# Why must Klein-Gordan equation describe spinless particles.

It's commonly said K-G equation describes spinless particles, but we know any solution of dirac eqn is a solution of K-G eqn, then can't we say K-G could possibly describe a Dirac spinor field? But if so, it's not spinless any more.

## Answers and Replies

Fredrik
Staff Emeritus
Science Advisor
Gold Member
Every component of every (non-interacting) quantum field is a solution of the K-G equation, but only scalar fields represent spin-0 particles. One way to find out what the spin is, is to use Noether's theorem to find the conserved quantities associated with the invariance of the Lagrangian under rotations in space. These will be the three spin operators. (Verify by checking their commutation relations). Now find the eigenvalue of (S1)2+(S2)2+(S3)2, set it equal to s(s+1) and solve for s≥0.

So the saying "K-G equation describes spinless particles" is not accurate?

jambaugh
Science Advisor
Gold Member
Right. The K-G equation is satisfied by all (massive) relativistic quanta (and in the massless case reduces to the wave equation.) For spin 1/2 quanta you can "take the square root" of the K-G equation yielding the the Dirac equation.

[EDIT] Actually saying "The K-G equation describes spinless particles" is totally accurate, it just isn't complete. Being described by the K-G equation doesn't imply anything about spin.

dextercioby
Science Advisor
Homework Helper
Historically, in 1926 about 6 people found the same equation which described a free relativistic particle with mass 'm'. At that time spin of the particle was not an issue. Even as we speak, no matter the spin, the equation has the same structural form: d'Alembertian, mass squared and <wavefunction>.

Later, when the theory of representations of the restricted Lorentz and Poincare groups was built, some spinor indices came on the <wavefunction> $\Psi$ to account for the spin, but both the d'Alembertian and the mass squared remained there.

So the equation has remained since 1926 essentially the same.

P.S. As is assumed in field theory and relativistic QM $c=1$ and $\hbar =1$.

Last edited:
I see, so the field equation can hardly characterize the field, instead we need the structure of the field (scalar, spinor, vector....) and how they transform under change of reference frame, to characterize the field, am I right? Then I'm a bit curious, what's the role of field equation in QFT, any comment?

Fredrik
Staff Emeritus
Science Advisor
Gold Member
The Lagrangian that the field equation is derived from can be used to explicitly construct an irreducible representation of the Poincaré group, which automatically has the right spin. I think this is done in most QFT books.

Also, I think the Hilbert space of the free theory can be taken to be the set of positive-frequency solutions of the classical field equation. (I probably just left out a bunch of technical details, because I don't have time to think about it now). There's an unpublished set of notes by Robert Geroch titled "Special topics in particle physics" that has some of the details. This is a direct link to the pdf.

dextercioby
Science Advisor
Homework Helper
I see, so the field equation can hardly characterize the field, instead we need the structure of the field (scalar, spinor, vector....) and how they transform under change of reference frame, to characterize the field, am I right? Then I'm a bit curious, what's the role of field equation in QFT, any comment?

The field equation fully characterizes the field. It encompasses both the spin and the mass, which are the 2 Casimir invariants of the Poincare algebra.

The field equation's solutions provide a start of the so-called <canonical quantization> approach. Likewise, the field equations give us the propagator used in the path-integral approach.

How does K-G equation alone characterize the field? Both scalar field and Dirac field satisfy K-G equation.

Demystifier
Science Advisor
Gold Member
It's commonly said K-G equation describes spinless particles, but we know any solution of dirac eqn is a solution of K-G eqn, then can't we say K-G could possibly describe a Dirac spinor field? But if so, it's not spinless any more.
The correct statement is: The ONE-COMPONENT KG equation describes spinless particles. Higher-spin wave functions have more than one component, even if each of them satisfies the KG equation.

Yeah I agree, but now it's a bit fuzzy to me "what's the role of field equation in QFT"

The Klein-Gordan Hamiltonian is the correct one needed for the partition function in statistical mechanics to describe the Boltzmann statistics of spinless particles. Therefore the Klein-Gordan equation describes spinless particles.

For example, using the Klein-Gordan Hamiltonian for your partition function, you can derive say the pressure of massless spinless particles, and you'll get that it is half that of the pressure from blackbody radiation. This is because massless spin 1 particles have two degrees of freedom due to spin, while spin 0 particles only have 1.

Fredrik
Staff Emeritus
Science Advisor
Gold Member
Yeah I agree, but now it's a bit fuzzy to me "what's the role of field equation in QFT"
"The field equation" is what you get from the Lagrangian. For the Dirac field, that's the Dirac equation. The field satisfies the field equation. Its components satisfy the Klein-Gordon equation, but since the field components aren't scalar fields, the method I described before won't give you a spin-0 representation.

Hmm, if that's all, it seems field equations do not really play a big role in QFT

Hmm, if that's all, it seems field equations do not really play a big role in QFT

Well you solve the field equation to find the Green's function (i.e., free-field propagator) by setting the RHS equal to a delta function.

These form the stems of your Feynman diagrams, and in principle knowledge of these stems, along with an interaction term, gives you scattering. The trick is that using perturbation theory along with Wick's theorem reduces solving interactions into solving for the free-field propagator and multiplying a bunch of them and adding them together.

What's amazing about these Green's functions for the Klein-Gordan propagator is that they propagate backwards and forwards in time, because no matter which direction you close the contour integral (dictated by the sign of t-t' as required by Jordan's lemma), there'll be a pole, so you get:

$$\Delta(x-x')=i\theta(t-t') \int \frac{d^3k}{2(2\pi)^3E_k}e^{ik(x-x')} +i\theta(t'-t) \int \frac{d^3k}{2(2\pi)^3E_k}e^{-ik(x-x')}$$

And this shows up in the field equation, and is not just something that comes from QFT.

jambaugh
Science Advisor
Gold Member
I see, so the field equation can hardly characterize the field, instead we need the structure of the field (scalar, spinor, vector....) and how they transform under change of reference frame, to characterize the field, am I right? Then I'm a bit curious, what's the role of field equation in QFT, any comment?

The field equation is[/i] the (free) dynamics, expressed as a space-time constraint, which also expresses in particular how the system transforms under time translation. It is thus used in QFT to construct the propagator for the quantized fields.

In QFT (and QM in general) we can resolve the dynamics in terms of free propagation and interactions. Free propagation is dictated by the appropriate wave equation and interactions by the gauge field coupling. These are succinctly expressed respectively by Feynman diagram lines and vertices.

Emm, now that Dirac field also satisfy K-G equation, is it possible to write down the propagator of Dirac field from K-G equation?

Demystifier
Science Advisor
Gold Member
Hmm, if that's all, it seems field equations do not really play a big role in QFT
They do, in the Heisenberg picture.