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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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    $\begingroup$ I recall that the Higgs is a spinless scalar field, so the Klein-Gordon equation describes it. $\endgroup$
    – resgh
    Commented Dec 2, 2012 at 16:06
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    $\begingroup$ Even without Higgs there are plenty of particles described by scalar fields. They are not fundamental particles, but nevertheless. $\endgroup$
    – Kostya
    Commented Dec 3, 2012 at 14:30
  • $\begingroup$ The Higgs field exists, as far as I know . P.S. @Dan: You stole my suggested edit and put it under your name. : ) ... $\endgroup$
    – Abhimanyu Pallavi Sudhir
    Commented Jul 27, 2013 at 7:16
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    $\begingroup$ @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. $\endgroup$
    – Dan
    Commented Jul 27, 2013 at 22:07
  • $\begingroup$ @Dan: I know, I was joking. $\endgroup$
    – Abhimanyu Pallavi Sudhir
    Commented Jul 28, 2013 at 1:57

4 Answers 4

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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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  • $\begingroup$ nice answer. Are all bosons described by spinless scalar fields? $\endgroup$
    – Mozibur Ullah
    Commented Dec 2, 2012 at 16:52
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    $\begingroup$ @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. $\endgroup$
    – Arnold Neumaier
    Commented Dec 2, 2012 at 17:24
  • $\begingroup$ 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. $\endgroup$
    – TMS
    Commented Dec 2, 2012 at 17:47
  • $\begingroup$ @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? $\endgroup$
    – Mozibur Ullah
    Commented Dec 2, 2012 at 17:52
  • $\begingroup$ @MoziburUllah: Yes you got it right, this because Spin can be represented mathematically only by spinors that has couple scalar quantities inside with a specific behavior. $\endgroup$
    – TMS
    Commented Dec 2, 2012 at 21:48
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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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