Free Scalar Field Theory
-
Upload
edwin-tan-pei-ming -
Category
Documents
-
view
240 -
download
0
Transcript of Free Scalar Field Theory
8/7/2019 Free Scalar Field Theory
http://slidepdf.com/reader/full/free-scalar-field-theory 1/25
Free Scalar Field Theory
1
8/7/2019 Free Scalar Field Theory
http://slidepdf.com/reader/full/free-scalar-field-theory 2/25
I. CANONICAL QUANTIZATION OF SCALAR FIELDS
A. The Klein-Gordon equation
In Chapter 1, we second quantized a nonrelativistic Schrodinger equation to obtain a
nonrelativistic quantum field theory. To find an interesting relativistic quantum field theory
along the same path as in Chapter 1, we need a relativistic Schrodinger equation to second
quantize. With little else to go by other than the nonrelativistic Schrodinger equation,
we can only guess at what an interesting relativistic equation is. This is what happened
historically. One natural relativistic generalization leads to the Klein-Gordon equation.
For a free particle of mass m, its energy E = p2/2m, where p is its momentum. By
substitutingE → i
∂
∂tand pi → −i
∂
∂xi, (1)
we obtain the nonrelativistic Schrodinger equation for a free particle,
i∂
∂tϕ(x, t) = − 1
2m2ϕ(x, t). (2)
The relativistic generalization of the above energy-momentum relation is E 2 = p2 + m2.
The same substitutions for E and pi above lead to Klein-Gordon equation,
∂ 2
∂t2ϕ(x, t) − 2ϕ(x, t) + m2ϕ(x, t) = 0. (3)
We note that Lorentz covariance requires that the field ϕ(x, t) transforms as a scalar under
the Lorentz group:
ϕ(xµ) → ϕ((Λ−1)µν xν ). (4)
Here, we recall that in nonrelativistic quantum mechanics the wave function ϕ(x, t) becomes
ϕ(x, t + δt) = ϕ(x, t) + δt∂ tϕ(x, t) = ϕ(x, t) − iδt(i∂ t)ϕ(x, t),
ϕ(x − δx, t) = ϕ(x, t) − δxi∂ iϕ(x, t) = ϕ(x, t) − iδxi(−i∂ i)ϕ(x, t)
under time evolution and space translation respectively. And, under anticloclwsie rotation
about say the z-axis by δθ,
ϕ(x, t) → {1 + iδθ[x(−i∂ y − y(−i∂ x))]}ϕ(x, t) = ϕ(x, t) + δθϕ(x − yδθ,y + xδθ,z,t).
2
8/7/2019 Free Scalar Field Theory
http://slidepdf.com/reader/full/free-scalar-field-theory 3/25
In analogy with the nonrelativistic Schrodinger equation,
ϕ∗(x, t)∂ 2
∂t2ϕ(x, t) − ϕ∗(x, t)2ϕ(x, t) + m2ϕ∗(x, t)ϕ(x, t) = 0,
ϕ(x, t) ∂
2
∂t2ϕ∗(x, t) − ϕ(x, t)2ϕ∗(x, t) + m2ϕ∗(x, t)ϕ(x, t) = 0.
Hence, like the nonrelativistic Schrodinger equation, the Klein-Gordon equation (3) defines
a conserved current.
∂ t(ϕ∗∂ tϕ − ϕ∂ tϕ∗) − · (ϕ∗ϕ − ϕϕ∗) = 0. (5)
Recalling the probability current density
J = −i
2m(ϕ∗ϕ − ϕϕ∗), (6)
we obtain the probability density
J 0 =i
2m(ϕ∗∂ tϕ − ϕ∂ tϕ
∗). (7)
Note that J 0 = 0 if ϕ∗ = ϕ.
Clearly, the plane wave
ϕ(x, t) = exp[−
i(ωt−
k·
x)] (8)
solves Eq.(3) if ω and k satisfy the mass-shell condition
ω2 − k · k = m2. (9)
That is,
ω = ±ωk, (10)
with
ωk ≡ +
k2 + m2. (11)
In this case, we have from Eq.(7),
J 0 =ω
m= ±ωk
m. (12)
3
8/7/2019 Free Scalar Field Theory
http://slidepdf.com/reader/full/free-scalar-field-theory 4/25
B. Difficulties of the Klein-Gordon equation
Historically, Eq.(3) was proposed as a relativistic wave equation by Schrodinger, Klein,
and Gordon in 1925 - 1927, but was abandoned (until 1934) due to difficulties in constructing
a one-particle theory based on it.
The first difficulty is that the Klein-Gordon equation (3) admits negative energy solutions.
For example, from Eq.(9), we could have ω = −ωk. Not only must we find an interpretation
of what a negative energy state is, but, in addition, the energy spectrum is not bounded
from below. That is, we could in principle extract an arbitrarily large amount of energy
from a one-particle system. For example, we could begin with a positive energy state and
apply some external influence that allows the particle to jump to a negative energy state.
The positive difference in energy between these two states could be used to do work. We
can repeat the procedure on the negative energy state, lowering its energy, and extract more
energy. Since the spectrum is not bounded from below, we could do this forever.
The second difficulty is the interpretation of ϕ(x, t) as a wave function, i.e., as a prob-
ability amplitude. To do so we must define some quantity based on ϕ(x, t) that has a
nonnegative norm that we can interpret as a probability density. This probability density
must be the time component of a conserved probability current so that the total proba-
bility is conserved with time. However, from Eq.(12), we could have J 0 = −ωk/m. The
use of the Klein-Gordon equation thus appears to exclude the possibility of a probability
interpretation.
These difficulties arise because we are trying to construct a one-particle theory from
Eq.(3). Our inability to do so is a consequence of the imposition of Lorentz covariance on
the quantum mechanical wave equation.
C. Real Klein-Gordon scalar field
To canonically quantize the field ϕ(x), we must find a Hamiltonian and quantum condi-
tions (equal-time commutation relations) such that the Klein-Gordon equation (3) becomes
an operator equation of motion,
∂ 2
∂t2ϕ(x, t) − 2ϕ(x, t) + m2ϕ(x, t) = 0. (13)
4
8/7/2019 Free Scalar Field Theory
http://slidepdf.com/reader/full/free-scalar-field-theory 5/25
This is accomplished with the aid of the classical action,
A[ϕ(x)] =
d4xL(x) =1
2
d4x[∂ µϕ(x)∂ µϕ(x) − m2ϕ2(x)]. (14)
Here, x = (x, t). We note that A[ϕ(x)] is Lorentz invariant if ϕ(x) transforms as a Lorentzscalar.
Equation (14) obviously yields Eq.(3):
∂ L∂ϕ
= −m2ϕ,
∂ L∂ (∂ µϕ)
= ∂ µϕ.
Therefore,
∂ L∂ϕ
− ∂ µ
∂ L
∂ (∂ µϕ)
= 0
⇒ ∂ µ∂ µϕ(x) + m2ϕ(x) = 0.
From the action, we find the conjugate momentum field
π(x) =∂ L
∂ (∂ tϕ)= ∂ tϕ. (15)
The Hamiltonian quickly follows,
H (t) =
d3x[π(x) · ∂ tϕ(x) − L]
=1
2
d3x[π2(x) + ϕ(x) · ϕ(x) + m2ϕ2(x)]. (16)
In analogy with non-relativistic quantum mechanics, we take as our quantum conditions:
[ϕ(x, t), π(y, t)] = iδ3(x − y), (17)
[ϕ(x, t), ϕ(y, t)] = 0, (18)
[π(x, t), π(y, t)] = 0. (19)
This choice will be shown to be consistent with microscopic causality as demanded by rela-
tivity.
5
8/7/2019 Free Scalar Field Theory
http://slidepdf.com/reader/full/free-scalar-field-theory 6/25
Using the Hamiltonian and commutators, we find
i[H (t), ϕ(x, t)] =i
2
d3y[π2(y, t) + ϕ(y, t) · ϕ(y, t) + m2ϕ2(y, t), ϕ(x, t)]
=i
2 d3y[π2(y, t), ϕ(x, t)]
= − i
2
d3y[ϕ(x, t), π2(y, t)]
=
d3yδ3(x − y)π(y, t)
= π(x, t). (20)
And,
i[H (t), π(x, t)] =i
2
d3y[π2(y, t) + ϕ(y, t) · ϕ(y, t) + m2ϕ2(y, t), π(x, t)]
= i2
d3y[ϕ(y, t) · ϕ(y, t) + m2ϕ2(y, t), π(x, t)]
= i
d3y[ϕ(y, t) · + m2ϕ(y, t)][ϕ(y, t), π(x, t)]
= −
d3y[ϕ(y, t) · + m2ϕ(y, t)]δ3(x − y)
= 2ϕ(x, t) − m2ϕ(x, t). (21)
It follows that we have the Klein-Gordon operator equation of motion,
∂
∂t π(x, t) =
∂ 2
∂t2 ϕ(x, t) = 2
ϕ(x, t) − m
2
ϕ(x, t) (22)
D. Particle interpretation
Expand ϕ(x, t) in terms of the classical solutions of the Klein-Gordon equation (3),
ϕ(x, t) = d3 k
(2π)31
2ωk
[exp(−ik · x)a(k) + exp(ik · x)b(k)]. (23)
Here,
k · x = gµν kµxν = ωkt − k · x. (24)
exp(−ik ·x) = exp[−i(ωkt− k ·x)] is a classical positive energy plane wave solution of Eq.(3),
and exp(ik · x) = exp[i(ωkt − k · x)] is a negative energy solution. These guarantee that
ϕ(x, t) satisfies the Klein-Gordon operator equation of motion (22). Classically ϕ(x, t) is a
real field, so ϕ(x, t) is a Hermitian operator and we demand that b(k) = a†(k). So,
ϕ(x, t) =
d3 k
(2π)31
2ωk
[exp(−ik · x)a(k) + exp(ik · x)a†(k)]. (25)
6
8/7/2019 Free Scalar Field Theory
http://slidepdf.com/reader/full/free-scalar-field-theory 7/25
It follows that
d3xϕ2(x, t)
=
d
3
x d3 k
(2π)3
1
2ωk d3 k
(2π)3
1
2ωk [e
−ik·x
a(k) + e
ik·x
a
†
(k)][e
−ik·x
a(k
) + e
ik·x
a
†
(k
)]
=
d3x d3 k
(2π)31
2ωk
d3 k
(2π)31
2ωk
[e−i(k+k)·xa(k)a(k) + ei(k+k)·xa†(k)a†(k)]
+
d3x
d3 k
(2π)31
2ωk
d3 k
(2π)31
2ωk
[e−i(k−k)·xa(k)a†(k) + ei(k−k)·xa†(k)a(k)]
=
d3 k
(2π)31
2ωk
d3 k
2ωk
δ3( k + k)[e−2iωkta(k)a(k) + e2iωkta†(k)a†(k)]
+ d3 k
(2π)3
1
2ωk
d3 k
2ωk
δ3( k
− k)[a(k)a†(k) + a†(k)a(k)] (26)
The normalization, 1/(2ωk), in the volume element above was chosen so as to be Lorentz
invariant.
d4k
(2π)42πδ(k2 − m2)θ(k0) =
d4k
(2π)3δ(k2
0 − k2 − m2)θ(k0)
=
d4k
(2π)3δ(k2
0 − ω2k)θ(k0)
= d4k
(2π)3
1
2ωk
[δ(k0−
ωk) + δ(k0 + ωk)]θ(k0)
=
d3 k
(2π)31
2ωk
. (27)
Here, θ is the Heaviside step function [θ(k0) = +1 if k0 > 0 and θ(k0) = 0 otherwise], and
we make use of the the following property of the δ-function.
δ(f (x)) =i
1
|f (xi)|δ(x − xi), (28)
where f
denotes df/dx and xi is a simple zero of f (x). To prove this relation, δ(f (x))dx =
δ(y)
1
|f (x)|dy =i
1
|f (xi)| . (29)
The expansion for π(x, t) follows from Eq.(25).
π(x, t) =∂
∂tϕ(x, t) =
d3 k
(2π)31
2ωk
[−iωk exp(−ik · x)a(k) + iωk exp(ik · x)a†(k)]. (30)
7
8/7/2019 Free Scalar Field Theory
http://slidepdf.com/reader/full/free-scalar-field-theory 8/25
It follows that
d3xπ2(x, t)
= − d
3
x d3 k
(2π)3
1
2ωk d3 k
(2π)3
1
2ωk ωkωk
[e
−ik·x
a(k) − e
ik·x
a
†
(k)][e
−ik·x
a(k
) − e
ik·x
a
†
(k
)]
= −
d3x d3 k
(2π)31
2ωk
d3 k
(2π)31
2ωk
ωkωk[e−i(k+k)·xa(k)a(k) + ei(k+k)·xa†(k)a†(k)]
+
d3x d3 k
(2π)31
2ωk
d3 k
(2π)31
2ωk
ωkωk [e−i(k−k)·xa(k)a†(k) + ei(k−k)·xa†(k)a(k)]
= −
d3 k
(2π)31
2ωk
d3 k
2ωk
δ3( k + k)ω2k[e−2iωkta(k)a(k) + e2iωkta†(k)a†(k)]
+ d3 k
(2π)31
2ωk
d3 k
2ωk
δ3( k − k)ω2k[a(k)a†(k) + a†(k)a(k)] (31)
Equations (26) and (31) together with
d3xϕ(x, t) · ϕ(x, t)
= −
d3x d3 k
(2π)31
2ωk
d3 k
(2π)31
2ωk
k · k[e−ik·xa(k) − eik·xa†(k)][e−ik·xa(k) − eik·xa†(k)]
= −
d3x
d3 k
(2π)31
2ωk
d3 k
(2π)31
2ωk
k · k[e−i(k+k)·xa(k)a(k) + ei(k+k)·xa†(k)a†(k)]
+
d3x
d
3 k(2π)3 12ωk
d
3 k
(2π)3 12ωk k · k[e−i(k−k
)·xa(k)a†(k) + ei(k−k
)·xa†(k)a(k)]
=
d3 k
(2π)31
2ωk
d3 k
2ωk
δ3( k + k) k2[e−2iωkta(k)a(k) + e2iωkta†(k)a†(k)]
+
d3 k
(2π)31
2ωk
d3 k
2ωk
δ3( k − k) k2[a(k)a†(k) + a†(k)a(k)] (32)
enable us to transform the Hamiltonian.
H =1
2 d3x[π2(x, t) +
ϕ(x, t)
· ϕ(x, t) + m2ϕ2(x, t)]
=1
2
d3 k
(2π)31
2ωk
ωk[a†(k)a(k) + a(k)a†(k)]. (33)
Using the expansion for ϕ( t, t) and π(x, t), we can solve for a(k).
ωkϕ(x, t) = d3 k
(2π)31
2ωk
[ωk exp(−ik · x)a(k) + ωk exp(ik · x)a†(k)],
iπ(x, t) =
d3 k
(2π)31
2ωk
[ωk exp(−ik · x)a(k) − ωk exp(ik · x)a†(k)],
8
8/7/2019 Free Scalar Field Theory
http://slidepdf.com/reader/full/free-scalar-field-theory 9/25
ωkϕ(x, t) + iπ(x, t) = d3 k
(2π)31
2ωk
[(ωk + ωk) exp(−ik ·x)a(k) + (ωk −ωk)exp(ik ·x)a†(k)].
It follows that
d3
x exp(−i k · x)[ωkϕ(x, t) + iπ(x, t)]
=
d3xe−i k·x d3 k
(2π)31
2ωk
[(ωk + ωk)e−ik·xa(k) + (ωk − ωk)eik·xa†(k)]
=
d3 k
2ωk
[δ3( k − k)(ωk + ωk) exp(−iωkt)a(k) + δ3( k + k)(ωk − ωk)exp(iωkt)a†(k)]
= exp(−iωkt)a(k). (34)
Therefore,
a(k) =
d
3
x exp(ik · x)[ωkϕ(x, t) + iπ(x, t)], (35)
a†(k) =
d3x exp(−ik · x)[ωkϕ(x, t) − iπ(x, t)]. (36)
The equal-time commutation relations for ϕ(x, t) and π(x, t) determine the algebra of
a(k) and a†(k).
[a(k), a†(k)]
=
d3x
d3y exp(ik · x − ik · y)[ωkϕ(x, t) + iπ(x, t), ωkϕ(y, t) − iπ(y, t)]
=
d3x
d3y exp(ik · x − ik · y)(ωk + ωk)δ(x − y)
= (2π)32ωkδ( k − k). (37)
Similarly,
[a(k), a(k)]
=
d3x
d3y exp(ik · x + ik · y)[ωkϕ(x, t) + iπ(x, t), ωkϕ(y, t) + iπ(y, t)]
= − d3
x
d3
y exp(ik · x + ik
· y)(ωk − ωk
)δ(x − y)= 0. (38)
[a†(k), a†(k)]
=
d3x
d3y exp(−ik · x − ik · y)[ωkϕ(x, t) − iπ(x, t), ωkϕ(y, t) − iπ(y, t)]
=
d3x
d3y exp(ik · x − ik · y)(ωk − ωk)δ(x − y)
= 0. (39)
9
8/7/2019 Free Scalar Field Theory
http://slidepdf.com/reader/full/free-scalar-field-theory 10/25
With the commutation relations for a(k) and a†(k) at hand, we observe that the Hamil-
tonian, Eq.(33), is the continuous sum of harmonic oscillator Hamiltonians, one for each k.
a†(k) is a creation (raising) operator, while a(k) is a destruction (lowering) operator.
[H, a†(k)] = 12
d3 k
(2π)31
2ωk
ωk[a†(k)a(k) + a(k)a†(k), a†(k)]
= ωka†(k), (40)
[H, a†(k)] =1
2
d3 k
(2π)31
2ωk
ωk[a†(k)a(k) + a(k)a†(k), a(k)]
= −ωka(k). (41)
The ground state or bare vacuum is the state in Fock space that satisfies
a(k)|0 = 0, (42)
and is normalized so that 0|0 = 1. The particle interpretation results from considering
a†(k) as an operator that creates a particle of energy ωk and momentum k, while a(k)
destroys such a particle. The state a†(k)|0 is a state containing one particle of energy ωk
and momentum k. (a†(k))2|0 contains two such particles, and so on.
The usual probability interpretation requires that all physical states be normalizable.
The state a†|0 however is not normalizable.
0|a(k)a†(k)|0 = 0|[a(k), a†(k)]|0= (2π)32ωkδ(0). (43)
This is not surprising, because a†(k) creates a particle of definite energy and momentum. By
the uncertainty principle, we have no idea where the particle is located. Its wave function
is a plane wave and such states are nonnormalizable if the volume containing the system isinfinite. To obtain a normalizable state, we build a wave packet by superposition:
d3 k
(2π)31
2ωk
f ( k)a†(k)|0, (44)
with d3 k
(2π)31
2ωk
|f ( k)|2 < ∞. (45)
10
8/7/2019 Free Scalar Field Theory
http://slidepdf.com/reader/full/free-scalar-field-theory 11/25
E. Normal ordering
The Hamiltonian, Eq.(33),
H =1
2 d3 k
(2π)31
2ωkωk[a†(k)a(k) + a(k)a†(k)]
=1
2
d3 k
(2π)31
2ωk
ωk{2a†(k)a(k) + [a(k), a†(k)]}
= d3 k
(2π)31
2ωk
ωka†(k)a(k) +1
2
d3 kωkδ3(0). (46)
It follows that the vacuum expectation value of H , the sum of zero point energies of all of
the oscillators, is divergent.
0|H |0 =1
2
d3 kωkδ3(0). (47)
We will dispose off this divergence by observing that the absolute energy cannot be measured.
We will merely redefine the zero point of the energy scale. The Hamiltonian, Eq.(33), is
redefined by subtracting this infinite constant. This is formally accomplished by normal
ordering the operators that appear in H . Normal ordering, denoted by placing colons on
both sides of the operator product, means that all creation operators are to appear to the
left of destruction operators.
: a(k)a†(k) := a†(k)a(k) =: a†(k)a(k) : (48)
The normal ordered Hamiltonian,
: H := H − 0|H |0 =
d3 k
(2π)31
2ωk
ωka†(k)a(k). (49)
From hereon, we will assume that the Hamiltonian is normal ordered. We note that we
have an explicitly positive Hamiltonian once we normal order the operators. In this fashion,
we have been able to handle the question of negative energy states for the Klein-Gordon
equation.
F. Lorentz invariance
We started with a classical field theory that is relativistic. In order to canonically quantize
the theory, we had to specify equal-time commutation relations. That is, we must choose a
11
8/7/2019 Free Scalar Field Theory
http://slidepdf.com/reader/full/free-scalar-field-theory 12/25
specific Lorentz frame. The equal-time commutation relations are thus not Lorentz covari-
ant. However, we want to get the same quantum theory no matter which frame we choose.
That is, we want the quantized field theory to remain relativisitic. One way to verify that
we do get a relativistic quantum field theory is to verify that the quantum operator forms
of the generators of the Lorentz algebra still satisfy the proper algebra after quantization.
To this end, we have to calculate the momentum
P i =1
2
d3x[π(x, t)∂ iϕ(x, t) + ∂ iϕ(x, t)π(x, t)]
= −1
2
d3x
d3 kd3 k
(2π)64ωkωk
(ωkki + kiωk)[e−ik·xa(k) − eik·xa†(k)][e−ik·xa(k) − eik·xa†(k)]
= −1
2 d3x d3 kd3 k
(2π)64ωkωk
(ωkki + kiωk)[e−i(k+k)·xa(k)a(k) + ei(k+k)·xa†(k)a†(k)]
+1
2
d3x
d3 kd3 k
(2π)64ωkωk
(ωkki + kiωk)[e−i(k−k)·xa(k)a†(k) + ei(k−k)·xa†(k)a(k)]
= −1
2
d3 k
(2π)31
2ωk
d3 k
2ωk
δ3( k + k)(ωkki + kiωk)[e−2iωkta(k)a(k) + e2iωkta†(k)a†(k)]
+1
2
d3 k
(2π)31
2ωk
d3 k
2ωk
δ3( k − k)(ωkki + kiωk)[a(k)a†(k) + a†(k)a(k)]
=1
2
d3 k
(2π)31
2ωk
ki[a†(k)a(k) + a(k)a†(k)]. (50)
From hereon, we will assume that the momentum is normal ordered.
: P i :=
d3 k
(2π)31
2ωk
kia†(k)a(k). (51)
Also, for
M µν = −1
2
d3x[π(xµ∂ ν ϕ − xν ∂ µϕ) + (xµ∂ ν ϕ − xν ∂ µϕ)π]
=1
2 d3x
d3 k
(2π)31
2ωk d3 k
2ωk
[ωk(xµkν − xν kµ) + ωk(xµkν − xν kµ)]
×[e−ik·xa(k) − eik·xa†(k)][e−ik·xa(k) − eik·xa†(k)]
= −1
2
d3 k
(2π)31
2ωk
(xµkν − xν kµ)[a†(k)a(k) + a(k)a†(k)] (52)
we consider
: M µν : = −
d3 k
(2π)31
2ωk
(xµkν − xν kµ)a†(k)a(k)
= −i
d3 k
(2π)31
2ωk
a†(k)
kµ ∂
∂kν
− kν ∂
∂kµ
a(k). (53)
12
8/7/2019 Free Scalar Field Theory
http://slidepdf.com/reader/full/free-scalar-field-theory 13/25
It follows that
i[P µ, ϕ] = i
d3k
(2π)31
2ωk
kµ[a†(k)a(k), ϕ]
= d3 k
(2π)3
1
2ωk
[−
ikµ exp(−
ik·
x)a(k) + ikµ exp(ik·
x)a†(k)]
= ∂ µϕ (54)
And,
i[M µν , ϕ] = −i
d3k
(2π)31
2ωk
(xµkν − xν kµ)[a†(k)a(k), ϕ]
=
d3 k
(2π)31
2ωk
[i(xµkν − xν kµ) exp(−ik · x)a(k) − i(xµkν − xν kµ) exp(ik · x)a†(k)]
= −(xµ∂ ν − xν ∂ µ)ϕ (55)
For an infinitesimal translation δaµ,
(I + iδaµP µ)ϕ(x)(I − iδaµP µ)
= ϕ(x) + iδaµ[P µ, ϕ(x)]
= ϕ(x) + δaµ∂ µϕ(x)
= ϕ(x0 + δa0, xi − δai). (56)
And, for an infinitesimal Lorentz transformation δωµν ,
(I − i2
δωµν M µν )ϕ(x)(I + i2
δωµν M µν )
= ϕ(x) − i
2δωµν [M µν , ϕ(x)]
= ϕ(x) +1
2δωµν (xµ∂ ν − xν ∂ µ)ϕ(x)
= ϕ(x) − δωµν x
ν ∂ µϕ(x)
= ϕ(xµ − δωµν x
ν ). (57)
If we exponentiate the generators of translations and Lorentz transformations, we can cal-
culate how ϕ transforms under the Poincare group.
Last, we check for instance that
J i = −1
2imnM mn
=i
2imn
d3 k
(2π)31
2ωk
a†(k)
km ∂
∂kn
− kn ∂
∂km
a(k)
= −i
d3 k
(2π)31
2ωk
a†(k)
imnkm ∂
∂kn
a(k) (58)
13
8/7/2019 Free Scalar Field Theory
http://slidepdf.com/reader/full/free-scalar-field-theory 14/25
satisfies
[J i, J j ] = −
d3 k
(2π)31
2ωk
d3 k
(2π)31
2ωk
[a†(k)
imnkm ∂
∂kn
a(k), a†(k)
jpqk p
∂
∂k q
a(k)]
= iijk J k (59)
since
[a†(k)
imnkm ∂
∂kn
a(k), a†(k)
jpqk p
∂
∂k q
a(k)]
= a†(k)[a†(k)
imnkm ∂
∂kn
a(k),
jpqk p
∂
∂k q
a(k)]
+ [a†(k)
imnkm ∂
∂kn
a(k), a†(k)]
jpqk p
∂
∂k q
a(k)
= a†(k)[a†(k),
jpqk p ∂ ∂k q
a(k)]
imnkm ∂
∂kn
a(k)
+ a†(k)[
imnkm ∂
∂kn
a(k), a†(k)]
jpqk p
∂
∂k q
a(k)
= a†(k)
jpqk p
∂
∂k q
[a†(k), a(k)]
imnkm ∂
∂kn
a(k)
+ a†(k)
imnkm ∂
∂kn
[a(k), a†(k)]
jpqk p
∂
∂k q
a(k)
= −a†
(k
)
jpqk p ∂
∂k q
[(2π)3
2ωk
δ3
( k −
k
)]
imnkm ∂
∂kn
a(k)
+ a†(k)
imnkm ∂
∂kn
[(2π)32ωkδ3( k − k)]
jpqk p
∂
∂k q
a(k)
= [(2π)32ωkδ3( k − k)]a†(k)
jpmimnk p
∂
∂kn
a(k)
− [(2π)32ωkδ3( k − k)]a†(k)
imn jnqkm ∂
∂k q
a(k)
= −[(2π)32ωkδ3( k − k)]a†(k)
(δijδnp − δinδ jp)k p
∂
∂kn
a(k)
+ [(2π)32ωkδ3( k − k)]a†(k)
(δijδmq − δimδ jq)km ∂
∂k q
a(k)
= −(2π)32ωkδ3( k − k)a†(k)
ki ∂
∂k j− k j
∂
∂k i
a(k)
= −(2π)32ωkδ3( k − k)a†(k)
ijkklmkl ∂
∂km
a(k) (60)
14
8/7/2019 Free Scalar Field Theory
http://slidepdf.com/reader/full/free-scalar-field-theory 15/25
G. Microscopic causality
No signal can travel faster than the speed of light. Two events separated by a space-like
distance cannot affect each other. The creation or destruction of a particle is such an event,
so if our field theory is to be relativistic, then the commutator [ϕ(x), ϕ(y)] should vanish
when the separation of the points x and y is space-like, (x − y)2 = (x0 − y0)2 − (x − y)2 < 0.
[ϕ(x), ϕ(y)]
=
d3 k
(2π)31
2ωk
d3 k
(2π)31
2ωk
×[exp(−ik · x)a(k) + exp(ik · x)a†(k), exp(−ik · y)a(k) + exp(ik · y)a†(k)]
= d3 k
(2π)3
1
2ωk
d3 k
(2π)3
1
2ωk
×[exp(−ik · x)exp(ik · y)[a(k), a†(k)] + exp(ik · x) exp(−ik · y)[a†(k), a(k)]]
=
d3 k
(2π)31
2ωk
{exp[−ik · (x − y)] − exp[ik · (x − y)]}≡ i∆(x − y). (61)
∆(x − y) is Lorentz invariant because of the invariant measure and the appearance of only
vector dot products. From the equal-time commutation relations, we know that ∆(x−y) = 0
at equal times. Since ∆(x
−y) is Lorentz invariant, it then follows that ∆(x
−y) = 0 for
space-like separations as desired.
Explicitly, let t = x0 − y0 and r = |x − y|. Then, in radial coordinates
d3 k
(2π)31
2ωk
{exp[−ik · (x − y)] − exp[ik · (x − y)]}
= −i
d3 k
(2π)31
ωk
sin[k · (x − y)]
=
−
i
(2π)3
2π
0
dφ ∞
0 | k
|2d
| k
| π
0
sin θdθsin(ωkt)cos(| k|r cos θ) − cos(ωkt)sin(| k|r cos θ)
ωk
= − i
(2π)2
∞0
d| k| | k|2 sin(ωkt)
ωk
π0
dθ sin θ cos(| k|r cos θ)
+i
(2π)2
∞0
d| k| | k|2 cos(ωkt)
ωk
π0
dθ sin θ sin(| k|r cos θ)
= − i
(2π)2
∞
0d| k| |
k|2 sin(ωkt)
ωk
| k|r−| k|r
cos z
| k|r dz − ∞0
d| k| | k|2 cos(ωkt)
ωk
| k|r−| k|r
sin z
| k|r dz
= − i
2π2r ∞
0d| k| |
k| sin(ωkt)sin(| k|r)
ωk
(62)
15
8/7/2019 Free Scalar Field Theory
http://slidepdf.com/reader/full/free-scalar-field-theory 16/25
Let | k| = m sinh κ, then ω2k = | k|2 + m2 = m2(1 + sinh2 κ) = m2 cosh2 κ and d| k| =
m cosh κdκ = ωkdκ. Therefore,
∆(x
−y) =
−
1
2π2
r
∞
0
d
| k
|
| k| sin(ωkt)sin(| k|r)
ωk
=1
2π2r
∂
∂r
∞0
dκ sin(mt cosh κ)cos(mr sinh κ)
=1
4π2r
∂
∂r
∞0
dκ{sin[m(t cosh κ + r sinh κ)] + sin[m(t cosh κ − r sinh κ)]}
=1
4π2r
∂
∂r
∞0
dκ{sin[m√
t2 − r2 cosh(κ + α) + sin[m√
t2 − r2 cosh(κ − α)]}
=1
2π2r
∂
∂r
∞0
dκ sin[m√
t2 − r2 cosh κ]
=1
4πr
∂
∂rJ 0(m
√t2 − r2)
= − 1
4πrJ 1(m
√t2 − r2) (63)
if √
t2 − r2 > 0 and t > r. Here, J 0 and J 1 are the Bessel functions, cosh α = t/√
t2 − r2,
and sinh α = r/√
t2 − r2. (Exercise: Consider the case when t < −r.) Otherwise, t < r.
sinh α = t/√
t2 − r2 and cosh α = r/√
t2 − r2,
∆(x − y) =1
4π2r
∂
∂r
∞0
dκ{sin[m√
r2 − t2 sinh(κ + α) − sin[m√
r2 − t2 sinh(κ − α)]}
= 0. (64)
Several other interesting properties of ∆ are that it is odd,
∆(x − y) = −i d3 k
(2π)31
2ωk
{exp[−ik · (x − y)] − exp[ik · (x − y)]}
= −i
d3 k
(2π)31
2ωk
{exp[ik · (y − x)] − exp[−ik · (y − x)]}
= i d3 k
(2π)31
2ωk
{exp[−ik · (y − x)] − exp[ik · (y − x)]}= −∆(y − x), (65)
that
∂
∂x0∆(x − y)
x0=y0
= − 1
2
d3 k
(2π)3{exp[−ik · (x − y)] + exp[ik · (x − y)]}
x0=y0
= −1
2
d3 k
(2π)3{exp[i k · (x − y)] + exp[−i k · (x − y)]}
=
−δ3(x
−y), (66)
16
8/7/2019 Free Scalar Field Theory
http://slidepdf.com/reader/full/free-scalar-field-theory 17/25
and that ∆(x) satisfies the Klein-Gordon equation.
∆(x) = −
d3 k
(2π)31
ωk
sin[k · (x − y)],
(∂ µ∂ µ + m2)∆(x) =
d3 k
(2π)31
ωk(k2 − m2) sin[k · (x − y)] = 0. (67)
The vanishing of the commutator of two fields at space-like separations is called micro-
scopic causality . The commutator should vanish no matter how close the two points are as
long as their separation is space-like. One implication of this is that since the commutator
vanishes, we can make precise measurement of the fields at the two points simultaneously
without one measurement disturbing the other. However, we show below that one cannot
measure the square of a field operator at one point.
Consider the vacuum expectation of two fields at two different points
0|ϕ(x)ϕ(y)|0 = d3 k
(2π)31
2ωk
d3 k
(2π)31
2ωk
exp(−ik · x)exp(ik · y)0|a(k)a†(k)|0
=
d3 k
(2π)31
2ωk
exp[−ik · (x − y)]
≡ ∆+(x − y). (68)
It follows that
∆+(0) = 0|ϕ(x)ϕ(x)|0 =
d
3 k(2π)3 12ωk
, (69)
which is divergent. This means that the vacuum fluctuations of ϕ(x, t) are infinite.
∆ϕ ≡
0|ϕ2(x)|0 − 0|ϕ(x)|0=
∆+(0), (70)
since 0|ϕ(x)|0 = 0. The quantum mechanical vacuum fluctuations of ϕ(x) make it impos-
sible to measure the square of the field at one point.
In addition to ∆(x − y) and ∆+(x − y), we can define
∆−(x − y) ≡
d3 k
(2π)31
2ωk
exp[ik · (x − y)], (71)
so that
i∆(x − y) = ∆+(x − y) − ∆−(x − y). (72)
The + on ∆+ refers to the positive energy (frequency) part of ∆, while the − on ∆− refers to
the negative energy part. The vanishing of the vacuum expectation value of the commutator
17
8/7/2019 Free Scalar Field Theory
http://slidepdf.com/reader/full/free-scalar-field-theory 18/25
of two fields at space-like separations as the two points approach each other is accomplished
by subtracting two infinite quantities!
The vacuum expectation value of the anticommutator of ϕ(x) and ϕ(y),
0|{ϕ(x), ϕ(y)}|0 = ∆+(x − y) + ∆−(x − y), (73)
which does not vanish for space-like separations - had we quantized ϕ(x) using anticom-
mutators, we would violate causality. Lorentz invariance therefore ultimately provides the
connection between spin and statistics.
II. CHARGED SCALAR FIELDS
We can generalize our discussion of the Klein-Gordon field by postulating the existence
of several scalar fields. In particular, we can arrange two independent scalar fields into a
single complex field.
ϕ(x) ≡ 1√2
[ϕ1(x) + iϕ2(x)],
ϕ∗(x) =1√
2[ϕ1(x) − iϕ2(x)]. (74)
The action then becomes
A[ϕ1(x), ϕ2(x)] =1
2
d4x[∂ µϕ1(x)∂ µϕ1(x) + ∂ µϕ2(x)∂ µϕ2(x) − m2ϕ2
1(x) − m2ϕ22(x)]
=
d4x[∂ µϕ∗(x)∂ µϕ(x) − m2ϕ∗(x)ϕ(x)]
= A[ϕ(x), ϕ∗(x)]. (75)
From the action, we find the conjugate momentum fields
πa(x) =∂ L∂ (∂ tϕa) = ∂ tϕa, (76)
where a = 1, 2. Or,
π(x) =∂ L
∂ (∂ tϕ)= ∂ tϕ
∗,
π∗(x) =∂ L
∂ (∂ tϕ∗)= ∂ tϕ. (77)
18
8/7/2019 Free Scalar Field Theory
http://slidepdf.com/reader/full/free-scalar-field-theory 19/25
The Hamiltonian quickly follows,
H =
d3x[πa(x) · ∂ tϕa(x) − L]
=1
2 d3x[π2
a(x) +
ϕa(x)
· ϕa(x) + m2ϕ2
a(x)]
=
d3x[π∗(x)π(x) + ϕ∗(x) · ϕ(x) + m2ϕ∗(x)ϕ(x)]. (78)
The action A[ϕ(x), ϕ∗(x)] possesses a global U (1) invariance, that is, it is invariant under
ϕ(x) → exp(−i)ϕ(x) =1√
2exp(−i)[ϕ1(x) + iϕ2(x)],
ϕ∗(x) → exp(i)ϕ∗(x) =1√
2exp(i)[ϕ1(x) − iϕ2(x)], (79)
where is a constant. It follows that the conserved current
J µ = ∂ L∂ (∂ µϕ)
∂ϕ∂
+ ∂ L∂ (∂ µϕ∗)
∂ϕ∗
∂
= −i[(∂ µϕ∗)ϕ − ϕ∗∂ µϕ], (80)
and the conserved charge
Q =
d3xJ 0
= i
d3x[ϕ∗∂ tϕ − (∂ tϕ∗)ϕ] (81)
We note that the action A[ϕ1(x), ϕ2(x)] must possess the same symmetry. It is invariantunder
ϕ1
ϕ2
→
cos sin
− sin cos
ϕ1
ϕ2
. (82)
This symmetry will only exist if the mass of type 1 particle is equal to the mass of type 2
particle, m1 = m2 = m.
To canonically quantize the fields ϕ1(x) and ϕ2(x) (or, ϕ(x) and ϕ∗(x)), we take as our
quantum conditions:
[ϕa(x, t), πb(y, t)] = iδabδ3(x − y), (83)
[ϕa(x, t), ϕb(y, t)] = 0, (84)
[πa(x, t), πb(y, t)] = 0. (85)
Equivalently,
[ϕ(x, t), π(y, t)] =1
2[ϕ1(x, t) + iϕ2(x, t), π1(x, t) − iπ2(x, t)]
= iδ3(x
−y), (86)
19
8/7/2019 Free Scalar Field Theory
http://slidepdf.com/reader/full/free-scalar-field-theory 20/25
[ϕ†(x, t), π†(y, t)] =1
2[ϕ1(x, t) − iϕ2(x, t), π1(x, t) + iπ2(x, t)]
= iδ3(x − y), (87)
with all other equal-time commutators vanishing.
A particle interpretation is introduced by Fourier transforming ϕa.
ϕ1(x, t) =
d3 k
(2π)31
2ωk
[a1(k) exp(−ik · x) + a†1(k)exp(ik · x)],
ϕ2(x, t) =
d3 k
(2π)31
2ωk
[a2(k) exp(−ik · x) + a†2(k)exp(ik · x)]. (88)
Similarly,
ϕ(x, t) =
d3 k
(2π)31
2ωk
[a(k) exp(−ik · x) + b†(k) exp(ik · x)],
ϕ†(x, t) =
d3 k
(2π)31
2ωk
[b(k)exp(−ik · x) + a†(k)exp(ik · x)], (89)
where
a(k) ≡ 1√2
[a1(k) + ia2(k)],
b(k) ≡ 1√2
[a1(k) − ia2(k)]. (90)
It follows that
a1(k) =
d3
x exp(ik · x)[ωkϕ1(x, t) + iπ1(x, t)],
a†1(k) =
d3x exp(−ik · x)[ωkϕ1(x, t) − iπ1(x, t)], (91)
and
a2(k) =
d3x exp(ik · x)[ωkϕ2(x, t) + iπ2(x, t)],
a†2(k) =
d3x exp(−ik · x)[ωkϕ2(x, t) − iπ2(x, t)], (92)
satisfy
[aa(k), a
†
b(k
)] = (2π)3
2ωkδabδ( k −
k
), (93)
[aa(k), ab(k)] = 0 = [a†a(k), a†b(k)].
Similarly,
a(k) =1√
2[a1(k) + ia2(k)]
=
d3x exp(ik · x){ωk
1√2
[ϕ1(x, t) + iϕ2(x, t)] + i1√
2[π1(x, t) + iπ2(x, t)]}
= d3x exp(ik · x)[ωkϕ(x, t) + iπ†(x, t)], (94)
20
8/7/2019 Free Scalar Field Theory
http://slidepdf.com/reader/full/free-scalar-field-theory 21/25
a†(k) =
d3x exp(−ik · x)[ωkϕ†(x, t) − iπ(x, t)], (95)
and
ˆb(k) =
1
√2[a1(k) − ia2(k)]
=
d3x exp(ik · x){ωk
1√2
[ϕ1(x, t) − iϕ2(x, t)] + i1√
2[π1(x, t) − iπ2(x, t)]}
=
d3x exp(ik · x)[ωkϕ†(x, t) + iπ(x, t)], (96)
b†(k) =
d3x exp(−ik · x)[ωkϕ(x, t) − iπ†(x, t)], (97)
satisfy
[a(k), a†
(k
)] =
d3
x
d3
yeik·x
e−ik·y
[ωkϕ(x, t) + iπ†
(x, t), ωkϕ†
(y, t) − iπ(y, t)]
= (2π)32ωkδ( k − k), (98)
[b(k), b†(k)] =
d3x
d3yeik·xe−ik·y[ωkϕ†(x, t) + iπ(x, t), ωkϕ(y, t) − iπ†(y, t)]
= (2π)32ωkδ( k − k), (99)
with all others vanishing.
The ground state or bare vacuum satisfies
a1(k)|0 = 0 = a2(k)|0, (100)
or equivalently
a(k)|0 = b(k)|0 = 0. (101)
It follows from Eqs.(93), (98), and (99) that the operator
ˆN 1 =
d3 k
(2π)31
2ωk a
†
1(k)a1(k) (102)
counts the number of type 1 particle, while
N 2 = d3 k
(2π)31
2ωk
a†2(k)a2(k) (103)
does the same for type 2 particle. Similarly,
N + =
d3 k
(2π)31
2ωk
a†(k)a(k) (104)
21
8/7/2019 Free Scalar Field Theory
http://slidepdf.com/reader/full/free-scalar-field-theory 22/25
is the number operator for + type particles and
N − =
d3 k
(2π)31
2ωk
b†(k)b(k) (105)
for
−type particles.
Clearly, N + + N − = N 1 + N 2. N 1, N 2, N +, and N − all commute with the Hamiltonian so
they are conserved. We may choose our Fock space to be eigenstates of N 1 and N 2 or N +
and N −.
Last, the conserved charge
: Q : = i
d3x : [ϕ†(x)π†(x) − π(x)ϕ(x)] :
=
d3 k
(2π)31
2ωk
[a†(k)a(k) − b†(k)b(k)]
= N + − N − (106)
since d3xϕ†(x)π†(x)
=
d3x
d3 k
(2π)31
2ωk
d3 k
(2π)31
2ωk
[b(k)e−ik·x + a†(k)eik·x][iωk b†(k)eik·x − iωk a(k)e−ik·x],
(107)
d3xπ(x)ϕ(x)
=
d3x
d3 k
(2π)31
2ωk
d3 k
(2π)31
2ωk
[iωka†(k)eik·x − iωkb(k)e−ik·x][a(k)e−ik·x + b†(k)eik·x].
(108)
Therefore, the + type particles, which are created by ϕ† and destroyed by ϕ have charge
+1, while the − type particles, which are created by ϕ and destroyed by ϕ†, carry charge
−1. The net effect of ϕ is to lower the charge by one unit, either by destroying a +1 charge
or creating a −1 charge, while operating with ϕ† raises the total charge by one unit.
III. TIME-ORDERING AND PROPAGATORS
Consider the propagation of charge in the charged scalar field theory. The state corre-
sponding to a particle of charge +1 at x is
ϕ†(x)|0 =
d3 k
(2π)31
2ωk
exp(ik · x)a†(k)|0, (109)
22
8/7/2019 Free Scalar Field Theory
http://slidepdf.com/reader/full/free-scalar-field-theory 23/25
while the state corresponding to the particle at x is
ϕ†(x)|0 = d3 k
(2π)31
2ωk
exp(ik · x)a†(k)|0. (110)
Thus, the quantum mechanical amplitude to transport the charge from x to x
is
0|ϕ(x)ϕ†(x)|0
=
d3 k
(2π)31
2ωk
d3 k
(2π)31
2ωk
θ(t − t)exp[−i(k · x − k · x)]0|a(k)a†(k)|0
= d3 k
(2π)31
2ωk
θ(t − t) exp[−ik · (x − x)]. (111)
Apparantly, we can interpret the propagation of charge as the creation of a particle of +1
charge out of the vacuum at x, the transport from x to x
, and the reabsorption of theparticle into the vacuum at x. Since we can’t absorb the particle before it is created, this
process only makes sense if t > t.
Equation (111) is not the total amplitude for propagation of +1 charge. The total ampli-
tude is the sum of all of the amplitudes of different processes that give equivalent physical
results. In the process above, the charge at x was increased by one unit, while the charge
at x was lowered by one unit. We can accomplish the same thing by creating a particle of
−1 charge at x
:ϕ(x)|0 =
d3 k
(2π)31
2ωk
exp(ik · x)b†(k)|0, (112)
and transporting it to x:
ϕ(x)|0 = d3 k
(2π)31
2ωk
exp(ik · x)b†(k)|0, (113)
then destroying it:
0|ϕ†(x)ϕ(x)|0 = d3 k
(2π)31
2ωkθ(t − t)exp[ik · (x − x)]. (114)
As before, we can’t destroy a particle before it is created, so this process only makes sense
if t > t.
The total amplitude for propagation, G(x, x), is the sum of the two amplitudes,
G(x, x) = θ(t − t)0|ϕ(x)ϕ†(x)|0 + θ(t − t)0|ϕ†(x)ϕ(x)|0= 0|T [ϕ(x)ϕ†(x)]|0. (115)
23
8/7/2019 Free Scalar Field Theory
http://slidepdf.com/reader/full/free-scalar-field-theory 24/25
Here, we introduce the Dyson time-ordering operator T .
T [ϕ(x)ϕ†(x)] =
ϕ(x)ϕ†(x) if t > t
ϕ†(x)ϕ(x) if t > t(116)
T orders operators by their time-ordering. Operators that occur at later times appear to
the left of operators that occur at earlier times.
From Eqs.(111) and (113), it follows that
G(x, x) = d3 k
(2π)31
2ωk
{θ(t − t) exp[−ik · (x − x)] + θ(t − t) exp[ik · (x − x)]}. (117)
To cast G(x, x) into a covariant form , we recall an integral representation for the Heaviside
step function,
θ(t) = lim→0+
dω
2πi
exp(iωt)
ω − i. (118)
Subtracting i displaces the pole above the Re(ω) axis. If t > 0, we close the contour above
the Re(ω) axis enclosing the pole. If t < 0, we close the contour in the bottom half-plane,
missing the pole, so the integral vanishes. Here, we make use of the Cauchy’s integral
formula . If f (z) is analytic on and inside a simple closed curve C , the value of f (z) at a
point z = a inside C is given by the following contour integral along C :
12πi
f (z)z − a
dz = f (a). (119)
Inserting the integral into Eq.(117), we have
G(x, x) = lim→0+
dω
2πi
d3 k
(2π)31
2ωk
1
ω − i[eiω(t
−t)e−ik·(x−x) + eiω(t−t)eik·(x−x)]
= lim→0+
dω
2πi
d3 k
(2π)31
2ωk
1
ω − i[ei(ω−ωk)(t
−t)ei k·( x− x) + e−i(ω−ωk)(t
−t)e−i k·( x− x)]
= − lim→0+ dk0
2πi d3 k
(2π)3
1
2ωk
1
ωk − k0 − i e
−ik0(t−t)
e
i k·( x−x)
− lim→0+
dk02πi
d3 k
(2π)31
2ωk
1
ωk + k0 − ie−ik0(t−t)ei
k·( x−x)
= i lim→0+
d4k
(2π)4e−ik·(x−x)
2ωk
1
ωk − k0 − i+
1
ωk + k0 − i
= i lim→0+
d4k
(2π)4e−ik·(x−x)
2ωk
2ωk − 2i
(ωk − i)2 − k20
= i lim→0+
d4k
(2π)4e−ik·(x−x)
ω2k
−2iωk
−k20
24
8/7/2019 Free Scalar Field Theory
http://slidepdf.com/reader/full/free-scalar-field-theory 25/25
= i lim→0+
d4k
(2π)4e−ik·(x−x)
k · k + m2 − k20 − i
= i lim→0+
d4k
(2π)4e−ik·(x−x)
−k · k + m2 − i
= −i lim→0+
d4k
(2π)4e−ik·(x
−x)
k · k − m2 + i
≡ −i∆F (x − x) (120)
In this form, it is clear that ∆F (x − x) satisfies
∂
∂xµ
∂
∂xµ
+ m2
∆F (x − x) = lim
→0+
d4k
(2π)4(−k2 + m2)e−ik·(x−x)
k · k − m2 + i
= −δ4(x − x). (121)
So, G(x, x) is a Green’s function for the Klein-Gordon equation.
The i prescription in
∆F (x − x) = lim→0+
d4k
(2π)4e−ik·(x−x)
k · k − m2 + i
implements the boundary conditions, namely, that ∆F describes the propagation of a particle
from x to x when t > t, and the propagation of an antiparticle from x to x when t > t.
i accomplishes this by displacing the positive energy pole (k0 = ωk) below the k0 axis and
the negative energy pole (k0 = −ωk) above the axis.
∆F (x − x) = − lim→0+
d4k
(2π)4e−ik·(x−x)
2ωk
1
ωk − k0 − i+
1
ωk + k0 − i
= lim→0+
d4k
(2π)4e−ik·(x−x)
2ωk
1
k0 − ωk + i− 1
k0 + ωk − i
For t > t, we must close the contour in the lower half-plane; thus only the positive energy
pole will be included. For t < t, the opposite is true. If we fix t > t, then ∆F will only
propagate positive energy states forward in time and negative energy states backwards in
time. From this observation, we can see that antiparticles of positive energy propagating
forward in time can be interpreted as negative energy partilces propagating backwards in
time.
25