March 2016
PX430: Gauge Theories for Particle Physics
Tim Gershon
(T.J.Gershon@warwick.ac.uk)
Handout 3: Gauge Invariance in Quantum Field Theory
The Free-Field Klein-Gordon Lagrangian
In our previous discussion of quantum field theory, we have considered a 1 dimensional “string”,
with (unquantised) Lagrangian density given by
1
L=T −V = ρ
2
∂φ
∂t
2
1
− ρc2
2
∂φ
∂x
2
,
(1)
where T and V are kinetic and potential energy terms, respectively, ρ is the mass per unit length,
and the field variable φ can be visualised as the displacement of the string as a function of distance
along the string x and time t; c is the speed of waves along the string. We can generalise this to
3 dimensions, take c = 1, and ignore ρ, which is a multiplicative constant of the whole Lagrangian
density, and therefore does not affect the solutions of the Euler-Lagrange equations. Doing so, we
find
1 ∂φ 2 1
1
L=
− (∇φ)2 = (∂µ φ) (∂ µ φ) .
(2)
2 ∂t
2
2
The Euler-Lagrange equations for this system give
∂L
∂L
− ∂µ
=0
−→
∂φ
∂ (∂µ φ)
∂µ ∂ µ φ = 0 ,
(3)
i.e. the massless Klein-Gordon equation.
In order to include the mass term in the Klein-Gordon equation we need to modify the Lagrangian
appropriately. It is not hard to show that the correct form is
L̂KG =
1
1
∂µ φ̂ ∂ µ φ̂ − m2 φ̂2 ,
2
2
(4)
where we have taken the liberty of quantising the fields. This is the free-field Klein-Gordon Lagrangian
density for a real valued scalar field.
Q1 Show that application of the Euler-Lagrange equations to the Lagrangian density of Eq. (4)
yields the free-field Klein-Gordon equation.
Recall that we can write the quantum field density φ̂ and its conjugate momentum Π̂ in terms of
their Fourier expansions:
Z +∞
h
i
d3 k
−ik.x
†
ik.x
φ̂(x, t) =
â(k)e
+
â
(k)e
,
3
−∞ (2π) (2ω)
Z +∞
h
i
dk
Π̂(x, t) =
(−iω) â(k)e−ik.x − ↠(k)eik.x ,
(5)
−∞ (2π)(2ω)
where k.x = kµ xµ = ωt − k.x. [As before, the dependence of ω = ω(k) on k – actually, on |k| – has
˙
not been made explicit.] Since the Hamiltonian density is given by Ĥ = Π̂φ̂ − L̂, we can show that
1 2
Π̂ + ∇φ̂.∇φ̂ + m2 φ̂2 .
2
Taking advantage of the commutation relation â(k), ↠(k0 ) = (2π)3 δ 3 (k − k0 ), we obtain
ĤKG =
Z
(6)
+∞
d3 xĤKG ,
ĤKG =
−∞
d3 k
1 †
†
â
(k)â(k)
+
â(k)â
(k)
ω,
(2π)3 (2ω) 2
Z
d3 k
†
â
(k)â(k)
ω,
:ĤKG: =
(2π)3 (2ω)
Z
=
(7)
where :ĤKG: is the normal ordered form, i.e. commutation relations are used to move all annihilation
operators (â(k)) to the right, and the (infinite) zero point energy term has been dropped.
Q2 Starting with the Klein-Gordon Lagrangian density of Eq. (4), derive Eq. (7).
Interacting Fields
With the number operator n̂(k) = ↠(k)â(k) as before, it is clear that ĤKG =
R
d3 k
n̂(k)ω,
(2π)3 (2ω)
and that [ĤKG , n̂(k)] = 0, i.e. that the Hamiltonian commutes with the number density operator,
and therefore that the number density is a constant of the motion. So although we have operators
that create and annihilate quanta, they cannot be actually doing anything in this situation, since
the number of quanta is conserved. To put this another way, imagine two plane waves that are
solutions of the Klein-Gordon equation given by the above Lagrangian density. If these waves start
off travelling towards each other, they will simply pass through each other and emerge unscathed on
the other side – i.e. there is no interaction.
The reason that this occurs has its roots early in our discussion of quantum field theory. We
started by considering a simple harmonic oscillator. This is a rather special situation, the solutions
of which can be superposed. However, the addition of further (non-linear) terms to the potential
will result in the system becoming anharmonic. As should be familiar from classical mechanics,
these anharmonic terms often appear as the consequence of approximations made in the harmonic
treatment (e.g. sin θ ∼ θ has corrections of O(θ3 )).
For a general harmonic oscillator, we have for the Hamiltonian, in terms of the generalised (and
quantised) coordinate q̂ and its conjugate momenta p̂,
Ĥ0 =
1 2 1
p̂ + mω 2 q̂ 2 ,
2m
2
(8)
to which we can add an interaction term
Ĥ0 7→ Ĥ = Ĥ0 + Ĥint ,
Ĥint = λq̂ 3 =
3
λ
†
â
+
â
,
(2mω)3/2
(9)
where λ determines the interaction strength. [Note that this choice of interaction term is more-orless arbitrary; we will encounter different, and more useful, forms for interaction terms later.] With
the addition of Ĥint , the Hamiltonian will no longer, in general, commute with the number density
operator.
h
i
Q3 For Ĥint = λq̂ 3 , what is Ĥint , n̂ ?
We cannot, in general, solve for this kind of situation exactly, and therefore we need to resort to
approximate methods. Two such methods are particularly noteworthy:
• perturbation theory is effective when the interaction strength is weak – the interaction can be
considered as a small perturbation on the stable, harmonic, system. The use of perturbation
theory should be familiar from previous modules, and will be discussed again later when it
comes to calculating amplitudes, for example in quantum electrodynamics (QED);
• computational methods can be seen as a brute force approach to be applied when nothing
else will work, although this is not to say that the techniques involved are without elegance.
Within particle physics, the perturbative approach fails for most calculations related to quantum chromodynamics (QCD), i.e. the strong interaction. To calculate QCD matrix elements,
a technique called lattice QCD is employed, wherein the four dimensional space-time is divided
into a discrete lattice, and the necessary quantities must be calculated at each lattice point.
The physical solution is obtained by extrapolating to infinitesimal lattice spacing.
Complex Scalar Fields
The discussion above related to the free-field Klein-Gordon Lagrangian gave a solution φ̂ that is
a real valued scalar field. There is no associated gauge symmetry (a change of phase would make φ̂
no longer real valued). These solutions describe neutral spin-0 particles, with the absence of gauge
symmetry representing the fact that neutral particles do not undergo electromagnetic interactions.
Rather few such fundamental particles have been discovered to date: neutral spinless particles such
as neutral pions (π 0 ) have been shown to be composed of quarks and antiquarks; the quantum field
description of the photon, which is a neutral boson, but spin-1 instead of spin-0, is somewhat more
complicated (as we will see later). However, in 2012 a particle with properties consistent with those
expected of the Higgs boson, a fundamental spinless neutral particle, was observed. If indeed this is
a fundamental particle (rather than a composite as predicted in some theories) then it seems that
nature does contain at least one real valued scalar field. (As a side comment, at high-energies the
Higgs field has a different Lagrangian density to that of Eq. (4), as will be discussed later. So it still
seems that nature does not contain any Klein-Gordon fields, for reasons that are not understood.)
Q4 The behaviour of the π 0 meson should be invariant under parity (inversion of spatial coordinates) and charge conjugation (replacement of particle by antiparticle) transformations. What
discrete symmetries of the Lagrangian in Eq. (4) do these correspond to?
√
¯
Q5 The quark content of the neutral pion is π 0 = (uū − dd)/
2. How is the π 0 affected by local
phase transformations of the u and d fields? What other symmetries does the π 0 respect, and
what are their physical significances?
We now turn our attention to the Lagrangian for two independent real valued scalar free-fields,
labelled (suggestively) by φ̂R and φ̂I . We can obtain the total Lagrangian from Eq. (4), by simply
adding the terms for each:
1
1
1
1
L̂ =
∂µ φ̂R ∂ µ φ̂R − m2 φ̂2R +
∂µ φ̂I ∂ µ φ̂I − m2 φ̂2I .
(10)
2
2
2
2
If we consider these as the real and imaginary parts of a complex scalar field φ̂ (n.b. beware that
we are reusing the φ̂ notation), we can write
1 1 φ̂ = √ φ̂R − iφ̂I ,
φ̂† = √ φ̂R + iφ̂I ,
(11)
2
2
and then rearrange Eq. (10) to give
L̂ = ∂µ φ̂† ∂ µ φ̂ − m2 φ̂† φ̂ .
(12)
Note that φ̂ and φ̂† are treated as independent fields, and that we have dropped the overall factor of
1
2 , which does not affect the physics.
Q6 Show explicitly that Eq. (10) can be rearranged to give Eq. (12).
Q7 Show that the Lagrangian of Eq. (12) is invariant under phase transformations.
Consider phase transformations of the form
φ̂ 7→ φ̂0 = e−iα φ̂ ,
φ̂† 7→ φ̂†0 = e+iα φ̂† .
(13)
Here we can assume that the phase transformation is global, i.e. α does not vary with position
in space-time;, though in fact we get the same outcome for local transformations. We can restrict
ourselves to infinitesimal values of the rotation α
,
φ̂ 7→ φ̂0 = φ̂ − iφ̂ = φ̂ + δ φ̂ ,
φ̂† 7→ φ̂†0 = φ̂† + iφ̂† = φ̂† + δ φ̂† ,
(14)
and identify δ φ̂ = −iφ̂ and δ φ̂† = iφ̂† .
Since the Lagrangian is invariant under such transformations, we can write
∂ L̂ ∂ L̂ ∂ L̂
∂ L̂
δ ∂µ φ̂ + δ ∂µ φ̂† +
(15)
δ φ̂ +
δ φ̂† .
0 = δ L̂ = †
∂ φ̂
∂ φ̂
∂ ∂µ φ̂
∂ ∂µ φ̂†
This looks similar to expressions found in the derivation of the Euler-Lagrange equations


∂ L̂
∂ L̂ 
= 0,
− ∂µ  ∂ φ̂
∂ ∂µ φ̂
(16)
and in fact we can now use the Euler-Lagrange equation for φ̂ to replace the third term in Eq. (15),
and that for φ̂† to replace the fourth, and obtain a total divergence:


∂ L̂
∂ L̂
δ φ̂ + δ φ̂†  .
0 = δ L̂ = ∂µ  (17)
†
∂ ∂µ φ̂
∂ ∂µ φ̂
Much of this derivation has been quite general, but if we now return
case
under scrutiny,
to the
†
using the Lagrangian of Eq. (12) and inserting the expressions for δ φ̂ and δ φ̂ that are relevant
for infinitesimal phase transformations, we find
0 = ∂µ ∂ µ φ̂† −iφ̂ + ∂ µ φ̂ iφ̂†
,
(18)
and hence
0 = i∂µ φ̂† ∂ µ φ̂ − φ̂∂ µ φ̂† .
(19)
Since this is true for arbitrary values of , we can write
0 = ∂µ N̂ µ ,
(20)
where the symmetry current N̂ µ = i φ̂† ∂ µ φ̂ − φ̂∂ µ φ̂† . The result that there is a symmetry current
associated with the symmetry of the Lagrangian is the quantum field theoretic version of Noether’s
theorem: if the Lagrangian is invariant under a continuous symmetry operation, there will be an
associated symmetry current.
Q8 Show that for a Lagrangian density involving several fields, φ̂1 , φ̂2 , . . ., invariant under φ̂i 7→
P
φ̂i + δ φ̂i , there is a conserved Noether current Jˆµ = i ∂ L̂ δ φ̂i .
∂(∂µ φ̂i )
We will return to the symmetry current momentarily. First let us consider the Fourier expansion
of the fields. Following Eq. (5), we can write
Z +∞
h
i
d3 k
†
−ik.x
ik.x
φ̂R =
â
(k)e
+
â
(k)e
,
R
R
3
−∞ (2π) (2ω)
Z +∞
h
i
d3 k
†
−ik.x
ik.x
φ̂I =
â
(k)e
+
â
(k)e
,
I
I
3
−∞ (2π) (2ω)
and hence
Z
+∞
φ̂ =
φ̂† =
−∞
+∞
Z
−∞
h
i
d3 k
−ik.x
†
ik.x
â(k)e
+
b̂
(k)e
,
(2π)3 (2ω)
h
i
d3 k
−ik.x
†
ik.x
,
b̂(k)e
+
â
(k)e
(2π)3 (2ω)
(21)
where
1
â(k) = √ (âR (k) − iâI (k))
2
1
b̂(k) = √ (âR (k) + iâI (k))
2
1 ↠(k) = √ â†R (k) + iâ†I (k) ,
2
1
b̂† (k) = √ â†R (k) − iâ†I (k) .
2
(22)
(n.b. beware that we are reusing the â, ↠notation.) The treatment of φ̂ and φ̂† as independent
fields results in the existence of distinct mode operators â(†) hand b̂(†) .
i
(†)
Since the âR,I operators obey the commutation relations âi (k), â†j (k0 ) = δij (2π)3 (2ω)δ(k − k0 ),
h
i with â†i (k), â†j (k0 ) = âi (k), âj (k0 ) = 0, it can be shown that the â(†) , b̂(†) operators obey
h
with
i h
i
â(k), ↠(k0 ) = b̂(k), b̂† (k0 ) = (2π)3 (2ω)δ(k − k0 ) ,
(23)
h
i h
i i
h
â(k), b̂† (k0 ) = b̂(k), ↠(k0 ) = â(k), â(k0 ) = b̂(k), b̂(k0 ) = 0 .
(24)
Using these, the (normal-ordered) Hamiltonian can be found to be
Z
d3 k
†
†
Ĥ =
â
(k)â(k)
+
b̂
(k)
b̂(k)
ω.
(2π)3 (2ω)
(25)
Q9 Show that the â(†) , b̂(†) operators obey the commutation relations given above.
Q10 Show that the Hamiltonian of Eq. (25) is obtained from the Lagrangian of Eq. (12).
Finally, we return, as promised, to the symmetry current. We have
0 = ∂µ N̂ µ =
∂ N̂ 0
+ ∇.N̂ .
∂t
(26)
If we integrate this over a large volume, which we will allow to tend to infinity, we can use Stokes’
theorem on the second term to obtain
Z
Z
∂ N̂ 0 3
0=
d x+
N̂ .dS ,
(27)
V →∞ ∂t
S→∞
with notation that should be familiar. We assume that N̂ drops off fast enough at infinity that we
can neglect the second term, and hence
Z
∂
N̂ 0 d3 x = 0 ,
(28)
∂t ∞
R
0 3
µ
i.e.
the symmetry
operator N̂ = ∞ N̂ d x is a constant of the motion. Moreover, since N̂ =
i φ̂† ∂ µ φ̂ − φ̂∂ µ φ̂† , we can show that
Z
+∞
N̂ =
0 3
N̂ d x =
−∞
Z
d3 k
†
†
â
(k)â(k)
−
b̂
(k)
b̂(k)
.
(2π)3 (2ω)
Q11 Using the Fourier expansion of the fields given by Eq. (21), show that N̂ =
given by Eq. (29).
h
i
h
i
h
i
Q12 What is N̂ , Ĥ ? What is N̂ , φ̂ ? What is N̂ , φ̂† ?
(29)
R +∞
−∞
N̂ 0 d3 x is as
Let us briefly recount some of the expressions we have encountered in the case of complex scalar
quantum fields:
h
i
R +∞
3
Field density
φ̂ = −∞ (2π)d3k(2ω) â(k)e−ik.x + b̂† (k)eik.x
i
h
R +∞
3
Conjugate field density
φ̂† = −∞ (2π)d3k(2ω) b̂(k)e−ik.x + ↠(k)eik.x
Lagrangian density
L̂ = ∂µ φ̂† ∂ µ φ̂ − m2 φ̂† φ̂
R
3
Hamiltonian
Ĥ = (2π)d3k(2ω) ↠(k)â(k) + b̂† (k)b̂(k) ω
symmetry current
N̂ µ = i φ̂† ∂ µ φ̂ − φ̂∂ µ φ̂†
R
3
symmetry operator
N̂ = (2π)d3k(2ω) ↠(k)â(k) − b̂† (k)b̂(k)
Several points are worthy of remark
• the operators ↠(k) and b̂† (k) commute – they are creation operators for different “particles”;
similarly â(k) and b̂(k) are annihilation operators for different “particles”;
• consequently, a state that contains two particles is symmetric under the interchange of those
particles – |k1 k2 i = ↠(k1 )↠(k2 )|0i = ↠(k2 )↠(k1 )|0i = |k2 k1 i – i.e. these particles follow
Bose-Einstein statistics;
• the Hamiltonian contains the sum of ↠(k)â(k) and b̂† (k)b̂(k) terms, thus both ↠(k) and b̂† (k)
create “particles” with positive energy; similarly, both â(k) and b̂(k) annihilate “particles”
with positive energy;
• the symmetry operator counts the difference of ↠(k) and b̂† (k) quanta; i.e. they carry a
conserved quantum number that takes opposite value in the two cases; they are particle and
antiparticle (and the quantum number is the electric charge);
• the Fourier decomposition of the field density operator φ̂ contains a positive frequency part
(â) that annihilates a particle, and a negative frequency part (b̂† ) that creates an antiparticle;
similarly, φ̂† creates a particle (through ↠) or annihilates an antiparticle (through b̂);
• the problem of negative energy solutions to the Klein-Gordon equation is solved in quantum
field theory!
[n.b. The reason “particles” is written in quote marks in the first two bullet points above is that it
is intended to include both particles and antiparticles.]
While the case of real scalar fields was related to neutral spinless particles, this discussion of complex scalar fields gives the quantum field theoretic description of charged spinless particles. Again,
we find that (perhaps surprisingly) nature does not appear to have endowed us with an abundance
of such objects – indeed, there are no fundamental charged spinless objects in the Standard Model
of particle physics. [In several extensions of the Standard Model, including supersymmetry, charged
Higgs particles are present, but there is no experimental evidence supporting the existence of such
things at present.] However, if we restrict our considerations to energies such that the quark substructure of pions is not an issue, we can use the above formulation to describe the electromagnetic
interactions of charged pions.
The Free-Field Dirac Lagrangian
Following the previous discussion, it is natural to be curious about the situation for Dirac particles.
It is fairly easy to show that the Dirac Lagrangian density operator is
L̂D = ψ̄ˆ (iγ µ ∂µ − m) ψ̂ ,
(30)
where the adjoint spinor ψ̄ˆ = ψ̂ † γ 0
Q13 Apply to Euler-Lagrange equations for i) ψ̄ˆ and ii) ψ̂ to Eq. (30).
The generalised momentum densities are
Π̂ =
∂ L̂D
= ψ̄ˆiγ 0 ,
˙
∂ ψ̂
ˆ = ∂ L̂D = 0 ,
Π̄
˙
∂ ψ̄ˆ
(31)
and so the Hamiltonian density is given by
˙
ĤD = Π̂ψ̂ − L̂D = −ψ̄ˆiγ.∇ψ̂ + mψ̄ˆψ̂ = ψ̂ † (−iα.∇ + βm) ψ̂ ,
(32)
where the α and β matrices will be familiar to those with good memories. Recall that we do not
expect to be able to write the Hamiltonian in a Lorentz invariant form.
Next we insert the Fourier expansions for ψ̂ and ψ̄ˆ, which are given by
R
P d3 k
−ik.x + dˆ† (k)v (k)e+ik.x ,
ĉ
(k)u
(k)e
ψ̂ =
s
s
s
s
s
(2π)3 (2ω)
(33)
R
P ˆ
†
ˆ
d3 k
−ik.x
ψ̄ =
d (k)v̄ (k)e
+ ĉ (k)ū (k)e+ik.x ,
(2π)3 (2ω)
s
s
s
s
s
where us e−ik.x and vs e+ik.x are positive and negative energy solutions of the Dirac equation, respectively (s labels the spin orientation, which can be either ↑ or ↓, and k.x = kµ xµ as before):
(−iα.∇ + βm) us e−ik.x = ωus e−ik.x ,
(−iα.∇ + βm) vs e+ik.x = ωvs e+ik.x .
ˆ?
Q14 What are the Fourier expansions for Π̂ and Π̄
In Eq. (33) above,
• ĉs (k) is an annihilation operator for a Dirac particle with spin s and momentum k;
• ĉ†s (k) is a creation operator for a Dirac particle with spin s and momentum k;
(34)
• dˆs (k) is an annihilation operator for a Dirac antiparticle with spin s and momentum k;
• dˆ†s (k) is a creation operator for a Dirac antiparticle with spin s and momentum k.
We can now find the Dirac Hamiltonian
Z
Z
X
d3 k
3
†
ˆs (k)dˆ† (k) ω .
ĤD = d xĤD =
ĉ
(k)ĉ
(k)
−
d
s
s
s
(2π)3 (2ω) s
(35)
Meanwhile, the Dirac Lagrangian of Eq. (30) is invariant under phase transformations
ψ̂ 7→ e−i ψ̂ ≈ ψ̂ − iψ̂ ,
ψ̄ˆ 7→ e+i ψ̄ˆ ≈ ψ̄ˆ + iψ̄ˆ ,
(36)
and therefore has an associated Noether current
N̂ µ = ψ̄ˆγ µ ψ̂ .
(37)
Q15 Show that that Noether current associated with the phase invariance of Eq. (30) is given by
Eq. (37).
Finally, we can show that in this case the symmetry operator is given by
R
R 0 3
N̂ d x = ψ̂† ψd3 x ,
N̂ =
R
P
†
d3 k
ˆs (k)dˆ†s (k) .
(k)ĉ
(k)
+
d
ĉ
=
s
s
3
s
(2π) (2ω)
(38)
Q16 Derive Eq. (38).
There appears to be a problem: while the quantum field theoretic treatment solved the problem
of negative energy solutions for the Klein-Gordon equation, it appears that we get a negative contribution to the energy from antiparticles in Eq. (35). Meanwhile, the symmetry operator of Eq. (38)
also appears to be counting antiparticles with the wrong sign. However, we might note that in both
Eq. (35 and Eq. (38) the antiparticle contributions are not in normal ordered form. It appears we
have a way out of the conundrum: if dˆs (k)dˆ†s (k) = −dˆ†s (k)dˆs (k) + const., the problems will be solved.
Can we justify using it?
A very important distinction, so far not discussed, between the spin-0 solutions to the KleinGordon equation and the spin- 12 solution to the Dirac equation, is that we know that wavefunctions
containing pairs of the former should be symmetric under interchange of particles, while those containing pairs of the latter should be antisymmetric. That is, they should follow Fermi-Dirac statistics.
Defining the vacuum |0i such that ĉs (k)|0i = 0 ∀ k and s =↑, ↓, a two-fermion state can be
written
|k1 , s1 ; k2 , s2 i = ĉ†s1 (k1 )ĉ†s2 (k2 )|0i .
(39)
In order for this wavefunction to be antisymmetric, we require
|k2 , s2 ; k1 , s1 i = ĉ†s2 (k2 )ĉ†s1 (k1 )|0i ,
= −|k1 , s1 ; k2 , s2 i
= −ĉ†s1 (k1 )ĉ†s2 (k2 )|0i .
(40)
Therefore, to obtain the required antisymmetry, we require the commutation relations of the spin-0
case to be replaced by anticommutation relations
n
n
o
o
ĉs (k), ĉ†s0 (k0 ) = (2π)3 (2ω)δss0 δ 3 (k − k0 ) = dˆs (k), dˆ†s0 (k0 ) ,
(41)
while
n
o
n
o n
o
ĉs (k), ĉs0 (k0 ) = ĉ†s (k), ĉ†s0 (k0 ) = 0 = dˆs (k), dˆs0 (k0 ) = dˆ†s (k), dˆ†s0 (k0 ) ,
(42)
and in addition the anticommutator of any c-type operator with any d-type operator vanishes.
With these anticommutation relations, we obtain the normal ordered forms
Z
X
d3 k
†
ˆ† (k)dˆs (k) ω ,
ĤD =
ĉ
(k)ĉ
(k)
+
d
(43)
s
s
s
(2π)3 (2ω) s
and
Z
N̂ =
X
d3 k
†
†
ˆ
ˆ
ĉ
(k)ĉ
(k)
−
d
(k)
d
(k)
,
s
s
s
s
(2π)3 (2ω) s
(44)
as expected.
To summarise,
• solutions to the Klein-Gordon equation are bosons, which satisfy commutation relations, and
obey Bose-Einstein statistics;
• solutions to the Dirac equations are fermions, which satisfy anticommutation relations, and
obey Fermi-Dirac statistics.
The manner in which quantum field theory enforces the connection between spin and statistics
via (anti)commutation relations is called the spin-statistics theorem. In case it appears somewhat
arbitrary as we have derived it here, note the use of the word “theorem” – this is, in fact, the only
way to obtain self-consistent quantum field theories.
Q17 The commutation of the creation and annihilation
were derived from the
h operators for bosons
i
0
3
fundamental commutator of quantum mechanics, φ̂(x, t), Π̂(x , t) = iδ (x−x0 ). This relation
is intimately connected to Heisenberg’s uncertainty principle. How do the anticommutation
relations of the fermions fit into this picture?
It should be clear by now that this description is appropriate for charged spin- 21 particles. In this
case, we do find an abundance of such particles in nature: all of the quarks (u, d, s, c, b and t) and
charged leptons (e, µ and τ ) fall into this category. Note, however, that quarks are never found as
isolated quantities, since they are bound by the strong interaction into hadrons.
Q18 As discussed, the formalism above is appropriate for charged spin- 21 particles. What is the
appropriate formalism for neutral spin- 21 particles? Do such particles exist in nature?
Quantum Electrodynamics
We wish to promote the global phase rotations discussed above to local transformations (i.e.
gauge symmetries). We start by considering the spinless case.
We know from our previous discussion that the physics described by quantum fields will be
invariant under
φ̂ 7→ φ̂0 = e−iqχ̂ φ̂ ,
(45)
provided we make the minimal substitution
∂ µ 7→ D̂µ = ∂ µ + iq µ ,
(46)
µ 7→ µ0 = µ + ∂ µ χ̂ .
(47)
with
The free-field Klein-Gordon Lagrangian density is modified by the introduction of the minimal substitution thus
L̂KG = ∂µ φ̂† ∂ µ φ̂ − m2 φ̂† φ̂ 7→ L̂KG + L̂INT = D̂µ φ̂† D̂µ φ̂ − m2 φ̂† φ̂ ,
(48)
from which the interaction Lagrangian is found to be
L̂INT = −iq φ̂† (∂ µ φ̂) − (∂ µ φ̂† )φ̂ µ + q 2 µ µ φ̂† φ̂ .
(49)
Recalling that a Lagrangian can be considered as consisting of kinetic and potential terms, L̂ =
T̂ − V̂, this can be interpreted as a potential: L̂INT = −V̂KG . We will return to the Klein-Gordon
potential shortly.
Similarly, introducing the minimal substitution to the Dirac Lagrangian density gives
L̂D = ψ̄ˆ (iγ µ ∂µ − m) ψ̂ 7→ L̂D + L̂INT = ψ̄ˆ iγ µ D̂µ − m ψ̂ ,
(50)
with interaction Lagrangian
L̂INT = −V̂D = −q ψ̄ˆγ µ ψ̂ µ .
(51)
The Dirac potential thus takes the form “ρem A0 − j em .A” with “ρem ” = q ψ̂ † ψ̂ and “j em ” = q ψ̂ † αψ̂.
Clearly, we can identify the electromagnetic current operator, which looks very similar to the symmetry current discussed previously
µ
ĵem
= q ψ̄ˆγ µ ψ̂ = q N̂ µ .
(52)
Indeed, it is not surprising to obtain the same symmetry current, since we are discussing the same
gauge invariance – previously, we did not specify the form of the transformation (local or global).
Therefore we find
µ
L̂INT = −V̂D = −ĵem
µ .
(53)
Since we know that Lagrangians contain both kinetic and potential terms, we might wonder if
there is a kinetic term associated with the µ . Indeed, something must be missing, since nothing in
the argument so far constrains the form of µ . However, we expect µ to represent photons, and
therefore we should expect to be able to derive Maxwell’s equations from the full Lagrangian. It
can be shown that this can be achieved by adding a kinetic term to the Lagrangian density: L̂em =
− 14 F̂µν F̂ µν , where the quantum field theory version of the field strength tensor F̂ µν = ∂ µ Âν − ∂ ν µ .
Q19 Consider adding a kinetic term to the Lagrangian density so that L̂ = L̂D + L̂INT + T̂ . What
constraints on T̂ do the Euler-Lagrange equations for µ give?
Q20 Show that the choice T̂ = L̂em = − 14 F̂µν F̂ µν yields Maxwell’s equations.
It is worth pointing out that the Lagrangian we have obtained is the full Lagrangian of quantum
electrodynamics (QED) for the interactions of fermions with photons, i.e. we have everything in
place to make a full quantum field theoretic treatment of electromagnetism. This theory, which
has been tested with quite exquisite precision, resulted in the 1965 Nobel Prize being awarded to
Tomonaga, Schwinger and Feynman.
with
L̂QED = L̂D + L̂INT + L̂em
(54)
L̂D = ψ̄ˆ (iγ µ ∂µ − m) ψ̂ ,
L̂INT = −q ψ̄ˆγ µ ψ̂ µ ,
L̂em = − 14 F̂µν F̂ µν .
(55)
Let us briefly return to the Klein-Gordon situation. In that case, the electromagnetic current is
given by
∂ L̂INT
µ
ĵem
=−
(56)
= iq φ̂† (∂ µ φ̂) − (∂ µ φ̂† )φ̂ − 2q 2 µ φ̂† φ̂ .
∂ µ
This has a more complicated form, and cannot be interpreted as simply eN̂ µ as for the Dirac case,
due to the presence of the q 2 term. However, recall that we have freedom to choose the gauge, and
can choose Â0 = 0, in which case the symmetry operator is unaffected by the additional term. The
full scalar QED Lagrangian is completed by adding the L̂em term as before.
Fourier Expansion of the Electromagnetic Field
There is one more piece of the jigsaw that we ought to fill in before progressing, namely the
Fourier expansion of µ . This is not entirely straightforward, since the four components of the
vector field are not independent. We therefore have to figure out how to quantise a constrained
system. This is an interesting topic, but one that we will skip – the interested reader can find details
in a good textbook. If will suffice for us to proceed by choosing a gauge, which is conventionally done
with the Coulomb gauge, which is the Lorenz gauge, ∂µ µ = 0, complemented with the requirement
that Â0 = 0, so that ∇. = 0. Then we have two independent polarisations each of which can be
quantised like massless real Klein-Gordon fields as in Eq. (5),
Z
i
d3 k X h
∗µ +ik.x
†
µ −ik.x
µ
,
(57)
e
 =
(k)
e
+
α̂
α̂
(k)
λ
λ
λ
λ
(2π)3 2ω
λ
where µλ are the polarisation vectors and α̂λ (k) and α̂λ† (k) are annihilation and creation operators, respectively. The Coulomb gauge imposes 0 = .k = 0, so there are only two independent
0
polarisations, that must satisfy µλ ∗µ
λ0 = −δλλ .
Dimensionality
The Lagrangian of Eq. (54) is the most general Lagrangian involving a fermion field that respects
both local gauge invariance and Lorentz invariance. (Specifically, the relevant gauge invariance is that
under U(1) symmetry; we will explain what this means a bit more when looking at more complicated
gauge symmetries later on.) An interesting way of demonstrating this is with dimensional analysis.
Since we are using natural units, we can put everything in
of mass. The Lagrangian
R dimensions
4
4
density must have dimensions of [M ] , since the action S = Ld x must be dimensionless. (If this is
not obvious, note that in the so-called path integral formulation of quantum field theory, the action
appears in an exponent, eiS which weights the contribution of each path to the total amplitude.)
3
Thus, from Eq. (55) we can see that [ψ̂] = [M ] 2 , [µ ] = [M ], and no other terms are possible. An
3
alternative way of seeing that [ψ̂] = [M ] 2 is to note that the symmetry operator of Eq. (38) should
be dimensionless.
Q21 Apply dimensional analysis to L̂KG and V̂KG . What are the dimensions of φ̂? Do you get a
consistent answer by considering the Klein-Gordon symmetry operator? Are any additional
terms possible in the Lagrangian density for scalar QED?