22 Quantum field theory 2 Quantization of the scalar field Commutator relations. The strategy to quantize a classical field theory is to inter- pret the fields Φ(x) and Π(x) = Φ(_ x) as operators which satisfy canonical commutation relations. This is completely analogous to the transition from classical to quantum me- chanics for discrete systems, where qi and pi are promoted to self-adjoint operators that satisfy [qi; pj] = iδij; [qi; qj] = [pi; pj] = 0 : (2.1) These relations hold in the Schr¨odinger picture where the time dependence is carried by the states alone; in the Heisenberg picture the operators are time-dependent and the commutation relations are imposed at equal times. In the following we will always work in the Heisenberg picture, so we demand that for equal times Φ(x); Π(y) = iδ(3)(x y) ; x0=y0 − (2.2) Φ(x); Φ(y) x0=y0 = Π(x); Π(y) x0=y0 = 0 : Despite appearances, this does not destroy Lorentz covariance because x and y are separated by a spacelike distance (x y)2 < 0 which is preserved under a Lorentz − transformation. By virtue of the Dirac delta function, Φ(x) and Π(x) are now operator- valued distributions; to arrive at well-defined expressions one should in principle `smear' them with smooth test functions. The wave functions in quantum mechanics are also fields Φ(x) that satisfy (relativistic or non- relativistic) wave equations, but there they are interpreted as single-particle wave functions in some Hilbert space by imposing an appropriate scalar product. (Unfortunately, already for relativistic Klein- Gordon particles the scalar product is not positive definite, so we lost the probability interpretation). In quantum field theory we impose instead an operator structure on Φ(x), which is why the procedure is often called `second quantization'. Since we really only quantize the field Φ(x) once, the correct term should be ‘field quantization'. Fourier expansion. We write the Fourier expansion for solutions of the free Klein- Gordon equation as 1 Z d3p Φ(x) = a(p) e−ipx + ay(p) eipx ; (2.3) 3=2 0 (2π) 2Ep p =Ep so the Fourier coefficients (from now on we abbreviate a(p) ap) will inherit the oper- ≡ ator structure. In the following we will often encounter the Lorentz-invariant integral R 3 measure d p=(2Ep) that is obtained by restricting the four-momentum integration to the positive-energy mass shell (which is a Lorentz-invariant condition): Z Z 0 0 Z 3 δ(p Ep) + δ(p + Ep) d p d4p Θ(p0) δ(p2 m2) = d4p Θ(p0) − = : (2.4) − 2Ep 2Ep 3 0 Consequently, also the combination 2Ep δ (p p ) is Lorentz-invariant. Upon inserting − y the Fourier expansion into Eq. (2.2) we obtain the commutation relations for ap, ap0 : y (3) 0 y y [ap; a 0 ] = 2Ep δ (p p ) ; [ap; ap0 ] = [a ; a 0 ] = 0 : (2.5) p − p p 2 Quantization of the scalar field 23 This can be shown in several ways. For example, let's write Z −iEpt y iEpt ! 1 3 ap e + a−pe ip·x Φ(x) = 3=2 d p e ; (2π) 2Ep (2.6) −iEpt y iEpt ! 1 Z ap e − a e Π(x) = φ_(x) = d3p −p eip·x ; (2π)3=2 2i and abbreviate the two brackets (the three-dimensional Fourier transforms) by 1 y 1 y Φep(t) = ap(t) + a−p(t) ; Πe p(t) = ap(t) − a−p(t) : (2.7) 2Ep 2i It follows that y Φep(t) = Φe−p(t) ; ap(t) = Ep Φep(t) + i Πe p(t) ; (2.8) y y y y Πe p(t) = Πe −p(t) ; ap(t) = Ep Φep(t) − i Πe p(t) : Now insert this into the commutator: Z 3 3 0 d p d p i(p·x−p0·y) y ! (3) Φ(x); π(y) 0 0 = e Φep(t); Πe 0 (t) = iδ (x − y) : (2.9) x =y (2π)3 p y Here we have changed the integration variable from p ! −p and used Πe −p(t) = Πe p(t). Hence, the commutator for the Fourier transformed quantities must be also a δ−function (the time dependence cancels), y (3) 0 Φep(t); Πe p0 (t) = iδ (p − p ) ; (2.10) y and we can extract the commutator relation for ap and ap0 : y y y 0 y y (3) 0 [ap; ap0 ] = [ap(t); ap0 (t)] = −iEp Φep(t); Πe p0 (t) + iEp Φep(t); Πe p0 (t) = 2Ep δ (p − p ) : (2.11) Another way to arrive at this result is to use Eq. (1.27) for the Fourier coefficients and calculate the commutator directly: Z $ Z $ a = hf ; Φi = i d3x f ∗(x) @ Φ(x) ; ay = −hf y; Φi = −i d3x f (x) @ Φ(x) : (2.12) p p p 0 x0=t p p p 0 x0=t For equal times x0 = y0 = t we can insert the commutator relations (2.2), so that $ $ Z Z y 3 3 ∗ @ @ ::: [a ; a ] = d x d y f (x) f 0 (y) Φ(x); Φ(y) = p p0 p p @x0 @y0 Z Z h i 3 3 ∗ _ _∗ 3 (2.13) = d x d x fp (x) fp0 (y) − fp (x) fp0 (y) iδ (x − y) Z $ 3 ∗ 3 = i d x fp (x) @0 fp0 (x) = hfp; fp0 i = 2Ep δ (p − q) : Hamilton and momentum operator. To proceed, we derive the Fourier decomposi- tion for the Hamiltonian (1.15) of the free scalar field theory. The form of the Hamilto- nian already resembles that of a collection of harmonic oscillators at each point x, but the term ( Φ)2 couples the degrees of freedom at x and x + δx. We can diagonalize r it in momentum space by inserting the relations (2.6{2.7); in that way it becomes the sum of decoupled harmonic oscillators with frequencies Ep: Z Z 3 1 2 2 2 2 3 1 h y 2 y i H = d x Π + ( Φ) + m Φ = d p Πe (t) Πe p(t) + E Φe (t) Φep(t) : (2.14) 2 r 2 p p p To arrive at this result, use Z Z Z 3 3 0 3 2 3 d p d p 0 i(p+p0)·x d x (rΦ) = d x Φep(t) Φep0 (t)(−p · p ) e (2π)3 Z Z (2.15) 3 2 3 y 2 = d p Φep(t) Φe−p(t) p = d p Φep(t) Φep(t) p 24 Quantum field theory and similarly Z Z Z Z 3 2 3 y 3 2 3 y d x Φ = d p Φep(t) Φep(t) ; d x Π = d p Πe p(t) Πe p(t) : (2.16) Inserting the decomposition (2.7) finally yields the result Z 3 y y d p ap ap + ap ap H = Ep : (2.17) 2Ep 2 Unfortunately this expression is divergent because it contains the sum of the zero-point energy of all oscillators: y y ap ap + ap ap y 1 y y 3 = a ap + [ap; a ] = a ap + Ep δ (0) : (2.18) 2 p 2 p p The Dirac delta is proportional to the volume; had we studied the system in a finite box, we would write (2π)3 δ3(0) V . (This is an infrared divergence.) However, p 2 !2 for large p we have Ep p + m p and the integral still diverges. If we regulate ∼ ' j j the divergence by integrating only up to a cutoff p Λ, the energy density of the j j ≤ vacuum becomes Λ Z 3 Z Evac 1 d p 3 4 ρvac = = Ep dp p Λ : (2.19) V 2 (2π)3 ∼ ∼ This is a first example of an ultraviolet divergence which we will frequently encounter later. Since (in a theory without gravity) we can only measure energy differences, we can simply discard it so that the vacuum energy is zero. This is formally called normal ordering or Wick ordering: we obtain the normal-ordered form : : of some operator O O by moving all creation operators to the left of all destruction operators. Later when we discuss renormalization we will see how UV divergences can be systematically removed from the theory; for the time being we interpret all operators as being normal-ordered. Hence, the Hamilton operator becomes Z Z 3 3 1 2 2 2 2 d p y H = d x : Π + ( Φ) + m Φ : = Ep ap ap : (2.20) 2 r 2Ep We can repeat the procedure to obtain the spatial momentum operator P . We identify it with the classical charge P i = R d3x T 0i in Eq. (1.50) that follows from the invariance under spatial translations. The analogous calculation gives Z Z Z 3 3 3 y d p y P = d x :Π Φ:= i d p p :Πe p(t) Φep(t): = p ap ap ; (2.21) − r 2Ep so that we can combine Eqs. (2.20) and (2.21) into the covariant four-momentum operator Z Z 3 µ 3 0µ d p µ y P = d x :T : = p a ap : (2.22) p 0 2Ep p =Ep 2 Quantization of the scalar field 25 Fock space. What is the Hilbert space on which the four-momentum operator acts? Since P µ is self-adjoint it has eigenstates with real eigenvalues. Let k be such an µ µ j i 0 eigenstate with P k = k k , so that k is the momentum of the state and k = Ek j i j i its energy.