1 Free real scalar field

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1 1 Free real scalar field The Hamiltonian is H = d 3 xh = 1 d 3 x p(x) +( φ) + m φ Let us expand both φ and p in Fourier series: d 3 p φ(t, x) = ω(p) φ(t, x)e ip x, p(t, x) = ω(p) p(t, x)eip x. where ω(p) = p + m.then: H = 1 p(p) + (π) 3 (ω(p)) φ(p) ω(p). This is a Hamiltonian for an infinite collection of harmonic oscillators labeled by p R 3 and energy ω(p). Introduce creation-annihilation operators: a(p) = p(p) iφ(p) ω(p), a (p) = p (p)+iφ (p) ω(p). Then: H = (π) 3 ω(p) ω(p) a (p)a(p)+ 1 The last term in parentheses can be dropped (divergent vacuum energy). The operators a, a satisfy: [a(p),a (p )] = (π) 3 ω(p)δ 3 (p p ), [a(p),a(p )] = 0. The expression ω(p)δ 3 (p p )islorenz-invariant,sothisisanaturalnormalization of creation-annihilation operators in a relativistic theory. So, as expected, the free scalar field describes noninteracting spinless bosonic particles with a relativistic energy-momentum relation E(p) = p + m. Free complex scalar field Commutation relations: 1

2 [φ(t, x),p(t, y)] = iδ 3 (x y), [φ(t, x),p(t, y) ]=iδ 3 (x y), [φ(t, x),p(t, y)] = 0, [φ(t, x),p(t, y) ]=0, [φ(t, x),φ(t, y)] = 0, [φ(t, x),φ(t, y) ]=0, [φ(t, x),φ(t, y) ]=0, [p(t, x),p(t, y)] = 0, [p(t, x),p(t, y) ]=0, [p(t, x),p(t, y) ]=0. Here p = φ,p = φ. Hamiltonian: H = d 3 x pp + i φ i φ + m φ φ. Let us show that these equations describe the bosonic Fock space for relativistic particles (with E p = p + m ). Let us Fourier transform the scalar field φ: φ(t, x) = E p (π) φ(t, p)e ip x. 3 The Klein-Gordon equation ( 0 + m )φ =0 gives an ordinary differential equation for φ(t, p): The general solution is φ t = (p + m ) φ. φ(t, p) =e iept a(p)+e iept c(p).

3 It will be convenient to rename c(p) =b( p).then φ(t, x) = ap e ip x + b(p) e ip x. E p (π) 3 Similarly φ(t, x) = E p (π) 3 bp e ip x + a(p) e ip x. We can invert these formulas and express a, b, a,b in terms of φ, φ and φ, φ. (This is an exercise). Then the commutation relations of a, a,b,b turn out [a(p),a (q)] = (π) 3 E p δ 3 (p q), (1) [b(p),b (q)] = (π) 3 E p δ 3 (p q), () with all other commutators vanishing. Thus it is natural to postulate the existence of the vacuum state 0, annihilatedbyalla(p) andb(p). Then the Hilbert space is the bosonic Fock space built on the sum of two copies of L (R 3 ). Why two copies? We expected only one! Resolution: we have an additional quantum number which distinguishes b-particles from a-particles. The b-particles are actually anti-particles of a-particles! (see below). Hamiltonian becomes H = 1 E p (π) E 3 p a (p)a(p)+a(p)a (p)+b (p)b(p)+b(p)b (p). Let us normal-order it: H = V (π) 3 E p +... Thus the vacuum energy density is divergent. If we cut off the integral at p =Λ,wefind E 0 = Λ4 8π This is the simplest example of an ultraviolet divergence. 3 Noether s theorem (Reading: section, pp ). 3

4 Noether s theorem says that for every continuous symmetry of the action there is a current j µ (vector-valued function made of fields and their derivatives) which satisfies µ j µ =0. This implies that Q = d 3 xj 0 (t, x) is time-independent. I.e. it is a conserved charge. In the Hamiltonian formalism this is expressed as Q, H =0,whichuponquantizationbecomes [Q, H] =0. Let us derive the Noether theorem for a theory of scalar fields with a Lagrangian L(φ a ). Suppose the infinitesimal symmetry transformation is given by δφ a = v a (φ). Consider now the same transformation, but with afunctionofx. Since the action is of first order in derivatives of φ, thevariationoftheactionmustbe of the form δs = d 4 xj µ µ, for some j µ independent of. But on equations of motion this must vanish, for arbitrary. Therefore µ j µ =0. Let us apply this procedure to the complex scalar field φ and the transformation δφ = iφ, δφ = iφ. The variation of the action is δs = i d 4 x µ φ µ φ + µ φ φ. Hence the current is j µ = i φ µ φ µ φ φ. What is the meaning of the corresponding charge, in terms of particles? d 3 k Q = a (k)a(k) b (k)b(k). (π) 3 E k I.e. it is the number of particles minus the number of anti-particles. 4

5 Let me consider another example: translational symmetry. Here δφ = µ µ φ. Note that here has a vector index. Thus we expect δs = d 4 x ν µ Tµ ν for some tensor T. (It is called the stress-energy tensor). Let us determine T.Forconstant we have δs = d 4 x µ µ L. This indeed vanishes for constant (by integration by parts), but does not vanish for nonconstant. But for nonconstant we also get other terms in the variation: δs = d 4 x µ µ L + µ ν L µ φ νφ. Hence For the free scalar field, we get For example: T µ ν = L µ φ νφ δ µ ν L. T µ ν = µ φ ν φ + µ φ ν φ δ µ ν L. T 0 0 = 0 φ 0 φ + φ φ + m φ φ. The corresponding charge is the energy (i.e. the Hamiltonian). Similarly, T 0 i = 0 φ i φ + i φ 0 φ. The corresponding charge is minus the momentum. Indeed, after expressing in terms of a and b get d 3 xti 0 d 3 k = k (π) 3 i a (k)a(k)+b (k)b(k). E k 5

6 Starting from a symmetry, one can get a conserved charge. Conversely, starting from a conserved charge Q, onecantrytogetasymmetrytransformation, by letting δf = {Q, F }. Then δh =0,andδ commutes with time translations. One can show directly that Q is the generator of symmetry transformations: Q = d 3 xp i δφ i, {Q, φ j } = δφ j. In quantum theory: A finite transformation is [Q, φ j ]= iδφ j. φ U 1 φu, U =exp( itq). In relativistic field theory, we are interested in translations and Lorenz transformations. Together they form Poincare group: x Λx + a. Generators of translations are momenta P µ = d 3 xt 0 µ. Lorenz transformations act by φ (x) =φ(λ 1 x). Infinitesimal transformation Λ = 1 + ω gives δφ = 1 (x µ ν x ν µ )φ. ω µν We can achieve this by letting M µν = d 3 x x µ T 0ν x ν T 0µ. This suggests that the conserved current for the Lorenz transformations is It is conserved because T µν = T νµ. L ρµν = x µ T ρν x ν T ρµ. 6

7 It is interesting to compute Poisson brackets or commutator of all these generators. For example: [M µν,m ρσ ]=i(g µρ M νσ (µ ν)) (ρ σ). This algebra characterizes infinitesimal Lorenz transformations. Infinitesimal rotations are J i = 1 ijkm jk, infinitesimal boosts are K i = M i0.intermsofj and K we have [J i,j j ]=i ijk J k, [J i,k j ]=i ijk K j, [K i,k j ]= i ijk J k. The other commutators are [P µ,m ρσ ]=i(g µσ P ρ (ρ σ)). 4 The spin-statistics relation Let us compute the commutator of φ(x) andφ(y) (intherealcase). It vanishes outside the light-cone. Now let us try to construct a similar theory based on the fermionic Fock space. The anticommutator comes out to be nonvanishing outside the lightcone,so this is unacceptable. 5 Scattering theory First: d 3 xe ikx φ(x) = 1 ω a(k)+ 1 ω eiωt a ( k), d 3 xe ikx 0 φ = i a(k)+ i eiωt a ( k). Hence a(k) = d 3 xe ikx (i 0 φ + ωφ) =i d 3 xe ikx 0φ. 7

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