4Spinors
III Quantum Field Theory
4.7 Solutions to Dirac’s equation
Degrees of freedom
If we just look at the definition of a Dirac spinor, it has 4 complex components,
so 8 real degrees of freedom. However, the equation of motion imposes some
restriction on how a Dirac spinor can behave. We can compute the conjugate
momentum of the Dirac spinor to find
π
ψ
=
∂L
∂
˙
ψ
= iψ
†
.
So the phase space is parameterized by
ψ
and
ψ
†
, as opposed to
ψ
and
˙
ψ
in the
usual case. But
ψ
†
is completely determined by
ψ
! So the phase space really
just has 8 real degrees of freedom, as opposed to 16, and thus the number of
degrees of freedom of
ψ
itself is just 4 (the number of degrees of freedom is in
general half the number of degrees of freedom of the phase space). We will see
this phenomenon in the solutions we come up with below.
Plane wave solutions
We want to solve the Dirac equation
(i
/
∂ − m)ψ = 0.
We start with the simplest ansatz (“guess”)
ψ = u
p
e
−ip·x
,
where
u
p
is a constant 4-component spinor which, as the notation suggests,
depends on momentum. Putting this into the equation, we get
(γ
µ
p
µ
− m)u
p
= 0.
We write out the LHS to get
−m p
µ
σ
µ
p
µ
¯σ
µ
−m
u
p
= 0.
Proposition. We have a solution
u
p
=
√
p · σξ
√
p · ¯σξ
for any 2-component spinor ξ normalized such that ξ
†
ξ = 1.
Proof. We write
u
p
=
u
1
u
2
.
Then our equation gives
(p · σ)u
2
= mu
1
(p · ¯σ)u
1
= mu
2
Either of these equations can be derived from the other, since
(p · σ)(p · ¯σ) = p
2
0
− p
i
p
j
σ
i
σ
j
= p
2
0
− p
i
p
j
δ
ij
= p
µ
p
µ
= m
2
.
We try the ansatz
u
1
= (p · σ)ξ
0
for a spinor ξ
0
. Then our equation above gives
u
2
=
1
m
(p · ¯σ)(p · σ)ξ
0
= mξ
0
.
So any vector of the form
u
p
= A
(p · σ)ξ
0
mξ
0
is a solution, for A a constant. To make this look more symmetric, we choose
A =
1
m
, ξ
0
=
√
p · ¯σξ,
with ξ another spinor. Then we have
u
1
=
1
m
(p · σ)
√
p · ¯σξ =
√
p · σξ.
This gives our claimed solution.
Note that solving the Dirac equation has reduced the number of dimensions
from 4 to 2, as promised.
Example. A stationary, massive particle of mass m and p = 0 has
u
p
=
√
m
ξ
ξ
for any 2-component spinor
ξ
. Under a spacial rotation, we have a transformation
ξ 7→ e
iφ·σ/2
ξ,
which rotates ξ.
Now let’s pick
φ
= (0
,
0
, φ
3
), so that
φ · σ
is a multiple of
σ
3
. We then pick
ξ =
1
0
.
This is an eigenvector of
σ
3
with positive eigenvalue. So it is invariant under
rotations about the
x
3
axis (up to a phase). We thus say this particle has spin
up along the
x
3
axis, and this will indeed give rise to quantum-mechanical spin
as we go further on.
Now suppose our particle is moving in the x
3
direction, so that
p
µ
= (E, 0, 0, p
3
).
Then the solution to the Dirac equation becomes
u
p
=
√
p · σ
1
0
√
p · ¯σ
1
0
=
p
E − p
3
1
0
p
E + p
3
1
0
In the massless limit, we have E → p
3
. So this becomes
√
2E
0
0
1
0
.
For a spin down field, i.e.
ξ =
0
1
,
we have
u
p
=
p
E + p
3
0
1
p
E − p
3
0
1
→
√
2E
0
1
0
0
.
Helicity operator
Definition
(Helicity operator)
.
The helicity operator is a projection of angular
momentum along the direction of motion
h =
ˆ
p · J =
1
2
ˆp
i
σ
i
0
0 σ
i
.
From the above expressions, we know that a massless spin up particle has
h = +
1
2
, while a massless spin down particle has helicity −
1
2
.
Negative frequency solutions
We now try a different ansatz. We try
ψ = v
p
e
ip·x
.
These are known as negative frequency solutions. The
u
p
solutions we’ve found
are known as positive frequency solutions.
The Dirac solution then becomes
v
p
=
√
p · ση
−
√
p · ¯ση
for some 2-component η with η
†
η = 1.
This is exactly the same as
u
p
, but with a relative minus sign between
v
1
and v
2
.
A basis
It is useful to introduce a basis given by
η
1
= ξ
1
=
1
0
, η
2
= ξ
2
=
0
1
We then define
u
s
p
=
√
p · σξ
s
√
p · ¯σξ
s
, v
s
p
=
√
p · ση
s
−
√
p · ¯ση
s
These then form a basis for the solution space of the Dirac equation.