Free Scalar Field Theory

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Free Scalar Field Theory

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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).

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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)

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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)

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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.

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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)

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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)

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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)],

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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)

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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)

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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



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)

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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



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



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



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



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



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



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



[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



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



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



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



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


2ωk

2ωk − 2i

(ωk − i)2 − k20

= i lim→0+

d4k


ω2k

−2iωk

−k20

24



= i lim→0+

d4k


k · k + m2 − k20 − i

= i lim→0+

d4k


−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


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


2ωk

1

ωk − k0 − i+

1

ωk + k0 − i

= lim→0+

d4k


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

Free Scalar Field Theory

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