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4 Canonical Quantization 



We will begin now the discussion of our main subject of interest: the role of quan- 
tum mechanical fluctuations in systems with infint(^ly many degrees of freedom. 
We will begin with a brief overview of quantum mechanics of a single particle. 



4.1 Elementary Quantum Mechanics 

Elementary Quantum Mechanics describes the quantum dynamics of systems 
with a finite number of degrees of freedom. Two key ingredients are involved 
in the standard procedure for quantizing a classical system. Let L{q,q) be the 
Lagrangian of an abstract dynamical system described by the generalized coor- 
dinate q. In chapter two, we recalled that the canonical formalism of Classical 
Mechanics is based on the concept of canonical pairs of dynamical variables. So, 
the canonical coordinate q has for partner the canonical momentum p: 

In this approach, the dynamics of the system is governed by the classical Hamil- 
tonian 

H{q,p)=pq-L{q,q) (2) 

which is the Legendre transform of the Lagrangian. In the canonical (Hamilto- 
nian) formalism the equations of motion are just Hamilton's Equations, 

dH . dH 

The dynamical state of the system is defined by the values of the canonical 
coordinates and momenta at any given time t. As a result of these definitions, 
the coordinates and momenta satisfy a set of Poisson Bracket relations 

{q,p}pb = 1 {q, q}pB = {p,p}pb = (4) 

where 

SAB} ^dAdB_dAdB 
dq dp dp dq 

In Quantum Mechanics, the primitive (or fundamental) notion is the concept 
of a physical state. A physical state of a system is a represented by a vector in 
an abstract vector space, which is called the Hilbert space H of quantum states. 
The space H is a. vector space in the sense that if two vectors \'i>) E H and 
1$) € 7i represent physical states, then the linear superposition ja^* + 6$) = 
a|\I') + &|$), where a and b are two arbitrary complex numbers, also represents a 
physical state and thus it is an element of the Hilbert space i.e., |avl' + 6i>) g H. 
Thus, the Superposition Principle is an axiom of Quantum Mechanics. 

In Quantum Mechanics, the dynamical variables, i.e., q,p,H, etc. , are 
represented by operators which act linearly on the Hilbert space of states. (In 



1 



this sense, Quantum Mechanics is hncar, even though the obscrvablcs obey non- 
linear equations of motion.) Let us denote by A an arbitrary operator acting 
on H. The result of acting on the state \^) G H with the operator A, i.e., a 
measurement, is another state |$) gH, 

m = I*) (6) 

The Hilbert space H is endowed with an inner product. An inner product is an 
operation which assigns a complex number ($|^') to a pair of states |$) e W 
and I*) G H. 

Since W is a vector space, there exists a set of linearly independent states 
{|A)}, called a basis, which spans the entire Hilbert space. Thus, an arbitrary 
state has the expansion 

|*) = ^*a|A) (7) 

A 

which is unique for a fixed set of basis states. The basis states can be chose to 
be orthonormal with respect to the inner product, i.e., 

(A|/^) = (8) 

In general if and are normalized states 

(*|*) = ($1$) = 1 (9) 

the action of ^4 on l^*) is merely proportional to |$) 

= a|$) (10) 

The coefficient a is a complex number which depends on the pair of states and 
on A. This coefficient is the matrix element of A between the state and |$), 
which we write with the notation 

a = {<^\A\<f) (11) 

Operators which act on a Hilbert space do not generally commute with 
each other. One of the ax;ioms of Quantum Mechanics is the Correspondence 
Principle which states that the classical limit, h ^ 0, the operators should 
become numbers, i. e. they commute in the classical limit. 

The procedure of canonical quantization consists in demanding that to the 
classical canonical pair {q,p), which satisfies the Poisson Bracket {q,p}pB = 1, 
we associate a pair of operators q and p, both acting on the Hilbert space of 
states H, which are required to obey the canonical commutation relations 

[q,p]=ih [q,q] = [p,p]=0 (12) 

where [A, B] is the commutator of the operators A and B, 

[A, b] = Ab- bA (13) 



2 



In particular, two operators that do not commute with each other cannot be 
diagonahzed simultaneously. Hence it is not possible to measure both observ- 
ables with arbitrary precision in the same physical state. This is the statement 
of the Uncertainty Principle. 

To the classical Hamiltonian H[q,p), which is a function of the variables 
q and p, we assign an operator H{q,p) which is obtained by replacing the dy- 
namical variables with the corresponding operators. Other classical dynamical 
quantities are similarly associated with quantum operators. All operators asso- 
ciated with classical physical quantities are Hermitian operators relative to the 
inner product defined in the Hilbert space H. Namely, if A is an operator and 
At is the adjoint of A 

= ($|i^f)* (14) 

then A is Hermitian if A = A^ . 

The quantum mechanical state of the system at time t, \'^{t)), is the solution 
of the Schrodinger Equation 

ih^^\m)=mp)\-^{t)) (15) 

The state |^'(t)) is uniquely determined by the initial state |^'(0)). 

It is always possible to choose a basis in which a particular operator is 
diagonal. For instance, if the operator is the canonical coordinate q, the basis 
states are labelled by q and are its eigenstates, i.e., 

q\q) = q \q) (16) 

The state vector can be expanded in an arbitrary basis. If the basis of states 
is {\q)} , the expansion is 

/ + 00 
dq^{q)\q) (17) 
t, -°° 

The coefficients ^{q) of this expansion 

*(?) = (18) 

are (the values of) the wave function associated with the state j^*) in the co- 
ordinate representation. Here we have used that the states \q) are orthonormal 
and complete, i. e. 

{q\q')=6iq-q') 1 = j dq\q){q\ (19) 

Since the canonical momentum p does not commute with q, it is not diagonal 
in this representation. In fact, just as in Classical Mechanics, the momentum 
operator p is the generator of infinitesimal displacements. Consider the states 



3 



\q) and exp{—^ap) \q). It is easy to prove that the latter is the state \q + a) 



since 



gexp(-^ap)|g)^g^- ^ \q) (20) 

71=0 ■ ^ ' 

Using the commutation relation [q, p] = ih is easy to show that 

=i?lnp"-^ (21) 

Hence, we can write 



Thus, 



q exp( -^apj \q) = (g + a) exp ( --ap ) \q) (22) 



exp(^-^apj\q) = \q + a) (23) 
We can now use this property to compute the matrix element 

(5|exp(^^ap^ |*) = *(g + a) (24) 

For a infinitesimally small, it can be approximated by 

^{q + a)^^{q) + ^a{q\p\^) + ... (25) 

We find that the matrix element for p has to satisfy 

, S *(g + a)-*(fl) 

{q\p\^) = - lim ' !^ (26) 

I a-+0 a 

Thus, the operator p is represented by a differential operator 

It is easy to check that the coordinate representation of the operator 

t oq 

and the coordinate operator q satisfy the commutation relation [q,p] = ih. 

4.2 Canonical Quantization in Field Theory 

We will now apply the axioms of Quantum Mechanics to a Classical Field The- 
ory. The result will be a Quantum Field Theory. For the sake of simplicity 
we will consider first the case of a scalar field (p{x). We have seen before that, 



4 



given a Lagrangian density 9^0), the Hamiltonian can be found once the 
canonical momentum Il{x) is defined, i.e., 

n(ar) = 7^7^ (29) 

On a given time surface xo, the classical Hamiltonian is 

H = j d^x [n{x, xo)do(t>{x, xo) - C{(f>, d^ct))] (30) 

We quantize this theory by assigning to each dynamical variable of the Classical 

theory, a Hcrmitian operator which acts on the Hilbcrt space of the quantum 
states of the system. Thus, the field (/>(x) and the canonical momentum n(x) 
are operators which act on a Hilbert space. These operators obey canonical 
commutation relations 

[cb{x),u{y)]=m£-y) (31) 

In the field representation, the Hilbert space is the vector space of wave functions 
^ which are functional of the field configurations {^(^)}, i.e., 

In this representation, the field is a diagonal operator 

The canonical momentum Il{x) is not diagonal in this representation but it acts 
like a functional differential operator, i.e., 

(W|n(f)|*) = ^^*(W) (34) 

What we just described is the Schrddinger Picture of QFT. In this picture, as 
usual, the operators arc time- independent but the states are time-dependent and 
satisfy the Schrodinger Equation 

ih^^'^m,t) = H^i{<l>},t). (35) 
For the particular case of a scalar field (p with the classical Lagrangian £ 

C=l{d^cl>r~V{cl>) (36) 
the quantum mechanical Hamiltonian operator H is 

H = Jd'x |in2(f) + livmr + V{${x))^ (37) 

The stationary states of the system are the eigenstates of H. While it is 
possible to proceed further with the Schrodinger picture, the manipulation of 



5 



wave functionals becomes very cumbersome rather quickly. For this reason an 
alternative approach has been devised. This is the Heisenberg Picture. 

In the Schrodinger Picture the information on the time evolution of the 
system is encoded in the time dependence of the states. In contrast, in the 
Heisenberg Picture the operators are time dependent while the states are not. 
The operators of the Heisenberg Picture obey quantum mechanical equations of 
motion. 

Let A be some operator which acts on the Hilbert space of states. Let us 
define Ah{xq), the Heisenberg operator at time xq, by 



(38) 



for a system with a time-independent Hamiltonian H. It is straightforward to 
check that Ah(xo) obeys the equation of motion 



ihdoAH{xo) = [Ah{xo),H] 



(39) 



Notice that in the classical limit, the dynamical variable A{xq) obeys the clas- 
sical equation of motion 



doA{xo) = {A{xo),H}pB 



(40) 



where it is assumed that all the time dependence in A comes from the time 
dependence of the coordinates and momenta. 

In the Heisenberg picture both (j){x,xo) and Il{x,xo) are time dependent 
operators which obey the equations of motion 



ihdocj) = 



(41) 



and 



ihdofl = [H, H] 



(42) 



The Heisenberg field operators and n (I will omit the subindex "iJ" from now 
on) obey equal-time commutation relations 



{x,xo),lliy,xo) =ihS{x-y) 



(43) 



4.3 Quantized elastic waves in a solid: Phonons 

Let us consider the problem of the quantum dynamics of an array of atoms. We 
will see below that this problem is closely related to the problem of quantization 
of a free scalar field. We will consider the simple case of a one-dimensional 
array of atoms, a chain. Each atom has mass M and their classical equilibrium 
positions are the regularly spaced lattice sites x'^^ = nao ,{n= 1 , . . . , n) where Oq 
is the lattice spacing. I will assiimc that we have a system with N atoms and, 
therefore that the length L of the chain is L = Nao- To simplify matters, I will 
assume that the chain is actually a ring and, thus, the N + 1 — st atom is the 



6 




n-1 n n+1 n+2 



Figure 1: A model of an elastic one- dimensional solid. 



same as the 1st atom. The dynamics of this system can be specified in terms of 
a set of coordinates which represent the position of each atom relative to 
their classical equilibrium positions a;° i.e., the actual position Xn of the n**^ 
atom is a;„ = + 

The Lagrangian of the system is a function of the coordinates and of 

their time derivatives In general the Lagrangian L it will be the difference 

of the kinetic energy of the atoms minus the potential energy, i.e., 

N_ 

M ( dUn 



L{{u„},{un})= \^ Yyir) 

We will be interested in the study of the small oscillations of the system. Thus, 

the potential will have a minimum at the classical equilibrium positions {m„ = 0} 
(which we will assume to be unique). For small oscillations V{{un}) can be 
expanded in powers 

nK})= E [§K+l-«n)' + f < + ...] (45) 
n=-f + l 

Here D is an elastic constant {i.e., spring constant!) which represents the 
restoring forces that keep the crystal together. The constant X is a measure of 
the strength of an external potential which favors the placement of the atoms at 
their classical equilibrium positions. For an isolated system, K = but D ^ 0. 
This must be the case since an isolated system must be translationally invariant 
and, therefore, V must not change under a constant, uniform, displacement of 
all the atoms by some amount a, u„ u„ + a. The term proportional to 
breaks this symmetry, although it does not spoil the symmetry n ^ n + m. 

Let us now proceed to study the quantum mechanics of this system. Each 
atom has a coordinate «„(*) and a canonical momentum Pn{t) which is defined 
in the usual way 

dL 

Pr^it) = = Miinit) (46) 



7 



The quantum Hamiltonian for a chain with an even number of sites N is 



(47) 



+1 

where the coordinates and momenta obey the commutation relations 

[Un,Pm\=ifiSn,m (48) 

and 

[Un, Um] = \pn,Pm] = 0. (49) 

For this simple system, the Hilbert space can be identified as the tensor product 
of the Hilbert spaces of each atom. Thus, if |\E')„ denotes an aribtrary state in 
the Hilbert space of the n*^ atom, the states of the chain |'J') can be written in 
the form 

I*) = O • • • O |*)„ O • • • O |*)jv = 1*1, . . . , *jv) (50) 

For instance, a set of basis states can be constructed by using the coordinate 
representation. Thus, if the state is an eigenstate of u„ with eigenvalue m„ 

Un\Un) = Un\Un) (51) 

we can write a set of basis states 

\ui,...,un) (52) 

of the form 

\ui,...,um) = \ui) \un) (53) 

In this basis, the wave functions are 

N N 

*(ui,...,ujv) = (m,...,Mjvi*) = n^^"!*") = n *"(^") (54) 

n=l n=l 

By inspecting the Hamiltonian it is easy to recognize that it represents a set of A'^ 
coupled harmonic oscillators. Since the system is periodic and invariant under 
lattice shifts n + m (m integer), it is natural to expand the coordinates u„ 
in a Fourier series 

Un = ^J2^ke'''' (55) 
fe 

where fc is a label (lattice momentum) and Uk are the Fourier components of 
Un- The fact that we have imposed periodic boundary conditions (PBC's) {i.e., 
the chain in a ring) means that 

U„ = Un+N (56) 



8 



This relation can hold only if the labels k satisfy 

e''^^ = 1 (57) 
This condition restricts the values of k to the discrete set 

fcm = 27r— rn = -y + (58) 

where I have set the lattice constant to unity, ao = 1 . Thus the expansion of u„ 
is 

N_ 

m=-f+l 

The spacing Ak between two consecutive values of k, k^ and k^+i, is 

27r 

Ak = fc„+i -km = — (60) 

which vanishes as -/V ^ cxd. In particular, the momentum label A;„ runs over the 
range (— Y + l)^<fc„<Y- Thus, in the limit N ^ oo the momenta fill up 
densly the interval (— tt.tt]. 

We then conclude that, in the thermodynamic limit N oo, the momentum 
sum converges to the integral 



1\_ 

^ E u.^e^'-- = f_^^me^'- (61) 

T — — ' - 



i+1 

Since u„ is a real Hermitian operator, the Fourier components u{k) must satisfy 

ut(fc) = u{-k) (62) 

The Fourier component u{k) can be written as a linear combination of operators 
Un of the form 

2 

u(A;)= ^ w„e-*'=" (63) 

n=-f +1 

where I have used the periodic Dirac delta function, defined by 

2 +00 

J2 e'('=-«)"= E 27r5(fc-g + 27rm) = 27r5p(fc-9) (64) 



m= — oo 



which is defined in the thermodynamic limit. 

The momentum operators p„ can also be expanded in Fourier series. Their 
expansions are 

— p(fc)e''=" p{k)= E (65) 

"=-2TT 



9 



They satisfy 

p^{k)^p{-k) (66) 

The transformation {uniPn) {u{k),p{k)) is a canonical transformation. In- 
deed, the Fourier ampHtudes u{k) and p{k) obey the commutation relations 



-+i 



n,n' = -f-i 

N_ 

= ^n ± 



Hence, we find 
and 



(67) 

[u{k),p{k')] = ih 27r 6p{k + k') (68) 

[u{k).P{k')] = m),p(k')]=Q (69) 

We can now write H in terms of the Fourier components u{k) and p{k). We 
find 

^ = /I S [^^^^(^)^(^) + ^'^\k)u\k)u{k)] (70) 

where ijj'^{k) is 

9, ^ K AD y,k^ , , 

^^-M^M^^ (71) 

Thus, the system decouples into its normal modes. The frequency ujik) is shown 
in the figure 2. It is instructive to study the long-wave length limit, — > 0. For 
-fC = (i.e., no external potential), a;(fc) goes to zero linearly as A; — > 0, 

^{k) « [K = 0) (72) 

However, for non-zero we get ( again in the limit fc ^ 0) 



If we now restore a lattice constant oq ^ 1, fc = fcoo we can write a;(fc) in the 
form 



ijj(U) \j ra^vl + fc2 (74) 
where is the speed of propagation of sound in the chain. 



^ao = \l^ (75) 



10 



co(k) 




Figure 2: The dispersion relation, 
where p is the density. The "mass" fh is 



m=^ (76) 



where wq = Thus, the waves which propagate on this chain behave hke 

"relativistic" particles with mass m and "speed of light" Vs- Indeed, the long 
wavelength limit [k — > 0) the discrete Lagrangian of eq (4.3.1) can be written 
in the form of an integral 



Thus, as ao — > and N ^ oo, the sums converge to an integral 

for a system of total length £. Apart from the overall factor of p, the mass 
density, wc sec that the Lagrangian for the linear chain is, in the long wavelength 
limit (or continuum limit) the same as the Lagrangian for the Klein-Gordon 
(KG) field u{x) in one-space dimension. The last term of this Lagrangian is 



2 



11 



precisely the mass term for the Lagrangian of the KG field. This explains the 
choice of the symbol m. Indeed, upon the change of variables 

Xq = Vgt Xi= X (79) 

and by defining the rescaled field 

if = ^pVgU (80) 
we see immediately that the Lagrangian density L is 

>C = ^ {d^^f - \ {d^^f - \fh\l^^ (81) 

which is the Lagrangian density for a free scalar field in one spacial dimension. 

Returning to the quantum theory, we seek to find the stationary states of 
the normal- mode Hamiltonian. Let a^{k) and a{k) be the operators defined by 



These operators satisfy the commutation relations 

[a{k),a{k')] = [d^k),a^k')] = 
[a{k),d^k')] = 2n5p{k + k') 



(82) 



(83) 



Up to normalization constants, the operators a^{k) and a{k) obey the algebra 
of creation and anihilation operators. 

In terms of the creation and anihilation operators, the momentum space 
oscillator operators u{k) and p{k) are 



p{k) = ^/2MhJ(k) ^. {a{k) - at(-fc)) 

(84) 

Thus, the coordinate space operators u„ and p„ have the Fourier expansions 



* ' ^ (a(fc)e''=" + at(fc)e-^'=") 



27r y 2Mw(fc) 
dk 



/ Hk 1 
— ^/2Mhuj{k) - (a(fc) e*'^" - a^{k) 6-^*=") 

(85) 



12 



The normal-mode Hamiltonian has a very simple form in terms of the creation 
and anihilation operators 

^ = /I S ^ ^,t(fc)a(fc) + a{k)aHk)) (86) 

It is customary to write H in such a way that the creation operators always 
appear to the left of anihilation operators. This procedure is called normal 
ordering. Given an arbitrary operator A, we will denote hy : A: the normal 
ordered operator. We can see by inspection that H can be written as a sum of 
two terms: a normal ordered operator : H : and a complex number. The complex 
number results from using the commutation relations. Indeed, by operating on 
the last term of eq (4.3.39), wc get 

a{k)aHk) = [a(fc), a^(fc)] + a^(fc)a(fc) (87) 

The commutator [a(fc),d^(fc)] is the divergent quantity 

+00 

[a(k),aUk)] = lim 27r5p(A; - fc') = lim 27r V 6(k-k' +2Trm) 
k'^k k'^k ^ 



= lim e'^*'-'^ = iV 

^*'n=-f +1 



which diverges in the thermodynamic limit, N ^ 00. 
Using these results, we can write H in the form 



(88) 



H=:H:+Eo (89) 
where :H: is the normal-ordered Hamiltonian 

/'^ rlh 
— hio{k) at(fc)a(fc) (90) 

and the real number Eq is given by 

We will see below that Eq is the ground state energy of this system. The linear 
divergence of Eq (as N 00) is natural since the ground state energy has to 
be an extensive quantity, i.e., it scales like the length (volume) of the chain 
(system) . 

We are now ready to construct the spectrum of eigenstates of this system. 

1. Ground state: 

Let |0) be the state which is anihilated by all the operators a(fc), 

o(A;)|0) = (92) 



13 



This state is an eigenstate with eigenvalue Eq since 

H\0)=:H:\0) + Eo\0) = Eo\0) (93) 

where we have used the fact that |0) is anihilated by the normal-ordered 
Hamilitonian -.H:. This is the ground state of the system since the energy 
of all other states is higher. Thus, Eq is the energy of the ground state. 
Notice that Eq is the sum of the zero-point energy of all the oscillators. 

The wave function for the ground state can be constructed quite easily. 
Let V^o {{u{k)}) — ({?i(fc)}|0) be the wave function of the ground state. 
The condition that |0) be anihilated by all the operators a{k) means that 
the matrix element ({u(A;)}|a(A;)|0) has to vanish. The definition of a{k) 
yields the condition 

= Muj{k) {{u{k)}\u{k)\0) + i{{u{k)}\p{k)\0) (94) 

The commutation relation 



u 



{k),p{k)]^ih2n dp{k + k') (95) 



implies that, in the coordinate representation p(A;) must bet the functional 
differential operator 

{{mmm = 7 ({^(fc)}) (96) 

Thus, the wave functional \l/o must obey the differential equation 

Muj{k) u*{k) *o {{u{k)}) + 2Trh *o {{u{k)}) = (97) 

for each value of A; € [— tt, it]. Clearly has the form of a product 

-^o{{u{k)})^\{^o,k{u{k)) (98) 
fe 

where the wave function "io^k {u{k)) satisfies 

Mw(fc)M*(fc)*o,fe {u{k)) + 27rft^-|^«'o,fe {u{k)) = (99) 

The solution of this equation is the ground state wave function for the 
fc-th oscillator 

*„,(.W)^^(.)e.p(-(^ifl)£efi) ,100) 

where J\f(k) is a normalization factor. The total wave function for the 
ground state is 

T /r-/,x.x .r r T dk f Muj{k)\ \u\^ (k) , , 

MMk)}) =^ exp[- - j ^] (101) 

where Af is another normalization constant. Notice that this wave function 
is a functional of the oscillator variables {u{k)}. 



14 



2. One-Particle States: 

Let the ket \lk) denote the one-particle state 

life) = at(fc)|0) (102) 

This state is an eigenstatc of H with eigenvalue Ei{k) 

H\lk)=:H:\lk)+E^\lk) (103) 

The normal-ordered term now does give a contribution since 



■.H: life) = (^j^ ^ nuj{k') a\k')a{k')^ a\k)\0} 



-L 



—tkj{k ){a^k')[a{k'),a'^{k)]\0)+a'^{k')a''ik)aik')\0)} 

ZTT 

(104) 

The result is 

— hLo{k') a^k') 2TT6p{k' - k) |0) (105) 

since the last term vanishes. Hence 

■.H: life) = njj{k) a\k) |0) = huj{k) |lfe) (106) 
Therefore, we find 

H\lk) ^ ihw(k) + Eo) \lk) (107) 
Let e{k) be the excitation energy 

s{k) = Ei{k) - Eo{k) = huj{k) (108) 

Thus the one-particle states represent quanta with energy tujj{k) above 
that of the ground state. 

3. Many Particle States: 

If we define the occupation num,ber operator n(fc) by 

h{k) = a){k)a{k) (109) 

i.e., the quantum number of the fc-th oscillator, we see that the most 
general eigenstate is labelled by the set of oscillator quantum numbers 
{n{k)}. Thus, the state |{?^(fc)}) defined by 

|{n(fe)})=n ^"'^'^ 10) (110) 

has energy E[{n{k)}] 

^ nik)huj{k) + Eo (111) 

It is clear that the excitations behave like free particles since the energies 
are additive. These excitations are known as phonons. They are the 
quantized fluctuations of the array of atoms. 



15 



4.4 Quantization of the Free Scalar Field Theory. 

We now return to the problem of quantizing a scalar field (f){x). In particular, 
we will consider a free real scalar field </> whose Lagrangian density is 

C = l{d^ct>){d'4>) - Im'^' (112) 

This system can be studied using methods which are almost identical to the 
ones we used in our discussion of the chain of atoms. 

The quantum mechanical Hamiltonian H for a free real scalar field is 



(113) 



where (f) and fl satisfy the equal-time commutation relations (in units with 

h = c=l) 

[4>{x, xq), U{y, xo)] =i6{x-y) (114) 

1. Equations of Motion: 

In the Hcisenberg representation, cj) and 11 arc time dependent operators 
while the states are time independent. The field operators obey the equa- 
tions of motion 

ido4>{x,xo) = [(l>{x,xo), H] 

idoIl{x,xo) = [Il{x,Xo),H] (115) 

These are operator equations. After some algebra, we get 

do4)ix,xo) = Il{x,xo) (116) 
i9on(.f,a;o) = \/'^4>{x,xo) - m'^(j){x,xo) (117) 
{D + m^)4,{x) = (118) 

Thus, the field operators 4>{x) satisfy the Klein-Gordon equation. 

2. Field Expansion: 

Let us solve this equation by Fourier Transforms. Let us write 4>{x) in the 
form 

^ci>{k,x^)e'^-^ (119) 

where (j){k,XQ) are the Fourier amplitudes of (t>{x). We now demand that 
the 4>{x) satisfies the KG equation we find that (j){k, xq) should satisfy the 
condition 

dl4>{k,xn) + {P + m^)^{k,XQ)=0 (120) 
Also, since (/)(x) is a real Hermitian field, (j){k,xo) must satisfy 

,xo) = 4>{-k,XQ) (121) 



16 



The time dependence of (t){k,Xo) is trivial. Let us write ^{k,Xo) as the 
sum of two terms 

${k,xo) = (^+(fc)e*'^(^)^« + ^_(fc)e-*'"(^)^° (122) 

The operators 4>+{k) and <^+(fc) are not independent since the reality 
condition imphes that 

4>+ik) = 4>U-k) 4>Uk) = 4>-{-k) (123) 

This expansion is a solution of the equations of motion if w(fc) is given by 

w{k) = \Jp+m? (124) 

Let us define the operators a{k) and its adjoint a^(fc) by 

a{k) = 2oj{k)(i)-{k) a\k) = 2oj{k)^l{k) (125) 

The operators a^k) and a{k) obey the (generalized) creation-anihilation 
operator algebra 

[a{k), a\k')] = {2TTf2u;{k) 6^{k - k') (126) 
In terms of the operators a\k) and a{k) field operator is 



Sk 



{2nf2oj{k) 



(127) 



We have chosen to normalize the operators in such a way that the phase 
space factor takes the Lorentz invariant form -^-4-. 

The canonical momentum also can be expanded in a similar way 

n{x) = -i J ^^^|^'^^'^^^^ a>(fc)[a(fc)e-^"<^)"°+'^-"-a^(fc)e'"^^)e''"(^)"°''^-"] 

(128) 

Notice that, in both expansions, there are terms with positive and neg- 
ative frequency and that the terms with positive frequency have creation 
operators d^(fc) while the terms with negative frequency have anihilation 
operators a{k). This observation motivates the notation 

0(x) =0+(a;)+0_(x) (129) 

where are the positive frequency terms and are the negative fre- 
quency terms. This decomposition will turn out to be very useful. 



17 



3. Hamiltonian: 

We will now follow the same approach that we used for the problem of 
the linear chain and write the Hamiltonian in terms of the operators a{k) 
and a){k). The result is 

This Hamiltonian needs to be normal-ordered relative to some ground 
state which we will now define. 

4. Ground State: 

Let |0) be state which is anihilated by all the operators a{k), i.e., 

a(fc)|0) = (131) 

Relative to this state, that we will call the vacuum state, the Hamiltonian 
can be written on the form 

H=:H:+Eo (132) 

where : H : is normal ordered relative to the state |0). In other words, 
in : _ff : all the destruction operators appear the right of all the creation 
operators. Therefore :H: anihilatcs the vacuum 

:^:|0)=0 (133) 

The real number Eq is the ground state energy. In this case it is equal to 

Eo = J d^fc^(5(0) (134) 

when S{0) is the infared divergent number 

6{0) = lim 6^{p) = lim / e'^'^ = (135) 

where V is the (infinite) volume of space. Thus, Eq is extensive and can 
be written as Eq = sqV, where sq is the ground state energy density. We 
find 

^°=y (2^-^"2y (2^v^'+-' 

5. Divergence: 



Eq. 136 is the sum of the zero-point energies of all the oscillators. This 
quantity is formally divergent since the integral is dominated by the con- 
tributions with large momentum or, what is the same, short distances. 



18 



This is an ultraviolet divergence. It is divergent because the system has 
an infinite number of degrees of freedom even if the volume is finite. We 
will encounter other examples of similar divergencies in field theory. It is 
important to keep in mind that they are not artifacts of our scheme but 
that they result from the fact that the system is in continuous space-time 
and thus it is infinitely large. 

It is interesting to compare this issue in the phonon problem with the 
scalar field theory. In both cases the ground state energy was found to 
be extensive. Thus, the infrared divergence in Eq was expected in both 
cases. However, the ultraviolet divergence that we found in the scalar field 
theory is absent in the phonon problem. Indeed, the ground state energy 
density Eq for the linear chain with lattice spacing a is 

p dknuj(k) /■» dkl [¥k 4Dh^ . 2.A:a, 

= 1, - J_, 2^2 Vir + -M- ^^^^t) (1^^) 

Thus integral is finite because the momentum integration is limited to 
the range \k\ < ^. Thus the largest momentum in the chain is ^ and it 
is finite provided that the lattice spacing is not equal to zero. In other 
words, the integral is cut off by the lattice spacing. However, the scalar 
field theory that we are considering does not have a cut off and hence the 
energy density blows up. 

We can take two different points of view with respect to this problem. 
One possibility is simply to say that the ground state energy is not a 
physically observable quantity since any experiment will only yield infor- 
mation on excitation energies and in this theory, they are finite. Thus, 
we may simply redefine the zero of the energy by dropping this term off. 
Normal ordering is then just the mathematical statement that all energies 
are measured relative to that of the ground state. As far as free field 
theory is concerned, this subtraction is sufficient since it makes the the- 
ory finite without affecting any physically observable quantity. However, 
once interactions are considered, divergencies will show up in the formal 
computation of physical quantities. This procedure then requires further 
subtractions. An alternative approach consists in introducing a regulator 
or cut off. The theory is now finite but one is left with the task of proving 
that the physics is independent of the cut off. This is the program of 
the Renormalization group. Although it is not presently known if there 
should be a fimdamcntal cut off in these theories, i.e., if there is a more 
fundamental description of Nature at short distances and high energies, it 
is clear that if these theories are to be regarded as effective hydrodynamic 
theories valid below some high energy scale, then a cut off is actually 
natural. 

6. Hilbert Space: 



19 



We can construct the spectrum of states by inspection of the normal or- 
dered Hamiltonian 



u{k) a\k)a{k) (138) 



(27r)32w(/c) 

This Hamiltonian commutes with the total momentum P 

P= d^xIl{x,xo)vH^,xo) (139) 

Jxo fixed 

which, up to operator ordering amibiguities, is the quantum mechanical 
version of the classical linear momentum , 

P^ = j d^xT°^= [ d^xU{x,xo)S7^ <(>{x,xo) (140) 

In Fourier space P becomes 

P= / 'i^^k aUkWk) (141) 

J (27r)32a;(fc) 

P has an operator ordering ambiguity which wc will fix below by normal 
ordering. By inspection wc sec that P commutes with H. 

: H : also commutes with the oscillator occupation number h{k), defined 

by 

h{k) = a''{k)aik) (142) 

Since {h{k)} and H commute with each other, we can use a complete set 
of eigenstates of {n{k)} to span the Hilbert space. Since we will regard the 
excitations counted by h{k) as particles, this Hilbert space has an indefi- 
nite number of particles and it is called Pack space. The states {|{?i(fc)})}, 
defined by 

\Mk})^^M{k)[a\k)r^~'^\0) (143) 
fe 

(with J\f{k) normalization constants) are eigenstates of the operator h{k) 

n{k)\{n{k)} >= {2nf2uj{k)n{k)\{n{k) > (144) 

These states are the occupation number basis of the Fock space. 
The total number operator N 

_fk 
{2TTf2uj{k) 

commutes with the Hamiltonian H and it is diagonal in this basis i.e., 

N\{n{k)}) = j d^kn{k) \{n{k)}) (146) 



^ n(k) (145) 

(27r)32a;(A;) 



20 



The energy of these states is 



H\{n{m 



£k n{k)Lu{k) + Eo 



\{n{k)}) 



(147) 



Thus, the excitation energy s{{n{k)}) of this state is J d^k n{k)uj{k). 

The operator P has an operator ordering ambiguity. It will be fixed 
by requiring that the vacuum state |0) be translationally invariant, i.e., 
P^\Q) = 0. In terms of creation and anihilation operators we get 



d^k 



k^h{k) 



(27r)32w(fc) 

Thus, P^ is diagonal in the basis |{'^(^)}) since 



Pi\{n{k)}) 



I' 



k k^n{k) 



\Mk)}) 



(148) 



(149) 



The state with lowest energy, the vacuum state |0) has n{k) = 0, for all 
k. Thus the vacuum state has zero momentum and it is translationally 
invariant. 

The state defined by 

\k) = aHfc)|0) (150) 

have excitation energy uj{k) and total momentum k. Thus, the states \k) 
are particle-like excitations which have an energy dispersion curve 



E 



(151) 



which is characteristic of a relativisitc particle of momentum k and mass m. 
Thus, the excitations of the ground state of this field theory are particle- 
like. From our discussion wc can sec that these particles are free since 
their energies and momenta are additive. 



7. Causality: 



The starting point of the quantization procedure was to impose equal- 
time commutation relations among the canonical fields (f){x) and momenta 
II(a;). In particular two field operators on different spacial locations com- 
mute at equal times. But, do they commute at different times? 

Let us calculate the commutator A(a; — y) 

iA{x-y) = [4>{x)A{y)] (152) 



21 



where 4'{x) and (t>{y) arc Hciscnbcrg field operators for space-time points 
X and y respectively. From the Fourier expansion of the fields we know 
that the field operator can be split into a sum of two terms 

(t>{x) ^ 4>+ix) + 4>^{x) (153) 

where contains only creation (anihilation) operators and positive 

(negative) frequencies. Thus the commutator is 

iA{x -y) = [4>+ (x) ,4>+{y)] + [4>^{x),$^{y)] 

+[Mx),4'-{y)] + [4'-ix),My)] 

(154) 

The first two terms always vanish since the 0+ operators commute among 
themselves and so do the operators (f>-. Thus, we get 

iA{x -y) = [4>+ {x) , 4>- (y)] + [4>- (x) , 4>+ (y)] = 

+[a{k), a}(k')\ exp {iuj{k)xo —ik-x — iu){k'), yo + ik' ■ y^} 

(155) 

where 

[dk= [ ^-^^ (156) 

J J (27r)32w(fc) 

By using the commutation relations, we get 

iA{x ~ ~ J [e''^(^)(^°~''°)~'^'(^~f — g-i'^i^){xo-yo)+ik-{x-v)-^ 

(157) 

With the help of the function e{k^), defined by 

e(fcO) = ^^sign(fcO) (158) 

we can write A(a; — y) in the manifestly Lorentz invariant form 

iA{x -y)= [ 7^<^(fc^ - m2)e(fc°)e-''=-(^-^) (159) 
J (27r)'^ 

The integrand vanishes unless the mass shell condition k^ — vr? = is 
satisfied. Notice that A(a; — y) satisfies the initial condition 

d^A\^,=y,=-b\x-y) (160) 



22 



At equal times xo = yo the commutator vanishes, 

A(f-y,0) = (161) 

Furthermore, it vanishes if the space-time points x and y are separated by 
a space-like interval, {x — y)"^ < 0. This must be the case since A(a: — y) 
is manifestly Lorentz invariant. Thus if it vanishes at equal times, where 
{x — yY = {xQ—yo)'^ — {x — y)'^ = {x — y^Y < 0; it must vanish for all events 
with the negative values of {x — y)^. This implies that, for events x and y, 
which are not causally connected A(a; — y) = and that A(a; — y) is non- 
zero only for causally connected events, i.e., in the forward light-cone. 



time 




Figure 3: The light-cone. 



4.5 Symmetries of the Quantum Theory 

In our discussion of Classical Field Theory we discovered that the presence of 

continuous global symmetries implied the existence of constants of motion. In 
addition, the constants of motion were the generators of infinitesimal symmetry 
transformations. It is then natural to ask what role do symmetries play in the 
quantized theory. 



23 



In the quantized theory all physical quantities are represented by operators 
which act on the Hilbert space of states. The classical statement that a quantity 
A is conserved if its Poisson Bracket with the Hamiltonian is zero 

A A 

— = {A,H}PB (162) 
becomes, in the quantum theory 

i^ = [AH,H] (163) 

and it applies to all operators in the Heisenberg representation. Then, the 
constants of motion of the quantum theory are operators which commute with 
the Hamiltonian. 

Therefore, the quantum theory has a symmetry if and only if the charge Q, 
which is a Hermitian operator associated with a classically conserved current 
j''(a;) via the correspondence principle, 

Q = I dPxf{x,xo) (164) 

J xq fixed 

commutes with H 

[Q,H]=Q (165) 

If this is so, the charges Q constitute a representation of the generators of the Lie 
group in the Hilbert space of the theory. The transformations U [a) associated 
with the symmetry 

U{a) = exp(iQ!Q) (166) 

are unitary transformations which act on the Hilbert space of the theory. 

For instance, we saw that for a translationally invariant system the classical 
energy-momentum four-vector P** 

= ( d^xT^f" (167) 

is conserved. In the quantum theory becomes the Hamiltonian operator H 

and P' the total momentum operator. In the case of a free scalar field we saw 
before that these operators commute with each other, [P^,H] = 0. Thus, the 
eigenstates of the system have well defined total energy and total momentum. 
Since P^ is the generator of infinitesimal translations of the classical theory, it 
is easy to check that its equal-time Poisson Bracket with the field (p{x) is 

{^{x,xo),P'}pB = di^ (168) 

In the quantum theory the equivalent statement is that the operators 4>{x) and 
P^ satisfy the equal-time commutation relation 

[^{x,xo),P']=idi^{S,xo) (169) 



24 



Consequently, ^{x^ + ,xo) and ^{x^,xo) are related by 

${x^ + a^,xo) = e'^i^^^ix, xo)e-'''^^^ (170) 

Translation invariancc of tlic ground state |0) implies that it is a state with zero 
total linear momentum, P-'IO) = 0. For a finite displacement a we get 

e*":'-^>) = |0) (171) 

which states that the state |0) is invariant and belongs to a one-dimensional 
representation of the group of global translations. 

Let us discuss now what happens to global internal symmetries. The simplest 
case that we can consider is the free complex scalar field (j>{x) whose Lagrangian 
C is invariant under global phase transformations. If </> is a complex field, we 
can decompose it into its real and imaginary parts 

<^=i=(</.i+i</)2) (172) 

The Classical Lagrangian for a free complex scalar field (j) is 

jC = d^(j)*d''(l)-m^<l)*(j) (173) 

now splits into two independent terms 

jC{<I>)=jC{<^^)+jC{cI>2) (174) 

where C{(j)i) and C{(j)2) are the Lagrangians for the free scalar real fields and 
(j}2- The canonical momenta n(a;) and n*(a;) decompose into 

n(a;) = ^ = ^(<^i - n\x) = -^(<^i + i^2) (175) 

In the quantum theory the operators (p and (j)^ no longer coincide with each other, 
and neither do II and 11'^. Still, the canonical quantization procedure tells us 
that (j) and 11 (and and fl^) satisfy the equal-time canonical commutation 
relations 

[4>ix, xo), U{y, xo)] = id\x - y) (176) 
The theory of a free complex scalar field is solvable by the same methods that 
we used for a real scalar field. Instead of a single creation anihilation algebra 
we must introduce now two algebras, with operators ai and a\,a2 and a\. Let 
a{k) and b{k) be defined by 



a{k) = 


71' 


^ai(fc) + ia2(fc)j 


d\k) = 


71 


(al(fc)-i4(fc)) 


b{k) = 


71 


(^ai(fc) - W2(fc)) 


b\k) = 


71 


(al(fc)+i4(fc)) 



(177) 



25 



which satisfy the algebra 

[a{k),a''{k')] = [b{k),b''{k')] = {2Trf2uj{k) 5^{k-k') (178) 
while all other commutators vanish. The Fourier expansion for the fields now is 

(179) 



where ui{k) = + rm? and ko = ui{k). The normal ordered Hamiltonian is 
■■H:=J {&Hk)a{k) + b\k)b{k)) (180) 



and the total momentum P is 

, , (fk 

pj 



{a){k)d{k) + b\k)b{k)j (181) 



(27r)32a;(fc) 

we see that there are two types of quanta, a and b. The field (j) creates 6-quanta 
and it destroys a-quanta. The vacuum has no quanta. 

The one-particle states have now a two-fold degeneracy since the states 
a\k)\Q) and 6^(fc)|0) have one particle of type a and one of type b respectively 
but these states have exactly the same energy, uj^k), and the same momentum 
k. Thus for each value of the energy and of the momentum we have a two 
dimensional space of possible states. This degeneracy is a consequence of the 
symmetry: the states form multiplcts. 

What is the quantum operator which generates this symmetry? The classi- 
cally conserved current is 

= iri<l> (182) 

In the quantum theory becomes the normal-ordered operator : :. The 
corresponding global charge Q is 

Q = : [ d^xi ($^do4> - do 



= : j d^x i 



d^k 
{2Trf2oj{k) 

(183) 



where Na and N), are the number operators for quanta of type a and b respec- 
tively. Since [Q,H] =0, the difference Na — Nb is conserved. Since this property 



26 



is consequence of a symmetry, it is expected to hold in more general theories 
than the simple non-interacting case that we are discussing here, provided that 
[Q, H] = 0. Thus, although Na and Nf, may not be conserved separately in the 
general case, the difference Na — Nh will be conserved if the symmetry is exact. 

Let us now briefly discuss how is this symmetry realized in the spectrum of 
states. The vacuum state has Na = Nf, = 0. Thus, the generator Q anihilates 
the vacuum 

g|0) = (184) 
Therefore, the vacuum state is invariant (i.e., a singlet) under the symmetry, 

|0)' = e"5«|0) = |0) (185) 

Because the state |0) is always defined up to an overall phase factor, it spans 
a one-dimensional subspace of states which are invariant under the symmetry. 
This is the vacuum sector and, for this problem, it is trivial. 

There are two linearly-independent one-particle states, |-|-, fc) and \ — ,k) de- 
fined by 

\+,k) = a\k) \0) \-,k) = b^{k)\0) (186) 

Both states have the same momentum k and energy oj{k). The Q-quantum 
numbers of these states, which we will refer to as their charge, are 

Q\+,k) = (Na - Nb)aHk)\0 >= Na a\k)\0) = +\+,k) 
Q\-,k) ={Na-Nt,)b\k)\0) = -\-,k) 

(187) 

Hence 

Q\a,k) =a\a,k) (188) 

where cr = ±1. Thus, the state a^k)\0) has positive charge while w{k)\0) has 
negative charge. Under a finite transformation U{a) = exp{iaQ) they transform 
like 

|+, k)' = U{a) |+, k) = exp{iaQ) |+, k) = e'" |+, k) 
\-,ky = U{a) \-,k) =exp{iaQ)\-,k) \-,k) 

(189) 

The field 4>{x) itself transforms like 

0'(a;) = exp(-iaQ) 4>{x) exp(iaQ) = e'°'4>{x) (190) 

since 

[Q, ${x)] = -0(x) [Q, $^ix)] = $^ix) (191) 

Thus the one-particle states are doubly degenerate, and each state transforms 
non-trivially under the symmetry group. By inspecting the Fourier expansion 
for the complex field (j), we see that </> is a sum of two terms: a set of positive 



27 



frequency terms, symbolized by and a set of negative frequency terms, 
In this case all positive frequency terms create particles of type b (which carry 
negative charge) while the negative frequency terms anihilate particles of type 
a (which carry positive charge). The states |±, fc) are commonly referred to as 
particles and antiparticles: particles have rest mass m, momentum k and charge 
+1 while the antiparticles have the same mass and momentum but carry charge 
— 1. This charge is measured in units of the electromagnetic charge — e (see the 
previous discussion on the gauge current). 

Let us finally note that this theory contains an additional operator, the 
charge conjugation operator C, which maps particles into antiparticles and vice 
versa. This operator commutes with the Hamiltonian, [C, H] = 0. This property 
insures that the spectrum is invariant under charge conjugation. In other words, 
for every state of charge Q there exists a state with charge —Q, all other quantum 
numbers being the same. 

Our analysis of the free complex scalar field can be easily extended to systems 
which are invariant under a more general symmetry group G. In all cases the 
classically conserved charges become operators of the quantum theory. Thus, 
there are as many charge operators as generators are in the group. The 
charge operators represent the generators of the group in the Hilbert (or Fock) 
space of the system. The charge operators obey the same commutation relations 
as the generators themselves do. A simple generalization of the arguments 
that we have used here tell us that the states of the spectrum of the theory 
must transform like the (irreducible) representations of the symmetry group. 
However, there is one important caveat that should be made. Our discussion 
of the free complex scalar field shows us that, in that case, the ground state is 
invariant under the symmetry. In general, the only possible invariant state is the 
singlet state. All other states are not invariant and transform non-trivially. But, 
shoiild the ground state always be invariant? In elementary quantum mechanics 
there is a theorem, due to Wigner and Weyl, which states that for a finite 
system, the ground state is always a singlet. However, there are many systems 
in Nature, such as magnets and many others, which have ground states which 
are not invariant under the symmetries of the Hamiltonian. This phenomenon, 
known as Spontaneous Symmetry Breaking, does not occur in simple free field 
theories but it does happen in non-linear or interacting theories. We will return 
to this important question later on. 



28