130_notes.dvi

the Schr ̈odinger equation. Writing that relation in the order termsappear in the Klein-Gordon
equation above we get (lettingc= 1 briefly).

−p^2 +E^2 −m^2 = 0

Its worth noting that this equation, unlike the non-relativistic Schr ̈odinger equation, relates the
second spatial derivative of the field to thesecond time derivative. (Remember the Schr ̈odinger
equation hasitimes the first time derivative as the energy operator.)

So far we have the Lagrangian and wave equation for a“free” scalar field. There areno sources
of the field (the equivalent of charges and currents in electromagnetism.) Lets assume the source
density isρ(xμ). The source term must be a scalar function so, we add the termφρto the Lagrangian.

L=−

1

2

(

∂φ

∂xν

∂φ

∂xν

+μ^2 φ^2

)

+φρ

This adds a term to the wave equation.

✷φ−μ^2 φ=ρ

Any source density can be built up from point sources so it is useful to understand thefield gen-
erated by a point sourceas we do for electromagnetism.

ρ(~x,t) =ρ(xμ) =Gδ^3 (~x)

This is a source of strengthGat the origin. It does not change with time so we expect a static field.

∂φ

∂t

= 0

The Euler-Lagrange equation becomes.

∇^2 φ−μ^2 φ=Gδ^3 (~x)

We will solve this for thefield from a point sourcebelow and get the result

φ(~x) =

−Ge−μr

4 πr

.

This solution should be familiar to us from the scalar potential for an electric point charge which
satisfies the same equation withμ= 0,∇^2 φ=−ρ=−Qδ^3 (~x). This is a field that falls off much
faster than^1 r.A massive scalar field falls off exponentiallyand the larger the mass, the faster
the fall off.(We also get a mathematical result which is useful in severalapplications. This is worked
out another way in the section on hyperfine splitting)∇^21 r=− 4 πδ^3 (~x).

Now we solve for the scalar field from a point source by Fourier transforming the wave equation.
Define the Fourier transforms to be.

φ ̃(~k) =^1

(2π)

(^32)

∫

d^3 x e−i

~k·~x

φ(~x)

φ(~x) =

1

(2π)

(^32)

∫

d^3 k ei

~k·~x ̃

φ(~k)