130_notes.dvi

ψ~p=u~pei(~p·~x−Et)/ ̄h

Plugging this into the equation, we get arelation between the momentum and the energy.

−p^2

̄h^2

+

E^2

̄h^2 c^2

−

(mc

̄h

) 2

= 0

−p^2 c^2 +E^2 −m^2 c^4 = 0

E^2 =p^2 c^2 +m^2 c^4

E=±

√

p^2 c^2 +m^2 c^4

Note that the momentum operator is clearly still ̄hi~∇and the energy operator is stilli ̄h∂t∂.

There is no coupling between the different components in this equation, but, we will see that (unlike
the equation differentiated again) theDirac equation will give us relations between the
components of the constant spinor. Again, the solution can be written as a constant spinor,
which may depend on momentumu~p, times the exponential.

ψ~p(x) =u~pei(~p·~x−Et)/ ̄h

We should normalize the state if we want to describe one particle per unit volume:ψ†ψ=V^1. We
haven’t learned much about what each component represents yet. We also have the plus or minus
in the relationE=±

√

p^2 c^2 +m^2 c^4 to deal with. The solutions for a free particle at rest will tell
us more about what the different components mean.

36.6.1 Dirac Particle at Rest

To study this further, lets take the simple case of the free particleat rest. This is just the~p= 0
case of the the solution above so the energy equation givesE=±mc^2. The Dirac equation can now
be used.
(
γμ

∂

∂xμ

+

mc

̄h

)

ψ= 0

(

γ 4

∂

∂(ict)

+

mc

̄h

)

ψ 0 e−iEt/ ̄h= 0

γ 4

−E

̄hc

ψ 0 =−

mc

̄h

ψ 0

±mc^2

̄hc

γ 4 ψ 0 =

mc

̄h

ψ 0

γ 4 ψ 0 =±ψ 0

This is a very simple equation, putting conditions on the spinorψ 0.

Lets take the case of positive energy first.

γ 4 ψ 0 = +ψ 0