January 2016
PX408: Relativistic Quantum Mechanics
Tim Gershon
(T.J.Gershon@warwick.ac.uk)
Selected worked answers
Handout 1: Revision & Notation
Q2 An electron travelling with energy 9.0 GeV in the +z direction collides with a positron travelling
with energy 3.1 GeV in the −z direction. What is the centre-of-mass energy of the system?
What are the Lorentz boost factors β and γ of this system in the laboratory frame?
−→ Add together the four-vectors (in the lab frame)
 
 

12.1
3.1
9.0
 0   0   0
 
 
Ptot = Pe− + Pe+ = 
 0 + 0 = 0
5.9
−3.1
9.0




(1)
where we have used the fact that both electron and positron are highly relativistic, and suppressed
the units of GeV (and taken c = 1).
The centre-of-mass energy squared is obtained from the Lorentz invariant
expression m2 = E 2 −p2
√
2
(in the centre-of-mass frame, p = 0, so Ecom = m) and thus Ecom = 12.1 − 5.92 GeV ≈ 10.56 GeV.
To boost into the centre-of-mass frame, we need to use Lorentz factors β and γ that satisfy
pz 7→ p0z = γ(pz − βEz ) = 0
(2)
and hence
β = pz /Ez = 5.9/12.1 ≈ 0.49 ,
p
γ = 1/ 1 − β 2 ≈ 1.15 .
(3)
Meanwhile, we can cross-check that the centre-of-mass energy is given by
Ez 7→ Ez0 = γ(Ez − βpz ) ≈ 10.56 GeV .
(4)
These are the parameters of the BaBar experiment, which collides asymmetric electron and
positron beams at energies tuned to produce the Υ(4S) particle, that in turn decays to produce pairs
of B mesons.
Q4 Calculate the elements of (Λµν )−1 (in its matrix representation). Show that the inverse Lorentz
transformation corresponds to a Lorentz boost with velocity −β.
−→ We are given that:


γ
−βγ 0 0
 −βγ
γ
0 0 
.
Λµν = 
 0
0
1 0 
0
0
0 1
(5)
It is easy to show, either by (i) explictly calculating the inverse, or (ii) showing that (Λµν )−1 Λµν = 1,
that


γ βγ 0 0
 βγ γ 0 0 
.
(6)
(Λµν )−1 = 
 0
0 1 0 
0
0 0 1
This is the same as Λµν with β −→ −β, and hence corresponds to a Lorentz boost with velocity −β.
Q6 Show that the Lorentz transformation can be written

cosh θ sinh θ

sinh θ cosh θ
Λµν = 
 0
0
0
0
0
0
1
0

0
0 

0 
1
(7)
where θ is called the “rapidity”. Write θ in terms of the Lorentz factors β and γ.
−→
We can simply identify γ = cosh θ. Then sinh2 θ = cosh2 θ − 1 = γ 2 − 1 = β 2 γ 2 , so sinh θ = ±βγ
and tanh θ = ±β (the sign can be chosen).
Q10 Show that F µν is antisymmetric.
Q11 How many independent elements does F µν have?
Q12 Show explicitly that

F µν

0
−Ex −Ey −Ez
 −Ex
0
−Bz By 

=
 −Ey Bz
0
−Bx 
−Ez −By Bx
0
(8)
−→
It’s trivial to show that F µν = −F νµ , either from its definition (F µν = ∂ µ Aν − ∂ ν Aµ ) or from
the explicit representation given.
Any antisymmetric 4 × 4 matrix has 6 independent components (the elements in its strict upper
triangle – i.e. excluding the diagonal components). Moreover, in the explicit representation given we
see that F µν can be written in terms of E and B, the components of which are all independent.
ν
∂A
ν
Consider µ = 0: F 0ν = ∂ 0 Aν − ∂ ν A0 = ∂A
∂t − ∂ φ so the space-like part gives ∂t + ∇φ = −E.
µν
(Diagonal elements are trivial.) Similarly for both µ and ν spacelike, F gives the elements of ∇×A
up to a sign.
Q15 At what energies/lengths are the following relativistic: (i) photons; (ii) electrons; (iii) protons?
−→
(i) Photons are massless and are always relativistic.
−34 Js−1
~
(ii) The Compton wavelength for electrons is λc ∼ mc
∼ 0.51.05×10
. To convert MeV
MeV×3×108 ms−1
2
to standard units, we need to use E = mc and multiply by the electron’s charge. We get λc ∼
4 × 10−13 m – a subatomic length scale.
(iii) Since the Compton wavelength is inversely proportional to mass, we can take the previous answer
and divide by mp /me ∼ 938/0.5, which gives λc ∼ 2 × 10−16 m – a subnuclear length scale.
Q16 If light can be treated as particles, why don’t these particles collide when two beams of light
are shone at each other?
−→ Somewhat surprisingly, we need quantum field theory to really answer this (apparently simple)
question properly, but in outline the answer is that photons only interact with electrically charged
particles, and since photons are neutral they do not self-interact (i.e. they do not “collide with each
other”).
It’s curious to note that in the full quantum field theory, due to higher-order processes (loop
diagrams), photons do in fact interact with each other, i.e. beams of photons do collide. Because these
are higher-order processes, their amplitudes are suppressed, so such collisions are rare, consistent with
our everyday experience. (n.b. It has been suggested that collisions of high energy photons could
be a convenient way to produce and study Higgs bosons, but there are no advanced plans for such a
facility.)
Q21 Prove the commutation relations [σi , σj ] = 2iijk σk (where ijk is the Levi-Civita symbol)
Q22 Prove the anti-commutation relations {σi , σj } = 2δij I
−→ These are best addresses by explicitly working through the possibilities. For example
i 0
0 −i
0 1
= iσ3 .
=
σ1 σ2 =
0 −i
i 0
1 0
(9)
Once all σi σj are worked out, it’s straightforward to show that
σi σj = δij + iijk σk =
1
({σi , σj } + [σi , σj ]) .
2
(10)
Q24 Verify the relation
(σ.a)(σ.b) = (a.b)I + iσ.(a × b)
(11)
where a and b are arbitrary vectors (of scalar quantities).
−→
(σ.a)(σ.b) = (σx ax + σy ay + σz az )(σx bx + σy by + σz bz )
=
σx2 ax bx
+
σy2 ay by
+
σz2 az bz
+
(12)
(13)
σx σy (ax by − bx ay ) + σy σz (ay bz − by az ) + σz σx (az bx − bz ax )
= a.b + iσz (a × b)z + iσx (a × b)x + iσy (a × b)y
(14)
= a.b + iσ.(a × b) ,
(15)
where in the second line we have used the anticommutation relations of the Pauli matrices, and in
the third we have used σx2 = σy2 = σz2 = 1. Note that the proof that (σ.a)2 = |a|2 follows trivially.
Q26 Give an expression for (σ.a)2 .
−→
(σ.a)2 = (σx .ax + σy .ay + σz .az )2
2
2
= (σx .ax ) + (σy .ay ) + (σz .az )
(16)
2
(17)
+ σx .ax σy .ay + σy .ay σx .ax + σy .ay σz .az + σz .az σy .ay + σz .az σx .ax + σx .ax σz .az
2
= |a|
(18)
where all the cross terms disappear due to the anticommutation relations of the Pauli spin matrices
(assuming here that the elements of a are scalar (or at least commuting) quantities.
Q27 Rearrange ...
−→ This question is answered in Handout 4 (under “Dirac’s explanation of g = 2”).
Handout 2: Relativistic Wave Equations
Q4 What is ∂µ (ψ ∗ ψ)?
Q5 Explicitly derive the Klein-Gordon current from the Klein-Gordon equation and its conjugate.
−→
This is simply the product rule for differentiation: ∂µ (ψ ∗ ψ) = (∂µ ψ ∗ )ψ + ψ ∗ (∂µ ψ).
ψ ∗ (∂µ ∂ µ + m2 )ψ = 0
(∂µ ∂ µ + m2 )ψ ∗ ψ = 0
∗
µ
µ
∗
ψ (∂µ ∂ ψ) − (∂µ ∂ ψ ) ψ = 0
∗
∗
µ
µ
µ
∗
∗
µ
ψ (∂µ ∂ ψ) + (∂µ ψ ) (∂ ψ) − (∂µ ∂ ψ ) ψ − (∂µ ψ ) (∂ ψ) = 0
∗ µ
(20)
(21)
(22)
µ
∗
(23)
µ
∗
(24)
∂µ (ψ ∂ ψ) − ∂µ (ψ∂ ψ ) = 0
∗ µ
(19)
∂µ (ψ ∂ ψ − ψ∂ ψ ) = 0
where the first two lines give the Klein-Gordon equation and its conjugate, left- and right-multiplied
appropriately, in the third line they have been subtracted, the scalar m2 term cancelling, in the
fourth line a term (∂µ ψ ∗ ) (∂ µ ψ) has been both added and subtracted, to make the step in the fifth
line, in which the product rule for differentiation is applied, more apparent. Finally, the result is put
in the form ∂µ J µ = 0 (a constant factor of i used in the usual Klein-Gordon current to make it real
is neglected).
Q9 Quantise and rearrange ...
−→ We start with
(E − qφ)2 = (p − qA)2 + m2 .
(25)
Quantising gives
(i
∂
− qφ)2 ψ = (−i∇ − qA)2 ψ + m2 ψ ,
∂t
(26)
which rearranges to
∂
(i − qφ)2 − (−i∇ − qA)2 ψ = m2 ψ
∂t
∂
∂
i − qφ + i∇ + qA
i − qφ − i∇ − qA ψ = m2 ψ
∂t
∂t
∂
∂
+ ∇ + iqφ − iqA
− ∇ + iqφ + iqA ψ = −m2 ψ
∂t
∂t
(28)
(∂µ + iqAµ )(∂ µ + iqAµ )ψ + m2 ψ = 0 .
(30)
(27)
(29)
or
Q10 Obtain the conserved current for the Klein-Gordon equation in an electromagnetic field.
−→ The Klein-Gordon equation in an electromagnetic field is
(∂µ + iqAµ )(∂ µ + iqAµ )ψ + m2 ψ = 0
so its complex conjugate is
(∂µ − iqAµ )(∂ µ − iqAµ )ψ ∗ + m2 ψ ∗ = 0
Left-multiply the first by ψ ∗ , right-multiply the second by ψ, and subtract:
ψ ∗ (∂µ ∂ µ ψ) − (∂µ ∂ µ ψ ∗ )ψ + iq (ψ ∗ Aµ (∂ µ ψ) + ψ ∗ (∂µ (Aµ ψ)) + Aµ (∂ µ ψ ∗ )ψ + (∂µ (Aµ ψ ∗ ))ψ) = 0
noting that the terms involving A2 and m2 cancel in the subtraction.
µ ∗
The first two terms give ∂µ (ψ ∗ ∂ µ ψ − (∂
ψ )ψ)as in the free-field case. The remaining terms can
be identified with 2iq∂µ (Aµ ψ ∗ ψ) = 2iq∂µ Aµ |ψ|2 .
So, after multiplying by i so that the conserved current is real for plane wave solutions , we obtain
for the current
j µ = i (ψ ∗ ∂ µ ψ − (∂ µ ψ ∗ )ψ) − 2q (Aµ ψ ∗ ψ) .
Q13 Show explicitly that this choice for α and β satisfies the requirements ...
−→
αi =
so
αi αj =
0 σi
σi 0
0 σi
σi 0
0
σj
β=
σj
0
=
1 0
0 −1
δij + iijk σk
0
0
δij + iijk σk
(31)
so the anticommutation relations of the Pauli matrices give {αi , αj } = 2δij .
Similarly
0 −σi
αi β =
= −βαi ,
σi
0
(32)
(33)
while β 2 = 1 is straightforward.
Q15 Show that any set of matrices α0 , β 0 related to the above forms by the transformation
αi0 = U αi U −1
β 0 = U βU −1
(34)
where U is any unitary 4 × 4 matrix, would also give satisfactory solutions to the requirements
...
−→
αi0 αj0 = U αi U −1 U αj U −1 = U αi αj U −1 ,
(35)
since U −1 U = 1. If i = j, then αi αj = 1, so αi0 αj0 = 1 too. If i 6= j, αi αj = −αj αi so
αi0 αj0 = −U αj αi U −1 = −U αj U −1 U αi U −1 = −αj0 αi0 .
This proves the anticommutation relations
similarly.
(36)
n
o
αi0 , αj0 . Those for {αi0 , β 0 } and (β 0 )2 are obtained
Q17 Write out ... explicitly, including all elements of the 4 × 4 matrices.
−→
i
∂
ψ = (−iα.∇ + βm)ψ
∂t
0 σ
1 0
=
−i
.∇ +
m ψ
σ 0
0 −1
 
∂
0
0
∂z
 
∂
∂
0
0
 
∂x + i ∂y
= −i 
∂
∂
∂
0
 
∂z
∂x − i ∂y
∂
∂
∂
− ∂z
0
∂x + i ∂y
(37)
(38)
∂
∂x
∂
− i ∂y
∂
− ∂z
0
0


m 0
0
0
 

0 
  0 m 0


+
 
ψ
0 0 −m 0 

0 0
0 −m

Q18 Show that if α and β satisfy ...
Q19 What are the commutation and anticommutation relations of the gamma matrices? Relate
{γ µ , γ ν } to g µν .
−→ These two questions are the same! As we showed above,
δij + iijk σk
0
.
αi αj =
0
δij + iijk σk
(39)
Similarly
γi γj =
0 σi
−σi 0
0
−σj
σj
0
=−
δij + iijk σk
0
0
δij + iijk σk
,
(40)
for i, j = 1, 2, 3. Hence, {γi , γj } = −2δij , while
[γi , γj ] = −2iijk
σk 0
0 σk
.
(41)
For µ = 0, (γ0 )2 = 1, while
γ0 γj =
1 0
0 −1
0
−σj
σj
0
0
σj
=
0
σj
σj
0
= −γj γ0 .
(42)
Hence {γ0 , γj } = 0 while
[γ0 , γj ] = 2
σj
0
.
(43)
Putting it all together, we obtain {γµ , γν } = 2gµν , while [γµ , γν ] cannot be written in any simple
form (this is related to the Clifford algebra of the gamma matrices discussed later).
Q22 For a generic covariant four vector aµ , write out explicitly the elements of a/ in the standard
representation.
−→


a
0
−a
−a
+
ia
0
3
1
2


a0 −σ.a
0
a0
−a1 − ia2
a3

a/ = γ µ aµ =
=
(44)


a3
a1 − ia2
−a0
0
σ.a −a0
a1 + ia2
−a3
0
−a0
1
Q25 Find solutions for χ, and consequently for Ψ, for plane-wave solutions with ψ ∝
and
0
0
with ψ ∝
. (Use the free-field form of the equations.)
1
−→ Put ψ =
1
0
e−ip.x , with p.x = pµ xµ and neglecting normalisation. Then
mχ = (i∂0 + iσ.∇)ψ
(45)
= (i(−iE) + iσ.(+ip))
Hence
χ=
1
0
e−ip.x .
(46)
E − σp
ψ.
m
(47)
With this particular choice of ψ, we get
1
E − pz
−px + ipy
1
χ =
e−ip.x
E + pz
0
m −px − ipy
1
E − pz
=
e−ip.x .
m −px − ipy
For the other linearly independent choice of ψ we get
1
−px + ipy
χ=
e−ip.x .
E + pz
m
(48)
(49)
(50)
Hence

ψ=
ψ=
1
0
0
1
e−ip.x
e−ip.x

m + E − pz

1 
 −px − ipy  e−ip.x
−→ Ψ =
m  m − E + pz 
px + ipy


−px + ipy

1 
 m + E + pz  e−ip.x
−→ Ψ =

px − ipy 
m
m − E − pz
(51)
(52)
Q27 Show that γ 0µ satisfy the same anticommutation relations as γ µ .
−→
γ 0µ , γ 0ν
= S −1 γ µ SS −1 γ ν S + S −1 γ ν SS −1 γ µ S
= S
−1 µ ν
= S
−1
γ γ S+S
µ
(53)
−1 ν µ
γ γ S
ν
{γ , γ } S = S
−1
(54)
µν
2g S = 2g
µν
(55)
Q28 Show that γµ a//bγ µ = 4aµ bµ .
−→
γµ a//bγ µ = γµ γν aν γλ bλ γ µ
ν λ
= a b γµ γν γλ γ
(56)
µ
(57)
ν λ
= a b (2gµν − γν γµ )γλ γ
µ
ν λ
= a b (2γλ γν − γν (2gµλ − γλ γµ )γ
ν λ
= a b (2γλ γν − (2γν γλ − 4γν γλ ))
ν λ
= a b 2 {γλ , γν }
ν λ
(58)
µ
(59)
(60)
(61)
ν
= 4a b gλν = 4a bν
where we have exploited the fact that gµν = gνµ and used γµ γ µ = 4.
(62)
Q30 Use the equivalence transformation to show that Tr(γ µ ) = 0 is true for any representation of
the gamma matrices.
−→
Tr(γ 0µ ) = Tr(S −1 γ µ S) = Tr(SS −1 γ µ ) = Tr(γ µ ) = 0
(63)
where we have used the cyclic property of the trace: Tr(ABC) = Tr(BCA) = Tr(CAB).
Q33 Show that (γ 5 )2 = 1.
Q34 Show that γ 5 , γ µ = 0.
−→
Using only anticommutation relations and the known values of the squares of the gamma matrices:
(γ 5 )2 = (iγ 0 γ 1 γ 2 γ 3 )(iγ 0 γ 1 γ 2 γ 3 )
0 1 2 3 0 1 2 3
= (−1)γ γ γ γ γ γ γ γ
(65)
4
0 2 1 2 3 1 2 3
(66)
6
0 2
1 2 2 3 2 3
(67)
7
0 2
1 2
(68)
10
=1
= (−1) (γ ) γ γ γ γ γ γ
= (−1) (γ ) (γ ) γ γ γ γ
2 2
3 2
= (−1) (γ ) (γ ) (γ ) (γ )
= (−1)
(64)
γ 5 , γ µ = γ 5 γ µ + γ µ γ 5 = iγ 0 γ 1 γ 2 γ 3 γ µ + γ µ iγ 0 γ 1 γ 2 γ 3 .
(69)
(70)
To get factors of (γ µ )2 we will need to anticommute (3 − µ) times on the first term and µ times on
the second, so they will pick up factors of (−1)3−µ and (−1)µ respectively. One of these factors will
be odd and the other even, hence we will end up with an overall minus sign difference between the
two terms that are otherwise identical. So they subtract to leave zero.