First to Second Quantization

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I. QUANTUM MECHANICS

Quantum field theory is based on the same quantum mechanics that was invented by

Heisenberg, Born, Schrodinger, Pauli, and others in 1925-26. Therefore, for completeness

and continuity, we will review first quantization, that is, (ordinary) nonrelativistic quantum

mechanics, before going on to second quantization.

A. The postulates

In any quantum mechanical process, the object of interest to calculate is the probability

amplitude - a complex number. The square of the modulus of the amplitude is proportional

to the probability that the process will occur. The postulates of quantum mechanics are

rules on how to calculate the amplitude for any process.

For a given physical system, there is a complete set of states that describes all of the

possible configurations of the system. A state, denoted by |Ψ〉, is a vector in some Hilbert

space. c|Ψ〉, c a constant, represents the same physical state. The inner product of two

states, |Ψ1〉 and |Ψ2〉, is written as 〈Ψ2|Ψ1〉. The amplitude for the process that takes the

system from the state |Ψ1〉 to |Ψ2〉 is just 〈Ψ2|Ψ1〉. The probability for this process to occur

is |〈Ψ2|Ψ1〉|2.With every physical observable A, there is associated a Hermitian operator A that acts

on the Hilbert space containing the states. If a measurement is made of the observable A,

the result of the measurement must be one of the real eigenvalues of A - no other values

are possible. The expectation value of the physical quantity A in the state |Ψ〉 is given by

〈Ψ|A|Ψ〉, where the state vector is normalized to 〈Ψ|Ψ〉 = 1.

Last, quantum mechanics provides for time evolution by requiring the states of the system

to satisfy the Schrodinger equation,

ihd

dt|Ψ(t)〉 = H|Ψ(t)〉. (1)

H is called the Hamiltonian operator and generates infinitesimal time translations. Suppose

it is time-independent, the formal solution to the Schrodinger equation is

|Ψ(t)〉 = exp(− i

htH

)|Ψ(0)〉 ≡ U(t)|Ψ(0)〉. (2)

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U(t) is the time evolution operator. Substituting Eq.(2) into Eq.(1),

ihd

dt|Ψ(t)〉 = ih

d

dtU(t)|Ψ(0)〉

= HU(t)|Ψ(0)〉= H|Ψ(t)〉 (3)

we obtain

H = ih

[d

dtU(t)

]U †(t) = −ihU(t)

d

dtU †(t). (4)

If we have a classical Hamiltonian system (say, a particle) to quantize, we may obtain

the Hamiltonian operator H using its classical Hamiltonian H(x, p) as a guide. Since both

the position x and momentum p of a particle are observable quantities, upon quantization,

they become operators x and p. We can turn H(x, p) into H by replacing x with x and p

with p. To complete the quantization, we must specify quantum conditions for x and p:

[x, p] = ih. (5)

x and p are defined such that their commutators are equal to ih times the corresponding

classical Poisson bracket. Since x and p do not commute, potential operator ordering prob-

lems arise in constructing H from H(x, p). For example, operator products xp and px are

inequivalent while xp and px are the same in the classical system. In general, we choose a

symmetric ordering, xp → (xp + px)/2.

If the quantum mechanical system we are trying to describe has no classical analog, such

as a system with spin, then we must guess what the Hamiltonian operator is and what the

quantum conditions are.

B. Heisenberg picture

So far we have been describing the Schrodinger picture, where the states of the system

are time dependent while the operators are not. In the Heisenberg picture, the states are

time independent. The operators carry the time dependence. The transformation between

the two pictures is done with the time evolution operator U(t). We want to obtain the same

results in both representations. In particular, the expectation of any operator A should be

the same:

〈Ψ(t)|A|Ψ(t)〉 = 〈Ψ(0)|U †(t)AU(t)|Ψ(0)〉 ≡ 〈Ψ(0)|AH(t)|Ψ(0)〉 (6)

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The time dependent operators in the Heisenberg picture are related to the Schrodinger

picture operators by

AH(t) = U †(t)AU(t). (7)

Since the states are now time independent, they do not have to satisfy any Schrodinger

equation. The dynamics is locked into the time dependent operators. They must satisfy an

operator equation of motion,

d

dtAH =

(d

dtU †

)AU + U †A

d

dtU

= U †U

(d

dtU †

)AU + U †A

d

dtUU †U

=i

hU †HAU − i

hU †AHU

=i

hHHAH − i

hAHHH

=i

h[HH, AH] (8)

Note that if AH commutes with HH, it is a constant of the motion. If the operator AH has

an explicit time dependence, then we must add ∂AH/∂t to the right-hand side of Eq.(8).

Equation (8) may also be obtained from the classical system by replacing ih times the

Poisson bracket with a commutator, just as in the quantum conditions, Eq.(5).

In summary, to quantize a classical system in the Heisenberg picture, we elevate observ-

ables to operators, specify quantum conditions, choose an operator ordering, and specify

a Hilbert space containing the time independent state vectors. To compute the quantum

dynamics, we must solve the operator equation of motion. We note that since xH and pH

are time dependent, we specify the commutator at equal times,

[xH(t), pH(t)] = ih. (9)

C. Harmonic oscillator in the Heisenberg picture

The classical Hamiltonian for the harmonic oscillator is

H =1

2p2 +

1

2ω2x2 =

1

2hω

(− i√

hωp +

√ω

hx

) (i√hω

p +

√ω

hx

). (10)

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To quantize this, we make x and p the operators x and p, and demand them to satisfy the

commutation relation, [x, p] = ih. It follows that

H =1

2p2 +

1

2ω2x2, (11)

dx

dt=

i

h[H, x] =

i

2h[p2, x] = p, (12)

d2x

dt2=

dp

dt=

i

h[H, p] =

iω2

2h[x2, p] = −ω2x. (13)

We note that

x(t) = A cos ωt + B sin ωt (14)

solves Eq.(13). And,

p(t) = −ωA sin ωt + ωB cos ωt. (15)

It follows that

√ωx =

√ωA cos ωt +

√ωB sin ωt,

1√ω

p = −√ωA sin ωt +√

ωB cos ωt,

and

√ωx +

i√ω

p =√

ω(A + iB) exp(−iωt),

√ωx− i√

ωp =

√ω(A− iB) exp(iωt).

Alternatively, we could express

H = hω(a†a +

1

2

), (16)

where

a =1√2

(√ω

hx +

i√hω

p

),

a† =1√2

(√ω

hx− i√

hωp

), (17)

with

[a, a†] =1

2[

√ω

hx +

i√hω

p,

√ω

hx− i√

hωp] = 1, (18)

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and

d

dta =

i

h[H, a] = iω[a†a, a] = −iωa,

d

dta† =

i

h[H, a†] = iω[a†a, a†] = iωa†. (19)

Substituting x =√

h/2ω(a + a†) and p = −i√

hω/2(a− a†) into the Hamiltonian operator,

we find

H =1

2p2 +

1

2ω2x2

=1

4hω[−(a− a†)2 + (a + a†)2]

=1

2hω(aa† + a†a)

= hω(a†a +

1

2

)(20)

Clearly if a state |n〉 is an eigenvector of H, then it must be an eigenvector of the number

operator N = a†a and vice versa.

N |n〉 = n|n〉,H|n〉 =

(n +

1

2

)hω|n〉. (21)

Next, consider the norm of the state a|n〉,

〈n|a†a|n〉 = n〈n|n〉 ≥ 0 and 〈n|n〉 > 0 ⇒ n ≥ 0. (22)

We will normalize the states |n〉 such that 〈n|n〉 = 1. The ground state has n = 0. It is the

state of lowest energy hω/2 and |0〉 satisfies

a|0〉 = 0. (23)

The first excited state has n = 1, the second n = 2, etc. The spectrum of H is En =

(n + 1/2)hω. The energy levels are evenly spaced with separation hω. To reach the excited

states from the ground state |0〉, we observe that

[N , a†] = [a†a, a†] = a†[a, a†] = a†. (24)

Thus,

N a†|n〉 = (a†N + [N , a†])|n〉 = (n + 1)a†|n〉 = (n + 1)λ|n + 1〉. (25)

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Similarly, we have

[N , a] = [a†a, a] = [a†, a]a = −a, (26)

N a|n〉 = (aN + [N , a])|n〉 = (n− 1)a|n〉 = (n− 1)µ|n− 1〉. (27)

λ and µ are constants that are fixed by making sure the states generated are properly

normalized.

λ|n + 1〉 = a†|n〉⇒ λ2 = 〈n|aa†|n〉 = 〈n|(a†a + 1)|n〉 = n + 1

⇒ λ =√

n + 1 (28)

µ|n− 1〉 = a|n〉⇒ µ2 = 〈n|a†a|n〉 = n

⇒ µ =√

n (29)

It follows that all the eigenstates of H can be generated by applying a† consecutively to |0〉.In general,

|n〉 =1√n!

(a†)n|0〉. (30)

The operators a† and a are called the ladder operators, or raising and lowering operators,

because they make states that march up and down the ladder of excitations. As we shall soon

see, free quantum field theories reduce to a collection of independent harmonic oscillators,

one for each energy-momentum. The raising and lowering operators of the collection of

oscillators provide a particle interpretation. To find a particle interpretation of any quantum

field we look for operators similar to a† and a in the quantum field theory.

D. Coordinate representation

The position x of a particle is a physical observable. Let the associated operator be x.

The eigenvectors of x satisfy

x|x〉 = x|x〉. (31)

The eigenvectors are normalized so that

〈x′|x〉 = δ(x′ − x), (32)

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where δ is the Dirac distribution. They are assumed complete,∫

dx|x〉〈x| = 1, (33)

and form a basis for the Hilbert space of states. If we expand a state vector |Ψ(t)〉 in the

position basis |x〉, the component of |Ψ(t)〉 in the |x〉 direction is 〈x|Ψ(t)〉, which is just a

number Ψ(x, t) - the wave function representing the state |Ψ(t)〉.

|Ψ(t)〉 =(∫

dx|x〉〈x|)|Ψ(t)〉 =

∫dx|x〉〈x|Ψ(t)〉 =

∫dxΨ(x, t)|x〉 (34)

|x〉 is the state of the system where the particle is at position x. 〈x|Ψ(t)〉 = Ψ(x, t) is the

amplitude for the system in the state |Ψ(t)〉 to also be in the state |x〉. Thus, |Ψ(x, t)|2dx

is the probability that the particle will be found in (x− dx/2, x + dx/2) when the system is

in state |Ψ(t)〉. Since the particle must be found somewhere,

1 =∫

dx|Ψ(x, t)|2 =∫

dx〈Ψ(t)|x〉〈x|Ψ(t)〉 = 〈Ψ(t)|Ψ(t)〉. (35)

In the coordinate representation, the Schodinger equation (1) becomes

ih∂

∂t〈x|Ψ(t)〉 = 〈x|H|Ψ(t)〉,

ih∂

∂tΨ(x, t) =

∫dx′〈x|H|x′〉〈x′|Ψ(t)〉

=∫

dx′〈x|H|x′〉Ψ(x′, t). (36)

〈x|H|x′〉 is called the matrix element of the operator H in the position basis. To compute

〈x|H|x′〉, we must determine the matrix elements 〈x|x|x′〉 of x and 〈x|p|x′〉 of p. Since |x〉is an eigenvector of x,

〈x|x|x′〉 = x〈x|x′〉 = xδ(x− x′). (37)

Suppose p has eigenvectors |p〉, then

〈x|p|x′〉 =∫

dp〈x|p|p〉〈p|x′〉

=∫

dp p〈x|p〉〈p|x′〉

=∫

dp

(−ih

∂

∂x

)〈x|p〉〈p|x′〉

= −ih∂

∂x

∫dp〈x|p〉〈p|x′〉

= −ih∂

∂x〈x|x′〉

= −ih∂

∂xδ(x− x′) (38)

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If the Hamiltonian is

H =1

2p2 + V (x), (39)

then

〈x|H|x′〉 =

[− h2

2

∂2

∂x2+ V (x)

]δ(x− x′). (40)

Substituting Eq.(40) into Eq.(36), we recover the familiar Schrodinger equation for the wave

function.

ih∂

∂tΨ(x, t) = − h2

2

∂2

∂x2Ψ(x, t) + V (x)Ψ(x, t). (41)

We note that

ihΨ∗(x, t)∂

∂tΨ(x, t) = − h2

2Ψ∗(x, t)

∂2

∂x2Ψ(x, t) + V (x)Ψ∗(x, t)Ψ(x, t),

−ihΨ(x, t)∂

∂tΨ∗(x, t) = − h2

2Ψ(x, t)

∂2

∂x2Ψ∗(x, t) + V (x)Ψ∗(x, t)Ψ(x, t),

from which we derive

∂

∂t(Ψ∗Ψ)− ih

2

∂

∂x

(Ψ∗∂Ψ

∂x−Ψ

∂Ψ∗

∂x

)= 0. (42)

From hereon, we set h = 1 and replace Ψ(x, t) with ϕ(x, t).

II. SECOND QUANTIZATION

To second quantize the Schrodinger equation,

i∂

∂tϕ(x, t) = −1

2

∂2

∂x2ϕ(x, t) + V (x)ϕ(x, t), (43)

we want to make the first quantized wave function ϕ(x, t) an operator ϕ(x, t), a time-

dependent field operator (so we will be working in the Heisenberg picture). Equation (43)

will then become an operator equation of motion,

i∂

∂tϕ(x, t) = −1

2

∂2

∂x2ϕ(x, t) + V (x)ϕ(x, t). (44)

The normalized eigenfunctions ϕn(x), with eigenvalue en, of the first quantized Hamilto-

nian,

−1

2

∂2

∂x2+ V (x), (45)

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are assumed to form a complete set, thus any solution ϕ(x, t) to the Schrodinger equation

(43) can be expanded in terms of the ϕn’s:

ϕ(x, t) =∑n

an(t)ϕn(x). (46)

In the first quantized system, ϕ(x, t) and ϕn(x) are wave functions and an(t) is just a number

times exp(−ient). After second quantizing, ϕ(x, t) becomes an operator. Making an(t) an

operator and leaving ϕn(x) a function,

ϕ(x, t) =∑n

an(t)ϕn(x) (47)

clearly solves Eq.(44).

In quantum mechanics, operator equations of motion take the form of Eq. (8),

∂

∂tϕ(x, t) = i[H, ϕ(x, t)]. (48)

Therefore, we must find a field Hamiltonian H and quantum conditions (equal-time com-

mutation relations) such that Eq.(48) reproduces Eq.(44). The commutator is the field

equivalent of [x(t), p(t)] = i for the first quantized system. In this field theory, ϕ(x, t) plays

the role of x(t). To write down the quantum commutator involving ϕ(x, t), we must find

what the momentum field π(x, t) conjugate to ϕ(x, t) is. Our strategy will be to treat the

first quantized system as a classical field theory, find an action principle that yields the

field equation (43), and use the action as a crutch to find the mometum field π(x, t) and

Hamiltonian H.

III. CLASSICAL MECHANICS

A. Methods of Lagrange and Hamilton

A classical mechanical system is described by a set of generalized coordinates q1, · · · , qN ,

the associated velocities, q1, · · · , qN , and a Lagrangian L[qi(t), qi(t), t]. The dot denotes the

time derivative d/dt. The Lagrangian governs the dynamics and is, at most, a quadratic

function of qi. The time integral

A[qi(t)] =∫ tb

tadt L[qi(t), qi(t), t] (49)

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over the Lagrangian along an arbitrary path qi(t) is called the action of this path. The path

qi(t) that is actually chosen by the system as a function of time is called the classical orbit,

qcli (t). It has the property of extremizing the action in comparison with all neighboring paths

qi(t) = qcli (t) + δqi(t) (50)

with fixed end points: δqi(ta) = δqi(tb) = 0. To express this property formally, we consider

the variation of A[qi(t)],

δA[qi(t)] ≡ {A[qi(t) + δqi(t)]−A[qi(t)]}lin

=∫ tb

tadt {L[qi(t) + δqi(t), qi(t) + δqi(t), t]− L[qi(t), qi(t), t]}lin

=∫ tb

tadt

[∂L

∂qi

δqi(t) +∂L

∂qi

δqi(t)

]

=∫ tb

tadt

[∂L

∂qi

− d

dt

(∂L

∂qi

)]δqi(t) +

∂L

∂qi

δqi(t)

∣∣∣∣∣tb

ta

=∫ tb

tadt

[∂L

∂qi

− d

dt

(∂L

∂qi

)]δqi(t) (51)

Here, repeated indices are understood to be summed - Einstein’s summation convention. For

the classical orbit qcli (t), δA[qi(t)] = 0 (Hamilton’s principle of least action) and we obtain

the Euler-Lagrange equations:∂L

∂qi

− d

dt

(∂L

∂qi

)= 0. (52)

There is an alternative formulation of classical dynamics which is based on a Legendre

transformed function of the Lagrangian called the Hamiltonian

H[qi(t), pi(t), t] ≡ qi(t)∂

∂qi

L[qi(t), qi(t), t]− L[qi(t), qi(t), t]. (53)

Its value at any time is identified with the energy of the system. The natural variables in H

are no longer qi(t) and qi(t), but qi(t) and the canonically conjugate momenta pi(t) defined

by the equations

pi(t) ≡ ∂

∂qi

L[qi(t), qi(t), t]. (54)

In order to specify the Hamiltonian H[qi(t), pi(t), t] in terms of its proper variables qi(t) and

pi(t), the equations (54) for pi(t) have to be solved for qi(t),

qi(t) = vi[qi(t), pi(t), t], (55)

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and inserted into Eq.(53), giving

H[qi(t), pi(t), t] = vi[qi(t), pi(t), t]pi(t)− L[qi(t), vi[qi(t), pi(t), t], t]. (56)

In terms of H, we have the canonical form of the action, Eq.(49),

A[qi(t), pi(t)] =∫ tb

tadt {qi(t)pi(t)−H[qi(t), pi(t), t]}. (57)

The variation of A[qi(t), pi(t)],

δA[qi(t), pi(t)] =∫ tb

tadt

[δqi(t)pi(t) + qi(t)δpi(t)− ∂H

∂qi

δqi(t)− ∂H

∂pi

δpi(t)

]

=∫ tb

tadt

{[qi(t)− ∂H

∂pi

]δpi(t)−

[pi(t) +

∂H

∂qi

]δqi(t)

}(58)

since δqi(ta) = δqi(tb) = 0. For the classical orbit, qcli (t) and pcl

i (t), δA[qi(t), pi(t)] = 0 and

we obtain the Hamilton equations:

qi(t) =∂H

∂pi

,

pi(t) = −∂H

∂qi

. (59)

These agree with the Euler-Lagrange equations (52) via Eq.(54), as can easily be verified.

pi =∂L

∂qi

=∂L

∂vi

⇒ pi =d

dt

(∂L

∂qi

),

∂H

∂qi

=∂vj

∂qi

pj − ∂L

∂qi

− ∂L

∂vj

∂vj

∂qi

=∂vj

∂qi

pj − ∂L

∂qi

− ∂vj

∂qi

pj

= −∂L

∂qi

.

The 2N -dimensional space of all qi and pi is called a phase space. An arbitrary function

F [qi(t), pi(t), t] changes along an arbitrary path as follows

d

dtF [qi(t), pi(t), t] =

∂F

∂qi

qi +∂F

∂pi

pi +∂F

∂t. (60)

If the path is a classical orbit, we may insert Eq.(59) and find

d

dtF [qi(t), pi(t), t] =

∂F

∂qi

∂H

∂pi

− ∂F

∂pi

∂H

∂qi

+∂F

∂t

=∂F

∂qi

∂H

∂pi

− ∂H

∂qi

∂F

∂pi

+∂F

∂t

≡ {F, H}+∂F

∂t, (61)

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where we have introduced the Poisson bracket,

{A,B} ≡ ∂A

∂qi

∂B

∂pi

− ∂B

∂qi

∂A

∂pi

, (62)

again with the Einstein summation convention for the repeated index i. From Eq.(61), a

function F [qi(t), pi(t)] which has no explicit dependence on time and which commutes with

H, i.e., {F, H} = 0, is a constant of motion along the classical orbit. In particular, H itself

is often of this type, i.e., H = H[qi(t), pi(t)]. In this case, the energy is a constant of motion.

Obviously, the original Hamilton equations themselves are a special case of Eq.(61).

d

dtqi = {qi, H} =

∂qi

∂qj

∂H

∂pj

− ∂H

∂qj

∂qi

∂pj

=∂H

∂pi

, (63)

d

dtpi = {pi, H} =

∂pi

∂qj

∂H

∂pj

− ∂H

∂qj

∂pi

∂pj

= −∂H

∂qi

. (64)

We also note that at each time t,

{qi, pj} =∂qi

∂qk

∂pj

∂pk

− ∂pj

∂qk

∂qi

∂pk

= δij,

{qi, qj} =∂qi

∂qk

∂qj

∂pk

− ∂qj

∂qk

∂qi

∂pk

= 0,

{pi, pj} =∂pi

∂qk

∂pj

∂pk

− ∂pj

∂qk

∂pi

∂pk

= 0. (65)

The methods of Lagrange and Hamilton provide for an elegant and flexible description

of dynamical systems. They can be applied to any chosen set of generalized coordinates qi,

the only prerequisite being the knowledge of the Lagrangian or Hamiltonian.

B. Canonical quantization

In a quantum mechanical system we may identify a set of coordinate operators {qi}, acting

on a Hilbert space whose vectors are identified with states of the system. The eigenvectors

of the coordinate operators satisfy

qi|q1, q2, · · · , qi, · · ·〉 = qi|q1, q2, · · · , qi, · · ·〉, (66)

and are states in which cooridnate i has value qi. Operators pi corresponding to classical

conjugate momenta pi = ∂L/∂qi obey canonical commutation relations with the qi,

[qi, pj] = ihδij,

[qi, qj] = 0,

[pi, pj] = 0. (67)

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The commutation relations are closely related to the classical Poission bracket.

{A,B} → 1

ih[A, B]. (68)

In particular, we have

dA

dt= {A,H}+

∂A

∂t→ dA

dt=

i

h[H, A] +

∂A

∂t. (69)

IV. CLASSICAL FIELD THEORY

In field theory, the analog of the generalized coordinates qi, is a field ϕ(~x, t), in which

the discrete index i has been replaced by the continuous position vector ~x. The position ~x

is not a coordinate, but rather a parameter that labels the field coordinate ϕ at point ~x at

a particular time t. There may be more than one field at each point in space, in which case

the fields may carry a distinguishing subscript, as in ϕa(~x, t). To qualify as a mechanical

system, the fields must be associated with a Lagrangian

L(t) =∫

d3~xL[ϕa(~x, t), ∂µϕa(~x, t)], (70)

which determines their time development.

In Eq.(70), we employ the conventions

Aµ = (A0, ~A) = gµνAν ,

Aµ = (A0,− ~A) = gµνAν , (71)

for any vector Aµ, where the Minkowski metric

gµν =

+1 0 0 0

0 −1 0 0

0 0 −1 0

0 0 0 −1

. (72)

For derivatives with respect to the coordinate vector xµ = (x0, ~x), we use the notation

∂µϕa ≡ ∂ϕa

∂xµ,

∂µϕa ≡ ∂ϕa

∂xµ

. (73)

14

Here x0 ≡ ct, where c is the speed of light, so that all the xµ have dimensions of length. From

hereon, we set c = 1. Generally, we shall use lower case Greek letters (α, β, · · · , µ, ν, · · ·) for

space-time vector indices (0, 1, 2, 3), and lower case italic letters (i, j, · · ·) for purely spatial

vector indices (1, 2, 3).

Now, let us demand that the action

A[ϕa(~x, t)] =∫ t2

t1dt

∫

Rd3~xL[ϕa(~x, t), ∂µϕa(~x, t)], (74)

within some region R of space to be extremal:

δA[ϕa(~x, t)] = δ∫ t2

t1dt

∫

Rd3~xL[ϕa(~x, t), ∂µϕa(~x, t)]

=∫ t2

t1dt

∫

Rd3~xδL[ϕa(~x, t), ∂µϕa(~x, t)]

=∫ t2

t1dt

∫

Rd3~x

[∂L∂ϕa

δϕa +∂L

∂(∂µϕa)δ∂µϕa

]

=∫ t2

t1dt

∫

Rd3~x

{∂L∂ϕa

− ∂µ

[∂L

∂(∂µϕa)

]}δϕa

+∫

Rd3~x

∂L∂(∂0ϕa)

δϕa

∣∣∣∣∣t2

t1

+∫ t2

t1dt

∫

∂Rdσi

∂L∂(∂iϕa)

δϕa

= 0, (75)

and obtain the equations of motion at every point inside R for every time between t1 and

t2,∂L∂ϕa

− ∂µ

[∂L

∂(∂µϕa)

]= 0. (76)

Here, the variation is over all possible fields ϕa(~x, t) inside R,

ϕa(~x, t) → ϕa(~x, t) + δϕa(~x, t), (77)

with the functions δϕa(~x, t) satisfying

δϕa(~x, t1) = 0 = δϕa(~x, t2) (78)

for all ~x in R, and

δϕa(~y, t) = 0 (79)

for all ~y on the surface ∂R of R, but are otherwise arbitrary. But, Eq.(76) must hold at

every point in space-time since t1, t2 and R were chosen arbitrarily.

15

The field momentum conjugate to ϕa(~x, t) is

πa(~x, t) =∂L

∂(∂0ϕa(~x, t)). (80)

The field Hamiltonian is given by the Legendre transform of the Lagrangian,

H =∫

d3~x{πa(~x, t) · ∂0ϕa(~x, t)− L[ϕa(~x, t), ∂µϕa(~x, t)]}. (81)

If we are given a Lagrangian, we may quickly find the field equations from Eq.(76), the

Hamiltonian from Eq.(81), and quantize by making ϕa(~x, t) and its conjugate momentum

πa(~x, t) operators satisfying equal-time commutation relations,

[ϕa(~x, t), πb(~x′, t)] = iδabδ

3(~x− ~x′). (82)

In the absence of difficult operator ordering ambiguities, the field Hamiltonian operator and

quantum conditions, will reproduce the field equations as operator equations of motion.

A. Noether’s theorem

If the action A[ϕa(~x, t)] is unchanged by a re-parametrization of xµ and ϕa, i.e., is in-

variant under some group of transformations on xµ and ϕa, then there exist one or more

conserved quantities, i.e., combinations of fields and their derivatives which are invariant un-

der the transformations. This crucial result is known as Noether’s theorem. It accounts for

conservation of energy, momentum, angular momentum, and any other “quantum” number

which particles happen to carry, like charge, color, isospin, etc.

To prove Noether’s theorem, consider

δA[ϕa(~x, t)] = δ∫

Rd4xL[ϕa(~x, t), ∂µϕa(~x, t)]

=∫

Rd4xδL[ϕa(~x, t), ∂µϕa(~x, t)]

=∫

Rd4x

[∂L∂ϕa

δϕa +∂L

∂(∂µϕa)δ∂µϕa + ∂µLδxµ

]

=∫

Rd4x

{∂L∂ϕa

− ∂µ

[∂L

∂(∂µϕa)

]}δϕa +

∫

∂Rdσµ

[∂L

∂(∂µϕa)δϕa + δµ

νLδxν

]

=∫

∂Rdσµ

[∂L

∂(∂µϕa)δϕa +

∂L∂(∂µϕa)

∂νϕaδxν

]−

∫

∂Rdσµ

[∂L

∂(∂µϕa)∂νϕa − δµ

νL]δxν

=∫

∂Rdσµ

[∂L

∂(∂µϕa)∆ϕa − θµ

νδxν

],

16

where the total variation

∆ϕa ≡ δϕa + ∂νϕaδxν , (83)

and the enery-momentum tensor

θµν ≡

∂L∂(∂µϕa)

∂νϕa − δµνL. (84)

We will first discuss the isospin symmetry, where the fields ϕa vary according to some

small parameter δεa.

ϕa → ϕa + δϕa = ϕa +∂ϕa

∂εb

δεb (85)

That is, ∆ϕa = δϕa = (∂ϕa/∂εb)δεb. If the action A[ϕa(~x, t)] is invariant under this trans-

formation, then

δA[ϕa] =∫

∂Rdσµ

∂L∂(∂µϕa)

∆ϕa =∫

∂RdσµJ

µa δεa = 0. (86)

Here, the current

Jµa ≡

∂L∂(∂µϕb)

∂ϕb

∂εa

. (87)

It follows from ∫

∂RdσµJ

µa δεa =

∫

Rd4xδεa∂µJ

µa = 0 (88)

that

∂µJµa = 0. (89)

From this conserved current, we can also establish a conserved charge, given by the integral

over the time-component of the current:

Qa ≡∫

d3~xJ0a . (90)

Now let us integrate the conservation equation:

0 =∫

d3~x∂µJµa

=∫

d3~x∂0J0a +

∫d3~x∂iJ

ia

=d

dt

∫d3~xJ0

a +∫

dSiJia

=d

dtQa + surface term (91)

Let us assume that the fields appearing in the surface term vanish sufficiently rapidly at

infinity so that the last term can neglected. Then,

∂µJµa = 0 ⇒ d

dtQa = 0 (92)

17

In summary, the symmetry of the action implies the conservation of a current Jµa , which in

turn implies a conservation principle:

symmetry → current conservation → conservation principle (93)

The second case, when the action is invariant under the space-time symmetry of the

Lorentz and Poincare groups, will be discussed later.

B. Schrodinger field

The Lagrangian density of the Schrodinger field

L = iϕ∗∂tϕ− 1

2∂xϕ

∗∂xϕ− V (x)ϕ∗ϕ. (94)

Here, ∂t ≡ ∂/∂t and ∂x ≡ ∂/∂x. We can determine the equations of motion by treating ϕ

and ϕ∗ as independent objects, so that for instance

∂L∂(∂tϕ∗)

= 0,

∂L∂(∂xϕ∗)

= −1

2∂xϕ,

∂L∂ϕ∗

= i∂tϕ− V (x)ϕ,

give us the Schrodinger equation (43). On the other hand,

∂L∂(∂tϕ)

= iϕ∗,

∂L∂(∂xϕ)

= −1

2∂xϕ

∗,

∂L∂ϕ

= −V (x)ϕ∗,

give us the complex-conjugate of Eq.(43).

From Eq.(94), we find that the conjugate momentum field is

π(x, t) =∂L

∂(∂tϕ)= iϕ∗(x, t). (95)

Since the field conjugate to ϕ∗ is found to vanish, there are only two independent fields,

ϕ(x, t) and π(x, t).

18

The Hamiltonian is easily computed to be

H =∫

dx[π(x, t)∂tϕ(x, t)− L]

=∫

dx1

2∂xϕ

∗∂xϕ + V (x)ϕ∗ϕ

=∫

dxϕ∗[−1

2

∂2

∂x2+ V (x)

]ϕ (96)

From Eqs.(42) and (87), we have

J0 =∂L

∂(∂0ϕ)

∂ϕ

∂ε+

∂L∂(∂0ϕ∗)

∂ϕ∗

∂ε

= iϕ∗∂ϕ

∂ε

= ϕ∗ϕ. (97)

It follows that

ϕ(ε) = ϕ(0) exp(−iε). (98)

And,

J1 =∂L

∂(∂1ϕ)

∂ϕ

∂ε+

∂L∂(∂1ϕ∗)

∂ϕ∗

∂ε

=1

2∂xϕ

∗(iϕ)− 1

2∂xϕ(iϕ∗)

= − i

2(ϕ∗∂xϕ− ϕ∂xϕ

∗). (99)

V. SECOND QUANTIZATION (CONT’)

In canonical field quantization, the classical fields ϕ(x, t) and π(x, t) are replaced by

operators ϕ(x, t) and

π(x, t) = iϕ†(x, t), (100)

where ϕ∗(x, t) is replaced by the Hermitian conjugate field operator ϕ†(x, t). We impose on

these operators the equal-time commutation relations:

[ϕ(x, t), ϕ†(x′, t)] = δ(x− x′), (101)

[ϕ(x, t), ϕ(x′, t)] = 0, (102)

[ϕ†(x, t), ϕ†(x′, t)] = 0. (103)

19

Using the Hamiltonian

H(t) =∫

dx[1

2∂xϕ

†∂xϕ + V (x)ϕ†ϕ], (104)

and the above equal-time commutation relations, we find

[H(t), ϕ(x, t)] =∫

dy[1

2∂yϕ

†(y, t)∂yϕ(y, t) + V (y)ϕ†(y, t)ϕ(y, t), ϕ(x, t)]

=∫

dy{

1

2∂y[ϕ

†(y, t), ϕ(x, t)]∂yϕ(y, t) + V (y)[ϕ†(y, t), ϕ(x, t)]ϕ(y, t)}

= −∫

dy{

1

2[∂yδ(y − x)]∂yϕ(y, t) + δ(y − x)V (y)ϕ(y, t)

}

=1

2∂2

xϕ(x, t)− V (x)ϕ(x, t) (105)

Alternatively, we could use

[H(t), ϕ(x, t)] =∫

dy[ϕ†(y, t)(−1

2∂2

y − V (y))

ϕ(y, t), ϕ(x, t)]

=∫

dy[ϕ†(y, t), ϕ(x, t)](−1

2∂2

y − V (y))

ϕ(y, t)

= −∫

dyδ(y − x)(−1

2∂2

y − V (y))

ϕ(y, t)

=1

2∂2

xϕ(x, t)− V (x)ϕ(x, t)

Therefore,∂

∂tϕ(x, t) = i[H, ϕ(x, t)]

implies

i∂

∂tϕ(x, t) = −1

2∂2

xϕ(x, t) + V (x)ϕ(x, t).

We note that we have picked a particular ordering for the operators in Eq.(104), called

normal ordering, to avoid the vacuum energy divergence.

A. Particle interpretation

Recall the expansion (47),

ϕ(x, t) =∑

i

ai(t)ϕi(x).

It follows that

ϕ†(x, t) =∑

i

a†i (t)ϕ∗i (x). (106)

20

Substitute Eq.(47) into the above equal-time commutation relations (101),

δ(x− x′) = [ϕ(x, t), ϕ†(x′, t)]

= [∑

i

ϕi(x)ai(t),∑

j

ϕ∗j(x′)a†j(t)]

=∑

i,j

ϕi(x)ϕ∗j(x′)[ai(t), a

†j(t)] (107)

only if

[ai(t), a†j(t)] = δij. (108)

This is because the first quantized energy eigenfunctions are assumed to be complete,

∑

i

ϕi(x)ϕ∗i (x′) =

∑

i

〈x|ϕi〉〈ϕi|x′〉

= 〈x|(∑

i

|ϕi〉〈ϕi|)|x′〉

= 〈x|x′〉= δ(x− x′) (109)

Alternatively, we could express ai(t) and a†i (t) in terms of ϕ(x, t) and ϕ†(x, t) respectively.

ai(t) =∫

dxϕ∗i (x)ϕ(x, t),

a†i (t) =∫

dxϕi(x)ϕ†(x, t). (110)

It follows that

[ai(t), a†j(t)] =

∫dx

∫dyϕ∗i (x)ϕj(y)[ϕ(x, t), ϕ†(y, t)]

=∫

dx∫

dyϕ∗i (x)ϕj(y)δ(x− y)

=∫

dxϕ∗i (x)ϕj(x)

=∫

dx〈ϕi|x〉〈x|ϕj〉= 〈ϕi|ϕj〉= δij (111)

Similarly,

[ai(t), aj(t)] =∫

dx∫

dyϕ∗i (x)ϕ∗j(y)[ϕ(x, t), ϕ(y, t)] = 0,

[a†i (t), a†j(t)] =

∫dx

∫dyϕi(x)ϕj(y)[ϕ†(x, t), ϕ†(y, t)] = 0. (112)

21

Next, substitute the expansion (47) into the Hamiltonian operator.

H(t) =∫

dxϕ†[−1

2

∂2

∂x2+ V (x)

]ϕ

=∫

dx∑

i

ϕ∗i (x)a†i (t)

[−1

2

∂2

∂x2+ V (x)

] ∑

j

ϕj(x)aj(t)

=∑

i,j

ej a†i (t)aj(t)

∫dxϕ∗i (x)ϕj(x)

=∑

i

eia†i (t)ai(t) (113)

The time dependence of the operators ai(t) is determined by

d

dtai(t) = i[H(t), ai(t)] =

∑

j

iej[a†j(t)aj(t), ai(t)] = −ieiai(t), (114)

which is solved by

ai(t) = exp(−ieit)ai(0) ≡ exp(−ieit)ai. (115)

It follows that

a†i (t) = exp(ieit)a†i , (116)

and H =∑

i eia†i ai. We note that the use of the eigenfunction basis {ϕi(x)} has the effect

that the time dependence of the operators ai(t) becomes trivial, being characterized by a

simple phase factor.

For fixed i, we note that a†i and ai look identical to the raising and lowering operators of

the harmonic oscillator. The field Hamiltonian is just an infinite sum of harmonic oscillator

Hamiltonians. The expansion (47) has reduced the quantum field theory to an infinite set

of harmonic oscillators. Following the discussion given on the harmonic oscillator, we can

now develop a particle interpretation. In the following we will construct the state vectors,

making use of the time-independent operators ai. This is possible within the Heisenberg

picture where the operators are time dependent while the state vectors are constant.

The lowest energy state of H, the ground state or bare vacuum, is the one that is empty,

ai|0〉 = 0. (117)

The destruction operator for any i, ai, finds no particle (excitation) to annihilate in the

empty vacuum |0〉, so the result is the null vector.

a†i |0〉 is a state of energy ei. It describes one particle of energy ei, created by a†i , the

creation operator for mode i. a†j|0〉 is also a one-particle state except that the energy of the

22

particle is ej. We have only one field in this theory, so there is only one type of particle.

The difference in energy between the two one-particle states must be due to a difference in

momentum between the two.

a†i a†j|0〉 is a two-particle state with energy ei + ej. The collection of all the states spanned

by the states formed by operating on |0〉 with any number of creation operators for any

mode i is called a Fock space. In general,

1√n1!n2! · · ·

(a†1)n1(a†2)

n2 · · · |0〉 ≡ |n1, n2, · · ·〉. (118)

These are eigenstates of ni ≡ a†i ai:

ni|n1, n2, · · · , ni, · · ·〉 = ni|n1, n2, · · · , ni, · · ·〉,

and the particle-number operator

N ≡ ∑

i

ni =∑

i

a†i ai, (119)

N |n1, n2, · · · , ni, · · ·〉 =

(∑

i

ni

)|n1, n2, · · · , ni, · · ·〉 = N |n1, n2, · · · , ni, · · ·〉,

where N is the total number of particles.

Last, we note that ϕ(x, t) is expanded in terms of ai(t) only, while ϕ†(x, t) is expanded in

terms of a†i (t) only. Thus, ϕ(x, t) is a destruction operator and ϕ†(x, t) is a creation operator.

ϕ†(x, t)|0〉 is a one-particle state where the particle is located at position x at time t.

〈0|ϕ(x, t)ϕ†(x′, t)|0〉 =∑

i,j

ϕi(x)ϕ∗j(x′)〈0|ai(t)a

†j(t)|0〉

=∑

i,j

ϕi(x)ϕ∗j(x′)〈0|{[ai(t), a

†j(t)] + a†j(t)ai(t)}|0〉

=∑

i,j

ϕi(x)ϕ∗j(x′)δij

=∑

i

ϕi(x)ϕ∗i (x′)

= δ(x− x′) (120)

So,

|x; t〉 ≡ ϕ†(x, t)|0〉. (121)

23

B. Particles in a box

Consider the case where V (x) = 0 with 0 ≤ x ≤ L. The normalized eigenfunctions,

ϕn(x) =1√L

exp(i2nπx

L

), (122)

are plane waves or momentum eigenfunctions. The expansions (47) and (106) turn into

Fourier series,

ϕ(x, t) =1√L

∑n

exp(i2nπx

L

)an(t),

ϕ†(x, t) =1√L

∑n

exp(−i

2nπx

L

)a†n(t), (123)

and Eq.(110) becomes

an(t) =1√L

∫dx exp

(−i

2nπx

L

)ϕ(x, t),

a†n(t) =1√L

∫dx exp

(i2nπx

L

)ϕ†(x, t). (124)

It follows that the state

a†n(t)|0〉 =1√L

∫dx exp

(i2nπx

L

)ϕ†(x, t)|0〉. (125)

Since we are integrating over x, a†n(t) creates a particle via ϕ†(x, t) at every point x with

amplitude exp(i2nπx/L)/√

L. In other words, the state a†n(t)|0〉 is a one-particle state with

position probability amplitude at each point given by exp(i2nπx/L)/√

L. That is, a†n(t)

creates a particle with wave function exp(i2nπx/L)/√

L. The square of this wave function is

independent of x, so a†n(t) creates a one-particle state and we do not know where the particle

is. It has equal probability of being anywhere. The particle has a definite momentum and

we have no idea what its position is, in accordance with the uncertainty principle.

Likewise, ϕ†(x, t) creates a particle at x, a definite position. Since we know its position,

we have no idea what its momentum and energy are. To create a particle definitely located

at x, we need to use a†n(t) for every n. The particle has equal probability of having any

momentum and hence n.

24

C. First quantized one-particle system

Now consider the general case where V (x) 6= 0. From above, a state containing one

particle described by wave function f1(x1, t) is given by

∫dx1 f1(x1, t)ϕ

†(x1, t)|0〉. (126)

Suppose this state has a definite energy, then it must be an eigenstate of H.

H∫

dx1 f1(x1, t)ϕ†(x1, t)|0〉

=∫

dx∫

dx1 f1(x1, t)ϕ†(x, t)

[−1

2

∂2

∂x2+ V (x)

]ϕ(x, t)ϕ†(x1, t)|0〉

=∫

dx∫

dx1 f(x1, t)ϕ†(x, t)

[−1

2

∂2

∂x2+ V (x)

]δ(x− x1)|0〉

=∫

dxϕ†(x, t)∫

dx1 f(x1, t)

[−1

2

∂2

∂x2+ V (x)

]δ(x− x1)|0〉

=∫

dx1

[−1

2

∂2

∂x21

+ V (x1)

]f(x1, t)ϕ

†(x1, t)|0〉

=∫

dx1 E1f(x1, t)ϕ†(x1, t)|0〉

= E1

∫dx1 f(x1, t)ϕ

†(x1, t)|0〉

which is true provided

[−1

2

∂2

∂x21

+ V (x1)

]f1(x1, t) = E1f1(x1, t). (127)

So, we see how the first quantized one-particle system emerges from the quantum field theory.

D. First quantized two-particle system

A two-particle state with one particle at x1 and another at x2 is ϕ†(x1, t)ϕ†(x2, t)|0〉. If

the two particles have wave function f2(x1, x2, t), then the state is

∫dx1dx2f2(x1, x2, t)ϕ

†(x1, t)ϕ†(x2, t)|0〉. (128)

Applying H(t) to this state, we have

∫dx

∫dx1

∫dx2f2(x1, x2, t)ϕ

†(x, t)

[−1

2

∂2

∂x2+ V (x)

]ϕ(x, t)ϕ†(x1, t)ϕ

†(x2, t)|0〉

25

=∫

dx∫

dx1

∫dx2f2(x1, x2, t)ϕ

†(x, t)

[−1

2

∂2

∂x2+ V (x)

]δ(x− x1)ϕ

†(x2, t)|0〉

+∫

dx∫

dx1

∫dx2f2(x1, x2, t)ϕ

†(x, t)

[−1

2

∂2

∂x2+ V (x)

]ϕ†(x1, t)ϕ(x, t)ϕ†(x2, t)|0〉

=∫

dx ϕ†(x, t)∫

dx2

∫dx1f2(x1, x2, t)

[−1

2

∂2

∂x2+ V (x)

]δ(x− x1)ϕ

†(x2, t)|0〉

+∫

dx ϕ†(x, t)∫

dx1

∫dx2f2(x1, x2, t)

[−1

2

∂2

∂x2+ V (x)

]δ(x− x2)ϕ

†(x1, t)|0〉

=∫

dx1dx2

[−1

2

∂2

∂x21

+ V (x1)

]f2(x1, x2, t)ϕ

†(x1, t)ϕ†(x2, t)|0〉

+∫

dx1dx2

[−1

2

∂2

∂x22

+ V (x2)

]f2(x1, x2, t)ϕ

†(x1, t)ϕ†(x2, t)|0〉

This state will be an eigenstate of H if f2 satisfies

[−1

2

∂2

∂x21

− 1

2

∂2

∂x22

+ V (x1) + V (x2)

]f2(x1, x2, t) = E2f2(x1, x2, t). (129)

This is a first quantized two-body Schrodinger equation.

E. First quantized N-particle system

If we apply the field Hamiltonian operator to its N -particle eigenstate,

∫dx1 · · · dxNfN(x1, · · · , xN , t)ϕ†(x1, t) · · · ϕ†(xN , t)|0〉, (130)

we find that the N -body wave function must satisfy

[−1

2

N∑

i=1

∂2

∂x2i

+N∑

i=1

V (xi)

]fN(x1, · · · , xN , t) = ENfN(x1, · · · , xN , t). (131)

In this way, all of the N -body first quantized systems are contained in the corresponding

quantum field theory. The operators ϕ(x, t) and ϕ†(x, t) change the number of particles

present so we can treat physical processes where the particle number is changing in a unified

manner. One way to exactly solve a quantum field theory is to solve the N -body Schrodinger

equation (131) for general N . Then any amplitude for any process will amount to a multi-

dimensional integral over the initial and final state wave functions.

26

F. Bosons and fermions

The quantum conditions we have chosen specify that ϕ†(x, t) commutes with itself. It

follows that

∫dx1dx2f2(x1, x2, t)ϕ

†(x1, t)ϕ†(x2, t)|0〉

=∫

dx2dx1f2(x2, x1, t)ϕ†(x2, t)ϕ

†(x1, t)|0〉

=∫

dx1dx2f2(x2, x1, t)ϕ†(x1, t)ϕ

†(x2, t)|0〉

Therefore,

f2(x1, x2, t) = f2(x2, x1, t). (132)

f2(x2, x1, t) is symmetric under the exchange of x1 and x2. The symmetric exchange is due

to the use of commutators in the quantum conditions. By the Pauli principle, this exchange

symmetry implies that we are dealing with bosons.

The wave function for two identical fermions must be antisymmetric under the exchange

of coordinates:

ψ2(x1, x2, t) = −ψ2(x2, x1, t). (133)

It follows that

∫dx1dx2ψ2(x1, x2, t)ϕ

†(x1, t)ϕ†(x2, t)|0〉

=∫

dx2dx1ψ2(x2, x1, t)ϕ†(x2, t)ϕ

†(x1, t)|0〉

= −∫

dx1dx2ψ2(x1, x2, t)ϕ†(x2, t)ϕ

†(x1, t)|0〉

If ϕ†(x1, t) and ϕ†(x2, t) commute, then

∫dx1dx2ψ2(x1, x2, t)ϕ

†(x1, t)ϕ†(x2, t)|0〉 = 0.

To avoid this, we demand

ϕ†(x1, t)ϕ†(x2, t) = −ϕ†(x2, t)ϕ

†(x1, t),

or

ϕ†(x1, t)ϕ†(x2, t) + ϕ†(x2, t)ϕ

†(x1, t) ≡ {ϕ†(x1, t), ϕ†(x2, t)} = 0. (134)

27

From hereon, we will use braces, {}, to denote an anticommutator. Therefore, to quantize

the field ϕ(x, t) satisfying Eq.(44) and obtain fermions, we must use anticommutators for

the quantum conditions,

{ϕ(x, t), ϕ†(x′, t)} = δ(x− x′),

{ϕ(x, t), ϕ(x′, t)} = 0,

{ϕ†(x, t), ϕ†(x′, t)} = 0. (135)

VI. LORENTZ INVARIANCE

A. Lorentz transformations

A Lorentz transformation is a linear, homogeneous change of coordinates from xµ to x′µ,

xµ → x′µ = Λµνx

ν , (136)

that preserves the interval x2 between xµ and the origin:

x2 ≡ t2 − ~x2 ≡ xµxµ = gµνxµxν . (137)

It follows that

x′2 = gµνx′µx′ν

= gµνΛµρx

ρΛνσx

σ

= gµνΛµρΛ

νσx

ρxσ

= gρσxρxσ

= x2

and the matrix (Λ)µν ≡ Λµ

ν must obey

gµνΛµρΛ

νσ = gρσ. (138)

If we take the determinant of Eq.(138), we get (det Λ)2 = 1, which implies that det Λ = ±1.

Transformations with det Λ = +1 are proper, and transformations with det Λ = −1 are

improper. Equation (138) implies

gµνΛµ0Λ

ν0

28

= Λ00Λ

00 − Λi

0Λi0

= g00

= 1

This means that (Λ00)

2 − Λi0Λ

i0 = 1; thus, either Λ0

0 ≥ +1 or Λ00 ≤ −1. Transformations

with Λ00 ≥ +1 are orthochronous, while those with Λ0

0 ≤ −1 are non-orthochronous.

B. Lorentz group

The set of all Lorentz transformations forms a group. The product of any two Lorentz

transformations is another Lorentz transformation. The product is associative. There is an

identity transformation,

Λµν = δµ

ν .

Every Lorentz transformation has an inverse:

gµνΛµρΛ

νσ = gρσ

⇒ ΛνρΛνσ = gρσ

⇒ gραΛναΛνσ = gραgασ

⇒ Λ ρν Λν

σ = δρσ

It follows that

(Λ−1)ρν = Λ ρ

ν = gνµΛµαgαρ, (139)

since, by definition, (Λ−1)ρνΛ

νσ = δρ

σ.

C. Infinitesimal Lorentz transformations

For an infinitesimal Lorentz transformation, we can write

Λµν = δµ

ν + δωµν . (140)

δω with both indices down (or up) is antisymmetric: Eq.(138) implies

gµνΛµρΛ

νσ = gµν(δ

µρ + δωµ

ρ)(δνσ + δων

σ)

≈ gµν(δµρδ

νσ + δµ

ρδωνσ + δν

σδωµρ)

29

= gρσ + δωρσ + δωσρ

= gρσ

It follows that

δωρσ = −δωσρ. (141)

Thus there are six independent infinitesimal Lorentz transformations (in four spacetime

dimensions). These can be divided into three rotations and three boosts:

δωij = −εijknkδθ (142)

for a rotation by angle δθ about the unit vector n,

δωi0 = niδη (143)

for a boost in the direction n by rapidity δη. For example,

Rz(θ) =

cos θ − sin θ 0

sin θ cos θ 0

0 0 1

and

Rz(δθ) =

1 −δθ 0

δθ 1 0

0 0 1

And,

Bz(η) =

cosh η 0 0 − sinh η

0 1 0 0

0 0 1 0

− sinh η 0 0 cosh η

,

where cosh η = γ ≡ 1/√

1− v2/c2, sinh η = γv/c, and rapidity η = tanh−1(v/c). So,

Bz(δη) =

1 0 0 −δη

0 1 0 0

0 0 1 0

−δη 0 0 1

.

30

We observe that

Rz(δθ) =

1 −δθ 0

δθ 1 0

0 0 1

=

1 0 0

0 1 0

0 0 1

− iδθ

0 −i 0

i 0 0

0 0 0

= 1− iδθ(ζ1|ζ1〉〈ζ1|+ ζ0|ζ0〉〈ζ0|+ ζ−1|ζ−1〉〈ζ−1|)= (1− iζ1δθ)|ζ1〉〈ζ1|+ (1− iζ0δθ)|ζ0〉〈ζ0|+ (1− iζ−1δθ)|ζ−1〉〈ζ−1|,

where |ζ1〉, |ζ0〉, and |ζ−1〉 are eigenvectors of the Hermitian matrix with eigenvalues ζ1 = 1,

ζ0 = 0, and ζ−1 = −1 respectively. It follows that

limn→∞[Rz(δθ)]

n = limn→∞(1− iζ1δθ)

n|ζ1〉〈ζ1|+ |ζ0〉〈ζ0|+ limn→∞(1− iζ−1δθ)

n|ζ−1〉〈ζ−1|= exp(−iθ)|ζ1〉〈ζ1|+ |ζ0〉〈ζ0|+ exp(iθ)|ζ−1〉〈ζ−1|= Rz(θ)

Here, δθ = limn→∞ θ/n.

Not all Lorentz transformations can be reached by compounding infinitesimal ones. In-

finitesimal transformations of the form Λ = 1 + δω are clearly proper and orthochronous:

det Λ = εκλµνΛκ0Λ

λ1Λ

µ2Λ

ν3

= εκλµν(δκ0 + δωκ

0)(δλ1 + δωλ

1)(δµ2 + δωµ

2)(δν3 + δων

3)

≈ εκλµν(δκ0δ

λ1δ

µ2δ

ν3 + δωκ

0δλ1δ

µ2δ

ν3 + δκ

0δωλ1δ

µ2δ

ν3 + δκ

0δλ1δω

µ2δ

ν3 + δκ

0δλ1δ

µ2δω

ν3

= 1,

Λ00 = δ0

0 + δω00 = 1.

Since the product of any two proper orthochronous Lorentz transformations is proper or-

thochronous:

det Λ = 1, det Λ′ = 1 ⇒ det(ΛΛ′) = det Λ det Λ′ = 1,

Λ00 ≥ 1, Λ′00 ≥ 1 ⇒ (ΛΛ′)0

0 = Λ0µΛ′µ0 = Λ0

0Λ′00 + Λ0

iΛ′i0 ≥ 1,

any transformation that can be reached by compounding infinitesimal ones is proper or-

thochronous. Thus, the Lorentz transformations that can be reached by compounding in-

finitesimal ones are both proper and orthochronous, and they form a subgroup. Generally,

31

when a theory is said to be Lorentz invariant, this means under the proper orthochronous

subgroup only.

Two discrete transformations that take one out of the proper orthochronous subgroup

are parity and time reversal. The parity transformation is

P µν =

+1 0 0 0

0 −1 0 0

0 0 −1 0

0 0 0 −1

= (P−1)µν . (144)

It is orthochronous, but improper. The time-reversal transformation is

T µν =

−1 0 0 0

0 +1 0 0

0 0 +1 0

0 0 0 +1

= (T−1)µν . (145)

It is nonorthochronous and improper. From hereon, in this chapter, we will treat the proper

orthochronous subgroup only. Parity and time reversal will be treated separately in later

chapters.

D. Quantum Lorentz transformations

In quantum theory, symmetries are represented by unitary (or antiunitary) operators.

This means that we associate a unitary operator U(Λ) to each proper, orthochronous Lorentz

transformation Λ. These operators must obey the composition rule

U(Λ′Λ) = U(Λ′)U(Λ). (146)

For an infinitesimal transformation, we can write

U(1 + δω) = I +i

2δωµνM

µν , (147)

where Mµν = −M νµ is a set of Hermitian operators called the generators of the Lorentz

group.

Let Λ′ = 1 + δω′, then to linear order in δω′,

U(Λ)−1U(Λ′)U(Λ) = U(Λ−1Λ′Λ)

32

⇒ U(Λ)−1(I +

i

2δω′µνM

µν)

U(Λ) = U((Λ−1)µρ(δ

ρσ + δω′ρσ)Λσ

ν)

⇒ I +i

2δω′µνU(Λ)−1MµνU(Λ) = U(δµ

ν + (Λ−1)µρδω

′ρσΛσ

ν)

= U(1 + Λ−1δω′Λ)

Therefore,

I +i

2δω′µνU(Λ)−1MµνU(Λ) = I +

i

2(Λ−1δω′Λ)ρσM

ρσ

where

δω′µνU(Λ)−1MµνU(Λ) = (Λ−1)ρµδω′µνΛ

νσM

ρσ

= gµαΛαρδω

′µνΛ

αρM

ρσ

= δω′µνΛµρΛ

νσM

ρσ

Since δω′µν is arbitrary, we have

U(Λ)−1MµνU(Λ) = ΛµρΛ

νσM

ρσ (148)

We observe that each vector index on Mµν undergoes its own Lorentz transformation. This

is a general result: any operator carrying one or more vector indices should behave similarly.

For example, for the energy-momentum four-vector P µ, where P 0 is the Hamiltonian H and

P i are the components of the total three-momentum operator,

U(Λ−1)P µU(Λ) = ΛµνP

ν . (149)

Now, let Λ = 1 + δω in Eq.(148), then to linear order in δω,(I − i

2δωρσM

ρσ)

Mµν(I +

i

2δωρσM

ρσ)

= (δµρ + δωµ

ρ)(δνσ + δων

σ)Mρσ

Mµν − i

2δωρσM

ρσMµν +i

2δωρσM

µνMρσ = Mµν + δωµρδ

νσM

ρσ + δµρδω

νσM

ρσ

− i

2δωρσ[Mρσ, Mµν ] = gµαδωαρM

ρν + gναδωασMµσ

i

2δωρσ[Mµν , Mρσ] = gµρδωρσM

σν + gνσδωσρMµρ

=1

2(gµρδωρσM

σν + gµσδωσρMρν + gνσδωσρM

µρ + gνρδωρσMµσ)

=1

2δωρσ(gµρMσν − gµσMρν − gνσMµρ + gνρMµσ)

=1

2δωρσ(−gµρM νσ + gµσM νρ − gνσMµρ + gνρMµσ)

= −1

2δωρσ(gµρM νσ − gνρMµσ − gµσM νρ + gνσMµρ)

33

Therefore,

[Mµν , Mρσ] = i(gµρM νσ − gνρMµσ − gµσM νρ + gνσMµρ). (150)

These commutation relations specify the Lie algebra of the Lorentz group. We can identify

the components of the angular momentum operator as Ji ≡ −1/2εijkMjk, and the compo-

nents of the boost operator Ki ≡ M i0.

[Ji, Jj] =1

4εimnεjpq[M

mn, Mpq]

=i

4εimnεjpq(g

mpMnq − gnpMmq − gmqMnp + gnqMmp)

= − i

4(εimnεjmqM

nq − εimnεjnqMmq − εimnεjpmMnp + εimnεjpnMmp)

= − i

4(εminεmjqM

nq + εimnεjqnMmq + εinmεjpmMnp + εimnεjpnM

mp)

= −iM ij

= iεijkJk (151)

since

εijkJk = −1

2εkijεklmM lm = −1

2(δilδjm − δimδjl)M

lm = −M ij.

[Ji, Kj] = −1

2εimn[Mmn, M j0]

= − i

2εimn(gmjMn0 − gnjMm0 − gm0Mnj + gn0Mmj)

=i

2(εijnMn0 − εimjM

m0)

= iεijkMk0

= iεijkKk (152)

[Ki, Kj] = [M i0, M j0]

= i(gijM00 − g0jM i0 − gi0M0j + g00M ij)

= iM ij

= −iεijkJk (153)

Similarly, we can let Λ = 1 + δω in Eq.(149), then to linear order in δω,

(I − i

2δωρσM

ρσ)

P µ(I +

i

2δωρσM

ρσ)

= (δµν + δωµ

ν)Pν

34

P µ − i

2δωρσ[Mρσ, P µ] = P µ + gµαδωανP

ν

i

2δωρσ[P µ, Mρσ] = gµρδωρσP

σ

=1

2(gµρδωρσP

σ + gµσδωσρPρ)

=1

2δωρσ(gµρP σ − gµσP ρ)

Therefore,

[P µ, Mρσ] = i(gµσP ρ − gµρP σ). (154)

It follows that

[H, Ji] = −1

2εijk[P

0, M jk]

= − i

2εijk(g

0kP j − g0jP k)

= 0, (155)

[Pi, Jj] =1

2εjmn[P i, Mmn]

=i

2εjmn(ginPm − gimP n)

=i

2(−εjmiP

m + εjinP n)

= iεijkPk, (156)

[H, Ki] = [P 0, M i0]

= i(g00P i − g0iP 0)

= −iPi (157)

[Pi, Kj] = −[P i, M j0]

= −i(gi0P j − gijP 0)

= −iδijH (158)

Also, the components of P µ should commute with each other.

[P i, P j] = 0,

35

[P i, H] = 0. (159)

In summary,

[H, P i] = 0,

[P i, P j] = 0,

[Ji, Jj] = iεijkJk,

[Ji, Kj] = iεijkKk,

[Ki, Kj] = −iεijkJk,

[H, Ji] = 0,

[Pi, Jj] = iεijkPk,

[H, Ki] = −iPi,

[Pi, Kj] = −iδijH

form the Lie algebra of the Poincare group.

E. Noether’s theorem (cont’)

Now let us investigate the case when the action is invariant under the space-time sym-

metry of the Lorentz and Poincare groups. First, consider the translation

xµ → x′µ = xµ + aµ, (160)

where aµ is a constant. a0 represents time displacements, and ai represents space displace-

ments. From Eq.(84),

θµν ≡

∂L∂(∂µϕa)

∂νϕa − δµνL

is the energy-momentum tensor. By integrating the energy-momentum tensor, we can gen-

erate conserved quantities. Define

Pν ≡∫

d3~xθ0ν . (161)

Then

∂µθµν = 0

⇒ ∂0θ0ν + ∂iθ

iν = 0

⇒ d

dtPν = 0. (162)

36

We observe that

P0 =∫

d3~xθ00 =

∫d3~x

∂L∂(∂0ϕa)

∂0ϕa − L = H. (163)

Therefore, displacement in time (space) leads to the conservation of energy (momentum).

Here,

Pi =∫

d3~xθ0i =

∫d3~x

∂L∂(∂0ϕa)

∂iϕa. (164)

Next, consider the Lorentz transformation

xµ → x′µ = xµ + δωµνx

ν . (165)

It follows that

∂µ(θµνδω

ναxα) = 0

⇒ ∂µ(θµνδωναxα) = 0

⇒ ∂µ(θµνδωναxα + θµαδωανxν) = 0

⇒ δωνα∂µ(θµνxα − θµαxν) = 0

⇒ ∂ρ(θρµxν − θρνxµ) = 0 (166)

Define

Mµν ≡∫

d3~x(θ0µxν − θ0νxµ). (167)

Then

∂0(θ0µxν − θ0νxµ) + ∂i(θ

iµxν − θiνxµ) = 0

⇒ d

dtMµν = 0 (168)

37