March 2016
PX430: Gauge Theories for Particle Physics
Tim Gershon
(T.J.Gershon@warwick.ac.uk)
Selected worked answers
Handout 1: Revision & Notation
Q2 For a classical particle in a potential V = V (qi ), write down the kinetic energy T , and show
that the solution of the Euler-Lagrange equations gives Newton’s law, F = ma.
Q3 Consider further a particle in an electromagnetic field, with kinetic energy given by T =
2
1
2m (p − qA) . Solve the Euler-Lagrange equations for this case.
−→ T = mv 2 /2 = m (q˙i .q˙i ) /2, so
d
dt
∂L
∂ q̇i
−
∂L
d
∂V
=
(mq˙i ) −
=0
∂qi
dt
∂qi
and so
m
Since F =
∂V
∂qi ,
d2 qi
∂V
=
2
dt
∂qi
this is F = ma.
For the particle in an electromagnetic field, instead of getting
∂ (p − qA)
= ∇V
∂t
or
∂p
∂t
= ∇V we get
∂p
∂A
= ∇V + q
.
∂t
∂t
Q6 Write Eq. (13) in full, expanding the Lorentz invariant term into
−→
∂µ
∂L
∂(∂µ ψ)
∂L
∂
=
−
∂ψ
∂t
∂L
∂(∂ψ/∂t)
−∇
∂L
∂(∇ψ)
∂
∂t ,
−
∇, etc.
∂L
= 0.
∂ψ
ν implies all of Maxwell’s equations as well as the continuity equation.
Q9 Show that ∂µ F µν = jem
Q10 Show that F µν is antisymmetric. What are the elements of F µν ?
−→ These questions are addressed in the selected worked answers to questions in the handouts for
the Relativistic Quantum Mechanics module.
Q11 Satisfy yourself that both B and E are unaffected by the gauge transformation Aµ 7→ A0µ =
Aµ − ∂ µ χ
Q12 Show the the field strength tensor F µν is unaffected by the gauge transformation Aµ 7→ A0µ =
Aµ − ∂ µ χ
−→
B = ∇ × A 7→ B 0 = ∇ × A − ∇ × ∇χ = ∇ × A = B
∂A
∂ (A − ∇χ)
∂χ
∂A
E = −∇V −
−
7→ E 0 = −∇ V −
= −∇V −
=E
∂t
∂t
∂t
∂t
F µν = ∂ µ Aν − ∂ ν Aµ 7→ F µν 0 = ∂ µ (Aν − ∂ ν χ) − ∂ ν (Aµ − ∂ µ χ) = ∂ µ Aν − ∂ ν Aµ = F µν
Q13 Suppose that ∂µ Aµ = f 6= 0. Show that a gauge transformation can be applied to change to
A0 where ∂µ A0µ = 0. What is the form of the transformation (in terms of f )?
−→ Make a gauge transformation: Aµ 7→ A0µ = Aµ − ∂ µ χ. Now,
∂µ A0µ = ∂µ Aµ − ∂µ ∂ µ χ = f − ∂µ ∂ µ χ
So to satisfy ∂µ A0µ = 0 we require ∂µ ∂ µ χ = f .
Q15 The Coulomb gauge specifies ∇.A = 0. If the Coulomb gauge is satisfied, find an expression
for the Lorenz condition. How many independent components does Aµ have?
−→ If ∇.A = 0, then ∂µ Aµ = 0 only if ∂0 A0 = 0, which is satisfied with A0 = 0. There are only
two independent components of Aµ , corresponding to the two polarisation states of a massless vector
field.
∂
∂
Q17 Prove (−i∇ − qA0 )ψ 0 = eiqχ (−i∇ − qA)ψ and i ∂t
− qV 0 ψ 0 = eiqχ i ∂t
− qV ψ
−→
(−i∇ − qA0 )ψ 0 = (−i∇ − qA − q(∇χ))(eiqχ ψ)
The only non-trivial term is the first one, where we have to differentiate by parts. This results in
a term involving ∇χ which cancels with that from the transform of A, so that the result can be
rearranged:
−ieiqχ ∇ψ − i(iq∇χ)eiqχ ψ − qAeiqχ ψ − q(∇χ)eiqχ ψ = eiqχ (−i∇ − qA)ψ
For the time-like derivative, the working is just the same.
Handout 2: An Introduction to Quantum Field Theory
Q1 Rearrange Eqs. (4) to write q̂ and p̂ in terms of â and ↠.
−→
1 √
i
1 √
i
â = √ ( mω q̂ + √
p̂)
↠= √ ( mω q̂ − √
p̂)
mω
mω
2
2
√
√
√
so q̂ = â + ↠/ 2mω and p̂ = i mω ↠− â / 2.
Q3 Explicitly derive Eqs. (7) and (8).
−→ Using the results above we get
Ĥ =
2 1
2 1 1 2 1
1 1
mω †
p̂ + mω 2 q̂ 2 =
(−1)
â − â + mω 2
â + ↠=
↠â + â↠ω
2m
2
2m
2
2
2mω
2
Using in addition [â, ↠] = 1 this can be rewritten as (↠â + 12 )ω.
1
1
[Ĥ, â] = Ĥâ − âĤ = (↠â + )ωâ − â(↠â + )ω
2
2
†
move â through to the left on the first term, using [â, â ] = 1, we pick up an extra term of −ωâ that
is all that is left when everything else cancels, so [Ĥ, â] = −ωâ.
The working for [Ĥ, ↠] is just the same.
Q4 Choosing a normalisation such that hn|ni = 1, show that the state |ni is given by
1
|ni = √ (↠)n |0i
n!
n.b. It will help to know that, e.g. (↠|0i)† = h0|â.
−→ Take n = 2 for example. Consider unnormalised states |ñi = (↠)n |0i and we will work out the
necessary normalisation factor.
h2̃|2̃i = h0|(â)2 (↠)2 |0i = h0|â ↠â + 1 ↠|0i = h0|â↠â↠|0i + h0|â↠|0i
The second term gives
h0|â↠|0i = h0| ↠â + 1 |0i = 1
since h0|↠= â|0i = 0 and h0|0i = 1. The first term gives
h0| ↠â + 1 ↠â + 1 |0i = 1
so with this normalisation
h2̃|2̃i = 2
i.e. we need to define |2i = √12 (↠)2 |0i in order to have proper normalisation.
To extend to the general case one needs to count how many commutations are necessary.
Q5 What is [Ĥ, n̂]? What is the physical meaning of this result?
−→
1
1
[Ĥ, n̂] = (↠â + )ω↠â − ↠â(↠â + )ω = 0
2
2
(Commutation relations not required here – the terms just cancel.) The number operator commutes
with the Hamiltonian, therefore it is a conserved quantity. Quanta of oscillation cannot be created
or annihilated in a simple harmonic system.
Q6 Satisfy yourself of the derivation of:
N X
1 2 1
2 2
Ĥ =
p̂ + mωr q̂r
2m r 2
=
=
r=1
N
X
r=1
N
X
r=1
1
(â†r âr + )ωr
2
1
(n̂r + )ωr
2
(1)
What are the various commutations relations between â†r , âr , n̂r and Ĥ?
−→ Starting from the Lagrangian,
L̂ =
N X
1
r=1
2
mr q̂˙r2
1
− mr ωr2 q̂r2
2
the Hamiltonian is obtained as
Ĥ =
N
X
p̂r q̂˙r − L̂ =
r=1
where
p̂r =
so that
Ĥ =
∂ L̂
= mq̂˙r
∂ q̂˙r
N
X
1 2 1
p̂ + mωr2 q̂r2
2m r 2
r=1
With definitions of
i
1 √
p̂r )
âr = √ ( mω q̂r + √
mω
2
1 √
i
â†r = √ ( mω qˆr − √
p̂r )
mω
2
we find
1
1 2 i
âr â†r = mω q̂r2 +
p̂ + [p̂r , qˆr ]
2
2mω r 2
1
1 2 i
â†r âr = mω q̂r2 +
p̂ − [p̂r , q̂r ]
2
2mω r 2
so that [âr , â†r ] = i[p̂r , q̂r ] = 1 (in fact, [âr , â†s ] = δrs , while [âr , âs ] = [â†r , â†s ] = 0) and
Ĥ =
N
X
1
r=1
2
(â†r âr
+
âr â†r )ωr
N
X
1
=
(â†r âr + )ωr
2
r=1
Using the number operator n̂r = â†r âr (this question does not ask us to show that this operator is
indeed the number operator, though that is a simple extension), we finally obtain
Ĥ =
N
X
r=1
1
(n̂r + )ωr
2
To obtain the commutation relations, the key point is that the oscillators of each mode are
independent (as can be seen by the commutation relations given above). So, for example
[Ĥ, âs ] = Ĥâs − âs Ĥ
N X
1
1
†
†
=
(âr âr + )ωr âs − âs (âr âr + )ωr
2
2
=
=
r=1
N
X
r=1
N
X
(2)
(3)
(â†r âr âs − (â†r âs + δrs )âr )ωr
(4)
δrs âr ωr
(5)
r=1
= −ωs âs
Similarly
[Ĥ, â†s ]
=
ωs â†s
(6)
(7)
To obtain the commutator [Ĥ, n̂s ] you can use the handy relation [a, bc] = [a, b]c + b[a, c], so with
n̂s = â†s âs we find
[Ĥ, n̂s ] = [Ĥ, â†s ]âs + â†s [Ĥ, âs ] = ωs â†s âs − â†s ωs âs = 0
Therefore the number operator is a constant of motion for this system.
Q8 What is the zero point energy of this system?
−→ We are discussing the system with Hamiltonian
Ĥ =
N
X
r=1
1
(n̂r + )ωr
2
At the zero-point, n̂r |0i = 0 ∀r, so
N
Ĥ|0i =
1X
ωr
2
r=1
Q13 Prove
Z
+∞
Z
+∞
dxĤ =
Ĥ =
−∞
Z
−∞
+∞
=
−∞
dk
(2π)(2ω)

1
1
dx  Π̂2 +
2
2
∂ φ̂
∂x
!2 

1 †
†
â (k)â(k) + â(k)â (k) ω
2
(8)
−→ This involves quite a lot of maths, but here goes. We start with the Fourier expansions for φ̂,
∂ φ̂
∂x and Π̂
Z +∞
i
h
dk
(9)
φ̂(x, t) =
â(k)eikx−iωt + ↠(k)e−ikx+iωt
−∞ (2π)(2ω)
Z +∞
i
h
∂ φ̂
dk
(10)
(x, t) =
(ik) â(k)eikx−iωt − ↠(k)e−ikx+iωt
∂x
−∞ (2π)(2ω)
Z +∞
i
h
dk
Π̂(x, t) =
(11)
(−iω) â(k)eikx−iωt − ↠(k)e−ikx+iωt
−∞ (2π)(2ω)
We square both Π̂ and ∂∂xφ̂ in their Fourier expanded forms by performing integrals over different
dummy variables (k and k 0 ), to obtain

!2 
Z +∞
1
1
∂
φ̂

Ĥ =
dx  Π̂2 +
(12)
2
2 ∂x
−∞
Z
Z +∞
Z +∞
dk
dk 0
−1 +∞
=
dx
(ωω 0 + kk 0 ) ×
(13)
0)
2 −∞
(2π)(2ω)
(2π)(2ω
−∞
−∞
n
on
o
0
0
0
0
â(k)eikx−iωt − ↠(k)e−ikx+iωt
â(k 0 )eik x−iω t − ↠(k 0 )e−ik x+iω t
When we multiply this out, we will get four terms that involve â(k)â(k 0 ), ↠(k)↠(k 0 ), â(k)↠(k 0 )
and ↠(k)â(k 0 ).
0
Now, integrating over x first of all, allows us to reduce expressions like ei(k+k )x to delta functions
((2π)δ(k + k 0 ) in this case). So we find
Z
Z +∞
dk
dk 0
−1 +∞
(ωω 0 + kk 0 ) ×
(14)
Ĥ =
2 −∞ (2π)(2ω) −∞ (2π)(2ω 0 )
0
â(k)â(k 0 )(2π)δ(k + k 0 )e−i(ω+ω )t +
(15)
0
↠(k)↠(k 0 )(2π)δ(−k − k 0 )ei(ω+ω )t −
†
0
0
â(k)â (k )(2π)δ(k − k )e
†
0
0
−i(ω−ω 0 )t
â (k)â(k )(2π)δ(−k + k )e
i(ω−ω 0 )t
−
(16)
(17)
(18)
We now take advantage of the delta functions to do the integral over k 0 . In the first two terms, the
effect is to set k 0 = −k, in the latter two, the effect is to set k 0 = k. Note that ω(k) = ω(−k), so that
ω 0 = ω in both cases.
Z
−1 +∞
dk
Ĥ =
×
(19)
2 −∞ (2π)(2ω)2
(ω 2 − k 2 ) â(k)â(−k)e−2iωt + ↠(k)↠(−k)e2iωt −
(20)
(ω 2 + k 2 ) â(k)↠(k) + ↠(k)â(k)
(21)
Now it just remains to assert that this is a massless quantum field. This was not made explicit
in the question, but is indeed the case: the Klein-Gordon Hamiltonian density for a massive system
includes an additional term 21 m2 φ̂2 – if this term is included in the working above we find the mass
appears in exactly the right place to cancel out, leaving the same result. Anyway, for a massless
field, ω 2 − k 2 = 0, so we are left with
Z
dk
1 +∞
↠(k)â(k) + â(k)↠(k) ω
(22)
Ĥ =
2 −∞ (2π)(2ω)
Q15 What is [Ĥ, n̂(k)]? What is [:Ĥ:, n̂(k)]?
Q16 Defining the total number operator n̂tot =
R +∞
dk
−∞ (2π)(2ω) n̂(k),
what is [Ĥ, n̂tot ]?
−→ Let’s consider first of all the normal ordered form.
Z +∞
Z +∞
dk 0
dk 0
0
0
n̂(k )ω n̂(k) − n̂(k)
n̂(k )ω
[:Ĥ:, n̂(k)] =
0
0
−∞ (2π)(2ω )
−∞ (2π)(2ω )
Z +∞
dk 0
[:Ĥ:, n̂(k)] =
[n̂(k 0 ), n̂(k)]ω
0)
(2π)(2ω
−∞
Since [n̂(k 0 ), n̂(k)] = 0 (which follows since the commutation relations of the â and ↠operators are
only non-vanishing for k = k 0 , and n̂(k) obviously commutes with itself), we see that [:Ĥ:, n̂(k)] = 0.
Without normal ordering, Ĥ includes an additional (zero-point energy) term which is, however,
constant, and therefore commutes with any operator. Therefore [Ĥ, n̂(k)] = 0.
The total number operator is just a sum (or integral) of number operators for individual modes,
each of which commutes with the Hamiltonian. Therefore [Ĥ, n̂tot ] = 0.
Q17 Using the Fourier expansion for φ̂, and the commutation relations for â(k) and ↠(k), show
that h0|φ̂|ki = h0|φ̂↠(k)|0i = eikx−iωt
−→ The Fourier expansion for φ̂ is
Z +∞
φ̂(x, t) =
−∞
n
o
dk 0
0
0
â(k 0 )eik x−iωt + ↠(k 0 )e−ik x+iωt
(2π)(2ω)
(using curly brackets and the dummy variable k 0 to try to reduce possible confusion). The required
commutation relations are [â(k), ↠(k 0 )] = (2π)(2ω)δ(k − k 0 ) and [â(k), â(k 0 )] = [↠(k), ↠(k 0 )] = 0.
We wish to calculate h0|φ̂|ki. We start with |ki = ↠(k)|0i (↠(k) creates a quanta of mode k).
Thus
Z +∞
n
o
dk 0
0
0
(23)
â(k 0 )eik x−iωt + ↠(k 0 )e−ik x+iωt ↠(k)|0i
h0|φ̂|ki = h0|
−∞ (2π)(2ω)
Z +∞
n
o
dk 0
0
0
â(k 0 )↠(k)eik x−iωt + ↠(k 0 )↠(k)e−ik x+iωt |0i
(24)
= h0|
−∞ (2π)(2ω)
Since [↠(k), ↠(k 0 )] = 0 we can reverse the order of the operators in the second of the terms inside
the integral, and then bring ↠(k) out of the integral to the left. This leaves us with (for this part),
something that looks like h0|↠(k), which vanishes. We are left with the first term inside the integral,
on which we can use the commutation relation [â(k), ↠(k 0 )] = (2π)(2ω)δ(k − k 0 ):
Z +∞
n
0
o
dk 0
(25)
h0|φ̂|ki = h0|
↠(k)â(k 0 ) + (2π)(2ω)δ(k − k 0 ) eik x−iωt |0i
−∞ (2π)(2ω)
Now, ↠(k) can come out of the left of this integral, so that the first part will now vanish, leaving
only
Z +∞
dk 0
0
(2π)(2ω)δ(k − k 0 )eik x−iωt |0i
(26)
h0|φ̂|ki = h0|
−∞ (2π)(2ω)
= eikx−iωt
(27)
Handout 3: Gauge Invariance in Quantum Field Theory
Q1 Show that application of the Euler-Lagrange equations to the Lagrangian density of Eq. (4)
yields the free-field Klein-Gordon equation.
−→
∂LKG
∂ φ̂
so

= −m2 φ̂
∂LKG
= ∂ µ φ̂
∂ ∂µ φ̂

∂LKG 
= −m2 φ̂ − ∂µ ∂ µ φ̂ = 0
− ∂µ  ∂ φ̂
∂ ∂µ φ̂
∂LKG
⇒
∂µ ∂ µ + m2 φ̂ = 0
i.e. the free-field Klein-Gordon equation.
h
i
Q3 For Ĥint = λq̂ 3 , what is Ĥint , n̂(k) ?
−→ [n̂, â] = −â and [n̂, ↠] = ↠so [n̂, â + ↠] = −â + ↠. Then
3 λ
3
†
â + â
, n̂
[Ĥint , n̂] = [λq̂ , n̂] =
(2mω)3/2
Use the commutation relation
to rewrite the second term.
To simplify the notation, the following
uses a+ to denote â + ↠and a− to denote −â + ↠. Then n̂a+ = a+ n̂ + a− .
n̂(a+ )3 = (a+ n̂ + a− ) (a+ )2
= a+ (a+ n̂ + a− ) a+ + a− (a+ )2
= (a+ )2 (a+ n̂ + a− ) + a+ a− a+ + a− (a+ )2
= (a+ )3 n̂ + (a+ )2 a− + a+ a− a+ + a− (a+ )2
so the commutator
(a+ )3 , n̂ = −(a+ )2 a− − a+ a− a+ − a− (a+ )2
Now we need the commutator [a+ , a− ]
[a+ , a− ] = â + ↠−â + ↠− −â + ↠â + ↠= 2 â↠− ↠â = 2
so after a bit more rearranging, we find
2
(a+ )3 , n̂ = −6a+ − 3a− (a+ )2 = −6 â + ↠− 3 = â + ↠â + â†
and so finally
[Ĥint , n̂] = [λq̂ 3 , n̂] =
3λ
−q̂ + ip̂q̂ 2
mω
Q6 Show explicitly that Eq. (10) can be rearranged to give Eq. (12).
−→ The trick is to replace +1 with −i2 so that
1
1 µ
2
µ
2 2
21 2 2
L̂ =
∂µ φ̂R ∂ φ̂R − i ∂µ φ̂I ∂ φ̂I −
m φ̂R − i m φ̂I
2
2
2
=
∂µ φ̂† ∂ µ φ̂ − m2 φ̂† φ̂
(28)
with φ̂ and φ̂† defined as in Eq. (11).
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 )
−→ The deriviation of this result follows exactly the steps taken for a Lagrangian depending on a
single field. Invariance of the Lagrangian gives us


X
 ∂ L̂ δ ∂µ φ̂i + ∂ L̂ δ φ̂i 
0 = δ L̂ =
∂ φ̂i
∂ ∂µ φ̂i
i
where all the fields are taken to be independent, and real (it is straightforward to further generalise
this result for complex fields). Using the Euler-Lagrange equation, this reduces to


X
∂
L̂
 δ φ̂i 
0 = δ L̂ = ∂µ
∂
∂
φ̂
µ i
i
P
Hence there is a conserved Noether current Jˆµ = i
∂ L̂
δ φ̂i .
∂(∂µ φ̂i )
Q11 Using the Fourier expansion of the fields given by Eq. (21), show that N̂ =
given by Eq. (29).
−→ The Fourier transforms are
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)
so
∂ φ̂
∂t
∂ φ̂†
∂t
Z
+∞
=
−∞
+∞
Z
=
−∞
i
d3 k0 (−iω 0 ) h
0 −ik0 .x
† 0 ik0 .x
â(k
)e
−
b̂
(k
)e
(2π)3 (2ω 0 )
i
d3 k0 (−iω 0 ) h 0 −ik0 .x
† 0 ik0 .x
b̂(k
)e
−
â
(k
)e
(2π)3 (2ω 0 )
R +∞
−∞
N̂ 0 d3 x is as
and
!
†
∂
φ̂
∂
φ̂
=
N̂ 0 d3 x =
i φ̂†
d3 x
− φ̂
∂t
∂t
−∞
−∞
Z +∞
d3 k
d3 k0 (−iω 0 ) †
0
0
id3 x
=
â (k)eik.x â(k0 )e−ik .x − b̂(k)e−ik.x b̂† (k0 )eik .x +
3 (2ω) (2π)3 (2ω 0 )
(2π)
−∞
Z
N̂
+∞
Z
+∞
0
0
â(k)e−ik.x ↠(k0 )eik .x − b̂† (k)eik.x b̂(k0 )e−ik .x
where we have taken advantage of the fact that only the ↠â and b̂† b̂ terms will give non-zero
contributions. We can use the commutation relations of these terms to move all daggered operators
to the left – in doing so we pick up a factor of 2, as well as delta functions of k − k0 which cancel
since the a terms and the b terms have opposite sign. If we do the integration over x, we get threedimensional delta functions of k − k0 . These can then be exploited to do the integration over k0 .
Finally, we are left with
Z +∞
d3 k
†
†
â
(k)â(k)
−
b̂
(k)
b̂(k)
N̂ =
3
−∞ (2π) (2ω)
as stated in Eq. (29).
ˆ?
Q14 What are the Fourier expansions for Π̂ and Π̄
−→ The generalized momentum densities are
Π̂ =
∂ L̂D
= ψ̄ˆiγ 0
˙
∂ ψ̂
ˆ = ∂ L̂D = 0
Π̄
˙
∂ ψ̄ˆ
and the Fourier expansions for ψ̂ and ψ̄ˆ have been 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
3
s
(2π) (2ω)
R
P †
ˆ
d3 k
ψ̄ =
ĉs (k)us (k)e−ik.x + dˆs (k)vs (k)e+ik.x
3
(2π) (2ω)
(29)
s
So, it is fairly straightforward to put these together to obtain
R
P †
d3 k
0 e−ik.x + dˆ (k)v (k)γ 0 e+ik.x
Π̂ =
i
ĉ
(k)u
(k)γ
s
s
s
s
s
(2π)3 (2ω)
ˆ
Π̄ = 0
(30)
Note that us and vs are the spinors that are acted upon by γ 0 .
Q15 Show that that Noether current associated with the phase invariance of Eq. (30) is given by
N̂ µ = ψ̄ˆγ µ ψ̂.
−→ As with Eq. (17) for the Klein-Gordon case


∂
L̂
∂
L̂
δ ψ̂ + δ ψ̄ˆ 
0 = δ L̂ = ∂µ  ˆ
∂ ∂µ ψ̂
∂ ∂µ ψ̄
This simplifies since ∂ L̂ ˆ = 0. As given in Eq. (37), δ ψ̂ = −iψ̂ so
∂ ∂µ ψ̄
0 = δ L̂ = ∂µ ψ̄ˆiγ µ (−iψ̂) = ∂µ N̂ µ
where N̂ µ = ψ̄ˆγ µ ψ̂.
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.
−→ The Euler-Lagrange equations for µ are
∂ L̂
∂ µ

∂ L̂

 = 0
− ∂µ  ∂ ∂µ µ
µ
µ this gives
With L̂D = ψ̄ˆ (iγ µ ∂µ − m) ψ̂ and L̂INT = −ĵem


∂
T̂
∂
T̂
µ
 = 0
−ĵem
+
− ∂µ  ∂ µ
∂ ∂µ µ
With the choice T̂ = − 14 F̂µν F̂ µν ,
∂ T̂
∂ µ

= 0 and

∂ T̂
 = −∂µ F̂ µν
∂µ  ∂ ∂µ µ
ν , i.e. Maxwell’s equations.
so, we obtain ∂µ F̂ µν = ĵem