @Dan: You stole my suggested edit and put it under your name. : ) ... –  Dimensio1n0 Jul 27 '13 at 7:16
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@Dimension10: I reviewed your edit in the "suggested edit" review queue, and then noticed some things I wanted to change too. The way SE works made it count as one edit. –  Dan Jul 27 '13 at 22:07
    
@Dan: I know, I was joking. –  Dimensio1n0 Jul 28 '13 at 1:57
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4 Answers

First reason for that is that it is the simplest one, there is no need to manipulate indexes when dealing with scalar field, and also it's important to realize the problems that happens with energy when you have to deal with this field in Klein-Gordon equation framework, and how vector field solves it.

Finally as already mentioned, Higgs bosons are described by scalar field, so if we really found it in LHC as we think, then it is actually a real thing.

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nice answer. Are all bosons described by spinless scalar fields? –  Mozibur Ullah Dec 2 '12 at 16:52
    
Yes because spins needs spinors to be described, and the simple structure of scalar field will be not enough, and that is what Dirac did through his famous equation because he didn't liked that Klein-Gordon equation doesn't include spins. –  TMS Dec 2 '12 at 16:56
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@MoziburUllah: No. Gauge bosons are described by spin 1 vector fields. Similarly, the gravitons in quantum gravity are spin 2 bosons described by a tensor field. –  Arnold Neumaier Dec 2 '12 at 17:24
    
Sorry I miss understood the question in the comments, my answer was for "Are spineless bosons described by scalar fields", anyway Arnold got the point. –  TMS Dec 2 '12 at 17:47
    
@Neumaier: Thanks. So bosons can be described by spinless scalar fields, and spinful vector & tensor fields? I'm assuming that scalar fields fundamentally cannot carry spin, but tensor & vector fields can. Is that correct? –  Mozibur Ullah Dec 2 '12 at 17:52
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Pions are also scalars. They may not be fundamental particles, but can in many situations accurately be described by scalar point-like particles.

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Fermi fields and gauge fields can emerge from scaler fields on lattice. So maybe the scalar fields are most fundamental and we only need scalar fields. See Are elementary particles actually more elementary than quasiparticles? (quantum spin models or qubit models are described by scalar fields on lattice.)

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Computations of Feynman diagrams factorize in three parts in general:

  1. The denominator has the same structure, regardless if you are dealing with scalar, spinor or vector fields.
  2. The numerator is different, you need trace identities for gamma matrices and so on, but can be calculated independently.
  3. There is a group theory factor, if you are dealing with a non-abelian gauge theory.

Because of 1., you can learn almost all the necessary tools for calculations from the case of a scalar theory. Since 2. and 3. introduce additional complications, they can be treated separately.

Conceptually all the different (free) fields arise as different irreducible representations of the Poincare group, in particular translations are generated by an Operator $P_\mu$, whose square is invariant $P^2 = M^2$. So even the components of spinors, for example, satisfy a version of the scalar Klein-Gordon equation, which is the reason for 1. above.

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