2. Free Fields

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2. Free Fields 2. Free Fields “The career of a young theoretical physicist consists of treating the harmonic oscillator in ever-increasing levels of abstraction.” Sidney Coleman 2.1 Canonical Quantization In quantum mechanics, canonical quantization is a recipe that takes us from the Hamil- tonian formalism of classical dynamics to the quantum theory. The recipe tells us to a take the generalized coordinates qa and their conjugate momenta p and promote them to operators. The Poisson bracket structure of classical mechanics morphs into the structure of commutation relations between operators, so that, in units with ~ =1, a b [qa,qb]=[p ,p ]=0 b b [qa,p ]=iδa (2.1) In field theory we do the same, now for the field φa(~x )anditsmomentumconjugate ⇡b(~x ). Thus a quantum field is an operator valued function of space obeying the com- mutation relations a b [φa(~x ),φb(~y )] = [⇡ (~x ),⇡ (~y )] = 0 [φ (~x ),⇡b(~y )] = iδ(3)(~x ~y ) δb (2.2) a − a Note that we’ve lost all track of Lorentz invariance since we have separated space ~x and time t. We are working in the Schr¨odinger picture so that the operators φa(~x )and ⇡a(~x ) do not depend on time at all — only on space. All time dependence sits in the states which evolve by the usual Schr¨odinger equation | i d i | i = H (2.3) dt | i We aren’t doing anything di↵erent from usual quantum mechanics; we’re merely apply- ing the old formalism to fields. Be warned however that the notation for the state | i is deceptively simple: if you were to write the wavefunction in quantum field theory, it would be a functional, that is a function of every possible configuration of the field φ. The typical information we want to know about a quantum theory is the spectrum of the Hamiltonian H.Inquantumfieldtheories,thisisusuallyvery hard. One reason for this is that we have an infinite number of degrees of freedom — at least one for every point ~x in space. However, for certain theories — known as free theories —wecanfind awaytowritethedynamicssuchthateachdegreeoffreedomevolvesindependently –21– from all the others. Free field theories typically have Lagrangians which are quadratic in the fields, so that the equations of motion are linear. For example, the simplest relativistic free theory is the classical Klein-Gordon (KG) equation for a real scalar field φ(~x , t ), µ 2 @µ@ φ + m φ =0 (2.4) To exhibit the coordinates in which the degrees of freedom decouple from each other, we need only take the Fourier transform, 3 d p ip~ ~x φ(~x , t )= e · φ(p~ , t )(2.5) (2⇡)3 Z Then φ(p~ , t )satisfies @2 +(p~ 2 + m2) φ(p~ , t )=0 (2.6) @t2 ✓ ◆ Thus, for each value of p~ , φ(p~ , t )solvestheequationofaharmonicoscillatorvibrating at frequency 2 2 !p~ =+ p~ + m (2.7) We learn that the most general solution top the KG equation is a linear superposition of simple harmonic oscillators, each vibrating at a di↵erent frequency with a di↵erent am- plitude. To quantize φ(~x , t ) we must simply quantize this infinite number of harmonic oscillators. Let’s recall how to do this. 2.1.1 The Simple Harmonic Oscillator Consider the quantum mechanical Hamiltonian 1 2 1 2 2 H = 2 p + 2 ! q (2.8) with the canonical commutation relations [q, p]=i.Tofindthespectrumwedefine the creation and annihilation operators (also known as raising/lowering operators, or sometimes ladder operators) ! i ! i a = q + p, a† = q p (2.9) 2 p 2 − p r 2! r 2! which can be easily inverted to give 1 ! q = (a + a†) ,p= i (a a†)(2.10) p − 2 − 2! r –22– Substituting into the above expressions we find [a, a†]=1 (2.11) while the Hamiltonian is given by 1 H = 2 !(aa† + a†a) 1 = !(a†a + 2 )(2.12) One can easily confirm that the commutators between the Hamiltonian and the creation and annihilation operators are given by [H, a†]=!a† and [H, a]= !a (2.13) − These relations ensure that a and a† take us between energy eigenstates. Let E be | i an eigenstate with energy E,sothatH E = E E .Thenwecanconstructmore | i | i eigenstates by acting with a and a†, Ha† E =(E + !)a† E ,HaE =(E !)a E (2.14) | i | i | i − | i So we find that the system has a ladder of states with energies ...,E !, E, E + !,E +2!,... (2.15) − If the energy is bounded below, there must be a ground state 0 which satisfies a 0 =0. | i | i This has ground state energy (also known as zero point energy), H 0 = 1 ! 0 (2.16) | i 2 | i Excited states then arise from repeated application of a†, n 1 n =(a†) 0 with H n =(n + )! n (2.17) | i | i | i 2 | i where I’ve ignored the normalization of these states so, n n =1. h | i6 2.2 The Free Scalar Field We now apply the quantization of the harmonic oscillator to the free scalar field. We write φ and ⇡ as a linear sum of an infinite number of creation and annihilation oper- ators ap†~ and ap~ ,indexedbythe3-momentump~ , 3 d p 1 ip~ ~x ip~ ~x φ(~x )= a e · + a† e− · (2.18) (2⇡)3 2! p~ p~ Z p~ 3 h i d p !p~ ip~ ~x ip~ ~x ⇡(~x )= (pi) a e · a† e− · (2.19) (2⇡)3 − 2 p~ − p~ Z r h i –23– Claim: The commutation relations for φ and ⇡ are equivalent to the following com- mutation relations for ap~ and ap†~ [φ(~x ),φ(~y )] = [⇡(~x ),⇡(~y )] = 0 [a ,a ]=[a†,a†]=0 p~ q~ p~ q~ (2.20) (3) 3 (3) [φ(~x ),⇡(~y )] = iδ (~x ~y ) , [a ,a†]=(2⇡) δ (p~ ~q ) − p~ q~ − 3 (3) Proof: We’ll show this just one way. Assume that [a ,a†]=(2⇡) δ (p~ ~q ). Then p~ ~q − 3 3 d pdq ( i) !q~ ip~ ~x i~q ~y ip~ ~x +i~q ~y [φ(~x ),⇡(~y )] = − [a ,a†] e · − · +[a†,a ] e− · · (2⇡)6 2 ! − p~ ~q p~ q~ Z p~ 3 r ⇣ ⌘ d p ( i) ip~ (~x ~y ) ip~ (~y ~x ) = − e · − e · − (2.21) (2⇡)3 2 − − Z = iδ(3)(~x ~y ) − ⇤ The Hamiltonian Let’s now compute the Hamiltonian in terms of ap~ and ap†~ .Wehave 1 H = d3x⇡2 +( φ)2 + m2φ2 2 r Z 3 3 3 1 d xdpdq p!p~ !q~ ip~ ~x ip~ ~x iq~ ~x i~q ~x = (a e · a† e− · )(a e · a† e− · ) 2 (2⇡)6 − 2 p~ − p~ ~q − q~ Z 1 ip~ ~x ip~ ~x i~q ~x i~q ~x + (ip~ ap~ e · ip~ ap†~ e− · ) (i~q aq~ e · i~q aq†~ e− · ) 2p!p~ !q~ − · − 2 m ip~ ~x ip~ ~x i~q ~x i~q ~x + (a e · + a† e− · )(a e · + a† e− · ) 2 ! ! p~ p~ q~ ~q p p~ ~q 3 1 d p 1 2 2 2 2 2 2 † † † † = 3 ( !p~ + p~ + m )(ap~ a p~ + ap~ a p~ )+(!p~ + p~ + m )(ap~ ap~ + ap~ ap~ ) 4 (2⇡) !p~ − − − Z h i where in the second line we’ve used the expressions for φ and ⇡ given in (2.18)and (2.19); to get to the third line we’ve integrated over d3x to get delta-functions δ(3)(p~ ~q ) ± which, in turn, allow us to perform the d3q integral. Now using the expression for the 2 2 2 frequency !p~ = p~ + m ,thefirsttermvanishesandwe’releftwith 1 d3p H = ! a a† + a† a 2 (2⇡)3 p~ p~ p~ p~ p~ Z 3 h i d p 1 3 (3) = ! a† a + (2⇡) δ (0) (2.22) (2⇡)3 p~ p~ p~ 2 Z h i –24– Hmmmm. We’ve found a delta-function, evaluated at zero where it has its infinite spike. Moreover, the integral over !p~ diverges at large p. What to do? Let’s start by looking at the ground state where this infinity first becomes apparent. 2.3 The Vacuum Following our procedure for the harmonic oscillator, let’s define the vacuum 0 by | i insisting that it is annihilated by all ap~ , a 0 =0 p~ (2.23) p~ | i 8 With this definition, the energy E0 of the ground state comes from the second term in (2.22), 1 H 0 E 0 = d3p ! δ(3)(0) 0 = 0 (2.24) | i⌘ 0 | i 2 p~ | i 1| i Z The subject of quantum field theory is rife with infinities. Each tells us something important, usually that we’re doing something wrong, or asking the wrong question. Let’s take some time to explore where this infinity comes from and how we should deal with it. In fact there are two di↵erent ’s lurking in the expression (2.24). The first arises 1 because space is infinitely large. (Infinities of this type are often referred to as infra-red divergences although in this case the is so simple that it barely deserves this name). 1 To extract out this infinity, let’s consider putting the theory in a box with sides of length L.Weimposeperiodicboundaryconditionsonthefield.Then,takingthelimit where L ,weget !1 L/2 L/2 3 (3) 3 i~x p~ 3 · (2⇡) δ (0) = lim d xe p~ =0 =lim d x = V (2.25) L L/2 L L/2 !1 Z− !1 Z− where V is the volume of the box. So the δ(0) divergence arises because we’re computing the total energy, rather than the energy density .Tofind we can simply divide by E0 E0 the volume, E d3p 1 = 0 = ! (2.26) E0 V (2⇡)3 2 p~ Z which is still infinite.
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