0.1 Klein-Gordon Equation 0.2 Dirac Equation

[revised Oct 15, 2010]
0.1 Klein-Gordon Equation
The relativistic Schrödinger equation
∂ µ ∂ ν ψ − m 2 ψ = 0
became known as the Klein-Gordon equation after it was rejected by Schrödinger
as a relativistic one particle generalization of the NR Schrödinger equation.
To list three of its defects (1.) It has solutions of unbounded negative energy
so that it predicts continuous radiative transitions, (2.) The only conserved
current 4-vector ψ † ∂ ↔ µ ψ can have negative j 0 and cannot therefore be interpreted
as a probability density, and (3.) When generalized to include
interaction with the electromagnetic field, its predictions do not agree with
the observed spectrum of the H-atom. (1.) is easily fixed in a context where
ψ is viewed as a quantized field. (Dirac had a solution in a one particle
context.) (2.) can be fixed by interpreting the current as a charge-current
4-vector. (Dirac’s equation fixed it for spin 1 2 particles.) (3.) was fixed by
Dirac. (It can also be used for spin 0 particles.)
0.2 Dirac Equation
Dirac’s solution to the lack of a positive definite probability density was to
make the equation linear in the space and time derivatives. In doing so he
found an equation that included the electron spin and accounted correctly for
its magnetic moment. His idea was to express the energy as a linear form in
the momentum and mass by introducing non-commuting coefficients, realized
as matrices, and allowing ψ to become a multicomponent wave function.
Now requiring E 2 = p 2 + m 2 gives
E = β · p + αm.
β i β j + β j β i = 2g ij
αβ i + β i α = 0
α 2 = 1, (1)
1

i.e., α and β i are a set of 4 anticommuting matrices whose square is unity.
In terms of derivatives the Dirac equation is:
i∂ 0 ψ = −iβ i ∂ i ψ + αmψ
To give it a covariant appearence (deceptive because the γ’s are numerical
matrices that do not change under LT, but see below), introduce γ i = αβ i
and γ 0 = α, multiply the equation by α and obtain
The γ µ satisfy
−iγ µ ∂ µ ψ + mψ = 0 (2)
{γ µ , γ ν } = −2g µν (3)
where {} means anticommutator. A possible explicit form is
( )
( )
1 0
0 σ
γ 0 =
γ i i
=
0 −1
−σ i 0
Each entry is a 2 by 2 matrix, making γ 4 by 4. An equivalent explicit form
is
( )
( )
0 1
0 σ
γ 0 =
γ i i
=
1 0
−σ i (5)
0
The first form is the standard form, good for taking the NR limit. The
second form is the chiral form, good for neutrinos and things. Changing
involves reshuffling components of ψ. Both forms can be represented as
direct products of two sets of Pauli matrices τ i for the skeleton and σ i for
the entries: the first set is (τ 3 , iτ 2 σ i ) and the second is (τ 1 , iτ 2 σ i ). Both
forms (and others you can easily invent) satisfy Eq.(3). In both forms, the
hermitian conjugate of γ 0 is γ 0 and of γ i is −γ i . A very useful relation valid
for both forms is
γ µ† = γ 0 γ µ γ 0 .
0.3 Free Particle Solutions
Look for solution in the form of an eigenstate of the energy-momentum vector:
ψ = ψ 0 e ipx ,
(4)
2

where ψ 0 is a constant 4-component column vector (Dirac spinor). ψ 0 must
satisfy (γ µ p µ + m)ψ 0 = 0. If ψ 0 has upper components χ and lower components
χ ′ (both ordinary 2-component spinors), then the equation takes the
form ( ) ( )
−E + m ⃗σ · ⃗p χ
−⃗σ · ⃗p E + m χ ′ = 0 (6)
using the standard form for the γ-matrices and replacing p 0 by E. There is
a non-trivial solution if the determinant is zero. To evaluate things with σ
use the identity σ i σ j = δ ij + iɛ ijk σ k . The determinant is (−(E) 2 + ⃗p 2 + m 2 ) 2
which has double roots at E = ± ω with ω = + √ ⃗p 2 + m 2 , implying two
positive energy states and two negative energy states for each value of ⃗ k.
Negative energy states are still present, but Dirac eliminated them by saying
that they were full, so the exclusion principle forbade transitions to them.
A hole in the negative energy ’sea’ behaves like positive particle of the same
mass: positron.
DISCUSS: motion of holes.
A solution of Eq.(6) is obtained by taking χ to be an arbitrary twocomponent
spinor. The equation then requires that χ ′ = ⃗σ · ⃗pχ/(E + m).
Note that the equation implies that χ ′ ≈ vχ in the NR limit. (χ called large
components and χ ′ small components.) We will return to these states when
we treat the Dirac equation as a field equation.
The physical meaning of states and operators clearer if we multiply the
Dirac equation by γ 0 and identify p 0 with the Hamiltonian:
H = p 0 = γ 0 γ i p i + γ 0 m,
where p i can be interpreted either as the momentum operator or its eigenvalue.
Clearly the spatial momentum commutes with H, but what about the
angular momentum? Calculation shows that the orbital angular momentum
⃗r × ⃗p does not commute with H. By inspecting the commutator, we can find
that adding a term gives an operator that does commute with H:
J i = ɛ ijk r j p k + i 4 ɛ ijkγ j γ k
The explicit form of the second term is
( )
S i = 1 σi 0
2
0 σ i
3

which can be identified with the spin. The total angular momentum J i =
L i + S i commutes with H.
As in NR QM, The plane wave solutions above are not eigenstates of
any of the angular momentum operators. (For eigenstates of ⃗ J see the H-
atom below.) There is however an important operator that commutes with
H and can be well-defined, the helicity h = ⃗σ · ⃗p/|⃗p|. It has eigenvalues ±1,
and by choosing χ to be one of its eigenstates, the full ψ will also be an
eigenstate. The helicity is not Lorentz invariant, however, because a boost
in the direction of the momentum can bring the particle to rest and reverse
its momentum without changing its spin.
0.4 Transformation
If we do an LT, the components of ψ get mixed like the components of a
3-vector under rotation. The new wave function satisfies
ψ ′ (x ′ ) = S(Λ)ψ(Λ −1 x ′ )
by analogy with the similar equation for rotations. Note first that S must
be a representation of the Lorentz group up to a phase. Starting with the
Dirac equation in x, change variables to x ′ = Λx and multiply by S:
−iSγ µ S −1 (∂ µ x ′ν )∂ ′ νSψ(Λ −1 x) + mSψ(Λ −1 x) = 0,
where we used the chain rule and inserted S −1 S remembering that S acts in
the same space as the γ’s and may not commute with them. The result is
the Dirac equation in the new frame with the new wave function identified
as above, provided S(Λ) satisfies the condition
S −1 γ µ S = Λ µ ν γν (7)
Since S(Λ) is a representation of the Lorentz Group, we can find it by finding
its infinitesimal elements. Putting S = 1 + ɛ ab Σ ab ,
(1 − ɛ ab Σ ab )γ µ (1 + ɛ ab Σ ab ) = (δ µ ν + ɛ abg µν G ab
µν )γν
By ansatz, the solution is found to be Σ ab = − 1 2γ a γ b . It is interesting to note
the explicit forms of Σ for rotations
( ) ⃗σ
Σ ij = i k
0
2
0 ⃗σ k .
4

and for boosts
Σ 0i = − 1 2
( 0 ⃗σ
k
⃗σ k 0
The i for rotations but not for boosts means that S is unitary for rotations
but not for boosts. Good because spatial volume is preserved by rotations
but not by boosts because of Lorentz contraction, so ψ † a ψ a is preserved by
rotations but changed by boosts.
Thus ψ † ψ is not a scalar. However, it is easy to show using γ µ† = γ 0 γ µ γ 0
that ψ † γ 0 ψ does not change under an infinitesimal LT, and therefore under
any LT. It therefore satisfies the condition S ′ (x) = S(Λ −1 x) that defines it
as a scalar. This is important enough to warrant special notation:
ψ = ψ † γ 0 .
)
.
Using Eq.(7), you should be able to prove the following
ψψ is a scalar.
ψγ µ ψ is a vector.
ψγ µ γ ν ψ is a second rank tensor.
Other covariants can be expressed with the aid of the matrix
γ 5 = iγ 0 γ 1 γ 2 γ
( )
3
0 1
=
1 0
( ) −1 0
=
0 1
in standard form
in chiral form
You can prove
ψγ 5 ψ is a pseudoscalar
ψγ µ γ 5 ψ is a pseudovector (axial vector).
It is also true that covariants with 3 γ’s can be expressed in terms of the
others.
5

0.5 NR Limit
Nothing better shows the physics buried in the Dirac equation than taking
the NR limit. To understand, we must add the coupling of the Dirac electron
to the electromagnetic field (discussion below in Dirac Equation as a Field
Equation).
γ µ (−i∂ µ + eA µ ) ψ + mψ = 0 (8)
or in Hamiltonian form
H = −eA 0 ψ + γ 0 γ i (p i + eA i )ψ + mγ 0 ψ (9)
Thus Hψ = Eψ in two-component form with E moved to left is
( m − eA 0 − E ⃗σ · (⃗p + eA)
⃗ ) ( ) χ
⃗σ · (⃗p + eA)
⃗ −m − eA 0 − E χ ′ = 0.
Eliminate χ ′ in favor of χ, and A 0 = φ:
(
(m − E − eφ) + ⃗σ · ⃗Π 1
m + E + eφ
)
⃗σ · ⃗Π χ = 0 (10)
where we used the abbreviation ⃗ Π = ⃗p+e ⃗ A. To get to the NR limit, subtract
the rest energy from E to get the NR energy W = E − m and write the
denominator
1
2m + (W + eφ) ≈ 1
2m − 1
1
(W + eφ) +
4m2 8m (W + 3 eφ)2 + . . .
First term in () in Eq.(10) is O(mv 2 ); second is O(mv 2 ) plus m times higher
powers of v 2 . Using the commutator [p i + eA i , p j + eA j ] = −ieF ij and the
identity σ i σ j = g ij + iɛ ijk σ k , and keep terms up to order mv 2 , this becomes
(
−W − eφ + 1 Π
2m ⃗ 2 + e )
⃗σ
m 2 · ⃗B χ = 0. (11)
This is the Pauli equation for a spinning NR particle. The last term is
the interaction of the spin magnetic moment with the magnetic field. The
magnetic moment is (e/m) S ⃗ rather than the orbital (e/2m) L, ⃗ indicating the
g-factor of 2 put in by hand in pre-Dirac times and justified a postiori as the
Thomas precession.
6

In higher order, the equivalence of DE to Pauli equation is approximate
because the energy eigenvalue appears in the expansion of the second term
in Eq.(10) so that is does not have the same form as the Pauli equation.
Nonetheless, we can obtain an approximate Pauli equation from the next
term in the expansion above as follows:
− 1
4m 2 (⃗σ · ⃗Π)(W + eφ)(⃗σ · ⃗Π) = − 1
4m 2 (⃗σ · ⃗Π) 2 (W + eφ) − 1
4m 2 (⃗σ · ⃗Π)[eφ, ⃗σ · ⃗Π].
It is correct to order v 4 to replace W + eφ in the first term on the right by
(⃗σ · ⃗Π) 2 according to Eq.(11). In the absence of a magnetic field ⃗ A = 0, the
result is the order v 4 relativistic kinetic energy correction −p 4 /8m 3 . Taking
the commutator in the second term and using the σ i σ j identity yields
second =
ie
4m (δ 2 ij + iɛ ijk σ k )Π i ∂ j φ
in which the first term is known as the Darwin term and the second is the
spin-orbit term which was added phonomenologically to the Pauli equation
in the olden days. To appreciate the physical significance of these terms, let’s
specialize to A ⃗ = 0, whence the Darwin term becomes
Darwin = − ie
4m 2 (∇2 φ − ⃗ E · ⃗∇).
This spin independent term represents a relativistic correction to the effective
potential which has obscure experimental consequences. For the pure
coulomb field, the first term is a delta function which shifts the energy of H-
atom s-states. The significance of the spin-orbit term appears if we specialize
to central potentials φ(r) where it becomes
SO = −
e 1 ∂φ
⃗σ · ⃗L.
4m 2 r ∂r
COMMENT
A more systematic and elegant way of approaching the NR approximation
is the Foldy-Wouthuysen (FW) transformation. The idea is to make a unitary
transformation of the states and operators, ψ = Uψ and O F W = UOU † with
UU † = 1, such that the new Hamiltonian is block diagonal, decoupling the
upper and lower components and giving separate two-component equations
for each. For a free particle, this is accomplished by
√
m + ω
U =
2ω + γ i p
√ i
,
2ω(ω + m)
7

in which ω is the operator + √ ⃗p 2 + m 2 which commutes with ⃗p. It is easily
verified that UU † = 1, and a lengthy calculation shows that the FW
Hamiltonian reduces to
( )
ω 0
H F W =
.
0 −ω
Thus the upper components represent the positive energy solutions and the
lower components represent the negative energy solutions.
Although the momentum operator is not altered by the FW transformation,
since U commutes with ⃗p, the position operator is modified and
becomes a non-local operator in a way that is difficult to calculate in this
context. This is an indication of the fact that it is not possible to form a
state that is localized to a single point using positive energy solutions only.
At best, the paritcle can be localized to within a Compton wavelength m −1 .
It is possible to do a FW transformation for particle subject to potentials,
but it must be done sucessively for each higher order in v 2 . Omit it here.
0.6 H Atom
0.6.1 Hamiltonian
The relativistic treatment of the H-atom is a major success of the Dirac
theory. More generally, we treat a particle in a central field in the absence of
a magnetic field. In particular, we neglect the (weak) magnetic field due to
the nuclear magnetic moment. With the Dirac equation γ 0 (−i∂ 0 + eA 0 )ψ +
γ i ∂ i ψ + mψ = 0 as a starting point, we rewrite in the form of a Schrödinger
equation i∂ t ψ = Hψ with the Hamiltonian
H = γ 0 γ i p i − α r + γ0 m
( )
m −
α
⃗σ · ⃗p
=
r
⃗σ · ⃗p −m − α (12)
r
where α = e 2 /4π is the fine structure constant, value approximately 1/137.
−α/r can be replaced by any central potential −eφ(r) or V (r).
0.6.2 Conserved Quantities
As noted earlier, the total angular momentum ⃗ J is conserved. Therefore we
can have simultaneous eigenstates of J 2 , J z , and H. in addition, the parity
8

is conserved. The operator P defined by P ψ(⃗r) = ψ(−⃗r) does not commute
with the Hamiltonian because it changes the sign of the momentum operator.
However P = γ 0 P does commute. Note that P commutes with the spin
operator S i = i 4 ɛ ijkγ j γ k , so it does not change its value. This is proper
because the spin, like the orbital angular momentum, should be a good axial
vector and not change sign under space reflection. According to NRQM,
operators J 2 , L 2 , and J z have eigenstates, generally denoted by Ylj m. j can
be either l + 1 2 or l − 1 2 . We introduce the quantity ω with values ±1 such
that j = l + 1 2ω. The parity P of the eigenstate is (−1) j− ω 2 so that states
with different values of ω have different parity. We denote the eigenstates by
Yωj m and they satisfy
J 2 Y m ωj = j(j + 1)Y m ωj
J z Y m ωj = mY m ωj
L 2 Y m ωj = (j − ω 2 )(j − ω 2 + 1)Ym ωj
PY m ωj = (−1) j− ω 2 Y
m
ωj
Don’t need to know their explicit forms.
0.6.3 Eigenstates
Look for solutions that are simultaneous eigenstates of H, P, and angular
momentum. Since P = γ 0 P looks like
( )
P 0
P =
,
0 −P
the parity of the upper components is opposite to the parity of the lower
components. Our eigenstate must look like
ψ = 1 ( ) f(r)Y
m
ωj
r ig(r)Y−ωj
m , (13)
where the i and 1/r are put in for convenience. The following identities ease
the calculation along. Try to prove them, but proofs will be provided in class.
ˆr is the unit radial vector.
i.)
⃗σ · ⃗p = −i⃗σ · ˆr∂ r +
9
i⃗σ · ˆr
⃗σ ·
r
⃗L

ii.) ⃗σ · ˆr commutes with ⃗ J and changes parity, so you can prove
⃗σ · ˆrY m ωj = −Y m −ωj
iii.) Since ⃗σ · ⃗L = J 2 − L 2 − 1 4 ⃗σ2 , it is easy to get
⃗σ · ⃗L Y m ωj = (ωj + 1 2 ω − 1) Ym ωj
Putting the Hamiltonian and the wave function together, we obtain the
coupled equations
0.6.4 Solution
f ′ − ω(j + 1 2 )
r
g ′ + ω(j + 1 2 ) g −
r
At r = ∞ both f and g satisfy the equation
(
f − m + E + α )
g = 0 (14)
r
(
m − E − α )
f = 0 (15)
r
f ′′ − (m 2 − E 2 )f = 0
which has decaying solution f = e −√ m 2 −E 2r , suggesting the new dimensionless
radial variable
ρ = √ m 2 − E 2 r
Introducing this variable and new independent variables such that f = F e −ρ
and g = Ge −ρ brings the equations to the form
dF
dρ − F − βF ρ
− 1 ν G − αG ρ
dG
dρ − G + βG ρ
− νF + αF ρ
= 0
= 0
(16)
in which ν stands for √ (m − E)/(m + E) and β stands for ω(j+ 1 2 ). Examination
of the equations near ρ = 0 shows F and G must behave as ρ s where
s = √ β 2 − α 2 , a power which is less than |β| (≥1) by something of the order
of α 2 . Introducing
F (ρ) = ∑ a p ρ s+p
p=0
10

and a similar series for G with coefficients b p leads to the following recursion
relation
(s + p + 1)a p+1 − a p − βa p+1 − 1 ν b p − αb p+1 = 0
(s + p + 1)b p+1 − b p + βb p+1 − νa p + αa p+1 = 0
The series must terminate to avoid generating the exponentially increasing
solution. If p is the index of the highest power of ρ before termination, then
we can deduce that a p = −νb p from the fact that a p+1 and b p+1 are both zero
while a p and b p are non-zero. If we use this relation in the recursion relations
above with p → p − 1, we find that all the coefficients can be eliminated and
a quadratic equation for ν obtained:
The appropriate solution is positive:
αν 2 + 2(p + s)ν − α = 0.
ν = − p + s
α
√ (p ) 2 + s
+ + 1
α
It is then a matter of algebra to deduce from this that
E =
m
√
1 + α2
(p+s) 2
To make sense of this and to compare it to the NR result, we can expand it in
powers of α = e 2 /4π ≈ 1/137, noting the dependence of s = √ (j+ 1 2) 2 − α 2
on α. The final result is
E = m
(1 − α2
2n 2 − α4
2n 4 ( n
j+ 1 2
− 3 4
)
)
+ O(α 6 ) ,
in which n stands for p + j + 1 2. After the rest energy, the α 2 term is the
Rydberg energy calculated in NR QM and the α 4 term is the relativistic
correction. In all orders the energy depends only on j and n, so some but
not all of the ’accidental’ degeneracy of the H atom remains. DISCUSSION
11

0.7 Dirac Equation as a Field Equation
Although the Dirac equation is somewhat sucessful as a one particle equation,
the idea of a filled sea of negative energy states is bizarre, and is also
not applicable to particles of integral spin that do not satisfy the exclusion
principle. The integral spin particles also do not have a consistent quantum
treatment with positive definite probability density. There were other difficulties
such a the Klein paradox. Therefore the Dirac equation came to be
regarded as a field equation to be subjected to a form of canonical quantization
just like the scalar and electromagnetic fields. To realize this, we must
have a Lagrangian for the Dirac equation. A Lagrangian density that serves
is
L = iψγ µ ∂ µ ψ − mψψ (17)
It is easily verified that the equations of motion arising from this are the
Dirac equation and its conjugate:
i∂ µ ψγ µ + mψ = 0
When we come to construct the Hamiltonian, however, there are pathological
elements in the calculation. For example, the momentum conjugate to ψ is
identically zero, and the momentum conjugater to ψ is
Π = ∂L
∂∂ 0 ψ
= iψγ 0
Since the momentum conjugate to ψ is iψ † , a coordinate, we are dealing with
a constrained system. Though there are sophisticated methods to deal with
such constraints, we will follow a simple practical method and consider only
ψ to be a coordinate in the Lagrangian. When we construct the Hamiltonian
as
H = Π∂ 0 ψ − L,
We find that the resulting Hamiltonian does not involve time derivatives,
and that these could not be eliminated anyway as in the previous examples
because the canonical momentum does not contain time derivatives. Despite
these peculiarities, the Hamiltonian density derived above leads to the
Hamiltonian
∫
H = d 3 xψ ( † iγ 0 γ i ∂ i + γ 0 m ) ψ,
12

expressed in terms of the fields. Although this looks like the expectation
value of the Dirac Hamiltonian in the single particle theory, it now signifies
the field Hamiltonian with ψ and ψ interpreted as fields to be quantized later.
0.7.1 Dirac Field in Fourier Space
To formulate the theory in Fourier space, we begin by expressing ψ as a sum
of free particle solutions of the free Dirac equation. Denote the two positive
energy solutions by
ψ = u s ( ⃗ k)e i⃗ k⃗x e −iωt
and the two negative ones by
ψ = ṽ s ( ⃗ k)e i⃗ k⃗x e iωt
with the usual definition of ω and s = 1, 2. The tilde is used because we will
later wish to redefine the negative energy states to make them more easily
identifiable as antiparticle states. Because u s and ṽ s are eigenstates of the
Hermitian H belonging to different eigenvalues, they satisfy u † s (⃗ k)ṽ s ′( ⃗ k) = 0.
They can also be chosen orthogonal among themselves by appropriate choices
of the spinor χ in Eq.(??). This can be done simply by choosing the χ s ’s
to be orthogonal. However, unless the χ’s are eigenstates of the helicity
h = σ · ˆp which is conserved by the free H, the resulting u’s or v’s will not
be eigenstates of any component of the spin even though χ is always such an
eigenstate. In fact, the components of the spin are not conserved by the free
Dirac Hamiltonian unless they are in the direction of the momentum. We
normalize the states to satisfy the orthonormality relations
u † s( ⃗ k)u s ′( ⃗ k) = 2ωδ ss ′,
ṽ † s (⃗ k)ṽ s ′( ⃗ k) = 2ωδ ss ′,
u † s (⃗ k)ṽ s ′( ⃗ k) = 0.
Beginning with the superposition
∫ (
ψ(x) = a s ( ⃗ k)u s ( ⃗ k)e i⃗k·⃗x + ˜b s ( ⃗ k)ṽ s ( ⃗ )
k)e i⃗ k·⃗x
˜dk ∑ s
where the quantities with tildes will be modified in anticipation of their role
as antiparticle amplitudes. Since an antiparticle associated with a negative
13

energy state of momentum ⃗ k has momentum − ⃗ k, we change to − ⃗ k in the
integral over ˜b. Then we redefine ṽ s (− ⃗ k) to be v s ( ⃗ k) and ˜b s (− ⃗ k) to b † s (⃗ k). The
latter takes account of the fact that after quantization a and a † annihilate
and create particles respectively. Annihilation of a particle in a negative
energy state creates an antiparticle there. so the roles of the operator and
its conjugate are interchanged. The expansion for ψ now looks like
∫ (
ψ(x) = a s ( ⃗ k)u s ( ⃗ k)e i⃗k·⃗x + b † s( ⃗ k)v s ( ⃗ )
k)e −i⃗ k·⃗x
(18)
˜dk ∑ s
It is easy to substitute this expansion into the Hamiltonian to obtain
∫ (
H = ω a † s (⃗ k)a s ( ⃗ k) − b s ( ⃗ )
k)b † s (⃗ k)
˜dk ∑ s
(19)
using the fact that u s and v s are eigenstates of H D that satisfy the above
orthonormality relations. It should be possible to prove that the Lagrangian
has a form something like
∫ ( i
)
L = ˜dk (a † ȧ + bḃ† − ȧ † a − ḃb† ) − 1
2
2ω(a † a + bb † ) .
Perhaps you can prove it with the aid of the following identities involving the
solutions u s ( ⃗ k) and v s ( ⃗ k), which we also list for future reference. Involving
bars:
Involving ± ⃗ k:
u s ( ⃗ k)u s ′( ⃗ k) = 2mδ ss ′
v s ( ⃗ k)v s ′( ⃗ k) = −2mδ ss ′
u s ( ⃗ k)v s ′( ⃗ k) = 0
Involving γ i k i :
u † s (⃗ k)v s ′(− ⃗ k) = 0
v † s (⃗ k)u s ′(− ⃗ k) = 0 (20)
u s ( ⃗ k)γ i k i u s ′( ⃗ k) = 2 ⃗ k 2 δ ss ′
v s ( ⃗ k)γ i k i v s ′( ⃗ k) = 2 ⃗ k 2 δ ss ′
u s ( ⃗ k)γ i k i v s ′( ⃗ k) = 2ω√
⃗k 2
δ ss ′
(21)
14

This Lagrangian leads to solutions a( ⃗ k) = a 1 ( ⃗ k)e −iωt and b( ⃗ k) = b 1 ( ⃗ k)e −iωt
where a 1 ( ⃗ k) and b 1 ( ⃗ k) are constants.
The free Dirac field can also be written as a superposition of the time
dependent solutions:
∫
ψ(x) =
˜dk
(a s ( ⃗ k)u s ( ⃗ k)e ikx + b † s (⃗ k)v s ( ⃗ )
k)e −ikx
with 4-dimensional k and x in analogy to the similar forms for the scalar and
electromagnetic fields. The Hamiltonian has the same expression in terms of
the a’s and b’s as in the previous time independent expansion, With the aid
of the orthogonality relations between u’s and v’s, The expansion of ψ can
be inverted to express a s ( ⃗ k) and b † s (⃗ k) in terms of ψ(x):
∫
a s ( ⃗ k) = d 3 xu † s( ⃗ k)e −ikx ψ(x)
∫
b † s (⃗ k) = d 3 xv s † (⃗ k)e ikx ψ(x)
15