12

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?

Dan
  • 5,795
Mozibur Ullah
  • 14,713

4 Answers4

11

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.

TMS
  • 2,111
7

Pions are also scalars. They may not be fundamental particles, but can in many situations accurately be described by scalar point-like particles.

Vibert
  • 2,672
5

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.)

Community
  • 1
Xiao-Gang Wen
  • 13,725
2

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.

orbifold
  • 1,504