Quantum Field Theory
Example Sheet 2
Michelmas Term 2011
Solutions by:
Johannes Hofmann
Laurence Perreault Levasseur
Dave M. Morris
Marcel Schmittfull
[email protected]
[email protected]
[email protected]
[email protected]
Exercise 1
We are given the classical Hamiltonian for a string
H=
∞ X
1 2 1 2 2
pn + ω n q n
2
2
n=0
(1)
Upon quantization, the pn and qn are promoted to Hermitian operators, and we impose the canonical commutation
relations upon them
[qn , qm ] = [pn , pm ] = 0
, [qn , pm ] = iδnm .
We then introduce creation and annihilation operators a†n and an as a decomposition of pn and qn
r
ωn †
1
†
an − an
an + an
, pn = i
qn = √
2
2ωn
(2)
Or, inverting these,
r
an =
ωn
i
pn
qn + √
2
2ωn
,
a†n
r
=
ωn
i
pn
qn − √
2
2ωn
Compute
r
[an , am ]
=
=
=
r
ωn
ωn
i
i
qn + √
qm + √
pn ,
pm
2
2
2ωn
2ωm
r
r
√
ωn ωm
i ωm
i ωn
1
[qn , qm ] +
[pn , qm ] +
[qn , pm ] − √
[pn , pm ]
2
2 ωn
2 ωm
2 ωn ωm
i
i
+ (−iδnm ) + iδnm
2
2
=0.
Taking the adjoint of this latter commutator, we find [a†n , a†m ] = 0.
The remaining commutator is
[an , a†m ]
r
r
ωn
i
ωn
i
qn + √
pn ,
qm − √
pm
2
2
2ωn
2ωn
r
r
i ωm
−i ωn
[qn , pm ] +
[pn , qm ]
=
2
ωm
2 ωn
r
r
−i ωn
i ωm
=
(iδmn ) +
(−iδnm )
2
ωm
2 ωn
= δmn
=
1
(3)
Now, using (2), and plugging into the expression for the Hamiltonian, we get
r
∞ r
X
1
ω2
ωn †
ωn †
1
1
√
H =
an + a†n
an + a†n
i
an − an
an − an + n √
i
2
2
2
2
2ωn
2ωn
n=1
=
∞ h
X
i
ωn
ωn † †
−
an an − an a†n − a†n an + an an +
an an + a†n an + an a†n + a†n a†n
4
4
n=1
=
∞
X
ωn an a†n + a†n an
2
n=1
as required.
Suppose now that there is a ground state |0i satisfying ân |0i = 0 for every n. Using the commutation relation (3), we
can rewrite the Hamiltonian in normal order as follow:
H=
∞
∞
∞
X
X
X
ωn ωn
2an a†n + δnn =
δnn .
ωn an a†n +
2
2
n=1
n=1
n=1
The last, infinite term is due to the vacuum energy. Indeed, we find that the vacuum expectation value is given by
∞
X
ωn
δnn .
2
n=1
h0|H|0i =
Physical experiments are typically only sensitive to differences of energy, not absolute magnitudes. This justifies us
therefore in neglecting the vacuum energy. Performing this subtraction, we recover the normal ordered prescription
for the Hamiltonian
∞
X
: H :=
= ωn a†n an .
n=1
l
From this, we calculate the energy of the states |(n, l)i = a†n |0i containing l particles in the n-th mode. First, we
may prove by induction that
[H, (a†n )l ] = l ωn (a†n )l , l ≥ 1.
(4)
For l = 1, this is a straightforward application of the commutation relations (3)
[H, a†n ] =
∞
X
ωm a†m [am , a†n ] =
m=1
∞
X
ωm a†m δnm = ωn a†n .
m=1
Supposing now that (4) is valid for l ≤ k, we find
[H, (a†n )k+1 ]
[H, (a†n )k ]a†n + (a†n )k Ha†n − (a†n )k+1 H
k ωn (a†n )k+1 + (a†n )k [H, a†n ]
k ωn (a†n )k+1 + ωn (a†n )k+1
(k + 1) ωn (a†n )k+1 ,
=
=
=
=
as required. Having removed the vacuum energy, we now have H|0i = 0, so the energies of the states |(n, l)i are,
H|(n, l)i = H(a†n )l − (a†n )l H |0i = l ωn (a†n )l |0i = l ωn |(n, l)i
(5)
Using the identity (5), we can now move on to compute the energy of the fock states of the form
l 1 l2 l N
|l1 , l2 , ..., lN i =
a†1
a†2
... a†N
|0i.
Act on these with H,
H|l1 , l2 , . . . , lN i =
[H, (a†1 )l1 ] + (a†1 )l1 H |l2 , l2 , . . . , ln i
= l1 ω1 |l1 , l2 , . . . , lN i + (a†n )l1 (H|l2 , l3 , . . . , lN i) .
This latter relation may be applied itteratively, to find
!
N
X
H|l1 , l2 , . . . lN i =
li ωi |l1 , l2 , . . . , lN i +
N
Y
i=1
i.e. the energy of the multi-particle state |l1 , l2 , . . . , lN i is
!
(a†i )li
i=1
PN
i=1 li
2
H|0i =
N
X
i=1
ωi .
!
li ωi
|l1 , l2 , . . . , lN i,
Exercise 2
Given the real-space expressions and commutation relations for φ(x) and π(x), we derive the momentum-space commutation relations for a†p and ap . Hence we take the mode expansions in the Schrödinger picture
Z
φ(x)
=
π(x)
=
Z
and transform into Fourier space
Z
φ(k) = d3 x e−ik·x φ(x)
1
d3 p
p
ap eip·x + a†p e−ip·x
3
(2π)
2Ep
(6)
−i
d3 p
p
ap eip·x − a†p e−ip·x
3
(2π)
2Ep
(7)
1
d3 p
p
(2π)3 2Ep
Z
=
Z
ap
d3 x ei(p−k)·x + a†p
Z
d3 x e−i(p+k)·x
1 d3 p
3 (3)
†
3 (3)
p
a
(2π)
δ
(k
−
p)
+
a
(2π)
δ
(p
+
k)
p
p
(2π)3 2Ep
1 √
ak + a†−k
2Ek
r
Z
Z
Z
d3 p
Ep
3
i(p−k)·x
†
3
−i(p+k)·x
−i
a
d
xe
−
a
d
x
e
p
p
(2π)3
2
r
Z
Ep d3 p
3 (3)
†
3 (3)
a
(2π)
δ
(k
−
p)
−
a
(2π)
δ
(p
+
k)
−i
p
p
(2π)3
2
r
Ek
−i
ak − a†−k ,
2
Z
=
=
Z
π(k) =
d3 x e−ik·x π(x)
=
=
=
where we use that Ek = E−k =
(7) to find
p
q2 + m2 . These relations may now be used to invert the mode expansions (6) and
r
ak
=
Ek
2
Z
=
Z
−ik·x
3
d xe
r
−ik·x
3
d xe
1
φ(x) + i √
2Ek
Z
d3 x e−ik·x π(x)
!
1
Ek
φ(x) + i √
π(x)
2
2Ek
We may take the adjoint of this to find
a†k
Z
=
r
d3 xe+ik·x
!
Ek
1
φ(x) − i √
π(x)
2
2Ek
Therefore we can compute
[ak ,
a†l ]
!
r
Ek
i El
d xe
d ye
[φ(x), π(y)] +
[π(x), φ(y)]
El
2 Ek
!
r
r
Z
Z
Ek (3)
El
3
−ik·x
3
il·y i
(3)
−
iδ (x − y) +
(−i)δ (x − y)
d xe
d ye
2
El
Ek
!
r
r
Z
Ek
El
3
−i(k−l)·x i
d xe
−i
+ (−i)
2
El
Ek
!
r
r
1
Ek
El
(2π)3 δ (3) (k − l)
+
2
El
Ek
Z
=
=
=
=
=
3
−ik·x
Z
3
il·y
−i
2
r
(2π)3 δ (3) (k − l) ;
3
!
r
Ek
i El
d xe
d ye
[φ(x), π(y)] +
[π(x), φ(y)]
El
2 Ek
!
r
r
Z
Z
Ek (3)
El
3
−il·y i
3
−ik·x
(3)
d ye
d xe
iδ (x − y) +
(−i)δ (x − y)
2
El
Ek
!
r
r
Z
Ek
El
3
−i(k−l)·x 1
d xe
−
+
2
El
Ek
!
r
r
Ek
El
1
3 (3)
(2π) δ (k − l)
+
2
El
Ek
Z
[ak , al ]
=
=
=
=
=
3
−ik·x
Z
3
−il·y
i
2
r
0.
The remaining relation [a†k , a†l ] = 0 follws simply by taking the adjoint of [ak , al ] = 0.
Exercise 3
Consider the Lagrangian for a free real scalar field φ,
L=
1
1
∂µ φ∂ µ φ − m2 φ2 .
2
2
The stress-energy tensor for this theory is (Peskin & Schroeder p.19)
T µν =
1
∂L
∂ ν φ − g µν L = ∂ µ φ ∂ ν φ −
∂α φ ∂ α φ − m2 φ2 g µν .
∂(∂µ φ)
2
The conserved four-momentum is therefore
Z
Z
δ 0ν 2
µ
3
0µ
3
ν
2
2 2
P = d x T = d x φ̇ ∂ φ −
φ̇ − (∇φ) − m φ
2
We examine the cases µ = 0, i separately. For µ = 0, write P 0 in terms of the mode expansions of the Heisenberg
picture operators
Z
1 2
P0 =
d3 x
φ̇ + (∇φ)2 + m2 φ2
2
Z
Z
Z
1
d3 k
d3 p
1
p
=
d3 x
3
2
(2π)
(2π)3 4Ek Ep
h
(−Ek Ep − p · k) ak e−ik·x − a†k eik·x ap e−ip·x − a†p eip·x
i
+m2 ak e−ik·x + a†k eik·x ap e−ip·x + a†p eip·x
Z
Z
1
d3 k
d3 p
1
p
=
3
3
2
(2π)
(2π)
4Ek Ep
Z
−Ek Ep − p · k + m2 a−k a−p e−i(Ek +Ep )t + a†k a†p ei(Ek +Ep )t
d3 x e−i(k+p)·x
Z
+ Ek Ep + p · k + m2 a†k ap ei(Ek −Ep )t + a−k a†−p e−i(Ek −Ep )t
d3 x e−i(k−p)·x
Z
Z
1
d3 k
d3 p
1
p
=
3
2
(2π)
(2π)3 4Ek Ep
h
−Ek Ep − p · k + m2 a−k a−p e−i(Ek +Ep )t + a†k a†p ei(Ek +Ep )t (2π)3 δ (3) (k + p)
i
+ Ek Ep − p · k + m2 a†k ap ei(Ek −Ep )t + a−k a†−p e−i(Ek −Ep )t (2π)3 δ (3) (k − p)
Z
1
d3 k 1 h
† †
2
2
2
−i2Ek t
i2Ek t
=
−E
+
|k|
+
m
a
a
e
+
a
a
e
−k
k
k
k −k
2
(2π)3 4Ek
i
+ Ek2 + |k|2 + m2 a†−k a−k + ak a†k
4
However, there is the on-shell condition
k · k = Ek2 − |k|2 = m2
⇒
−Ek2 + |k|2 + m2 = 0 and Ek2 + |k|2 + m2 = 2Ek2
The first term in P 0 therefore vanishes, leaving
Z
i
d3 k 1 h 2 †
†
0
2E
a
a
+
a
a
P
=
−k
k
k
−k
k
(2π)3 4Ek
Z
i
0 h
3
d k p
a†k ak + ak a†k ,
=
(2π)3 2
where we change variables −k → k in the first term, recalling that E−k = Ek . Normal ordering, we finally obtain
Z
d3 k 0 †
p ak ak .
: P0 : =
(2π)3
We now proceed to evaluate the spatial components of the conserved four-momentum
Z
h
i
i
P =
d3 x φ̇ (−∂i φ)
Z
Z
Z
h
i
d3 k
d3 p
−1
p
=
d3 x
−iEk ipi ak e−ik·x − a†k eik·x ap e−ip·x − a†p eip·x
3
3
(2π)
(2π) 2 Ek Ep
Z
Z
Z
d 3 p − E k pi d3 k
† † i(Ek +Ep )t
−i(Ek +Ep )t
p
d3 x e−i(k+p)·x
=
−a
a
e
+
a
a
e
−k −p
k p
(2π)3
(2π)3 2 Ek Ep
Z
†
†
i(Ek −Ep )t
−i(Ek −Ep )t
3
−i(k−p)·x
− ak ap e
d xe
− a−k a−p e
Z
Z
d3 p − Ek pi h † † i(Ek +Ep )t
d3 k
−i(Ek +Ep )t
p
a
=
a
e
−
a
a
e
(2π)3 δ (3) (k + p)
−k
−p
k p
(2π)3
(2π)3 2 Ek Ep
i
+ a−k a†−p e−i(Ek −Ep )t − a†k ap ei(Ek −Ep )t (2π)3 δ (3) (k − p)
Z
i
d3 k 1 h i † † i2Ek t
†
†
−i2Ek t
i
k
a
a
e
−
a
a
e
=
−
k
a
a
−
a
a
−k
k
−k
k
k −k
−k
k
(2π)3 2
Since E−k = Ek and ak a−k = a−k ak etc., we see that the first term in the final expression above,
k i a†k a†−k ei2Ek t − a−k ak e−i2Ek t ,
is an odd function of k, so that its
Z
i
P =
Z
=
Z
⇒
: Pi : =
3-momentum integral vanishes. This leaves
i
d3 k k i h †
†
a
a
−
a
a
k
−k
−k
(2π)3 2 k
i
3
i h
d k k
†
†
a
a
+
a
a
(sending k → −k in the second term)
k
k
k
(2π)3 2 k
d3 k i †
k ak ak
(2π)3
Putting everything together, we obtain the required identity for all space-time components,
Z
d3 k µ †
: P µ :=
k ak ak .
(2π)3
Finally, we can compute the following commutation relation
Z
Z
i
d3 k µ
d3 p
1 h †
µ
−ip·x
† ip·x
p
[: P :, φ(x)] =
k
a
a
,
a
e
+
a
e
k
p
p
(2π)3
(2π)3 2Ep k
Z
Z
d3 k µ
d3 p
1 †
p
=
k
[ak ak , ap ]e−ip·x + [a†k ak , a†p ]eip·x
3
3
(2π)
(2π)
2Ep
Z
Z
3
3
d k µ
d p
1 †
†
−ip·x
†
ip·x
p
=
k
[a
,
a
]a
e
+
a
[a
,
a
]e
p
k
k
p
k
k
(2π)3
(2π)3 2Ep
Z
d3 k k µ √
=
−ak e−ik·x + a†k eik·x
3
(2π)
2Ek
µ
= −i∂ φ(x)
5
Exercise 4
Begin with the classical Hamiltonian system with Hamiltonian function
Z
1
1 2
1
2
2
2
3
π(y, t) + |∇φ(y, t)| + m φ(y, t)
H =
d y
2
2
2
This system is to be quantized by supposing that φ(x) and π(x) are operators satisfying the canonical equal time
commutation relations in the Heisenberg picture,
[φ(x, t), φ(y, t)] = [π(x, t), π(y, t)] = 0 , [φ(x, t), π(y, t)] = iδ (3) (x − y).
(8)
In the Heisenberg picture, all operators O are to satisfy the Heisenberg equations of motion
∂
O
∂t
= i[H, O].
Thus, the time derivative of φ is given by
φ̇(x)
= i[H, φ(x)]
Z
1
1
1
= i d3 y
[π(y, t)2 , φ(x, t)] + [|∇φ(y, t)|2 , φ((x), t)] + m2 [φ(y, t)2 , φ(x, t)]
2
2
2
Now, with the use of (8),
[π(y, t)2 , φ(x, t)]
= π(y, t)[π(y, t), φ(x, t)] + [π(y, t)2 , φ(x, t)]π(y, t)
= −2iπ(y, t)δ 3 (x − y),
and
[|∇φ(y, t)|2 , φ((x), t)]
= ∇φ(y, t) · [∇φ(y, t), φ(x, t)] + [∇φ(y, t), φ(x, t)] · ∇φ(y, t)
∂
∂
∂φ
∂φ
(y, t)
[φ(y, t), φ(x, t)] +
[φ(y, t), φ(x, t)]
(y, t)
=
∂yi
∂yi
∂yi
∂yi
= 0
and finally,
[φ(y, t)2 , φ(x, t)] = 0.
Altogether then,
= −i2
φ̇(x)
Z
d3 y π(y, t)δ (3) (x − y) = π(x, t).
Now for the equation of motion for π,
π̇(x) = i[H, φ(x)]
Z
1
1
1 2
3
2
2
2
=i d y
[π(y, t) , π(x, t)] + [|∇φ(y, t)| , π((x), t)] + m [φ(y, t) , π(x, t)]
2
2
2
This time round we have,
[π(y, t)2 , π(x, t)] = 0,
and
[|∇φ(y, t)|2 , π((x), t)]
= ∇φ(y, t) · [∇π(y, t), φ(x, t)] + [∇φ(y, t), φ(x, t)] · ∇π(y, t)
∂φ
∂
∂
∂φ
=
(y, t)
[φ(y, t), π(x, t)] +
[φ(y, t), π(x, t)]
(y, t)
∂yi
∂yi
∂yi
∂yi
=
2i∇φ(y, t) · ∇δ (3) (x − y).
Finally,
[φ(y, t)2 , π(x, t)]
= φ(y, t)[φ(y, t), π(x, t)] + [φ(y, t)2 , π(x, t)]φ(y, t)
= 2iφ(y, t)δ 3 (x − y).
6
(9)
In all,
= i2
π̇(x)
Z
n
o
d3 y ∇φ(y, t) · ∇δ (3) (x − y) + m2 π(y, t)δ 3 (x − y)
= ∇2 φ(x) − m2 φ(x),
(10)
where the first term in the integrand is integrated by parts to remove the derivative on the delta function.
Assembling (9) and (10) then,
∂
∂2
φ(x) = π(x) = ∇2 φ(x) − m2 φ(x),
2
∂t
∂t
implying
∂2
2
2
− ∇ + m φ(x) = ∂ 2 + m2 φ(x) = 0,
∂t2
which is precisely the Klein-Gordon equation. The equations of motion for the quantized system therefore coincide
with those of the classical system.
Exercise 5
The relativistically normalised one particle states are defined by
p
|pi = 2Ep a†p |0i.
With the mode expansion
Z
φ(x) =
1
d3 q
p
(aq e−iq·x + a†q eiq·x ),
(2π)3 2Eq
we find
Z
h0|φ(x)|pi =
d3 q
(2π)3
s
2Ep
h0| aq e−iq·x + a†q eiq·x a†p |0i,
2Eq
but the second term in the brackets vanishes because
†
0 = (ap |0i) = h0|a†p .
Therefore, upon commuting and using ap |0i = 0, we arrive at
s
Z
2Ep −iq·x
d3 q
e
h0|aq a†p |0i
h0|φ(x)|pi =
3
(2π)
2Eq
s
Z
2Ep −iq·x
d3 q
†
3 (3)
e
h0|
a
a
+
(2π)
δ
(p
−
q)
|0i
=
p
q
(2π)3 2Eq
= e−ip·x
Exercise 6
We calculate
d3 q
1
p
h0| ap e−ip·x + a†p eip·x , aq e−iq·y + a†q eiq·y |0i
3
(2π) 2 Ep Eq
Z
Z
3
d p
d3 q
1
p
=
h0| [ap , a†q ]eiq·y e−ip·x − [aq , a†p ]e−iq·y e+ip·x |0i
(2π)3
(2π)3 2 Ep Eq
Z
d3 p 1 −ip·(x−y)
ip·(x−y)
e
−
e
,
=
(2π)3 2Ep
Z
h0|[φ(x), φ(y)]|0i =
where p0 = Ep =
operators.
d3 p
(2π)3
Z
p
p2 + m2 and we used the by now familiar commutation relations for the creation and annhilation
7
The retarded Klein-Gordon propagator DR (x − y) is therefore given by the following 3-momentum integral
Z
d3 p 1 −ip·(x−y)
ip·(x−y) 0
0
0
0
e
−
e
θ(x − y )h0|[φ(x), φ(y)]|0i = θ(x − y )
0
(2π)3 2Ep
p =Ep
(11)
As a retarded propogator, DR (x − y) quite naturally vanishes if x0 < y 0 .
We seek a more covariant form of (11), i.e. re-express the above integral expression as a 4-momentum integral. In
particular, we demonstrate that
Z
Z
d3 p 1 −ip·(x−y)
d4 p
i
0
0
ip·(x−y) θ(x − y )
=
e−ip·(x−y)
e
−
e
(12)
4
2
0
(2π)3 2Ep
(2π)
p
−
m2
p =Ep
Γ
by evaluating the p0 integral
Z
Γ
0
0
0
i
dp0
e−ip (x −y )
2π p2 − m2
(13)
via a suitable contour Γ in the complex p0 -plane, employing the residue theorem together with the following perhaps
unfamiliar result from complex analysis
Lemma: (Jordan’s Lemma) Let f (z) be a function on the upper half plane. If there is an
r ≥ 0 such that f (z) is continuous whenever |z| ≥ r and
supθ∈[0,π] |f (Reiθ )| −→ 0 , as R −→ ∞,
then for any a > 0,
Z
f (z)eiaz dz −→ 0 , as R −→ ∞,
CR
where CR is the anti-clockwise contour in the upper half plane given by the semi-circle of
radius R
CR = Reiθ | θ ∈ [0, π] .
The standard application of this result is to situations where f (z) is holomorphic except at a finite number of poles
z1 , . . . zn all of which satisfy Im zi 6= 0 : Complete the contour CR to a closed contour ΓR by integrating along the
segment [−R, R] on the Re z axis; then we can evaluate various indefinite integrals using Jordan’s Lemma as follows
Z
lim
R→∞
f (z)eiaz dz = lim
Z
R→∞
ΓR
f (z)eiaz dz + lim
R→∞
CR
Z
+R
f (x)eiax dx =
−R
Z
+∞
f (x)eiax dx,
−∞
and as R gets indefinitely large, the contour ΓR encloses all the poles of f lying in the upper half plane, so that by
the residue theorem,
Z +∞
X
f (x)eiax dx = 2πi
Res f (z)eiaz , zi .
(14)
−∞
0
0 2
2
2 −1
In our case, f (p ) = ((p ) − |p| − m )
p2
Im zi >0
, and as
1
1
1
= 0
,
= 0 2
2
2
2
−m
(p ) − (p + m )
(p − Ep )(p0 + Ep )
we see that, unfortunately, the poles of the integrand in (13) are at p0 = ±Ep , which lie on the real p0 -axis. However,
we shall refuse to be phased by this, and perform a regularization by applying a perturbation that moves our poles
away from the real axis.
Now, how do we choose a regularization that will produce the retarded propagator? The important feature to notice
is that the coefficient of ip0 in the exponent in (13) changes sign as we go from x0 < y 0 to x0 > y 0 . When this
coefficient is positive (x0 < y 0 ), Jordan’s Lemma gives us a contour in the upper half plane to integrate along; when
it is negative (x0 > y 0 ), by symmetry we simply integrate in the lower half plane.
Now, DR (x − y) vanishes when x0 < y 0 , so from (14), we see that the perturbation should be such that the contour in the upper half plane encloses no poles. This gives us our prescription: perturb the poles into the lower half plane.
8
Figure 1: The pole prescription for evaluating the retarded propagator.
A perturbation that achieves this is p0 → p0 + i, for > 0: the poles are now at points with
(p0 + i)2 − Ep2 = 0 ⇒ p0 = ±Ep − i,
which have a strictly negative imaginary part. We summarize the above in figure 1.
In the limit → 0, the residues are
i
i ∓iEp (x0 −y0 )
−ip0 (x0 −y 0 )
0
lim Res
e
, p = ±Ep − i = ±
e
.
(p0 + i)2 − Ep2
2Ep
→0−
(15)
Now, we have described how this prescription gives 0 when x0 < y 0 . When x0 > y 0 the contour ΓR in the lower half
plane is traversed counter-clockwise, which contributes a factor of −1. Using (14) and sending → 0, we see then that
Z
dp0
i
i 1 −iEp (x0 −y0 )
i 1 +iEp (x0 −y0 )
−ip0 (x0 −y 0 )
e
= −2πi
e
−
e
.
2π p2 − m2
2π 2Ep
2π 2Ep
We may now assemble everything to find
Z
Z
d4 p
i
d3 p
1 −iEp (x0 −y0 )
1 +iEp (x0 −y0 ) +ip·(x−y)
−ip·(x−y)
0
0
e
= θ(x − y )
e
−
e
e
(2π)4 p2 − m2
(2π)3 2Ep
2Ep
Z
d3 p 1 −iEp (x0 −y0 )+ip·(x−y)
e
= θ(x0 − y 0 )
(2π)3 2Ep
0
0
−e+iEp (x −y )+ip·(x−y) ,
(16)
with the θ taking account of the vanishing of this integral when x0 < y 0 . The first exponential is
0
0
e−iEp (x −y )+ip·(x−y) = e−ip·(x−y) 0
,
p =Ep
whereas the second is
0
e+iEp (x
−y 0 )+ip·(x−y)
0
0
= eip ·(x−y) p00 =Ep0
0
where p = (E−p , −p). On making the substitution p → p in the second term in (16) and then relabling, we arrive
at the expression (11).
Had we left in the perturbation merely as a reminder of how to evaluate this integral,
Z
d4 p
i
DR (x − y) =
e−ip·(x−y) ,
4
0
2
(2π) (p + i) − |p|2 − m2
then the comparison with the Feynman propagator
Z
d4 p
i
DF (x − y) =
e−ip·(x−y)
(2π)4 p2 − m2 + i
is immediate. With the Feynman propagator, we push the pole at the positive energy up and that at the negative
energy down, so that now the above contours pick up only one contribution for each ordering of x0 , y 0 . Had we defined
an advanced propagator in the obvious way, the suitable prescription would be to perturb both poles into the upper
half plane, i.e. p0 → p0 − i, for > 0.
9
Exercise 7
We examine the following non-relativistic model
L = iφ∗ ∂0 φ −
1
∇φ∗ · ∇φ
2m
The equations of motion are
∂L
∂L
1 2 ∗
0 = ∂µ
−
=−
∇ φ + iφ̇∗ ⇒
∂µ φ
∂φ
2m
∂L
1 2
∂L
−
=−
∇ φ − iφ̇ ⇒
0 = ∂µ
∂µ φ∗
∂φ∗
2m
1 2 ∗
∇ φ = iφ̇∗ ,
2m
1 2
−
∇ φ = iφ̇
2m
(17)
Note that (17) looks exactly like the free Schrödinger equation of quantum mechanics. However, in this context there
is no probabilistic interpretation of φ.
The energy momentum tensor is
T µν =
∂L
∂L
∂ν φ +
∂ν φ∗ − δ µ ν L.
∂(∂µ φ)
∂(∂µ φ∗ )
The components are
T 00
T 0i
T i0
T ij
1
(∇φ∗ ) · (∇φ),
2m
= iφ∗ ∂i φ,
1
1
= −
(∂ i φ∗ )φ̇ −
(∂ i φ)φ̇∗ ,
2m
2m
1
1
= −
(∂ i φ∗ )(∂j φ) −
(∂ i φ)(∂j φ∗ ) − δ i j L.
2m
2m
=
The transformation
φ 7−→ eiα φ , α ∈ R
is clearly a symmetry of L, and, infinitesimally,
∆φ = iφ , ∆φ∗ = −iφ∗ , ∆L = 0.
The conserved Noether current is therefore given by
jµ =
∂L
∂L
iφ +
(−iφ∗ ).
∂(∂µ φ)
∂(∂µ φ∗ )
The components are
j0
ji
= −φφ∗ ,
i
(φ∗ ∂ i φ − φ∂ i φ∗ ),
=
2m
which, in quantum mechanics, would have the interpretation of a probability density and current, respectively.
The momenta conjugate to φ and φ∗ are
πφ =
∂L
= iφ∗ ,
∂ φ̇
πφ∗ =
∂L
= 0.
∂ φ̇∗
The classical Hamiltonian density is therefore
H = πφ φ̇ − L =
1
(∇φ∗ )(∇φ).
2m
To quantise this theory, promote the classical fields φ, π to Schrödinger picture operators φ(x) and π(x) with canonical
commutation relations
[φ(x), φ(y)] = [π(x), π(y)] = 0 , [φ(x), πφ (y)] = iδ (3) (x − y)
In the quantum theory, φ∗ is replaced by the adjoint φ† , so that πφ = iφ† in the quantum theory. This gives an
alternate form of the above relations
[φ(x)† , φ(y)† ] = 0 , [φ(x), φ(y)† ] = δ (3) (x − y).
10
Now, to mimic the quantum field theories met in lectures, we find a mode expansion for the field φ(x). To do this,
first look for plane wave (or mode) solutions to the equation of motion (17), i.e. those of the form
φ(x, t) ∼ e−i(Et−p·x) .
These provide solutions iff the non-relativistic dispersion relation
E=
p2
2m
(18)
holds.
Here we note a marked difference from Klein-Gordon or Dirac theory: the equations of motion being first order in
time gives only one kind of plane wave solution, i.e. those with positive frequency E > 0 given by (18). Contrast this
with Klein-Gordon and Dirac theory in which there are a number of differing independent solutions, namely
Positive Frequency
Negative frequency
e−ip·x
e+ip·x
Klein-Gordon
Dirac
with p0 = +
us (p)e−ip·x
s=± 21
vs (p)e+ip·x
s=± 21
p
|p|2 + m2 in all cases.
The up-shot of there being only one type of mode is that upon superposing these solutions, we see that we can only
create positive frequency particles in the non-relativistic theory,
Z
Z
d3 p † −ip·x
d3 p
ip·x
†
a
e
,
φ(x)
=
a e
,
φ(x) =
p
(2π)3
(2π)3 p
and so cannot have anti-particles.
Then
Z
d3 x φ(x)e−ip·x , a†p =
ap =
Z
d3 x φ(x)e+ip·x ,
so that the only non-trivial commutator is
[ap , a†q ]
Z
=
=
d3 x
Z
d3 y [φ(x), φ(y)† ] e−ip·x e+iq·y
(2π)3 δ (3) (x − y)
as expected.
Finally, we compute the energy of one particle states. First, write the Hamiltonian in terms of creation and annihilation
operators,
Z
Z
1
3
H = d xH =
d3 x ∇φ† · ∇φ
2m
Z
Z
Z
1
d3 p
d3 q i(p−q)x
3
=
d x
e
(−iq) · (ip) a†q ap
2m
(2π)3
(2π)3
Z
1
d3 p
=
|p|2 a†p ap ,
2m
(2π)3
where we pick up and evaluate a delta function in the penultimate line.
Defining one particle states by |pi = a†p |0i, with the ground state |0i being specified by ap |0i = 0, we find
1
H|pi =
2m
Z
d3 q
1
|p|2
2 †
†
2 †
|q|
a
[a
,
a
]|0i
=
|p|
a
|0i
=
|pi,
q
q
p
p
(2π)3
2m
2m
which is the expected energy of a non-relativistic free particle of mass m.
11
(19)
Exercise 8
The object is to show that the S µν defined below generate a representation of the Lie algebra so(1, 3), the Lie algebra
of the Lorentz group SO(1, 3).
[γ κ γ λ , γ µ γ ν ]
= γκγλγµγν − γµγν γκγλ
= γκ γλ, γµ γν − γκγµγλγν − γµγν γκγλ
= 2g λµ γ κ γ ν − {γ κ , γ µ } γ λ γ ν + γ µ γ κ γ λ γ ν − γ µ γ ν γ κ γ λ
= 2g λµ γ κ γ ν − 2g κµ γ λ γ ν + γ µ γ κ γ ν , γ λ − γ µ γ ν γ κ γ λ
= 2g λµ γ κ γ ν − 2g κµ γ λ γ ν + 2g µκ γ ν γ λ − γ µ {γ κ , γ ν } γ λ + γ µ γ ν γ κ γ λ − γ µ γ ν γ κ γ λ
= 2g λµ γ κ γ ν − 2g κµ γ λ γ ν + 2g µκ γ ν γ λ − 2g κν γ µ γ λ
Now, to perform the main computation of interest, write S µν = 41 [γ µ , γ ν ] in a slightly more convenient form,
S κλ =
1 κ λ
[γ , γ ]
4
=
=
=
1 κ λ
γ γ − γλγκ
4
1 κ λ λ κ
γ γ − γ , γ + γκγλ
4
1 κ λ
γ γ − g κλ .
2
This allows us to calculate that
[S κλ , S µν ]
1 κ λ µ ν
[γ γ , γ γ ]
2
1 λµ κ ν
=
g γ γ − g κµ γ λ γ ν + g µκ γ ν γ λ − g κν γ µ γ λ
2
1
1
1
1
= g λµ S κν + g λµ g κν − g µκ S λν − g µκ g λν + g λν S µκ + g λν g µκ − g κν S µλ − g κν g µλ
2
2
2
2
= g λµ S κν − g µκ S λν + g λν S µκ − g κν S µλ
=
(20)
The S µν thus define a representation of so(1, 3), called the spin representation - it is not irreducible (see next exercise).
Exercse 9
We use the brackets (20) worked out in the previous exercise, but, first, write
Si =
i
ijk γ j γ k
4
=
=
i
ijk γ j γ k − δ jk
4
i
ijk S jk ,
2
where skew symmetry of ijk allows us to insert the symmetric term δ ij . Thus
[S i , S j ]
1
= − ikl jmn [S kl , S mn ]
4
1
= − ikl jmn δ lm S kn − δ km S ln + δ ln S mk − δ kn S ml
4
1
= − ikl jln S kn − ikl ikn S ln + ikl jml S mk − ikl jmk S ml .
4
Relabeling indices in the final three terms and permuting in the alternating symbols, we have
[S i , S j ]
1
ikl jln S kn − ilk iln S kn + ikl jnl S nk − iln jnl S nk
4
−ikl jln S kn
ikl jnl S kn
(δij δkn − δin δkj ) S kn
−S ji
S ij ,
= −
=
=
=
=
=
12
where in the final two steps the obvious skew-symmetry S kn = −S nk was employed, so that, in the penultimate step,
S kn δkn = 0. To obtain the desired result, consider
1
1
iijk S k = − ijk kln S ln = ijk nlk S ln
2
2
1
(δim δjl − δil δjm ) S ln
2
1 ij
=
S − S ji
2
= S ij
=
Thus
[S i , S j ] = iijk S k ,
and we see that the S i furnish a representation of the Lie algebra of the rotation group in three dimensions,
so(3) ∼
= su(2).
Since γ 0 γ i = −γ i γ 0 ,
γ0Si =
i
i
i
ijk γ 0 γ j γ k = − ijk γ j γ 0 γ k = + ijk γ j γ k γ 0 = S i γ 0
4
4
4
i.e. [γ 0 , S i ] = 0.
Using the following identity, the same reasoning demonstrates that [γ 5 , S i ] = 0.
5 µ
γ ,γ = 0
(21)
This is straightforward to prove: since γ µ anti-commutes with γ ν whenever µ 6= ν, we can commute γ µ past the three
γ ν in γ 5 for which µ 6= ν and pick up a factor of (−1)3 = −1.
Mathematical Remark: In fact, the previous argument for [γ 5 , S i ] = 0 demonstrates that [γ 5 , S µν ] = 0,
which in turn tells us that the representation of so(1, 3), furnished by the S µν is reducible: since
(γ 5 )2 = 1 (see Exercise 10) it follows that the space of Dirac spinors (i.e. C4 ) splits into eigensubspaces of
γ 5 with eigenvalues ±1 (i.e. the spaces of positive and negative Weyl spinors); the identity [γ 5 , S µν ] = 0
shows that the action of the S µν preserves this eigenspace decomposition, i.e. sends all Dirac spinors of
γ 5 eigenvalue ±1 into a spinor with the same eigenvalue. These eigenspaces are thus proper invariant
subspaces, and so the representation is reducible.
More generally, this is a feature of even dimensions - in odd dimensions the representation defined here
is indeed irreducible. See Michelson & Lawson, Spin Geometry for full details.
Fixing i ∈ {1, 2, 3}, there are j, k ∈ {1, 2, 3} uniquely defined by j, k 6= i and j < k. Then, without any implicit
summations,
i
S i = ijk γ j γ k − γ k γ j .
4
2
Now, as ijk = ±1, (ijk ) = +1,
(S i )2 = −
1
γ j γ k γ j γ k + γ k γ j γ k γ j − γ j (γ k )2 γ j − γ k (γ j )2 γ k .
16
In the first two terms use that γ j γ k = −γ k γ j , whenever j 6= k, so that as (γ l )2 = −I, any l,
(S i )2 = −
1
1
1
−(γ j )2 (γ k )2 − (γ k )2 (γ j )2 − γ j (γ k )2 γ j − γ k (γ j )2 γ k = − (−4 I) = I
16
16
4
Consider the following 4-dimensional representation of the Clifford algebra
I2
0
0
σi
0
i
γ =
γ =
,
0 −I2
−σ i 0
where, for reference,
i
(σ ) =
0
1
1
0
0 −i
1 0
,
,
i 0
0 −1
and these satisfy
σ i σ j = δ ij I2 + iijk σ k ,
13
(22)
so that, in particular,
σ i , σ j = 2δ ij I2 , [σ i , σ j ] = 2iijk σ k .
(23)
i
The S have the following matrix representatives
Si =
i
i
ijk γ j γ k = ijk
4
4
−σ j σ k
0
0
−σ j σ k
.
According to (22),
ijk σ j σ k
= iijk jkl σ l = i (δil δjj − δij δjl ) σ l = 2iσ i .
So that
Si =
1
2
σi
0
0
σi
.
Then by (23),
1
[S , S ] =
4
i
j
[σ i , σ j ]
0
0
[σ i , σ j ]
1
=
4
2iijk σ k
0
0
2iijk σ k
k
1
σ
= iijk
0
2
0
σk
0
−σ 3
σ3
0
= iijk S k ,
as desired.
From (23), (σ i )2 = +1, so
(S i )2 =
1
4
(σ i )2
0
0
(σ i )2
=
1
I2
4
Now compute γ 5 ,
γ 5 = iγ 0 γ 1 γ 2 γ 3
σ1 σ2 σ3
⇒ γ5
Therefore
= i
I2
0
0
−I2
0
−σ 1
σ1
0
3
0
−σ 2
σ2
0
0
−σ 1 σ 2 σ
= i
1 2 3
−σ σ σ
0
0 1
0 −i
1 0
=
= i I2
1 0
i 0
0 −1
0 I2
=
I2 0
1
[γ , S ] =
2
5
i
0
σi
σi
0
−
0
σi
σi
0
=0
Likewise, [γ 0 , S i ] = 0.
Since so(3) (the Lie algebra of the rotation group, SO(3)) is a sub-algebra of the Lorentz algebra, these calculations
tell us that by restricting the above representation of so(1, 3) to a representation of so(3), we obtain two copies of
the usual spin half representation encountered in non-relativistic physics. In other words, Dirac spinors furnish the
(reducible) representation 12 ⊕ 12 of so(3), and thus have spin 12 .
Notice that although we have exhibited the repersentation of so(3) ≤ so(1, 3) as a sum of irreducibles, i.e. the matrices
i
1
σ
0
Si =
0 σi
2
generating so(3) are block diagonal, this does not show us explicitly that the representation of so(1, 3) here constructed
is reducible as claimed above since the S 0i are not, in fact, block diagonal in this matrix representation (additional
exercise: show this).
Following from the discussion above, we instead seek a representation in which γ 5 is block diagonal. The so-called
chiral representation achieves this for us,
0 I2
0
σi
γ0 =
γi =
,
I2 0
−σ i 0
then one would find all the S µν and γ 5 to be block diagonal (additional exercise: show this).
14
Exercise 10
As the identity (γ 5 )2 = I is used in numerous places throughout the following calculations, it is more logical to prove
this first. Picking up various factors of −1 from commuting factors, we have
(γ 5 )2
=
=
=
=
−γ 0 γ 1 γ 2 γ 3 γ 0 γ 1 γ 2 γ 3
−(−1)3 (γ 0 )2 γ 1 γ 2 γ 3 γ 1 γ 2 γ 3
+(−1)2 (γ 0 )2 (γ 1 )2 γ 2 γ 3 γ 2 γ 3
−(γ 0 )2 (γ 1 )2 (γ 2 )2 (γ 3 )2 ,
and, finally, using (γ 0 )2 = +I, (γ i )2 = −I, we see that (γ 5 )2 = I.
1. We prove instead that
Tr (γ µ1 · · · γ µ2k+1 ) = 0 , k ∈ N,
as this accounts for this exercise and Exercises 3 and 8 also.
We begin by inserting a factor of (γ 5 )2 = I into the far right of the trace and use the cyclicity of the trace to
move one of these factors of γ 5 to the far left,
Tr (γ µ1 · · · γ µ2k+1 ) = Tr γ µ1 · · · γ µ2k+1 (γ 5 )2
= Tr γ 5 γ µ1 · · · γ µ2k+1 γ 5 .
Now use the identity (21) to begin anti-commuting this γ 5 past each factor of γ µi to its right.
Tr (γ µ1 · · · γ µ2k+1 ) = (−1)Tr γ µ1 γ 5 γ 2 · · · γ µ2k+1 γ 5
= (−1)2k+1 Tr γ µ1 · · · γ µ2k+1 γ 5 γ 5
= (−1)2k+1 Tr γ µ1 · · · γ µ2k+1 (γ 5 )2
= −Tr (γ µ1 · · · γ µ2k+1 )
⇒ Tr (γ µ1 · · · γ µ2k+1 ) = 0
(24)
where in the penultimate step, we use (γ 5 )2 = I again.
2. Splitting the product γ µ γ ν into symmetric and skew-symmetric parts, we have
1 µ ν
1 µ ν
µ ν
Tr (γ γ ) = Tr
{γ , γ } + [γ , γ ]
2
2
1
1
Tr {γ µ , γ ν } + Tr [γ µ , γ ν ]
=
2
2
The second term vanishes since the identity Tr(XY ) = Tr(Y X) implies
0 = Tr(XY ) − Tr(Y X) = Tr (XY − Y X) = Tr [X, Y ].
Using the defining relation of the Clifford algebra, this then gives
Tr (γ µ γ ν )
= Tr (g µν I4 )
= 4g µν
(25)
5. Here, insert a factor of (γ 0 )2 = I and anti-commute as before
Tr γ 5
=
Tr γ 5 (γ 0 )2
= −Tr γ 0 γ 5 γ 0
= −Tr γ 5 (γ 0 )2
⇒ Tr γ 5
=
0
(26)
6. As was done previously, split the product γ µ γ ν into symmetric and skew-symmetric parts and then use the
Clifford relations,
1
pµ qν ({γ µ , γ ν } + [γ µ , γ ν ])
2
1
=
pµ qν (2g µν I + 4S µν )
2
= p · q I + 2S µν pµ qν .
p
//q =
15
This time simply anti-commute factors,
µ ν
p
//q = pµ qν γ γ
= pµ qν ({γ µ , γ ν } − γ ν γ µ )
= pµ qν (2g µν I − γ ν γ µ )
= 2p · q − /qp
/
7. Clearly, by (25),
µ ν
µν
Tr(p
//q) = pµ qν Tr (γ γ ) = 4pµ qν g = 4p · q
8. Clearly
1
2k+1
µ1
· · · γ µ2k+1 ) ,
Tr p
= p1µ1 · · · p2k+1
/ ···p
/
µ2k+1 Tr (γ
which vanishes by (24).
9. As before, we use cyclicity of the trace and then perform a sequence of anti-commutations to move a factor back
to its original position,
Tr (γ µ γ ν γ ρ γ σ ) = Tr (γ σ γ µ γ ν γ ρ )
=
=
=
⇒ Tr (γ µ γ ν γ ρ γ σ ) =
=
Tr ({γ σ , γ µ } γ ν γ ρ − γ µ γ σ γ ν γ ρ )
2g σµ Tr (γ ν γ ρ ) − Tr (γ µ {γ σ , γ ν } γ ρ ) + Tr (γ µ γ ν γ σ γ ρ )
2g σµ Tr (γ ν γ ρ ) − 2g σν Tr (γ µ γ ρ ) + Tr (γ µ γ ν {γ σ , γ ρ }) − Tr (γ µ γ ν γ ρ γ σ )
g σµ Tr (γ ν γ ρ ) − g σν Tr (γ µ γ ρ ) + g σρ Tr (γ µ γ ν )
4 (g σµ g νρ − g σν g µρ + g σρ g µν )
Contracting this identity with p1 , p2 , p3 , p4 gives the desired result.
10. First,
Tr γ 5 γ µ γ ν = −Tr γ µ γ 5 γ ν = −Tr γ 5 γ ν γ µ ,
so that Tr γ 5 γ µ γ ν is skew in µ ν. To calculate this trace we need therefore only look at the cases µ = i, ν = 0
and µ = i, ν = j, i 6= j. First, for µ = i, ν = 0,
Tr γ 5 γ 0 γ i
= iTr (γ 0 )2 γ 1 γ 2 γ 3 γ i
= iTr γ 1 γ 2 γ 3 γ i
As γ i appears somewhere in the product γ 1 γ 2 γ 3 , it may be commuted through the relevant factors until we
obtain a square (γ i )2 = −I. Thus, up to sign, we have, for j < k and j, k 6= i,
Tr γ 5 γ i γ 0 = ±iTr γ j γ k = ±4iδ jk = 0,
where the result (25) has been used.
When µ = i, ν = j, i 6= j, we similarly have, for k 6= i, j,
Tr γ 5 γ i γ j = ±iTr γ 0 γ k = ±2ig 0k = 0,
again by (25).
11. Again, perform anti-commutations and use the Clifford relations,
µ
ν σ µ
γµ p
/γ = γ γ γ gνµ pσ
= γ ν {γ σ , γ µ } gνµ pσ − γ ν γ µ γ σ gνµ pσ
1 ν µ
1 ν µ
ν σµ
= 2γ g gνµ pσ − gνµ
{γ , γ } + [γ , γ ] γ σ pσ
2
2
νµ
= 2p
/ − gνµ g p
/
= −2p
/,
where in the third line
gνµ [γ ν , γ µ ] = 0,
since gνµ is symmetric in ν µ yet [γ ν , γ µ ] is skew.
16
(27)
12. Using (27) in the second line,
µ
γµ p
/2 γ
/1 p
µ
µ
= γµ p
/2
/1 {p
/1 , γ } − γµ p
/1 γ p
=
=
=
=
µ
2γµ p
/2
/1 p2 + 2p
/1 p
2p
/2 p
/1 + 2p
/1 p
/2
2{p
/2 , p
/1 }
4p · q
(28)
13. Using (28),
µ
γµ p
/3 γ
/2 p
/1 p
14. We show first that
=
=
=
=
µ
µ
µ
γµ p
/3
/2 γ p
/2 {p
/3 , γ } − γ p
/1 p
/1 p
2p
/2 − 4p1 · p2 p
/3
/1 p
/3 p
2p
/1 − 4p1 · p2 p
/3
/2 p
/ 3 {p
/1 , p
/2 } − 2p
/3 p
p
.
p
−2p
/3 /2 /1
S µνρσ ≡ Tr γ 5 γ µ γ ν γ ρ γ σ = Tr γ 5 γ [µ γ ν γ ρ γ σ] ,
so that, as a totally skew contravariant 4-tensor in 4 dimensions, S µνρσ must be proportional to the alternating
tensor µνρσ .
We can show this for adjacent pairs of indices since, for instance,
1 µ ν
1 µ ν
5 µ ν ρ σ
5
ρ si
Tr γ γ γ γ γ
= Tr γ
{γ , γ } + [γ , γ ] γ γ
2
2
1
µν
5 ρ σ
= g Tr γ γ γ + Tr γ 5 [γ µ , γ ν ]γ ρ γ σ ,
2
the first term of which vanishes by the identity (26), leaving the term skew in µ ν. This then implies that S µνρσ
is totally skew since, for instance,
S ρνµσ = −S νρµσ = +S νµρσ = −S µνρσ .
Recall that the alternating tensor µνρσ is defined by
µνρσ = αβγδ g αµ g βν g γρ g δσ
where 0123 = +1. Therefore
0123 = αβγδ g α0 g β1 g γ2 g δ3 = det (g −1 ) = −1.
Thus
S µνρσ
= −µνρσ Tr γ 5 γ 0 γ 1 γ 2 γ 3
= iµνρσ Tr (γ 5 )2
= iµνρσ Tr(I4 )
= 4iµνρσ
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