February 2016
PX408: Relativistic Quantum Mechanics
Tim Gershon
(T.J.Gershon@warwick.ac.uk)
Handout 3: Solving the Dirac Equation
We would like to find, and investigate, the values of ψ which solve the Dirac equation:
(iγ µ ∂µ − m)ψ = 0 .
(1)
We start by looking for (positive energy) plane wave solutions of the form ψ ∼ ωe−ip.x . It is
convenient to use a form of the Dirac equation where space- and time-like parts have been separated
(i ∂ψ
∂t = (−iα.∇ + βm)ψ – refer to earlier handouts for definitions of α, β and σ). We then find the
momentum-space Dirac equation
Eω = (α.p + βm)ω ,
(2)
ω+
into which we can insert ω =
to give, in block form,
ω−
E
ω+
ω−
=
m σ.p
σ.p −m
ω+
ω−
,
(3)
which we can rewrite as coupled equations
(E − m)ω+ = σ.pω−
(4)
(E + m)ω− = σ.pω+ .
(5)
Thus we can choose to substitute either ω+ or ω− in our working. Choosing the latter, we find
ω+
ω=
.
(6)
σ.p
E+m ω+
Note also that both choices give (E + m)(E − m)ω± = (σ.p)2 = p2 , i.e. we recover (again) the
relativistic mass-energy relation E 2 = m2 +p2 . This implies the existence of negative energy solutions,
which we will discuss in more detail below.
From the above we can see that there are in fact two linearly independent solutions (corresponding
to the fact that ω+ is a two-component spinor). For a particle at rest we can choose orthogonal (but
unnormalised) solutions as
1
0
ω+ =
or
(7)
0
1
giving


1
 0  −imt

ψ=
 0 e
0

or

0
 1  −imt
 e
.
 0 
0
(8)
The Helicity Operator
There must be some operator that distinguishes the two linearly independent solutions to the
Dirac equation given above. Since they have the same energy, this operator must commute with the
Hamiltonian. (Recall from Heisenberg’s equation of motion
h
i
˙
 = −i Â, Ĥ ,
(9)
that any operator that commutes with the Hamiltonian represents a constant of motion.) The
Hamiltonian is given by H = (α.p + βm), and choosing the particle’s rest frame for simplicity, this
reduces to βm. We are therefore looking for an operator that commutes with β, while distinguishing
 
 
1
0
 0 
 1 
 
 .
from
(10)
 0 
 0 
0
0
It is fairly obvious that a possible choice is
Σz =
where we recall that
σz =
σz 0
0 σz
1 0
0 −1
(11)
.
(12)
It is worth noting that, based on the arguments above, we could have chosen the bottom right
element of the block form matrix in Eq. (11) to take any value. Hopefully the choice we made appears
sensible, but we will return to this point later.
Q1 Show that Σz commutes with β.
Q2 Show that Σz distinguishes the two solutions for ψ given in Eq. (8).
Generalising, we might expect our operator in an arbitrary frame to be
σ 0
Σ=
.
0 σ
(13)
This operator does not commute with (α.p + βm), and therefore is not the correct answer, but
let us briefly examine its properties anyway. Using knowledge of the Pauli spin matrices, we see
immediately
2
1
3
1
1
1
Σ = ,
Σj , Σk = ijkl Σl (j 6= k) .
(14)
2
4
2
2
2
Q3 Show explicitly that
2
1
2Σ
= 43 .
There are in fact many possible solutions for the operator that we are searching for, of which
several can be found in the literature, but the most convenient is the so-called “helicity operator”
given by
!
σ.p
0
|p|
h(p) =
.
(15)
σ.p
0
|p|
Q4 Show that Σ defined in Eq. (13) does not commute with (α.p + βm).
Q5 Show that h(p) does commute with the Hamiltonian.
Q6 Find the value of (h(p))2 , and state the eigenvalues of h(p).
Q7 Consider the solutions to the Weyl equations discussed in Handout 2. Show that these are
eigenstates of the helicity operators and give their eigenvalues.
Note that σ.p
|p| gives the projection of σ onto the direction of momentum p. Thus, h(p) can be
understood, physically, as the projection of Σ onto the direction of motion. Helicity is not a Lorentz
invariant quantity, except in the special case of massless particles, but is invariant under spatial
rotations. (The spin matrix Σ is Lorentz invariant, but is not a constant of the motion.)
The Helicity Operator – Second Derivation
It is fairly evident already that the operator we discovered above, called the helicity operator,
is strongly related to the quantum property of spin that appears inevitably in the solution of the
Dirac equation. To further illustrate this, an alternative derivation in which the connection is more
apparent is useful. This approach, following Feynman (QED, p60) also makes use of some of the
gamma matrix algebra we developed earlier.
Let us start by restating the problem: we have two solutions to
p/ω = mω ,
(16)
and therefore we are looking for an operator that commutes with p/.
Firstly, recall that γ 5 anticommutes with the slashed version of any four vector. Therefore
5 γ , p/ = 0 .
(17)
Let us now define an operator O such that Oµ pµ = 0. Then
{O/, p/} = 2Oµ pµ = 0 .
(18)
Now, define the eigenvalues of −γ 5 O/ as s, i.e.
− γ 5 O/ω = sω
(19)
so that
−γ 5 O/ −γ 5 O/ ω = −γ 5 γ 5 O/O/ω
µ
(20)
2
= −Oµ O ω = s ω
(21)
2
(since γ 5 , O/ = 0, γ 5 = 1 and O/O/ = Oµ Oµ ). So with the choice Oµ Oµ = −1, we find eigenvalues
s = ±1. (n.b. Although the choice −γ 5 O/ for the operator looks a bit unnatural, it is necessary to
get the sign right in the end.)
Note that the Lorentz invariant requirement Oµ pµ = 0, obliges us to put O0 = 0 in the particle’s
rest frame. Therefore, in the rest frame, Oµ Oµ = −1 reduces to |O| = 1.
Now, choosing the orientation to be along the z-axis, so that O/ = γ 3 , we find
−γ 5 O/ω = −γ 5 γ 3 ω
0 1 2 3 3
= −iγ γ γ γ γ ω
(22)
(23)
0 1 2
(24)
1 2 0
(25)
1 2
(26)
= iγ γ γ ω
= iγ γ γ ω
= iγ γ ω
where we have used the fact that γ 3 γ 3 = −1, used two anticommutation relations ((−1)2 = 1) to
bring γ 0 through to the right, and, in the last step, used γ 0 ω = ω in the particle’s rest frame (recall
that the two positive energy solutions in the rest frame are given by ω = (1 0 0 0)T , (0 1 0 0)T ).
Now,
0
σx
0
σy
−σx σy
0
σz 0
1 2
γ γ =
=
= −i
.
(27)
−σx 0
−σy 0
0
−σx σy
0 σz
and thus
− γ 5 O/ω = i(−i)Σz ω = sω ,
i.e.
(28)
−γ 5 O/
is the Σz operator.
The above derivation is still not very general, since we have chosen the particle to be in its rest
frame and moreover specified an axis with which to align ourselves. Nonetheless, we can at this stage
µ
state that the Dirac spinor ψ ∼ ωe−ipµ x , where ω satisfies p/ω = mω and −γ 5 O/ω = sω, represents a
particle moving with momentum pµ and with spin aligned either parallel with (s = +1) or antiparallel
to (s = −1) the O axis in the co-moving frame.
The generalisation to an arbitrary frame proceeds as before, to obtain the helicity operator of
Eq. (15). Since (σ.p)2 = p2 , it is clear that (h(p))2 = 1, and so the eigenvalues of h(p) must be ±1
(in fact +1 twice and −1 twice). To find the eigenstates, consider
!
σ.p
0
ω+
ω+
|p|
h(p)ω =
= ±ω = ±
.
(29)
σ.p
σ.p
σ.p
0
E+m ω+
E+m ω+
|p|
For the positive solution we find σ.p
|p| ω+ = ω+ , while for the negative solutions we find
which as before can be solved for p aligned with the z-axis as
0
1
.
or
ω+ =
1
0
σ.p
|p| ω+
= −ω+ ,
(30)
Q8 Find the eigenvectors of h(p) in the cases that (i) p is aligned along the x-axis; (ii) p is aligned
along the y-axis; (iii) p has a general direction.
Helicity and Chirality
As discussed above, the helicity operator h(p) distinguishes the solutions of the Dirac equation
(i.e. h(p) commutes with the Hamiltonian).
The term “chirality” is often confused with helicity, and therefore it is worthwhile to briefly
discuss its meaning at this stage. Hopefully this will reduce the possibility for potential confusion.
The chirality operator is actually just γ 5 . Recall that in block form, and in the standard representation, we write
0 1
γ 5 = iγ 0 γ 1 γ 2 γ 3 =
.
(31)
1 0
It is common to encounter the following operators that project out right- and left-handed chiral
states, respectively:
1
1
PR =
1 + γ5
PL =
1 − γ5
(32)
2
2
and we use these to define right- and left-handed spinors
ψR = PR ψ
ψ L = PL ψ
(33)
The importance of the chirality operators will be apparent to those who have studied particle physics
since the weak interaction operates only on the left-handed spinors. (Some of the profound consequences of this fact will be discussed in the Gauge Theories module.) Note that the chirality operator
is Lorentz invariant, as are the PR and PL projection operators.
Q9 Show that ψR and ψL are eigenstates of the chirality operator γ 5 , and give their eigenvalues.
Q10 Show that ψ = ψR + ψL .
Q11 Find expressions for PR ψR , PR ψL , PL ψR and PL ψL .
Q12 Using the standard convention, give matrix forms for PR and PL .
Q13 Using matrix representation, show that (PR )2 = PR , (PL )2 = PL and PR PL = PL PR = 0.
Now, the expressions for the helicity operator of Eq. (15) and the chirality operator of Eq. (31)
appear completely different, so it is not obvious how any confusion could possibly arise. However,
let us recall that a spinor that is a solution of the Dirac equation can be written
(E − m) ω+ = σ.p ω−
ω+
−ipµ xµ
(34)
ψ ∝ ωe
where
ω=
and
(E + m) ω− = σ.p ω+
ω−
Thus we can write
0 1
5
γ ω=
1 0
σ.p
E−m ω−
σ.p
E+m ω+
=
σ.p
E+m ω+
σ.p
E−m ω−
=
σ.p
E+m
0
0
σ.p
E−m
ω+
ω−
.
(35)
Consequently, when acting on solutions of the Dirac equation, the chirality operator can in fact be
represented as
σ.p
0
0 1
E+m
ω=
γ5ω =
ω.
(36)
σ.p
1 0
0
E−m
It should now be apparent how helicity and chirality operators can be confused. In particular, in the
highly relativistic limit (E m such that E − m ∼ E + m ∼ |p|), they tend to become equivalent.
In particular, for massless particles helicity and chirality are equivalent. Correspondingly, it is for
the nearly massless neutrinos where the largest confusion between the two operators tends to arise.
Nonetheless, in a general frame helicity and chirality are distinct quantities represented by different
operators. A nice summary of their similarities and differences can be found at
http://www.quantumfieldtheory.info/Chiralityvshelicitychart.htm.
Q14 Find the commutator of the chirality operator with the Hamiltonian.
The Dirac Current
As before, we want to find something that satisfies the continuity equation
∂ρDirac
= −∇.j Dirac
∂t
or
µ
∂µ jDirac
= 0.
(37)
It is tempting to try ρ = ψ † ψ, where ψ † is the Hermitian conjugate (or adjoint) of ψ. That is


ψa
 ψb 

if
ψ=
then
ψ † = (ψa∗ ψb∗ ψc∗ ψd∗ ) .
(38)
 ψc 
ψd
Then ρ = |ψa |2 +|ψb |2 +|ψc |2 +|ψd |2 is automatically positive definite, and looks like a good candidate
for the probability density. (n.b. We will discuss the normalisation of the Dirac spinors later, which
will cover possible concerns that the density should transform as the time-like component of a fourvector.)
To find j Dirac , we use a similar approach to that taken for the Klein-Gordon case, and start by
taking the conjugate of the Dirac equation. It is easiest to see how this works by explicitly writing
out the Lorentz invariant term which gives
(iγ µ ∂µ − m)ψ = 0
0
1
2
3
(39)
iγ ∂0 ψ + iγ ∂1 ψ + iγ ∂2 ψ + iγ ∂3 ψ − mψ = 0
(40)
† 0
† 1
† 2
† 3
−i∂0 ψ γ + i∂1 ψ γ + i∂2 ψ γ + i∂3 ψ γ − mψ
†
= 0
(41)
3
i∂0 ψ̄γ + i∂1 ψ̄γ + i∂2 ψ̄γ + i∂3 ψ̄γ + mψ̄ = 0
(42)
µ
(43)
0
1
2
i∂µ ψ̄γ + mψ̄ = 0
where between the second and third lines we have taken the Hermitian conjugate, and exploited
the fact that (γ 0 )† = γ 0 while (γ i )† = −γ i (i = 1, 2, 3), and between the third and fourth lines we
have right-multiplied by −γ 0 , exploited the anticommutation relations of the gamma matrices, and
introduced the conjugate spinor ψ̄ = ψ † γ 0 .
Q15 Work carefully through the steps above to derive Eq. (43).
Now, we left-multiply the Dirac equation of Eq. (39 by ψ̄, right-multiply the conjugate Dirac
equation of Eq. (43) by ψ, and add
ψ̄(iγ µ ∂µ − m)ψ = 0
µ
(i∂µ ψ̄γ + mψ̄)ψ = 0
µ
µ
iψ̄γ (∂µ ψ) + i(∂µ ψ̄)γ ψ = 0
µ
i∂µ (ψ̄γ ψ) = 0
(44)
(45)
(46)
(47)
µ
= ψ̄γ µ ψ, with components ρDirac = j 0 = ψ̄γ 0 ψ = ψ † ψ
allowing us to identify the Dirac current jDirac
i
i
and j Dirac = ψ̄γ ψ.
Normalisation of Dirac Spinors
µ
Since ρDirac = ψ † ψ is the time-like component of jDirac
, and must behave accordingly under
Lorentz transformations, an appropriate choice of normalisation is ψ † ψ = 2E (or often 2E particles
per unit volume, making the connection with density explicit). It should be noted that this is merely
a convention (albeit a very useful one), and other choices appear in the literature.
Let us write our solutions for the Dirac spinors as
µ
ω+
ω+
−ipµ xµ
−ipµ xµ
ψ = N ωe
=N
e
=N
e−ipµ x ,
(48)
σ.p
ω−
E+m ω+
where N is the normalisation factor to be found. Without loss of generality, we can choose ω+ to be
†
normalised such that ω+
ω+ = 1 (any other choice will just scale the resulting value of N ). We then
find
2 !
(E + m)2 + p2
2E
σ.p
†
= N2
= N2
= 2E ,
(49)
ψ † ψ = N 2 ω+
ω+ 1 +
2
E+m
(E + m)
E+m
where we have used (σ.p)2 = p2 and the √
relativistic mass energy relation E 2 = p2 + m2 . Clearly, to
normalise to ψ † ψ = 2E, we require N = E + m.
Finally, the (positive energy) solutions of the Dirac equation are
ψ = u(p, s)e−ipµ x
µ
(50)
where
1/2
φs
u(p, s) = (E + m)
and
σ.p
s
(E+m) φ
↑
φ =
1
0
↓
φ =
0
1
.
Until now we have only considered solutions of the Dirac equation of the form
µ
ω+
ω+
−ipµ xµ
−ipµ xµ
ψ = N ωe
=N
e
=N
e−ipµ x .
σ.p
ω−
E+m ω+
(51)
(52)
However, as can be seen from Eq. (4), we could equally well write
σ.p
µ
ω−
ω+
−ipµ xµ
−ipµ xµ
E−m
ψ = N ωe
=N
e
=N
e−ipµ x .
ω−
ω−
(53)
The reason that we have avoided this solution is that its behaviour in the case that |p| → 0 (i.e. E →
m) is not immediately apparent.
Q16 By giving the non-relativistic expression for E, find the limit of
σ.p
E−m
as |p| → 0.
To avoid such problems, we can “fix” the energy by writing Efix = −E. We then obtain
σ.pfix
σ.p
− Efix
ω−
ω
−
+m
E
+m
fix
ω=
=
,
ω−
ω−
(54)
where pfix = −p. We can write the solutions as
−ipµ xµ
ψ = N ωe
=N
σ.pfix
Efix +m ω−
ω−
µ
e+ipfix xµ ,
(55)
√
where the normalisation factor N can be found to be Efix + m. We can interpret these as negative
energy solutions. Alternatively, yet equivalently, we can drop the subscript “fix”, and consider these
solutions as corresponding to antiparticles with positive energy. We will discuss the interpretation
of these solutions further below (and more in the next handout).
The complete expression for the negative energy solutions is then
ψ = v(p, s)e+ipµ x
where
v(p, s) = (E + m)
1/2
σ.p
s
(E+m) χ
χs
µ
(56)
and
↑
χ =
0
1
↓
χ =
1
0
.
(57)
The two-component vectors φs and χs describe the spin-states of particles and antiparticles
respectively. As noted after Eq. (11), we have some freedom to choose how to define the relative
helicity of particles and antiparticles. (Specifically γ 0
h(p) would be a valid
alternative choice
for
the
1
0
0
helicity operator.) It is conventional to choose φ↑ =
and φ↓ =
, with χ↑ =
and
0
1
1
1
χ↓ =
. With this choice the absence of a “spin-up” negative-energy electron is equivalent to
0
the presence of a “spin-down” positive-energy positron. The utility of this equivalence will become
clear when we discuss the negative energy solutions in more detail below.
Q17 What are the values of ū(p, s)u(p, s) and v̄(p, s)v(p, s)?
Q18 What are the values of ū(p, s)γ 5 u(p, s) and v̄(p, s)γ 5 v(p, s)?
Lorentz Transformations of Dirac Spinors
We discussed the Lorentz transformation properties of gamma matrices earlier, showing that they
could be treated as invariant since the transformation is an equivalence transformation. Effectively,
we can choose to implement the Lorentz transformation in one of two ways:
• transform γ µ like a 4-vector, leaving ψ unchanged
,→ γ µ0 differs by an equivalence transformation
• leave γ µ unchanged
,→ ψ transforms according to an equivalence transformation
The question we would like to answer here is, exactly which equivalence transformation is required?
That is, what is the form of S such that S −1 γ µ S = Λµν γ ν , where Λµν is a Lorentz transformation?
To solve this, we start by considering a simpler transformation: that of a rotation in spatial
dimensions. For a rotation by θ about the x axis, we can write
 
 0  
 
x
x
1
0
0
x
cos θ sin θ   y  .
x= y 
7→
x0 =  y 0  =  0
(58)
0
z
z
0 − sin θ cos θ
z
Note also that p transforms in the same way.
Let us consider how a two component spinor φ transforms under this rotation. This question is
not trivial since the internal space described by the two components of the spinor are not related to
our spatial coordinates in any straightforward way. However, we can still progress by noting that
0
0 0
the helicity will be unchanged under the transformation,
i.e.
if h(p)φ+ = +φ+ then h(p )φ+ = +φ+ .
1
Then, if we start with a positive helicity spinor φ+ =
, we require
0
σ.p0 0
φ = φ0+ ,
|p0 | +
(59)
i.e. that φ0+ is an eigenvector of the helicity operator in the transformed space, with eigenvector +1.
Without loss of generality we can take p to be normalised to unity, and since rotation preserves
normalisation, p0 is also normalised to unity. Then
σ.p0 = σx px + σy (py cos θ + pz sin θ) + σz (−py sin θ + pz cos θ) .
(60)
Choosing p to be aligned with the z-axis, we then find
σ.p0 = σy sin θ + σz cos θ .
(61)
[Note that the choice of aligning p with the z-axis here is arbitrary, but simplifies the maths. Since
we are looking at a rotation about the x-axis it would be silly to align in that direction – nothing
would happen! One can choose to align along the y-axis, and the same solutions are obtained. If you
are feeling brave you can try to prove this, but it is not recommended.]
So, using explicit forms for σy and σz , Eq. (59) tells us
a
a
a
cos θ −i sin θ
=
with
φ0+ =
.
(62)
i sin θ − cos θ
b
b
b
Q19 Find the solution to Eq. (62).
hint: recall that cos 2θ = cos2 θ − sin2 θ = 1 − 2 sin2 θ = 2 cos2 θ − 1.
The solution to Eq. (62) is
φ0+
=
a
b
=
cos(θ/2)
i sin(θ/2)
,
(63)
where we have chosen normalisation such that (φ0+ )† φ0+ = φ†+ φ+ = 1.
Q20 Find a similar description for the transformation of the negative helicity state φ− .
Q21 Give the values of (φ0+ )† φ0− and (φ0− )† φ0+ .
The outcome of all this work is that we can show that
cos(θ/2) i sin(θ/2)
0
φ
7→
φ =
φ = eiσx θ/2 φ .
i sin(θ/2) cos(θ/2)
(64)
The last equality of Eq. (64) may require explanation. Recall that we can use a Taylor expansion
to define
1
eA = 1 + A + A2 + . . .
(65)
2
While we may be most familiar with the case where A is a scalar, we can use this definition for any
mathematical form of A for which A2 , A3 , etc. are defined (and have the same dimensionality as A).
This is the case where A is any square matrix. All that is required is to interpret “1” as the unit
matrix of the appropriate dimension. Thus
1
eiσx θ/2 = 1 + iσx θ/2 + (iσx θ/2)2 + . . .
2
1
= 1 + iσx θ/2 − (θ/2)2 + . . .
2
= cos(θ/2) + σx sin(θ/2) ,
(66)
(67)
(68)
where we have used the fact that σx2 = 1.
Q22 Find similar expressions to Eq. (64) for transformations corresponding to rotations about the
y-axis and the z-axis.
The generalisation to a rotation about an arbitrary axis n̂ is fairly obvious:
φ
7→
φ0 = eiσ.n̂ θ/2 φ .
(69)
Q23 Show that the transformation of Eq. (69) can be written φ 7→ φ0 = U φ, where U is a unitary
matrix.
There is a subtle but important consequence of this result. A rotation in three-dimensional space
can be represented by a 3 × 3 matrix (in the same 3D space), and can equivalently be represented by
a 2 × 2 matrix in bispinor space. This can be expressed mathematically in terms of isomorphisms
of groups (specifically SO(3) 7→ SU(2) – though knowledge of the meaning of these groups is not
necessary for this module).
We now wish to extend this discussion to Dirac spinors. Recall that we can write
√
√
√
µ
ω+
ω+
−ipµ xµ
−ipµ xµ
ψ = E+m ω e
= E+m
e−ipµ x .
(70)
e
= E+m
σ.p
ω
ω−
E+m +
Thus it is clear that the top and bottom two-component spinors transform in the same way under
rotations. Hence
ψ
7→
ψ 0 = eiΣ.n̂ θ/2 ψ ,
(71)
where
Σ=
σ 0
0 σ
(72)
Q24 Show that the Dirac probability density is invariant under rotations.
Q25 Show that the Dirac (3-)current transforms as a vector under rotations.
Now, at last, let us consider Lorentz transformations, specifically a Lorentz boost along the x-axis
with velocity β:
 
 0  
 
t
t
cosh θ sinh θ 0 0
t
0








x 
µ0
 x  =  sinh θ cosh θ 0 0   x  = Λµν xν ,
=
xµ = 
→
7
x
(73)
0
 y 
 y   0
0
1 0  y 
z
0
0
0 1
z0
z
where the rapidity θ is given by tanh θ = β (n.b. note that β here is the Lorentz boost factor, and
should not be confused with the gamma matrix – we will use the rapidity θ in what follows in the
hope of helping to avoid confusion).
As before we can proceed by looking at the transformation of specific spinors. Let us start with
a positive energy, positive helicity spinor in its rest frame

 

1
1
√
√
 0 

0 
,

(74)
7→
ω0 = E 0 + m 
ω = 2m 

 0 
 p0
0
x
E 0 +m
1
0
where we have used the fact that σx
1
0
=
0
1
.
Q26 Give the appropriate transformation for a positive energy, negative helicity spinor in its rest
frame.
Q27 Give the appropriate transformations for negative energy spinors (i.e. antiparticle spinors),
both positive and negative helicity.
It can be shown that the required transformation can be written as

cosh(θ/2)
0
0
sinh(θ/2)

0
cosh(θ/2)
sinh(θ/2)
0
ω
7→
ω0 = 

0
sinh(θ/2) cosh(θ/2)
0
sinh(θ/2)
0
0
cosh(θ/2)


ω,

(75)
i.e.
ω 0 = (cosh(θ/2) + αx sinh(θ/2)) ω = eαx θ/2 ω .
Q28 Show that cosh(θ/2) =
q
E 0 +m
2m
2
and sinh(θ/2) = √
hint: cosh 2θ = cosh2 θ + sinh θ.
p0x
.
2m(E 0 +m)
(76)
Q29 Show that the transformation of Eq. (75) correctly gives the result of Eq. (74).
Q30 Show that the transformation of Eq. (75) correctly gives the results obtained in Q26 and Q27.
Q31 Verify by expanding in power series that eαx θ/2 = cosh(θ/2) + αx sinh(θ/2).
Q32 Find similar expressions for eαy θ/2 and eαz θ/2
The generalisation is
ω
7→
ω 0 = Sω = eα.v̂ θ/2 ω ,
(77)
where v̂ is a unit vector in the direction of the boost (and the magnitude of the boost |v| = tanh θ).
Finally, we have found S that solves S −1 γ µ S = Λµν γ ν .
Q33 Show that (i) e−αx θ/2 γ 0 eαx θ/2 = γ 0 cosh θ + γ 1 sinh θ; (ii) e−αx θ/2 γ 1 eαx θ/2 = γ 0 sinh θ +
γ 1 cosh θ; (iii) e−αx θ/2 γ 2 eαx θ/2 = γ 2 ; (iv) e−αx θ/2 γ 3 eαx θ/2 = γ 3 .
Q34 Show that S is not a unitary matrix.
0 1
Q35 Show
q that, for a boost along the x-axis, S can be written S = a+ + a− γ γ , where a± =
± 12 (γ ± 1). (Note that γ in the parentheses corresponds to the Lorentz factor, not a gamma
matrix). Give a similar expression for S for an arbitrary Lorentz transformation.
µ
transforms as a four-vector under Lorentz transformations.
Q36 Show that the Dirac current jDirac
Finally, note that since S is a function of the particular Lorentz transformation being used (the
boost and the direction), we can write S = S(Λ). This is sometimes called the “spinor representation
of the Lorentz group” and can be expressed mathematically as an isomorphism between groups. [n.b.
The Lorentz group is the group of Lorentz transformations, including (as a subgroup) rotations of
the three spatial dimensions. It, in turn, is a subgroup of the Poincaré group.] Indeed, one can derive
much of relativistic quantum mechanics starting from this isomorphism: it is an equally valid, though
arguably more abstract, approach to that taken in this module. One interest aspect of that approach
is that it underlines the deep connection between the geometry of space-time and the solutions of the
Dirac equation. This seems to indicate that a theory of quantum gravity, i.e. a unification of general
relativity and quantum theory, may be developed; however, this has proved to be beyond the grasp
of Einstein, Dirac and everyone else who has attempted it since.
Discrete Symmetries – Parity and Charge Conjugation
Discrete symmetries play an important role in particle physics. Two are particularly important
and their relations to relativistic wave equations will be discussed below. These are parity, which
is the simultaneous inversion of all three spatial co-ordinates (x 7→ −x), and charge conjugation,
which transforms a particle into the corresponding antiparticle. [There is a third important discrete
symmetry, time reversal, which we will not discuss in detail here. It is perhaps worth mentioning
that, if we had (much) more time, supersymmetry could also be discussed in this context. But we
don’t.]
Parity
Let us start with the Dirac equation, which we write as
i
∂
ψ = −iα.∇ψ + βmψ .
∂t
(78)
The parity operation transforms
x 7→ x0 = −x
(79)
0
t 7→ t = t
(80)
ψ 7→ ψP .
(81)
We wish to find the form of ψP such that the transformed system satisfies the same equation, i.e.
i
∂
ψP (x0 , t0 ) = −iα.∇0 ψP (x0 , t0 ) + βmψP (x0 , t0 ) .
∂t0
(82)
It is obvious that ∇0 = −∇, and therefore if we left-multiply by β and take advantage of the
anticommutation relations between β and α, we find
i
∂
βψP (x0 , t) = −iα.∇ βψP (x0 , t) + βm βψP (x0 , t) ,
∂t
(83)
from which it is straightforward to identify that
βψP (x0 , t) = ψ(x, t) ,
(84)
ψP (x, t) = βψ(−x, t) .
(85)
or
[In fact, most generally we can allow the parity transformation to introduce an arbitrary phase.
However, we will neglect this minor complication.]
Let us now consider the effect of the parity operation on a general (positive energy) solution of
the Dirac equation:
φ
1/2
ψ(x, t) = (E + m)
e−iEt+ip.x ,
(86)
σ.p
(E+m) φ
so that
ψP (x, t) = βψ(−x, t) = (E + m)
1/2
φ
σ.p
− (E+m)
φ
e−iEt−ip.x ,
(87)
which is equivalent to p 7→ −p. Note that the top two components of the spinor (i.e. the “particle”
part) transform oppositely to the bottom two components (i.e. the “antiparticle” part). This is the
origin of the statement that particle and antiparticle have opposite parity.
Q37 Consider the effect of the parity operation on a general negative energy solution of the Dirac
equation.
Q38 Consider the effect of the parity operation on helicity eigenstates.
Q39 Consider the effect of the parity operation on chirality eigenstates.
Q40 Working through equivalent steps to those above, find an expression for the time-reversal
operation acting on solutions of the Dirac equation.
We could also consider the effect of parity on solutions of the Klein-Gordon equation. Here the
solution is straightforward since
(∂µ ∂ µ + m2 )φ = 0 ,
(88)
is invariant under parity, we simply have
φP (x, t) = φ(−x, t)
(up to an arbitrary phase).
(89)
Q41 Find the effect of the parity operation on (i) the time-like part and (ii) the space-like part of
the Klein-Gordon current i (φ∗ ∂ µ φ − φ∂ µ φ∗ ).
Q42 Find the effect of the parity operation on (i) the time-like part and (ii) the space-like part of
the Dirac current ψ̄γ µ ψ.
Q43 Find the effect of the parity operation on the bilinears (i) ψ̄γ 5 ψ, (ii) ψ̄γ µ γ 5 ψ, (iii) ψ̄σ µν ψ.
Charge conjugation
This time, let us start with the Klein-Gordon equation for a particle with electric charge q and
mass m in an electromagnetic field:
(∂µ + iqAµ )(∂ µ + iqAµ )φ + m2 φ = 0 .
(90)
We wish to find the appropriate equation for the corresponding antiparticle, which is a particle with
electric charge −q and mass m. [Aside: we know that antiparticles are a necessary consequence of
the Dirac equation, but there is no such requirement in RQM for the Klein-Gordon equation. Put
another way, from what we have covered so far, there is no requirement that the anti-Klein-Gordonparticle exists. But let’s assume that it does (and indeed, in quantum field theory it does turn out
to be necessary).] We can find the appropriate equation by simply taking the charge conjugate:
(∂µ − iqAµ )(∂ µ − iqAµ )φ∗ + m2 φ∗ = 0 .
(91)
Thus φC = φ∗ gives the relevant transformation.
Q44 How does the Klein-Gordon current transform under charge conjugation?
The situation for the Dirac equation is slightly (only slightly) more complicated. If we write the
Dirac equation as
γ µ (∂µ + ieAµ )ψ + imψ = 0 ,
(92)
then taking the complex conjugate transforms not only e 7→ −e (which we want) but also m 7→ −m
(which we don’t) as well as γ µ 7→ γ µ∗ .
γ µ∗ (∂µ − ieAµ )ψ ∗ − imψ ∗ = 0 .
(93)
Let us assume that the solution is given by ψC = Cψ ∗ , where C is some operator to be found. Then
ψ ∗ = C −1 ψC , so
γ µ∗ (∂µ − ieAµ )C −1 ψC − imC −1 ψC = 0 ,
(94)
and left-multiplying by C gives
Cγ µ∗ C −1 (∂µ − ieAµ )ψC − imCC −1 ψC = 0 .
(95)
Since CC −1 = 1, we can multiply through by −1 and find that we require
Cγ µ∗ C −1 = −γ µ .
(96)
The question is then: what is the form of C that satisfies this equation?
The solution is easier to spot if we rewrite the problem as Cγ µ∗ = −γ µ C, and consider first of
all the cases where γ µ is real (in the standard representation, this is all µ 6= 2). Thus we require
{C, γ µ } = 0 (µ 6= 2), which can be solved by C = γ 2 . Noting that γ 2∗ = −γ 2 , we find that this choice
also satisfies the requirement for µ = 2. In fact, any
solution, and it is conventional to choose C to be real,

0 0

0 0
C = iγ 2 = 
 0 −1
1 0
linear multiple of γ 2 provides a satisfactory
i.e.

0 1
−1 0 
.
(97)
0 0 
0 0
∗.
Thus ψC = iγ 2 ψ ∗ and ψ = iγ 2 ψC
Q45 How does the Dirac current transform under charge conjugation?
Q46 Find the effect of the charge conjugation operation on the bilinears (i) ψ̄γ 5 ψ, (ii) ψ̄γ µ γ 5 ψ, (iii)
ψ̄σ µν ψ.
Q47 Find the effect of the charge conjugation operation on a general solution of the Dirac equation.
Consider both positive and negative energy with both positive and negative helicity.
Majorana fermions
This is an appropriate stage to go back and scrutinise one of the assumptions that we made earlier.
We observed that the Dirac spinor has four components, and interpreted this as corresponding to
four degrees of freedom – (particle, antiparticle) × (positive helicity, negative helicity). If we consider
electrons as our prototype fermion, this is indeed a natural thing to do, since the electron and positron
have opposite charges and are clearly distinct, and similarly we know very well that a spin-1/2 particle
has two possible spin states.
However, what about if we have a fermion that is neutral – i.e. carries no quantum numbers that
distinguish particle from antiparticle (including, but not limited to, no electric charge)? Indeed, a
candidate for such a particle exists in nature – the neutrino. If this particle possesses a conserved
quantum number that distinguishes neutrino from antineutrino (i.e. a lepton number) then our
previous discussion is appropriate. But could it is also be possible that there is no such conserved
quantum number? If so, there would only be two degrees of freedom. So the question to ask is, can
we find a relativistic wave equation for massive fermions with only two degrees of freedom?
We earlier discussed such equations for massless fermions, i.e. Weyl spinors, which satisfy:
(D0 + σ.D) φ = 0 ,
(98)
(D0 − σ.D) χ = 0 .
(99)
The Weyl spinors φ and χ have different helicity eigenvalues – i.e. different “handedness” (“R” and
“L” respectively). (More technically, the spinors have different transformation properties under the
Lorentz group.) In terms of these, the Dirac equation can be written as two coupled equations
(switching to the free-field case since we will be considering uncharged particles):
(i∂0 + iσ.∇) φ = mχ ,
(100)
(i∂0 − iσ.∇) χ = mφ .
(101)
We can see that the effect of a non-zero mass term is to couple the left-and right-handed terms
together. Specifically
the operator (i∂0 + iσ.∇) transforms R into L ,
the operator (i∂0 − iσ.∇) transforms L into R ,
where the rate of transformation is governed by the mass, m (and is zero in the limit m → 0).
Now, we are trying to find a solution for a massive fermion with only two degrees of freedom, so
one way to ask this question is, can we write χ in terms of the elements of φ? This is quite hard to
solve in general (see questions below), but we can take an educated guess – since we expect that a
two-degree-of-freedom massive fermion has no distinction between particle and antiparticle, we might
try χ = φC = Cφ∗ , where C is now the 2 × 2 equivalent of the 4 × 4 version of Eq. (97), i.e. C = iσ2 .
Q48 Insert χ = iσ2 φ∗ into Eq. (100). Using complex conjugation (noting the properties of the Pauli
matrices) and anticommutation relations, show that Eq. (101) is obtained.
Q49 Show that both φ and χ satisfy the Klein-Gordon equation.
Q50 Give an expression for φ in terms of the elements of χ.
Q51 Consider a general two-component spinor that can be constructed from φ as
ΦA = (A0 + A.σ) φ .
(102)
Show that no choice of (A0 , A) can make ΦA transform like χ.
Q52 Consider a general two-component spinor that can be constructed from φ∗ as
ΦB = (B0 + B.σ) φ∗ .
(103)
Find a choice of (B0 , B) that makes ΦB transform like χ.
The solution to the equation
(i∂0 + iσ.∇) φ = imσ2 φ∗ ,
(104)
is therefore a valid relativistic two-component wavefunction for a massive fermion – a so-called
“Majorana fermion”. We can consider plane wave solutions
c
a
−ip.x
e+ip.x ,
(105)
e
+
φ=
d
b
where we can show, with the help of Eq. (100), that
∗ 1
c
a
=
(E + σ.p) (−iσ2 )
.
d
b∗
m
(106)
It is also possible to obtain a conserved current, with ρM = φ† φ and j M = φ† σφ, which satisfies
µ
∂µ JM
= 0.
(107)
Q53 Prove Eq. (106).
Q54 Obtain Eq. (107) from Eq. (100) and (101). hint: insert χ = iσ2 φ∗ , left-multiply Eq. (100) by
φ† and take the transpose of Eq. (101) before right-multiplying by φ.
µ
Q55 Show that ρM = χ† χ and j M = χ† σχ provides an alternative choice for JM
. How does this
choice differ from that given in terms of φ?
Q56 How do φ and χ transform under parity? How does the Majorana current transform under
parity?
A Dirac spinor can be written in terms of Weyl spinors as
φ
ψ=
.
χ
Thus a Majorana fermion can be written as a Dirac spinor:
φ
ψM =
.
iσ2 φ∗
We can now investigate the properties of Majorana fermions under charge conjugation:
0
iσ2
φ∗
−φ
∗
ψM C = iγ2 ψM =
=
= −ψM .
−iσ2 0
iσ2 φ
−iσ2 φ∗
(108)
(109)
(110)
Hence ψM C = −ψM – i.e. Majorana fermions are self-conjugate (up to a phase). This is known as
the “Majorana condition”.
We can also consider the action of charge conjugation on the “L” and “R” components of ψM
separately. It is easy to show that
ψM,L C
= −ψM,R ,
(111)
ψM,R C
= −ψM,L ,
(112)
i.e. charge conjugation swaps L ←→ R. This is the origin of the (potentially misleading) statement
that for Majorana neutrinos, the right-handed neutrino is the antiparticle of the left-handed neutrino.
Q57 Give plane wave solutions for Majorana fermions written as Dirac spinors.
Q58 Show that ψM,L C = −ψM,R and ψM,R C = −ψM,L .
Neutrinos – Majorana or Dirac?
As mentioned above, nature has provided us with a class of particle that may be a Majorana
fermion – the neutrino. Neutrinos are electrically neutral and do not interact with the strong interaction. They also have non-zero (though tiny) mass. It is significant that the mass is non-zero,
since for massless fermions the distinction between Majorana and Dirac particles is purely academic:
the “R” and “L” components are decoupled. A non-zero mass couples together the two handedness
states, and then the question of the nature of the neutrino (Majorana or Dirac) arises. Put another
way, we can ask does the neutrino have a conserved quantum number (i.e. a lepton number)? If it
does, then neutrino and antineutrino are distinct, and the neutrino is described by a Dirac spinor.
If not, then the Majorana formalism is appropriate.
The question of the nature of the neutrino is one of the most important in contemporary particle
physics for several reasons, some of which are briefly discussed below. Note that more detailed
discussions are beyond the scope of this module, but may be partially covered in some other modules,
including PX430 Gauge Theories for Particle Physics, PX445 Advanced Particle Physics and PX435
Neutrinos.
• Neutrinos are massless in the Standard Model, and therefore it is important to understand the
correct way to incorporate neutrino mass into a more advanced theory.
• If the neutrino is Majorana, then a fundamentally new type of wavefunction is realised in
nature. (In quantum field theory, we should say that a fundamentally new type of field is
realised.)
• If the neutrino is Majorana, then a new type of mass is possible. All fermions have a so-called
“Dirac mass” term, which originates from the couplings of the fermions to the Higgs boson.
However, Majorana particles can have an additional, “Majorana mass” term, which is not
associated to the Higgs. (Note the potentially confusing nomenclature: the correspondence between Dirac/Majorana fermions and Dirac/Majorana mass terms is not one-to-one. Explicitly,
a Dirac fermion can only have a Dirac mass term, while a Majorana fermion can have both
Dirac and Majorana mass. The reasons for this are discussed in the Gauge Theories module.)
• If the neutrino is Majorana, there is a possible explanation for the smallness of their masses
compared to the other fermions (quarks and charged leptons). This is called the “see-saw
mechanism”: if the Dirac mass is given by m, and the Majorana mass by M , then the physical
mass of the left-handed neutrinos we see in experiments is given by neither m nor M , but
(roughly) by m2 /M . Thus the Dirac mass can be comparable to other fermion masses (which
are pure Dirac terms), if the Majorana mass is very large (roughly 1015 GeV, give or take a few
orders of magnitude). (Recall that the mass of the proton is about 1 GeV, that of the electron
about 0.5 × 10−3 GeV and the physical mass of the neutrino is expected to be . 10−10 GeV.)
• If the neutrino is Majorana, and the seesaw mechanism is correct, then the Majorana mass
(roughly 1015 GeV) is approximately at the scale where “grand unification” (i.e. the unification
of the strong and electroweak couplings) is expected to occur. Thus there is a very exciting
possibility that the small neutrino mass is a hint of physics at very high energies, of which we
currently know very little, and that further studies of neutrinos may allow us to learn more
about this new physics.
• If the neutrino is Majorana, then interactions of the neutrinos that differ between particle and
antiparticle (so-called “CP violation”, where C represents charge conjugation and P represents
parity) could be responsible for the origin of the asymmetry between matter and antimatter
that we observe in the Universe. This scenario is known as “leptogenesis”, and has been a hot
topic in theoretical particle physics over the past few years.
From the above, it should be clear that there are reasons for a strong theoretical prejudice that
neutrinos should be Majorana particles. However, the nature of the neutrino is a question that
can only be settled experimentally. The most promising approach is the search for a phenomenon
called neutrinoless double beta decay. In certain radioactive nuclei, it is energetically favourable to
undergo simultaneous double beta decay (usually accompanied by the emission of two neutrinos).
However, if the neutrino is a Majorana particle, then the same decay can occur without neutrinos
being emitted. This process is illustrated in Fig. 1. It can be visualised as an electron being emitted
together with a neutrino at one beta decay site; the neutrino then being absorbed (i.e. acting as an
antineutrino) at the second beta decay site. This can only occur if the neutrino is the same particle
as the antineutrino, i.e. if it is a Majorana fermion. It is also straightforward to see that the process
violates the conservation of lepton number, since the initial state has zero lepton number, while the
final state has two electrons (and no neutrinos). Clearly this is not possible if the neutrino is a Dirac
fermion.
Experimental investigations of neutrinoless double beta decay have been continuing for many
years, with gradually improving sensitivity. Due to the smallness of the neutrino mass, neutrinoless
double beta decay is an extremely rare process (even for Majorana neutrinos), and therefore the
experiments need to be extraordinarily sensitive, looking for only a few decays in many tonnes of
material. Ongoing and forthcoming experiments will improve further the sensitivity to neutrinoless
double beta decay, and may be able to show that that neutrino is a Majorana particle within the
next few years – if so, this would be a dramatic discovery. On the other hand, if the neutrino is a
Dirac particle, it could unfortunately be very difficult to rule out the Majorana possibility.
p
n
n
eep
Figure 1: Diagram for neutrinoless double beta decay. The symbol ⊗ represents the Majorana mass
– the diagram can only occur if the neutrino is a Majorana fermion.