QUANTUM FIELD THEORY. Kenzo INOUE

Transcription

1 QUANTUM FIELD THEORY Kenzo INOUE September 0, 03

2

3 Contents Field and Lorentz Transformation. Fields Lorentz Transformation Lorentz Transformation of Fields Free Scalar Quantum Field 5. Classical Theory of Real Scalar Field Canonical Quantization Free Scalar Field T-Product and Propagator Complex Scalar Field Chapter-Project/Q Free Dirac Quantum Field 5 3. Statistics of Particles and Commutators of Fields Dirac Equation Free Dirac Field Mode Expansion Charge Conjugation T-Product and Propagator of Fermi Field Chapter-Project/F Interaction of Fields 4 4. Interacting Fields Green s Function Path Integral Path Integral Representation of Quantum Mechanics Path Integral Representation of Quantum Field Theory Perturbation Theory Path Integral of Fermi Field Connected Green s Function and Effective Action Connected Green s Function Effective Action Scattering Amplitude and Scattering Cross Section Transition Amplitude Scattering Cross Section One-Particle Wave Function i

4 ii CONTENTS 8 Renormalization Loop Expansion of Effective Action Renormalization Renormalization of λφ 4 Theory Symmetry and Effective Potential Symmetry of Effective Action Effective Potential Spontaneous Symmetry Breakdown Nambu-Goldstone Theorem Chapter-Project/T A Perturbation Theory Based on the Interaction Picture 03 A. Interaction Picture A. Time Development Operator A.3 Green s Function A.4 Generating Functional

5 Chapter Field and Lorentz Transformation. Fields The field φ(x is a set of dynamical degrees of freedom distributed all over the points x in the three-dimensional space. There are varieties of field in Nature. The familiar examples may be an electric field E(x and a magnetic field B(x. The field φ(x may change its value in accordance with the time variation of its surroundings. What is more, the field itself evolves its motion by its own force. Its time evolution φ(x, t is governed by the equation of motion of the field φ(x. One of the basic facts recognized at the microscopic level of Nature is that Any Dynamical Freedom Reveals Its Dynamics as Quantum Dynamics, where the dynamical freedom behaves as a quantum mechanical operator. The field φ(x behaves as a quantum field operator ˆφ(x. The astonishing consequence of this Quantum Principle is that the quantum field ˆφ(x describes, through its quantum excitation, a quantum mechanical particle specific to the field φ(x. Any of the quantum mechanical particles of species s is described by its own field φ s (x. The quantum electro-magnetic field describes a particle photon, which reveals itself at a macroscopic level as a light. One more basic fact in Nature is summarized as the Principle of Relativity. The principle asserts that The Law of Nature is Equally Realized in Any Inertial System. For the field φ(x, this assertion amounts to that the equation of motion of the field takes the same form in any inertial system. Precisely speaking, the form of the equation of motion written down in terms of the time t and the coordinate x of one inertial system is identical to the form written down in termes of the time t and the coordinate x of another inertial system. If the form of two expressions were different, we might be able to distinguish one inertial system from another inertial system by carrying out the same experiment in each inertial system. The clear experiment may be a measurement of a velocity of light. The velocity of light is a universal constant for any inertial system. The transformation which connects one inertial system to another inertial system is called Lorentz transformation. Therefore, the equation of motion of fields keeps its form unchanged under the Lorentz transformation.

6 CHAPTER. FIELD AND LORENTZ TRANSFORMATION. Lorentz Transformation The principle of relativity is nicely formulated by introducing the 4-dimensional coordinate system x µ (µ = 0,,, 3 which combines the time t and the space coordinate x as x µ = (x 0, x, x, x 3 (ct, x, y, z = (ct, x, (. where c is the velocity of light. The coordinate x µ specifies the point P in the 4-dimensional space. In this 4-dimensional space, 4-dimensional distance is introduced. The distance s from the origin O to the point P specified by the coordinate x µ is defined as where s (x 0 (x (x (x 3 = c t x 3 x 0 = ct = g µν x µ x ν, (. P µ,ν=0 x µ g µν = O x (.3 is called metric tensor of the 4-dimensional space which measures the distance s between O and P from the difference of their coordinates. When s > 0, the interval of O and P is called time-like. When s < 0, it is called space-like. The special case s = 0 is called light-like. Now, we write down the Lorentz transformation. Suppose one inertial system is described by the coordinate x µ and another one by x µ. Since both x µ and x µ are coordinate systems, their relation must be linear. Otherwise, the twice-length stick might not be measured as twice-length in another coordinate system. We assume the origin O of the 4-dimensional space is common for both systems. Therefore, the transformation should take a form x µ x µ = L.T. 3 Λ µ νx ν, (.4 with the constant 4 4 matrix Λ µ ν. In addition, it must preserve the 4-dimensional distance (.: s = 3 µ,ν=0 g µν x µ x ν = ν=0 3 µ,ν=0 g µν x µ x ν = s. (.5 This requirement comes from the motion of light. Suppose the light, passing through the origin O, arrives at the point P. For an observer in the inertial system with x µ, the distance s between O and P should be s = 0, because the light moves along the straight line with the velocity c and therefore x = ct. For another observer with x µ, this distance s should also be s = 0, because the velocity of light c is universal and he must observe x = ct. The condition (.5 assures s = s = 0. Once this equality is realized for zero distance s = 0, the equality for nonzero distance s = s 0 results from the linearity of the transformation (.4. The 4-component quantity A µ = (A 0, A, A, A 3 = (A 0, A which transforms in the same way as x µ is called Lorentz vector : A µ A µ = L.T. 3 Λ µ νa ν. (.6 ν=0

7 .. LORENTZ TRANSFORMATION 3 The square of A µ defined below is invariant under Lorentz transformation as the distance s is invariant : 3 3 A g µν A µ A ν = g µν A µ A ν A. (.7 µ,ν=0 The invariant quantity under Lorentz transformation is called Lorentz scalar. The scalar product of two Lorentz vectors A µ and B µ is defined by µ,ν=0 A B A 0 B 0 A B = 3 µ,ν=0 g µν A µ B ν. (.8 This is also Lorentz scalar because (A + B = A + B + A B is Lorentz scalar. It is convenient to introduce the subscript Lorentz vector A µ by A µ = (A 0, A, A, A 3 (A 0, A, A, A 3 = (A 0, A = g µν A ν, (.9 where, in the final expression, we have omitted the summation symbol Σ 3 ν=0. Hereafter, when the same Greek indices appear in one term simultaneously as superscript and subscript indices, we assume that the summation over 0,,, 3 is taken and we omit the symbol Σ. Thus, we write A µ = g µν A ν ; g µν = g µν = , (.0 A B = g µν A µ B ν = A µ B µ = A µ B µ = g µν A µ B ν. (. Let us investigate the Lorentz transformation of the subscript vector A µ. According to the definition, we have Therefore, A µ transforms as A µ = g µρ A ρ L.T. A µ = g µρ A ρ = g µρ Λ ρ σa σ = g µρ Λ ρ σg σν A ν. (. A µ L.T. A µ = Λ µ ν A ν ; Λ µ ν g µρ g νσ Λ ρ σ. (.3 Combining (.6 and (.3, we find that the scalar product of A µ and B µ transforms as This must be equal to A µ B µ L.T. A µb µ = Λ µ ρ Λ µ σa ρ B σ. (.4 A µ B µ = δ ρ σa ρ B σ ; δ ρ σ = Thus, we find the condition for the Lorentz transformation ; (.5 Λ µ ρ Λ µ σ = δ ρ σ and similarly Λ µ ρ Λ ν ρ = δ ν µ. (.6 Note that the inverse of the Lorentz transformation x µ = Λ µ νx ν is expressed as x ν = Λ µ ν x µ. (.7

8 4 CHAPTER. FIELD AND LORENTZ TRANSFORMATION Let us consider for a while a free motion of a particle with mass m. The dynamical quantities which characterize the motion of the particle are its energy E and momentum p. They compose a Lorentz vector (4-momentum p µ = (E/c, p. (.8 The square of the 4-momentum p µ is a Lorentz scalar and it is independent of the inertial system. From a dimensional consideration, it must be proportional to m c. In fact, it is p = E /c p = m c. (.9 This is the so-called Einstein s relation. This relation should be regarded as a definition of m. Suppose the particle is at rest (p=0 in the inertial system x µ (0 and sitting on the origin O at the time t (0 = 0. Its coordinate x µ (0 (O and 4-momentum pµ (0 are xµ (0 (O = 0 and pµ (0 = mc δµ 0. After a time interval t, the particle moves to the point P specified by the coordinate x µ (0 (P = c tδµ 0. In another inertial system, the particle is no longer at rest and all 4-vectors are transformed to p µ = Λ µ νp ν (0 = mcλ µ 0, x µ (O = Λ µ νx ν (0(O = 0, x µ (P = Λ µ νx ν (0(P = c tλ µ 0. (.0 Therefore, the velocity v of the particle is related to its energy E and momentum p by v i = xi (P x i (O (x 0 (P x 0 (O/c = c Λi 0 Λ 0 0 = c p i E/c, that is, v = c p E. (. Notice that the massless particle (m = 0 like photon moves with velocity v = c..3 Lorentz Transformation of Fields Let φ(x be the value of the field at the space-time point P described in one inertial system x µ, and φ (x be its value at the same point P but in another inertial system x µ. Scalar field does not change its value at any point P under the Lorentz transformation : φ(x L.T. φ (x = φ(x space-time point P x µ x µ (. Vector field changes values of its components according to the Lorentz transformation property : superscript 4-dim. vector A µ (x A µ (x = Λ µ L.T. νa ν (x, (.3 subscript 4-dim. vector A µ (x A L.T. µ(x = Λ ν µ A ν (x. (.4 The derivative of scalar field by the coordinate x µ transforms as φ(x φ (x x µ L.T. x µ = xν φ(x ν x µ = Λ φ(x x ν µ, (.5 x ν where we have used (.7. This expression invites us to introduce the notation µ φ(x = φ(x ( ( ; x µ µ = x, 0 x, x, x 3 x,. (.6 0 The differential µ transforms as a subscript Lorentz vector : µ L.T. µ = Λ µ ν ν ; µ = x µ. (.7 The contracted second derivative g µν µ ν φ(x is a Lorentz scalar. The differential operator is called d Alembertian. = g µν µ ν = c t (.8

9 Chapter Free Scalar Quantum Field. Classical Theory of Real Scalar Field Let φ(x be the scalar field which takes a real number as its value. Since φ(x represents the continuously distributed dynamical degrees of freedom, the Lagrangian L of the system consists of the space integral of the Lagrangian density L(x : ( L = d 3 x L φ(x, φ(x, φ(x. (. Suppose the function φ(x, t is an arbitrary time trajectory of φ(x which need not necessarily be the physical time development of φ(x. The action integral I[φ] of this trajectory is defined as a time integral of the Lagrangian L with φ(x replaced by φ(x, t ; ( I[φ] = dt L = dtd 3 x L φ(x, t, φ(x, t, φ(x, t. (. The basic principle of Nature recognized as the classical law of motion is summarized as the Action Principle. The principle states, for the system of the field, that The Classical Motion of the Field φ(x is Realized So That the Time Development φ(x, t Gives the Minimum Value of the Action Integral I[φ]. This means that the action integral I[φ] is stationary under any infinitesimal variation δφ of φ around physically realized trajectory φ(x, t, subject to the constraint δφ(x, t t, x, = 0, which is fixed by its boundary condition : φ(x, t φ(x, t + δφ(x, t ; { δφ(x, t t ± = 0 δφ(x, t x = 0 (.3 δi I[φ(x, t + δφ(x, t] I[φ(x, t] = 0 (.4 t t, δφ = 0 δφ(x, t 0 x x δφ = 0 t, δφ = 0 5

10 6 CHAPTER. FREE SCALAR QUANTUM FIELD Notice that the principle is stated in terms of the geometrical terms independent of any inertial system. This is essential for the realization of the Lorentz invariance of the dynamics. The Lorentz invariance is realized only when the action integral I[φ] is Lorentz invariant. Then the minimum point of I[φ] is common for any observer in any inertial system. Under the Lorentz transformation x µ x µ = Λ µ νx ν, the 4-dimensional volume cdtd 3 x transforms as cdtd 3 x = dx µ dx µ = det Λ µ ν dx µ. (.5 µ=0 µ=0 From (.3 and (.6, however, we find det Λ µ ν =, and then the 4-dimensional volume is Lorentz invariant. Therefore, the Lagrangian density L must be Lorentz invariant. From now on, we simply call L Lagrangian instead of Lagrangian density. The simplest form of Lorentz invariant Lagrangian L which consists of φ, φ and φ is L = φ c ( φ V (φ µ=0 = c gµν µ φ ν φ V (φ L(φ, µ φ, (.6 where V (φ is some polynomial of φ called potential. Let us derive the consequence of the action principle for the general form of the Lagrangian L(φ, µ φ. Starting from (.4 and utilizing partial integration, we obtain 0 = δi = = dtd 3 x dtd 3 x L φ(x δφ(x + L ( µ φ(x δ( } µφ(x {{} = µδφ(x ( L φ(x L µ δφ(x + [surface term]. (.7 ( µ φ(x }{{}}{{} arbitrary =0 δφ =0 The surface term resulting from the partial integration vanishes owing to the boundary condition δφ t, x = 0. Since the right-hand-side of (.7 is required to vanish for any space-time dependent variation δφ(x, the quantity in the parenthesis must be zero at all space-time points x µ : µ L ( µ φ(x L φ(x = 0. (.8 This equation is called Euler-Lagrange equation, which serves as an equation of motion of the field φ(x. For the Lagrangian (.6, we have c g µν µ ν φ(x + dv dφ = 0 or equivalently φ + c dv dφ = 0. (.9. Canonical Quantization For the quantization of the field φ(x, we start with giving a reminder of quantum mechanics. Reminder of quantum mechanics generalized coordinates : q i (i =,, N, (.0

11 .. CANONICAL QUANTIZATION 7 generalized velocities : q i, (. Lagrangian : L(q i, q i, (. canonical momentum : p i = L q i q i = q i (q i, p i, (.3 Hamiltonian : H(p i, q i = N p i q i L(q i, q i, (.4 i= canonical quantization : q i ˆq i, p i ˆp i, H Ĥ(ˆp i, ˆq i, (.5 ˆf represents the quantum operator corresponging to f. canonical commutator : [ˆq i, ˆp j ] = i δ ij, [ˆq i, ˆq j ] = [ˆp i, ˆp j ] = 0, (.6 Schrödinger equation : i d ψ(t = Ĥ ψ(t. (.7 dt Since the field φ(x is labeled by the continuous index x, instead of the discrete index i, we need some prescription to apply the above quantization procedure. Among variety of possible prescriptions, we adopt here the so-called lattice discretization method. We discretize the threedimensional space by the cubic lattice with infinitesimal lattice spacing ɛ. At each lattice point, we distribute the address α which consists of a set of three integers (α x, α y, α z. α i s run from to + by one unit successively in accordance with the ordering of the lattice points. Discretization of the space by lattice x α = (α x, α y, α z Lattice point α i =,,, 0,,, + ɛ (0,, 0 (, 0, 0 (0, 0, 0 (, 0, 0 (0,, 0 Under this preparation, we represent the continuous degrees of freedom of the field φ(x by the countable degrees of freedom φ α located on each lattice point α. The correspondence between the continuous description and the discrete description is as follows ; φ(x φ α φ(x φ α i φ(x ɛ (φ α φ α ni d 3 x α ɛ3 i = x, y, z n x = (, 0, 0 n y = (0,, 0 n z = (0, 0, (.8 Now, we are ready to apply the quantization procedure. The Lagrangian is expressed as [ ] L = d 3 x L(x = d 3 x φ(x c ( φ(x V (φ(x = [ ɛ 3 φ α c ( φα φ α ni V (φ α ]. (.9 ɛ α i=x,y,z

12 8 CHAPTER. FREE SCALAR QUANTUM FIELD The canonical momentum conjugate to φ α is p α L φ α = ɛ 3 φα ɛ 3 π α, (.0 where we have introduced π α as a rescaled quantity of p α so that it survives in the limit ɛ 0. The Hamiltonian is given by H = p α φα L α = [ ɛ 3 π α + c ( φα φ α ni + V (φ α ] ɛ α i=x,y,z [ ] = d 3 x π(x + c ( φ(x + V (φ(x d 3 x H(x, (. where we have returned to the continuous description, with an additional correspondence The field π(x is called canonical momentum density. π α π(x. (. Quantization : The quantization to ˆφ α and ˆp α is straightforward. The canonical commutators are [ ˆφ α, ˆp β ] = i δ 3 αβ, [ ˆφ α, ˆφ β ] = [ˆp α, ˆp β ] = 0 ; δ 3 αβ δ αx β x δ αy β y δ αz β z. (.3 From the relation ˆp α = ɛ 3ˆπ α, we have [ ˆφ α, ˆπ β ] = i ɛ 3 δ3 αβ. (.4 The limit ɛ 0 of the right-hand-side is expressed in terms of the Dirac s delta function δ 3 (x ; ɛ 3 δ3 αβ ɛ 0 δ 3 (x y. (.5 Thus, we obtain the continuous expression for the commutation relations : [ ˆφ(x, ˆπ(y] = i δ 3 (x y, (.6 [ ˆφ(x, ˆφ(y] = [ˆπ(x, ˆπ(y] = 0. (.7 Hamiltonian Operator : The Hamiltonian operator is obtained by simply replacing φ(x and π(x in (. with the corresponding quantum operators ˆφ(x and ˆπ(x ; [ ( Ĥ = d 3 x Ĥ(x = d 3 x ˆπ(x + c ˆφ(x ] + V ( ˆφ(x (.8 Remember that we are treating a classically real scalar field, that is, φ and π are real quantities. Therefore, the quantum operators ˆφ and ˆπ are hermitian operators ; ˆφ = ˆφ, ˆπ = ˆπ. (.9

13 .. CANONICAL QUANTIZATION 9 The Schrödinger equation i d ψ(t = Ĥ ψ(t (.30 dt determines the time development of the quantum state ψ(t of the dynamical system which consists of the field ˆφ and ˆπ. Momentum Operator : The operator ˆP, which realizes the commutation relations [ ˆP, ˆφ(x] = i ˆφ(x, [ ˆP, ˆπ(x] = i ˆπ(x, (.3 is called momentum operator. It is given by ˆP = d 3 x ˆπ(x ˆφ(x. (.3 We can easily confirm that ˆP given by (.3 reproduces the commutators (.3. For example, [ ˆP, ˆπ(x] = d 3 y [ˆπ(y (y ˆφ(y, ˆπ(x] = d 3 y ˆπ(y (y [ ˆφ(y, ˆπ(x] = d 3 y ˆπ(y (y i δ 3 (y x = i ˆπ(x ; (y y, (.33 where, in the second line, we have used a partial integration. The remarkable property of the momentum operator ˆP is that it realizes the commutation relation [ ˆP, Ô(x] = i Ô(x (.34 for any operator Ô(x which is an arbitrary product of ˆφ(x and ˆπ(x. Ô = ÔÔ, we have This is because, for [ ˆP, Ô] = [ ˆP, Ô]Ô + Ô[ ˆP, Ô] = (i ÔÔ + Ô(i Ô = i (ÔÔ. (.35 According to the formula of similarity transformation we have eâ ˆBe  = ˆB + [Â, ˆB] +! [Â, [Â, ˆB]] + 3! [Â, [Â, [Â, ˆB]]] +, (.36 e i a ˆP Ô(xe i a ˆP = Ô(x + a Ô(x +! (a Ô(x + = Ô(x + a (.37 for any constant vector a. Since the Hamiltonian Ĥ is a spacial integral of the Hamiltonian density Ĥ(x, we find e i a ˆP Ĥe i a ˆP = d 3 x Ĥ(x + a = d 3 x Ĥ(x = Ĥ. (.38 Owing to the arbitrariness of the vector a, this leads us to the important observation [ ˆP, Ĥ] = 0. (.39 For an arbitrary state ψ, let us define ψ = e i a ˆP ψ. Then we have ψ Ô(x ψ = ψ Ô(x + a ψ. (.40

14 0 CHAPTER. FREE SCALAR QUANTUM FIELD Ô(x a Ô(x + a ψ ψ This means that the state ψ is a state obtained by translating the state ψ by a distance a. Thus, ˆP is also called a space translation operator. Heisenberg Picture : The integral of the Schrödinger equation is given by ψ(t = e i Ĥt ψ(0. (.4 Suppose the states ψ(t and ψ (t are the solutions of the Schrödinger equation specified by the initial states ψ(0 and ψ (0, respectively, at t = 0. The matrix element of any operator Ô between ψ(t and ψ (t is expressed as ψ (t Ô ψ(t = ψ (0 Ô(t ψ(0. (.4 This defines the operator Ô(t called Heisenberg operator : Ô(t = e i Ĥt Ôe i Ĥt. (.43 The left-hand-side of (.4 is viewing the time development of the matrix element through the time evolution of the state vectors. This point of view is called Schrödinger picture. The righthand-side on the other hand is viewing the same matrix element through the motion of the operator Ô(t, with the state vectors fixed at t = 0. This point of view is called Heisenberg picture. The Heisenberg operator of a product of two operators Ô and Ô is a product of the corresponding Heisenberg operators ; Ô O (t = Ô(tÔ(t. (.44 Equal Time Commutation Relation : The operators ˆφ(x, t and ˆπ(x, t in the Heisenberg picture satisfy the equal time commutation relations [ ˆφ(x, t, ˆπ(y, t] = i δ 3 (x y, (.45 [ ˆφ(x, t, ˆφ(y, t] = [ˆπ(x, t, ˆπ(y, t] = 0. (.46 Since Ĥ(t = Ĥ, we have Ĥ(t = [ ( d 3 x ˆπ(x, t + c ˆφ(x, ] t + V ( ˆφ(x, t = Ĥ. (.47 The commutation relation (.39 also states ˆP(t = ˆP, and we have ˆP(t = d 3 x ˆπ(x, t ˆφ(x, t = ˆP. (.48 These two equations represent the conservation of energy and momentum. Heisenberg s Equation of Motion : Taking a time derivative of the Heisenberg operator (.43, we obtain the equation of motion for the Heisenberg operator Ô(t ; Ô(t = i [Ô(t, Ĥ(t]. (.49

15 .3. FREE SCALAR FIELD For the local operator Ô(x, t, this equation and the equation (.34 represented in terms of the Heisenberg operators are combined to the Lorentz covariant equation ; i µ Ô(x, t = [Ô(x, t, ˆP µ ], ˆP µ = (Ĥ/c, ˆP, µ = ( / x 0, (.50 Note the derivative by the coordinate is originally defined as a subscript vector µ = ( / x 0,. Let us derive the equations of motion for ˆφ(x, t and ˆπ(x, t. Noticing [ ˆφ(x, t, ˆφ(y, t] = [ˆπ(x, t, ˆπ(y, t] = 0, we have ˆφ(x, t = i [ ˆφ(x, t, d 3 y ] ˆπ(y, t, (.5 ˆπ(x, t = i [ ( ] c ˆπ(x, t, d 3 y ˆφ(y, (y t (y ˆφ(y, t + V ( ˆφ(y, t. (.5 The identity [ ˆX, Ŷ Ẑ] = [ ˆX, Ŷ ]Ẑ + Ŷ [ ˆX, Ẑ] gives us the commutators [ ˆφ(x, t, ˆπ(y, t ] = [ ˆφ(x, t, ˆπ(y, t]ˆπ(y, t + ˆπ(y, t[ ˆφ(x, t, ˆπ(y, t] = i δ 3 (x yˆπ(y, t,(.53 ] [ˆπ(x, t, (y ˆφ(y, t (y ˆφ(y, t = i (y δ 3 (y x ˆφ(y, t. (.54 For the potential V ( ˆφ(y, t = N v ˆφ(y, N t N, the identity [ ˆX, Ŷ N ] = N[ ˆX, Ŷ ]Ŷ N, which holds when [ ˆX, Ŷ ] is a c-number, gives [ˆπ(x, t, V ( ˆφ(y, t] = N v N [ˆπ(x, t, ˆφ(y, t N ] = N v N N( i δ 3 (y x ˆφ(y, t N = i δ 3 (y x dv ( ˆφ(y, t dφ. (.55 Gathering all together, we obtain the Heisenberg s equations of motion for ˆφ(x, t and ˆπ(x, t ; ˆφ(x, t = ˆπ(x, t, ˆπ(x, t = c ˆφ(x, t dv ( ˆφ(x, t dφ. (.56 Combining these two equations, we arrive at the equation ˆφ(x, t = ˆπ(x, t = c ˆφ(x, t dv ( ˆφ(x, t dφ. (.57 This reproduces the original Euler-Lagrange equation ˆφ + c d ˆV dφ = 0 in terms of ˆφ(x, t..3 Free Scalar Field From now on, we use the Natural unit system c =, =. Klein-Gordon Equation : Field φ whose equation of motion is linear in φ is called free field. Thus the potential V is quadratic V = m φ. (.58 The equation of motion ( + m ˆφ(x, t = 0 (.59

16 CHAPTER. FREE SCALAR QUANTUM FIELD is called Klein-Gordon equation. m expresses the mass of the particle the field ˆφ describes. Fourier transform : Let us express the field operator ˆφ(x, t in the Fourier integral in terms of the plane-wave function exp(ik x ; d ˆφ(x, 3 k t = (π ˆφ 3/ k (t e ik x. (.60 The operator property of ˆφ(x, t is inherited by ˆφ k (t. The hermiticity of ˆφ is translated into ˆφ k (t = ˆφ k (t. (.6 Substituting ˆφ(x, t in the Klein-Gordon equation (.59 by the expression (.60, we have 0 = ( + m ˆφ(x, t d 3 k ( = ˆφk (t + (k + m (π ˆφ 3/ k (t e ik x. (.6 Since the plane-wave functions form a complete set of orthonormal functions, this requires ˆφ k (t + (k + m ˆφ k (t = 0. (.63 This is nothing but an equation of motion of the harmonic oscilator with angular frequency The solution which satisfies the hermiticity condition (.6 is ω k = k + m. (.64 ˆφ k (t = ωk (â k e iω kt + â k eiω kt, (.65 where we have introduced a normalization constant / ω k for the later convenience. Thus, we obtain d ˆφ(x, 3 k t = (â (π 3/ k e i(ωkt k x + â k ei(ω kt k x, (.66 ωk ˆπ(x, t = ˆφ(x, d 3 k ωk ( t = iâ (π 3/ k e i(ωkt k x + iâ k ei(ω kt k x, (.67 where, in the terms originally containing â k, we have replaced the integration variable k by k. Commutator of â k and â k : The commutation relations (.45 and (.46 of ˆφ and ˆπ determine the commutation relations of â k and â k. The result is [â k, â k ] = δ 3 (k k, (.68 [â k, â k ] = [â k, â k ] = 0. (.69 We can easily confirm that this really reproduces (.45 and (.46. For example, [ ˆφ(x, d 3 kd 3 k ωk { t, ˆπ(y, t] = i[â (π 3 k, â k ]e i((ω k ω k t k x+k y ω k } i[â k, â k ]ei((ω k ω k t k x+k y d 3 k ( = ie ik (x y + ie ik (x y (π 3 d 3 k = i eik (x y (π 3 = iδ 3 (x y. (.70

17 .3. FREE SCALAR FIELD 3 Let us express the Hamiltonian Ĥ(t = ( d 3 x ˆπ(x, t + ( ˆφ(x, t + m ˆφ(x, t (.7 in terms of â k and â k. By substituting the expressions for ˆφ(x, t and ˆπ(x, t, we have Ĥ(t = d 3 x d 3 kd 3 k (π 3 ω k ω k [ ( ( ω k ω k iâ k e i(ωkt k x + iâ k ei(ω kt k x iâ k e i(ω k t k x + iâ k e i(ω k t k x ( ( +k k iâ k e i(ωkt k x iâ k ei(ω kt k x iâ k e i(ω k t k x iâ k e i(ω k t k x ( ( ] +m â k e i(ωkt k x + â k ei(ω kt k x â k e i(ω k t k x + â k e i(ω k t k x = d 3 kd 3 k d 3 x ω k ω k (π [ ( 3 (ω k ω k + k k + m â k â k e i(ω k ω t k e i(k k x + â kâk ei(ω k ω t k e i(k k x ( ] +( ω k ω k k k + m â k â k e i(ω k+ω t k e i(k+k x + â kâ k e i(ω k+ω t k e i(k+k x = d 3 k [ ( (ωk + k + m â k â k ω + â kâk k ( ] +( ωk + k + m â k â k e iωkt + â kâ k eiω kt, (.7 where we have used the formula d 3 x (π 3 ei(k k x = δ 3 (k k. (.73 Since ω k = k + m, the second term in the bracket vanishes, and we find Ĥ(t = d 3 k ω k (â kâk + â k â k = d 3 k ω k â kâk + [const.]. (.74 The term [const.] in the final expression appears due to the interchange of â k and â k. Obviously, it is a collection of the zero-point energies of the harmonic oscillators â k. It is expressed as [const.] = d 3 k ω k δ3 (k k = d 3 k ω k d 3 x (π 3 ei(k k x = V d 3 k ω k 6π 3, (.75 where V is the total volume of the three-dimensional space. Thus, the space posseses the energy ɛ vac = d 3 k ω k /(6π 3 per unit volume. Since ω k is given by ω k = k + m, the energy density ɛ vac is quarticaly divergent. This is the first divergence we encounter in the dynamical system of quantum field. As a matter of fact, [const.] plays no significant role in quantum field theory, because it is merely a constant, though divergent. We can always shift the origin of energy so that [const.]=0 keeping intact any physical content of the system. So, hereafter we omit it.

18 4 CHAPTER. FREE SCALAR QUANTUM FIELD Next, let us turn to the momentum operator ˆP. We leave the details of the calculation for the exercise. The result is ˆP(t = d 3 xˆπ(x, t ˆφ(x, t = d 3 k k (â kâk + â k â k = d 3 k k â kâk. (.76 At this time, we do not have [const.] term because the integrand contains an odd quantity k. Notice that t dependence disapeared in the final expressions of Ĥ(t and ˆP(t as it should be. Creation and Annihilation Operator : The Hamiltonian Ĥ and momentum ˆP contain the product â kâk. The commutation relations of this operator with â k and â k are [â kâk, â k ] = [â k, â k ]â k + â k [â k, â k ] = δ 3 (k k â k, (.77 [â kâk, â k ] = [â k, â k ]â k + â k [â k, â k ] = δ 3 (k k â k. (.78 From this result, we find the following commutation relations : [Ĥ, â k] = ω k â k, [Ĥ, â k ] = ω kâ k, (.79 [ ˆP, â k ] = kâ k, [ ˆP, â k ] = kâ k. (.80 These commutation relations play the fundamental role as a bridge between fields and particles. Suppose a state E, P is an eigenstate of Ĥ and ˆP : Ĥ E, P = E E, P, ˆP E, P = P E, P. (.8 Then we have ( Ĥâ k E, P = [Ĥ, â k] + â k Ĥ E, P = (E ω k â k E, P, (.8 ([Ĥ, Ĥâ k E, P = â k ] + â kĥ E, P = (E + ω k â k E, P. (.83 Similarly, ˆPâ k E, P = (P kâ k E, P, ˆPâ k E, P = (P + kâ k E, P. (.84 That is, â k, when operated to a state, decreases and â k increases the energy and momentum of the state by ω k and k, respectively. The energy ω k and the momentum k satisfy the Einstein s relation ωk k = m (.85 of a particle with mass m. This shows that â k annihilates and â k creates the particle with energy ω k and momentum k. Thus, â k and â k are called creation and annihilation operator, respectively. Space of Quantum States (Fock Space : The state which plays a basis of the system of quantum field is the normalized ground state 0 of the Hamiltonian Ĥ ; Ĥ 0 = E 0 0, 0 0 =. (.86 Evidently, it should be a state where no particle exists. That is, no particle can be annihilated by any of the annihilation operator â k ; â k 0 = 0. (.87

19 .3. FREE SCALAR FIELD 5 The state 0 is called the vacuum of the quantum system. keeping in mind of our convention [const.]=0, From (.74 and (.76, we find, Ĥ 0 = 0, ˆP 0 = 0. (.88 Any state other than the vacuum 0 is obtained by operating the creation operators â k The one particle state is on 0. k = â k 0, (.89 Ĥ k = ω k k, ˆP k = k k, ωk = k + m. (.90 Owing to the commutator (.68 and the vacuum condition (.87, k satisfies the orthonormality The two particle state is k k = 0 â k â k 0 = 0 [â k, â k ] + â k â k 0 = δ 3 (k k. (.9 k, k = â k â k 0, (.9 Ĥ k, k = (ω k + ω k k, k, ˆP k, k = (k + k k, k, ω ki = k i + m.(.93 Due to the commutability (.69 of â k and â k, the state k, k is symmetric under the interchange of k and k ; k, k = â k â k 0 = â k â k 0 = k, k. (.94 The normalization of k, k is k, k k, k = 0 â k â k â k â k 0 = 0 â k [â k, â k â k ] 0 = 0 â k [â k, â k ]â k + â k â k [â k, â k ] 0 = δ 3 (k k δ 3 (k k + δ 3 (k k δ 3 (k k. (.95 It is straightforward to extend these results to the general N particle state ; k,, k N = â k â k N 0 = k,, k j,, k i,, k N, i j (.96 N N Ĥ k,, k N = ω ki k,, k N, ˆP k,, k N = k i k,, k N, (.97 i= k,, k N k,, k N = σ(permutation i= δ 3 (k k σ δ 3 (k N k σ N. (.98 Thus, we find the remarkable property of the particle described by the scalar field ˆφ ; State vector is completely symmetric under interchange of any set of two particles. This property of the particle is called Bose-Einstein statistics. statistics is called boson. The particle subject to this Problem (.3- Prove the equal time commutabilities [ ˆφ(x, t, ˆφ(y, t] = 0, [ˆπ(x, t, ˆπ(y, t] = 0, (.99

20 6 CHAPTER. FREE SCALAR QUANTUM FIELD based on the commutation relations [â k, â k ] = δ 3 (k k, [â k, â k ] = [â k, â k ] = 0. (.00 Also derive the following expression of ˆP by carrying out the space integration. ˆP = d 3 k k (â kâk + â k â k. (.0 Problem (.3- Prove the N particle state vector (.96 satisfies the ortho-normality condition k, k,, k N k, k,, k N = δ 3 (k k σ δ 3 (k N k σn. (.0 σ(permutation The general N-particle state ψ (N is expressed in terms of this vector as ψ (N = d 3 k d 3 k N k,, k N c(k,, k N, (.03 N! where c(k,, k N is a completely symmetric function of k i (i =,, N. Show ψ (N ψ (N = d 3 k d 3 k N c (k,, k N c(k,, k N. (.04 Problem (.3-3 Prove that the identity operator of the quantum states is expressed as ˆ = N= N! by showing, for any state ψ, the identity equation d 3 k d 3 k N k, k,, k N k, k,, k N, (.05 ˆ ψ = ψ. (.06 Problem (.3-4 Suppose the real scalar field is interacting with the classical matter distribution density ρ(x. Hamiltonian is given by ( Ĥ = d 3 x ˆπ(x + ( ˆφ(x + m ˆφ(x + gρ(x ˆφ(x. (.07 Let us define the unitary operator Û by Û = e ig R d 3 xd 3 yˆπ(xf(x yρ(y, (.08 f(x y e m x y 4π x y ; ( + m f(x y = δ 3 (x y. (.09 Check the following unitary transformations of ˆφ(x and ˆπ(x ; Û ˆφ(xÛ = ˆφ(x + g d 3 yf(x yρ(y, (.0 Û ˆπ(xÛ = ˆπ(x. (.

21 .4. T-PRODUCT AND PROPAGATOR 7 According to this fact, show that the eigenstate n of Ĥ (Ĥ n = E n n is represented in terms of the eigenstate n 0 of the free scalar field (Ĥ0 n 0 = E 0 n n 0 described by the Hamiltonian as Ĥ 0 = ( d 3 x ˆπ(x + ( ˆφ(x + m ˆφ(x (. n = Û n 0 (.3 E n = En 0 g d 3 xd 3 yρ(xf(x yρ(y. (.4 Derive the expectation value of ˆφ(x with respect to the ground state 0 of Ĥ..4 T-Product and Propagator T-Product : Imagine that there is an interaction between particles. Interactions of particles are described by the annihilations and creations of particles at various space-time points x i. In quantum field theory, this is realized by the operation of the products of various operators Ô(x i on the state vector ψ. In general, operators are non-commutative, [Ô(x i, Ô(x j] 0. Therefore, the ordering of the operators, which act on the state vector, is significant. t x 3 x x x Intuitively, the natural ordering will be Each operator Ô i (x i stands from right to left along the ordering of each time x 0 i. This is called T -product (time-ordered product ; T [Ô(x Ôn(x n ]. (.5 For example, T [Ô(x Ô(x ] = { Ô (x Ô(x x 0 > x 0 Ô (x Ô(x x 0 > x 0 = θ(x 0 x 0 Ô(x Ô(x + θ(x 0 x 0 Ô(x Ô(x, (.6

22 8 CHAPTER. FREE SCALAR QUANTUM FIELD where θ(x 0 is the Heviside s θ function ; θ(x 0 { x 0 > 0 0 x 0 < 0. (.7 Propagator : Let us express the field operator (.66 in slightly compact form d ˆφ(x 3 k ( = â (π 3/ k e ik x + â eik x k, (.8 ωk k x k µ x µ = k 0 x 0 k x, k µ = (k 0 = ω k, k, ω k = k + m, (.9 and take the matrix element of T [ ˆφ(x ˆφ(x ] between vacuum 0 and 0 ; 0 T [ ˆφ(x ˆφ(x ] 0 i F (x x. (.0 The functon F (x x is called Feynman propagator. Due to the property of the vacuum â 0 = 0 and 0 â = 0, F (x x expresses the amplitude of the process where the particle is created at a space-time point x and annihilated at x when x 0 > x 0, and particle is created at x and annihilated at x when x 0 > x 0. t x x x t x x x x 0 > x 0 x 0 > x 0 Let us derive the explicit form of F (x x : i F (x x = θ(x 0 x 0 d 3 k 0 â (π 3/ k e ik x ωk + θ(x 0 x 0 d 3 k 0 â (π 3/ k e ik x ωk = d 3 k â (π 3/ k e ik x 0 ωk d 3 k â eik x (π 3/ k 0 ωk d 3 k ( θ(x 0 (π 3 ω x 0 e ik (x x + θ(x 0 x 0 e ik (x x. (. k We show that this is expressed in a compact form as a 4-dimensional integration d 4 k i i F (x x = e ik (x x, k (k 0 k, d 4 k dk 0 d 3 k, (. (π 4 k m + iɛ where ɛ is a positive infinitesimal quantity. Be careful that k 0 is an integration variable. Since k m = (k 0 k m = (k 0 ω k (k 0 + ω k, (.3 the right-hand-side is expressed as [ d 3 k ] dk 0 i r.h.s. = (π 3 π (k 0 ω k (k 0 + ω k + iɛ e ik0 (x0 x0 e ik (x x. (.4

23 .4. T-PRODUCT AND PROPAGATOR 9 Due to the presence of the factor exp( ik 0 (x 0 x 0, when x 0 x 0 < 0, the integration over k 0 can be extended to the closed contour circling from + to on the upper complex k 0 plane, where exp( ik 0 (x 0 x 0 vanishes. x 0 x 0 < 0 complex k 0 -plane ω k + iɛ ω k iɛ x 0 x 0 > 0 The contour encloses the pole at k 0 = ω k + iɛ. According to the residue theorem, the k 0 integral gives [ ] = πi i e iωk(x0 x0. (.5 π ω k Replacing in (.4 the integration variable k by k, this gives the second term of i F (x x. When x 0 x 0 > 0, the k 0 integration contour can be taken to circle the lower complex k 0 plane, which encloses the pole at k 0 = ω k iɛ. Taking account of the minus sign due to the opposite integration direction, the k 0 integral gives this time [ ] = πi i e iωk(x0 x0, (.6 π ω k which reproduces the first term of i F (x x. Thus, we have proved the formula d 4 k F (x x = e ik (x x. (.7 (π 4 k m + iɛ F (x satisfies the following formulas ; F (x = F ( x, (.8 d ( + m 4 k F (x = e k x = δ(x 0 δ 3 (x δ 4 (x. (.9 (π 4 Problem (.4- Show that the 4-dimensional (non-equal-time commutator of the free scalar field ˆφ(x is represented in the form [ ˆφ(x, ˆφ(x d 4 k ] i (x x = (π 3 δ(k m ɛ(k 0 e ik (x x, (.30 { + k ɛ(k 0 = 0 > 0 k 0. (.3 < 0 Next, replace all of the integration variable k µ of this expression with k µ and define the new k µ in terms of the Lorentz transformation matrix Λ µ ν by k µ = Λ µ νk ν. Evidently, we have k = k and d 4 k = d 4 k. Derive the expression i (x x = d 4 k (π 3 δ(k m ɛ(k 0 e ik (x x, (.3

24 0 CHAPTER. FREE SCALAR QUANTUM FIELD where x i Λ µ νx ν. Confirm that the sign of k 0 is preserved under Lorentz transformation when 4-vector k µ is timelike (k > 0 which is ensured by δ(k m. This verifies the Lorentz invariance of the function (x x = (x x. When the interval of x and x is spacelike, that is, (x x < 0, it is possible to Lorentz transform two points x and x to the equal time points, x 0 = x 0. The equal-time commutator was [ ˆφ(x, t, ˆφ(x, t] = 0. Thus the Lorentz invariance of (x x implies that the spacelikely separated two fields ˆφ(x, ˆφ(x are commutative, [ ˆφ(x, ˆφ(x ] = 0. The consequence of this fact is that, any two operations separated by the spacelike interval give no influence on each other. This is a reflection of the fact that the causality is accurately maintained in the quantum field theory. Problem (.4- Show that the vacuum expectation value of the T -product of four free scalar fields is expressed as 0 T [ ˆφ(x ˆφ(x ˆφ(x 3 ˆφ(x 4 ] 0 = i F (x x i F (x 3 x 4 + i F (x x 3 i F (x x 4 + i F (x x 4 i F (x x 3. (.33.5 Complex Scalar Field Suppose there are two free real scalar fields φ (x and φ (x with the equal mass m. The Lagrangian is a sum of each Lagrangian ; L = ( µφ µ φ + µ φ µ φ m Each quantum field ˆφ i (i =, satisfies the Klein-Gordon equation ( φ + φ. (.34 ( + m ˆφ i (x = 0 i =,. (.35 Thus, they are expanded in terms of each creation and annihilation operators â ik and â ik ; ˆφ i (x = d 3 k ( â (π 3/ ik e ik x + â eik x ik ωk, (.36 [â ik, â jk ] = δ ij δ 3 (k k, other [, ] = 0. (.37 Conserved Current and Charge : This system has a following remarkable property. Let us define the Lorentz vector operator ĵ µ (x, called current, by ĵ µ (x ˆφ (x µ ˆφ (x ˆφ (x µ ˆφ (x. (.38 Due to the Klein-Gordon equation, ĵ µ (x satisfies the continuity equation µ ĵ µ = ˆφ ˆφ ˆφ ˆφ = ˆφ m ˆφ + ˆφ m ˆφ = 0. (.39 The current ĵ µ which satisfies µ ĵ µ = 0 is in general called conserved current. The space integral of its time component ĵ 0 is called charge ˆQ ; ( ˆQ = d 3 x ĵ 0 (x = d 3 x ˆφ ˆφ ˆφ ( ˆφ = i d 3 k â kâk â kâk. (.40

25 .5. COMPLEX SCALAR FIELD Due to the current conservation, ˆQ is a conserved quantity : ˆQ = d 3 x tĵ0 (x = d 3 x ĵ = x Another Viewpoint : Let us combine φ and φ to form a complex field φ ; d S ĵ = 0 (.4 φ = (φ + iφ, L = µ φ µ φ m φ φ. (.4 Then, the quantum field ˆφ is expressed as ˆφ(x = ( ˆφ + i ˆφ (x = ˆφ (x = ( ˆφ i ˆφ (x = d 3 k ( â (π 3/ (+k e ik x + â eik x ( k ωk d 3 k ( â (π 3/ ( k e ik x + â eik x (+k ωk, (.43, (.44 where â (±k (â k ± iâ k, â (±k (â k iâ k. (.45 They satisfy the commutation relations [â (±k, â (±k ] = δ 3 (k k, (.46 [â (±k, â ( k ] = 0, [â, â] = 0, [â, â ] = 0. (.47 The Hamiltonian Ĥ, momentum ˆP and charge ˆQ turn out to be simply a sum of two sectors (+ and ( ; Ĥ = d 3 k ω k (â (+kâ(+k + â ( kâ( k (.48 ( ˆP = d 3 k k â (+kâ(+k + â ( kâ( k (.49 ( ˆQ = d 3 k â (+kâ(+k â ( kâ( k (.50 The commutators of ˆQ with â (±k and â (±k are (.5 [ ˆQ, â (±k ] = â (±k, [ ˆQ, â (±k ] = ±â (±k. (.5 Thus, the operators â (± annihilate and â (± create a particle with charge ±. If we call the particle created by â (+ particle, the particle created by â ( is called antiparticle : From (9.49, (.44 and (.5, we find [ ˆQ, ˆφ] = ˆφ, [ ˆQ, ˆφ ] = ˆφ. (.53 These commutators show that, ˆφ operated on a state ψ decreses the charge of the state by one unit by annihilating a particle or creating an antiparticle. ˆφ works in a contrary way. Fock Space : The vacuum 0 is originally defined by â k 0 = â k 0 = 0. Thus, â (+k 0 = â ( k 0 = 0. (.54

26 CHAPTER. FREE SCALAR QUANTUM FIELD The one-particle state is â (+k 0 particle with momentum k, (.55 â ( k 0 anti-particle with momentum k. (.56 Propagator : Let us first examine 0 T [ ˆφ(x ˆφ (x ] 0 ; 0 T [ ˆφ(x ˆφ (x ] 0 = ( 0 T [ ˆφ (x + i ˆφ ( (x ˆφ (x i ˆφ (x ] 0 = ( 0 T [ ˆφ (x ˆφ (x + ˆφ (x ˆφ (x ] 0 = i F (x x (.57 This propagator represents the processes where, a particle is created at x and annihilated at x when x 0 > x 0, and an anti-particle is created at x and annihilated at x when x 0 > x 0. When figuring these processes, we add an arrow pointing from ˆφ (x to ˆφ(x. Then the particle moves along the arrow but the antiparticle moves opposite to the arrow. t t x x particle anti-particle x x x x x 0 > x 0 x 0 > x 0 Owing to the charge conservation, other propagators vanish ; 0 T [ ˆφ(x ˆφ(x ] 0 = 0 T [ ˆφ (x ˆφ (x ] 0 = 0. (.58 Problem (.5- Derive the final expression of ˆQ presented in (.40 ( ˆQ = i d 3 k â kâk â kâk. (.59 by carrying out the space integration. Problem (.5- Check the results (.58 based on the real fields ˆφ and ˆφ..6 Chapter-Project/Q We have completed the framework for the description of the motion of the free quantum scalar fields. Though it treats only free scalar fields, the framework is applicable to the quantum fields interacting with the classical external source as we found in the problem (.3-4. Here, we slightly modify the model and extract physically significant information on the particle creation through the interaction with the classical source. Suppose the real scalar field ˆφ(x interacts, for the time interval 0 < t < T, with the classical matter density ρ(x through the interaction Hamiltonian Ĥ int = d 3 x gρ(x ˆφ(x = d 3 k ω k (f kâ k + f k â k : f k = g ω k ωk d 3 x ρ(x e ik x (π 3/, (.60

27 .6. CHAPTER-PROJECT/Q 3 where g represents the strength of the interaction. We treat this system in the Schrödinger picture. The total Hamiltonian Ĥ for 0 < t < T is a sum of the free Hamiltonian Ĥ 0 = d 3 k ω k â kâk (.6 and the interaction Hamiltonian Ĥint : [ ] Ĥ = Ĥ0 + Ĥint = d 3 k ω k (âk + f k (â k + f k fkf k, 0 < t < T. (.6 Let us assume the state ψ(t for t < 0 is the vacuum 0 of Ĥ 0, and calculate the number of particles N T created through the interaction. Since the system is free at t > T, the particle number operator ˆN commutes with Ĥ0 and it is given by ˆN = d 3 k â kâk : ˆN k,, k N = N k,, k N. (.63 Therefore, N T evaluated by the state ψ(t > T = e Ĥ0(t T e iĥt 0 is expressed as N T = ψ(t ˆN ψ(t = 0 e iĥt ˆNe iĥt 0, t > T. (.64 For the calculation of the matrix element (.64, it is useful to introduce the unitary operator Û = exp d 3 k (fkâ k f k â k : Û = Û (.65 which realizes, through the similarity transformation formula (.36, Û â k Û = â k + f k and then (â k + f k (â k + f k = Û â kâkû. (.66 Thus, we have Ĥ = Û Ĥ 0 Û d 3 k ω k fkf k and Û ˆNÛ = d 3 k (â k f k (â k f k. (.67 The second term in Ĥ cancels in (.64. Consequently, N T is expressed as N T = d 3 k 0 Û e iĥ0t Û ˆNÛ e iĥ0t Û 0 = d 3 k 0 Û e iĥ0t (â kâk fkâ k f k â k + f kf k e iĥ0t Û 0 = d 3 k 0 Û (â kâk fkâ k e iωkt f k â k eiω kt + fkf k Û 0 = d 3 k 0 ( (â k + f k (â k + f k fk(â k + f k e iωkt f k (â k + f k e iωkt + fkf k 0. (.68 Taking the matrix elements, we obtain the formula N T = d 3 k fkf k ( cos ω k T. (.69

28 4 CHAPTER. FREE SCALAR QUANTUM FIELD From this expression, we understand that in the limit T, N T approaces to the value N T = d 3 k fkf k (.70 because the second term vanishes in this limit due to the rapid sign change of the integrand. This situation represents the limit where the number of particles created from the source is balanced with the number of particles absorbed by the source. Let us examine the more detailed analysis. Suppose the matter distribution in (.60 is the normalized Gaussian function ρ(x = ( /D e x πd 3 : d 3 x ρ(x =. (.7 Then, the Fourier component f k is given by f k = g ω k ωk (π 3/ e D k /4 (.7 and N T is expressed as d N T = g 3 k cos ω k T e D k /. (.73 (π 3 ωk 3 In the case of massless scalar field (m = 0, it takes a compact expression N T (m = 0 = g π 0 dx cos rx e x : r x T D. (.74 This is a monotonically increasing function of T, and it diverges in the limit T. If we limit the interaction time T to an infinitesimal time interval T, N T is expressed as N T = ( T g π D 0 dx x x + m D / e x + O ( ( T 4. (.75 This diverges in the localized limit D 0 of ρ(x. In the case of massless scalar field, we obtain N T = ( T g 4π D + O ( ( T 4. (.76

29 Chapter 3 Free Dirac Quantum Field 3. Statistics of Particles and Commutators of Fields Quantum Statistics : Quantum mechanics gives a stringent constraint on the quantum state containing the same kind of particles, which is called quantum statistics. The allowed statistics is only two kind : Bose-Einstein statistics : State vector is completely symmetric : Boson under interchange of particles. Fermi-Dirac statistics : State vector is completely anti-symmetric : Fermion under interchange of particles. Description of Particles : The description of particles in quantum field theory is based on the vacuum 0 which serves, by a requirement â k 0 = 0, a ground state of the Hamiltonian Ĥ = d 3 k ω k â kâk, Ĥ 0 = 0, (3. and the commutation relations of Ĥ with the annihilation and creation operators The boson system realizes these commutation relations by based on the identity relation [Ĥ, â k] = ω k â k, [Ĥ, â k ] = ω kâ k. (3. [â k, â k ] = δ 3 (k k, [â k, â k ] = 0, [â k, â k ] = 0, (3.3 [â k â k, â k ] = â k [â k, â k ] + [â k, â k ]â k. (3.4 Owing to the commutativity â k i â k j = â k j â k i, the multi-particle state â k â k N 0 is completely symmetric under the interchange of any set of â k i and â k j. The commutators (3.3 came from the commutators of the fields [ ˆφ(x, ˆπ(y] = iδ 3 (x y, [ ˆφ(x, ˆφ(y] = 0, [ˆπ(x, ˆπ(y] = 0. (3.5 The key to the description of fermion system is the identity relation alternative to (3.4, which is a trivial miracle of mathematics, [â k â k, â k ] = â k {â k, â k } {â k, â k }â k, (3.6 5

30 6 CHAPTER 3. FREE DIRAC QUANTUM FIELD where, {A, B} AB + BA is called anti-commutator. We find the anticommutators {â k, â k } = δ 3 (k k, {â k, â k } = 0, {â k, â k } = 0 (3.7 also realize (3.. The anti-commutativity â k i â k j = â k j â k i ensures the complete anti-symmetry of the multi-particle state â k â k N 0. Thus, the quantization prescription for fermionic field is { ˆφ(x, ˆπ(y} = iδ 3 (x y, { ˆφ(x, ˆφ(y} = 0, {ˆπ(x, ˆπ(y} = 0. (3.8 Question : Can we quantize scalar field by fermi statistics? The answer is No!! The Hamiltonian was ( Ĥ = d 3 x ˆπ + ( ˆφ + m ˆφ = d 3 k ω ( k â kâk + â k â k. (3.9 The anti-commutator leads to Ĥ = d 3 k ω kδ 3 (k k = [constant]. Thus, [Ĥ, ˆφ] = 0, [Ĥ, ˆπ] = 0, and Ĥ cannot work as a quantum mechanical operator. Therefore, the scalar particle, which has no spin degree of freedom, must be subject to the Bose-Einstein statics. 3. Dirac Equation The Dirac spinor field ψ(x is a 4-component complex field ; ψ (x ψ(x = ψ (x ψ 3 (x. (3.0 ψ 4 (x Whenever the field ψ(x represents a particle with mass m, its energy ω k and momentum k should satisfy the Einstein s relation ωk k = m. Therefore, ψ(x must be subject to the Klein-Gordon equation ( + m ψ(x = 0. If we assume that the Klein-Gordon equation is realized as the Euler-Lagrange equation of ψ(x, its Lagrangian takes a form L = µ ψ µ ψ m ψ ψ. Since this Lagrangian merely represents four species of complex scalar fields, the resulting particles are bosons. Dirac Equation : If ψ(x represents the particle subject to the Fermi-Dirac statistics, its equation of motion must be stronger than the Klein-Gordon Equation. It should be the first-order equation with respect to the derivative. The equation is called Dirac equation ; (iγ µ µ mψ(x = 0 ; µ x µ, (3. where γ µ (µ = 0,,, 3 are the 4 4 constant matrices which act on the four components of ψ(x. The term m also tacitly contains the 4 4 unit matrix. The condition for the solution of this equation to satisfy the Klein-Gordon equation is obtained by the requirement 0 = ( iγ ν ν m(iγ µ µ mψ(x = (γ ν γ µ ν µ + m ψ(x ( = {γµ, γ ν } µ ν + m ψ(x because µ ν = ν µ ( + m ψ(x. (3.

31 3.. DIRAC EQUATION 7 That is, the matrices γ µ must satisfy {γ µ, γ ν } = g µν : (γ 0 =, (γ i = (i =,, 3. (3.3 For the later convenience, we require the (anti-hermiticity (γ 0 = γ 0, (γ i = γ i : γ µ = γ 0 γ µ γ 0. (3.4 Standard Representation : Dirac composed γ µ satisfying (3.3 and (3.4 in the form ( ( σ γ σ = 0 σ 0, γ i i = σ i i =,, 3, (3.5 0 where σ 0 is a unit matrix, and σ i are the Pauli matrices ( ( ( 0 0 i 0 σ =, σ =, σ 3 = 0 i 0 0. (3.6 It will be an instructive exercise to confirm that there is no possibility for the smaller matrices than 4 4 ones to satisfy (3.3 and (3.4. Unitary Equivalence : Suppose the two sets of gamma matrices γ µ and γ µ both satisfy the conditions (3.3 and (3.4. Then, they are always connected by the unitary transformation γ µ = Uγ µ U, UU =. (3.7 We skip here the proof of this Pauli s fundamental theorem which is tedious though straightforward. The theorem (3.7 states that, when ψ(x is the solution of Dirac equation with γ µ, the solution with γ µ is given by ψ(x Uψ(x. That is, ψ(x and ψ(x represent the same physical situation using different convention for the labeling of the four components of ψ(x. Lorentz Transformation of ψ(x : Under the Lorentz transformation x µ x µ = Λ µ νx ν, ψ(x should also recieve a transformation ψ(x L.T. ψ (x = Sψ(x. (3.8 S is the 4 4 transformation matrix, which we are going to investigate. The principle of relativity states that, the equation of motion takes the same form in any inertial frame. Therefore, when ψ(x satisfies the Dirac equation (iγ µ µ mψ(x = 0, ψ (x should also satisfy the Dirac equation Using the relation (.7, µ = Λ µ ν ν, we obtain Thus, only when S satisfies the requirement (iγ µ µ mψ (x = 0, µ x µ. (3.9 (iγ µ µ mψ (x = S(iS γ µ SΛ µ ν ν mψ(x. (3.0 S γ µ SΛ µ ν = γ ν, (3. (3.9 is realized ; (iγ µ µ mψ (x = S(iγ ν ν mψ(x = 0. (3.

32 8 CHAPTER 3. FREE DIRAC QUANTUM FIELD Owing to the condition of the Lorentz transformation Λ µ ρλ µ ν = δ ν ρ, (3. is expressed in the form S γ µ S = Λ µ ργ ρ. (3.3 The solution to this equation S(Λ µ ν is uniquely determined in terms of Λ µ ν under the requirement of the group property of the Lorentz transformation ; S(δ µ ν =, S(Λ µ ρ Λρ ν = S(Λµ ν S(Λµ ν. (3.4 The latter represents the multiplicative nature of two successive Lorentz transformations by Λ µ ν and Λ µ ν. The four component quantity ψ(x which transforms according to (3.8 is called Lorentz spinor. Taking a hermitian conjugate of (3.3, we have By the relations γ µ = γ 0 γ µ γ 0 and (γ 0 =, this is expressed as S γ µ (S = Λ µ ργ ρ (3.5 (γ 0 S γ 0 γ µ (γ 0 S γ 0 = Λ µ ργ ρ. (3.6 Comparing this expression with (3.3, we obtain, from the uniqueness of S, the impotant nature of the transformation matrix S : γ 0 S γ 0 = S. (3.7 Dirac Conjugate : Let us define the four component row spinor by Under the Lorentz transformation, ψ(x transforms as ψ(x ψ (xγ 0. (3.8 ψ(x L.T. ψ (x ψ (x γ 0 = ψ (xs γ 0 = ψ (xγ 0 γ 0 S γ 0 = ψ(xs. (3.9 This determines the Lorentz transformation property of the bi-linear quantities : ψψ ψγ µ ψ L.T. ψ ψ = ψs Sψ = ψψ : Lorentz scalar (3.30 L.T. ψ γ µ ψ = ψs γ µ Sψ = Λ µ ψγ ρ ρ ψ : Lorentz vector (3.3 Taking the dagger of the Dirac equation (iγ µ µ mψ(x = 0, we have ψ (x( iγ µ µ m = 0, where µ means that the differential operates on ψ (x. Inserting γ 0 γ 0 = just after ψ (x and multiplying γ 0 from the right, we obtain the Dirac equation for ψ(x : ψ(x( iγ µ µ m = 0. (3.3 Problem (3.- Let us express the infinitesimal version of the Lorentz transformation matrices Λ µ ν and Λ µ ν which satisfy the condition (.6 as Λ µ ν = δ µ ν + ɛ µ ν, Λ µ ν = δ ν µ ɛ ν µ : ɛ µ ν. (3.33