x ● Consider waves on a string, mass per unit length ρ, tension T , displacement φ(x, t).
∂φ 2 K.E. T = 1 ρ dx 2 ∂t ! " # ∂φ 2 ∂φ 2 P.E. V = 1 T dx = 1 ρc2 dx 2 ∂x 2 ∂x ! " # ! " # where wave velocity c = T/ρ.
$ 1 ● Lagrangian L = T V = dx with Lagrangian density − L % ∂φ 2 ∂φ 2 = 1 ρ c2 L 2 ∂t − ∂x &" # " # ' For brevity, write ∂φ ∂φ = φ˙ , = φ! ∂t ∂x
1 2 2 2 = ρ φ˙ c φ! ⇒ L 2 − ( )
● Equations of motion given by least action δS = 0 where S = L dt. Now ∂ ∂ ∂ % δS = Lδφ + L δφ! + Lδφ˙ dx dt ∂φ ∂φ! ∂φ˙ ! " # But + ∂ ∂ ∂ ∂ ∞ ∂ ∂ L δφ! dx = L δφ dx = L δφ L δφ dx ∂φ! ∂φ! ∂x ∂φ! − ∂x ∂φ! ! ! * +−∞ ! " # 2 We can assume the boundary terms vanish. Similarly ∂ ∂ ∂ Lδφ˙ dt = L δφ dt ∂φ˙ − ∂t ∂φ˙ ! ! " # and so ∂ ∂ ∂ ∂ ∂ δS = L L L δφ dx dt ∂φ − ∂x ∂φ! − ∂t ∂φ˙ ! * " # " #+
● This has to vanish for any δφ(x, t), so we obtain the Euler-Lagrange equation of motion for the field φ(x, t): ∂ ∂ ∂ ∂ ∂ L L L = 0 ∂φ − ∂x ∂φ − ∂t ˙ " ! # " ∂φ #
1 2 2 2 ● For the string, = ρ φ˙ c φ! . Hence L 2 − ( ) ∂φ ∂φ˙ ρc2 ! ρ = 0 ∂x − ∂t
3 which gives the wave equation ∂2φ ∂2φ c2 = 0 ∂x2 − ∂t2
● We shall also need the Hamiltonian
H = dx H ! Recall that for a single coordinate q we have L = L(q˙, q) and H = pq˙ L − where p is the generalized momentum ∂L p = ∂q˙
● Similarly, for a field φ(x, t, ) we define momentum density ∂ π(x, t) = L ∂φ˙
4 Then (π, φ) = πφ˙ H − L
● For the string π = ρφ˙ (as expected) and
2 π 1 π 1 2 2 = π ρ + ρc φ! H ρ − 2 ρ 2 " # " # 1 ∂φ 2 = π2 + 1 ρc2 2ρ 2 ∂x " #
5 Klein-Gordon Field ● We choose the Lagrangian density (! = c = 1) ∂φ 2 ∂φ 2 = 1 1 1 m2φ2 L 2 ∂t − 2 ∂x − 2 " # " # to obtain the Klein-Gordon equation of motion: ∂ ∂ ∂ ∂ ∂ L L L = 0 ∂φ − ∂x ∂φ − ∂t ˙ " ! # " ∂φ # ∂2φ ∂2φ m2φ + = 0 ⇒ − ∂x2 − ∂t2 ● The momentum density is ∂ ∂φ π = L = ∂φ˙ ∂t and so the Klein-Gordon Hamiltonian density is ∂φ 2 = 1 π2 + 1 + 1 m2φ2 H 2 2 ∂x 2 " #
6 ● In 3 spatial dimensions ∂φ/∂x φ, → ∇ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ L L L L L = 0 ∂φ − ∂x ∂(∂φ/∂x) − ∂y ∂(∂φ/∂y) − ∂z ∂(∂φ/∂z) − ∂t ˙ " # " # " # " ∂φ # ∂2φ m2φ + 2φ = 0 ⇒ − ∇ − ∂t2
● Covariant notation:
= 1 g (∂µφ) (∂ν φ) 1 m2φ2 L 2 µν − 2 Euler-Lagrange equation: ∂ ∂ L ∂µ L = m2φ ∂µ (g ∂ν φ) = 0 ∂φ − ∂(∂µφ) − − µν " # i.e. µ 2 ∂ ∂µφ + m φ = 0
7 ● Note that the Lagrangian density and the action S = d3r dt = d4x L L L are scalars (invariant functions), like φ. % % ● On the other hand the momentum density π = ∂φ/∂t and the Hamiltonian density = 1 π2 + 1 ( φ)2 + 1 m2φ2 H 2 2 ∇ 2 are not: time development frame dependence. ⇒
8 Fourier Analysis ● We can express any real field φ(x, t) as a Fourier integral:
ikx iωt ikx+iωt φ(x, t) = dk N(k) a(k)e − + a∗(k)e− ! , - where N(k) is a convenient normalizing factor for the Fourier transform a(k). The frequency ω(k) is obtained by solving the equation of motion: KG equation ω = +√k2 + m2. ⇒ ● The Hamiltonian 1 2 1 2 1 2 2 H = 2 π + 2 φ! + 2 m φ dx ! takes a simpler form in terms o.f the Fourier amplitud/es a(k). We can write e.g.
2 φ = dk N(k) [. . .] dk! N(k!) [. . .] ! ! and use
i(k k!)x dx e ± = 2π δ(k k!) ± ! 9 to show that
2 i(ω+ω!)t φ dx = 2π dk dk! N(k) N(k!) a(k)a(k!)δ(k + k!)e− ! ! i(ω,+ω!)t i(ω ω!)t + a∗(k)a∗(k!)δ(k + k!)e + a(k)a∗(k!)δ(k k!)e− − − i(ω ω!)t + a∗(k)a(k!)δ(k k!)e − − ● Noting that ω( k) = ω(k) and choosing N(k-) such that N( k) = N(k), this − − gives
2 2 2iωt φ dx = 2π dk [N(k)] a(k)a( k)e− − ! ! 2iωt + a∗(k)a∗( k)e , + a(k)a∗(k) + a∗(k)a(k) − ● Similarly -
2 2 2iωt φ! dx = 2π dk [kN(k)] a(k)a( k)e− − ! ! 2iωt + a∗(k)a∗( k)e ,+ a(k)a∗(k) + a∗(k)a(k) − - 10 while
2 2 2iωt φ˙ dx = 2π dk [ω(k)N(k)] a(k)a( k)e− − − ! ! +2iωt a∗(k)a∗( k)e ,+ a(k)a∗(k) + a∗(k)a(k) − − and hence, using k2 = ω2 m2, - − 2 H = 2π dk [N(k)ω(k)] [a(k)a∗(k) + a∗(k)a(k)] ! or, choosing 1 N(k) = 2π 2ω(k) ·
1 H = dk N(k) 2 ω(k) [a(k)a∗(k) + a∗(k)a(k)] ! i.e. the integrated density of modes N(k) times the energy per mode ω(k) a(k) 2. | | Each normal mode of the system behaves like an independent harmonic ⇒ oscillator with amplitude a(k).
11 ● In 3 spatial dimensions we write
3 ik r iωt ik r+iωt φ(r, t) = d k N(k) a(k)e · − + a∗(k)e− · ! , - and use
3 i(k k!) r 3 3 d r e ± · = (2π) δ (k k!) ± ! Therefore we should choose 1 N(k) = (2π)3 2ω(k) to obtain an integral with the usual relativistic phase space (density of states) factor: d3k H = ω(k) a(k) 2 (2π)3 2ω(k) | | !
12 Second Quantization ● First quantization was the procedure of replacing classical dynamical variables q and p by quantum operators qˆ and pˆ such that
[qˆ, pˆ] = i (! = 1)
● Second quantization is replacing the field variable φ(x, t) and its conjugate momentum density π(x, t) by operators such that
[φˆ(x, t), πˆ(x!, t)] = i δ(x x!) − N.B. x and x! are not dynamical variables but labels for the field values at different points. Compare (and contrast)
[qˆj, pˆk] = i δjk (j, k = x, y, z)
● The wave function φ satisfying the Klein-Gordon equation is replaced by the field operator φˆ, satisfying the same equation.
13 ● The Fourier representation becomes
ikx iωt ikx+iωt φˆ(x, t) = dk N(k) aˆ(k)e − + aˆ†(k)e− ! , - i.e. φˆ is hermitian but the Fourier conjugate operator aˆ is not. ● Keeping track of the order of operators, the Hamiltonian operator is
ˆ 1 H = dk N(k) 2 ω(k) [aˆ(k)aˆ†(k) + aˆ†(k)aˆ(k)] !
● Comparing this with the simple harmonic oscillator,
ˆ 1 HSHO = 2 ω (aˆaˆ† + aˆ†aˆ)
we see that aˆ†(k) and aˆ(k) must be the ladder operators for the mode of wave number k. They add/remove one quantum of excitation of the mode. These quanta are the particles corresponding to that field:
aˆ†(k) = creation operator , aˆ(k) = annihilation operator ⇒ for Klein-Gordon particles.
14 ● Ladder operators of simple harmonic oscillator satisfy
[aˆ, aˆ†] = 1
The analogous commutation relation for the creation and annihilation operators is
N(k) [aˆ(k), aˆ†(k!)] = δ(k k!) − [aˆ(k), aˆ†(k!)] = 2π 2ω(k) δ(k k!) ⇒ · − or in 3 spatial dimensions
3 3 [aˆ(k), aˆ†(k!)] = (2π) 2ω(k) δ (k k!) · − On the other hand
[aˆ(k), aˆ(k!)] = [aˆ†(k), aˆ†(k!)] = 0
15 ● The commutators of the creation and annihilation operators correspond to the field commutation relation
3 3 [φˆ(r, t), πˆ(r!, t)] = d k d k! N(k) N(k!)[ iω(k!)] − × ! ik x ik x ik! x! ik! x! aˆ(k)e− · + aˆ†(k)e · , aˆ(k!)e− · aˆ†(k!)e · − 0 1 3 ik (x x!) ik (x x!) = i d k N(k)ω(k) e− · − + e · − ! 0 1 µ µ where x = (t, r) and x! = (t, r!). Hence
3 [φˆ(r, t), πˆ(r!, t)] = i δ (r r!) − as expected. On the other hand
[φˆ(r, t), φˆ(r!, t)] = [πˆ(r, t), πˆ(r!, t)] = 0
16 ● The fact that the field operator has positive- and negative-frequency parts now appears quite natural: ik r iωt ❖ positive frequency part aˆ(k)e · − annihilates particles ik r+iωt ❖ negative frequency part aˆ†(k)e− · creates particles
● !ω is the energy released/absorbed in the annihilation/creation process. ± N.B. The hermitian field describes particles that are identical to their antiparticles, e.g. π0 mesons. ● More generally (as we shall see shortly) the negative-frequency part of φˆ creates antiparticles.
17 Single Particle States ● If 0 represents the state with no particles present (the vacuum), then | ' aˆ†(k) 0 is a state containing a particle with wave vector k, i.e. momentum !k. | ' More generally, to make a state with wave function φ(r, t) where
3 ik r iωt φ(r, t) = d k φ˜(k)e · − ! we should operate on the vacuum with the operator
3 d k φ˜(k)aˆ†(k) !
3 ● Writing this state as φ = d k φ˜(k)aˆ†(k) 0 we can find the wave function | ' | ' using the relation % 0 φˆ(r, t) φ = φ(r, t) ( | | '
● Thus the field operator is an operator for “finding out the wave function” (for single-particle states).
18 Two Particle States ● Similarly, we can make a state φ of two particles: | 12' 3 3 φ = d k d k φ˜(k , k )aˆ†(k )aˆ†(k ) 0 | 12' 1 2 1 2 1 2 | ' ! Since the two particles are identical bosons, this state is symmetric in the
labels 1 and 2 even if φ˜ isn’t, because aˆ†(k1) and aˆ†(k2) commute:
3 3 φ = d k d k φ˜(k , k )aˆ†(k )aˆ†(k ) 0 | 12' 1 2 1 2 1 2 | ' ! 3 3 = d k d k φ˜(k , k )aˆ†(k )aˆ†(k ) 0 1 2 2 1 2 1 | ' ! = φ | 21' ● The 2-particle wave function is φ(r , r , t) = 0 φˆ(r , t)φˆ(r , t) φ 1 2 ( | 1 2 | 12' = φ(r2, r1, t) Thus quantum field theory is also a good way of dealing with systems of many identical particles. 19 Number Operator
● Recall that for the simple harmonic oscillator the operator aˆ†aˆ tells us the number of quanta of excitation:
aˆ†aˆ φ = n φ | n' | n' when φ is the nth excited state. | n' ● Similarly, in field theory the operator
ˆ = dk N(k) aˆ†(k)aˆ(k) N ! counts the number of particles in a state. For example,
ˆ φ = dk dk dk N(k)φ˜(k , k )aˆ†(k)aˆ(k)aˆ†(k )aˆ†(k ) 0 N | 12' 1 2 1 2 1 2 | ' !
20 where
N(k)aˆ†(k)aˆ(k)aˆ†(k1)aˆ†(k2)
= N(k)aˆ†(k)aˆ†(k )aˆ(k)aˆ†(k ) + aˆ†(k)aˆ†(k )δ(k k ) 1 2 2 − 1 = N(k)aˆ†(k)aˆ†(k )aˆ†(k )aˆ(k) + aˆ†(k)aˆ†(k )δ(k k ) + aˆ†(k)aˆ†(k )δ(k k ) 1 2 1 − 2 2 − 1 However, aˆ(k) 0 = 0 (by definition of the vacuum) and so | ' ˆ φ = 0 + φ + φ = 2 φ N | 12' | 12' | 12' | 12' N.B. In general, states do not need to be eigenstates of ˆ : they need not N contain a definite number of particles.
21 Electromagnetic Field ● Each component of Aµ (in Lorenz gauge) is quantized like a massless Klein-Gordon field (i.e. with ω = k ): | | d3k Aˆµ(r, t) = (2π)3 2ω P2=L,R ! µ ik r iωt µ ik r+iωt εP (k)aˆP (k)e · − + εP∗ (k)aˆP† (k)e− · 0 1 where aˆP (k) annihilates a photon with momentum !k and polarization P and
aˆP† (k) creates one. µ ● εL,R(k) are the left/right-handed circular polarization 4-vectors for wave vector k and
3 3 aˆP (k), aˆ† (k!) = (2π) 2ω(k) δP P δ (k k!) P ! ! − N.B. Different po0larizations com1mute. ● The Hamiltonian for the e.m. field is
ˆ 1 3 2 2 H = 2 d r E + B ! 22 . / where (in A0 = 0 gauge) from Eˆ = ∂Aˆ /∂t, Bˆ = Aˆ we have − ∇ × 3 d k ik x +ik x Eˆ = iω ε aˆ e− · ε∗ aˆ† e · (2π)3 2ω P P − P P 2P ! 0 1 3 d k ik x +ik x Bˆ = ik ε aˆ e− · ε∗ aˆ† e · (2π)3 2ω × P P − P P 2P ! 0 1
● Using k ε = iωε , k ε = iωε [recall that for k along the z-axis we × L L × R − R have εµ = (0, 1, i, 0)/√2] we find, as expected, that L,R ∓ 3 d k 1 Hˆ = ω(k) aˆ (k)a† (k) + aˆ† (k)a (k) (2π)3 2ω 2 P P P P 2P ! 0 1
23 Vacuum Energy and Normal Ordering ● Using the above expression for the e.m. Hamiltonian and the commutation relation for the photon annihilation and creation operators, we find that the energy of the vacuum is given by
3 3 Hˆ 0 = d k ω(k) lim δ (k k!) 0 | ' k! k − | ' ! →
● In a finite volume V , we should interpret 3 3 d r i(k k!) r V ! lim δ (k k ) = lim 3 e − · = 3 k! k − k! k (2π) (2π) → → !V The energy density of the vacuum is thus U where Hˆ 0 = U V 0 0 | ' 0 | ' d3k U = ω(k) = !! ⇒ 0 (2π)3 ∞ !
1 ● This is due to the zero-point fluctuations of the e.m. field ( 2 !ω per mode). In reality, we don’t understand physics at very short distances (very large wave
24 numbers) and presumably the integral gets cut off at some very large k Λ. | | ∼ We shall see that quantities we can measure don’t depend much on Λ or the form of the cut-off. ● For most purposes we can throw away the term in Hˆ that gives rise to the vacuum energy, which corresponds to measuring all energies relative to the vacuum. This means writing d3k Hˆ = ω(k)aˆ† (k)aˆ (k) (2π)3 2ω P P 2P ! which is called the normal-ordered form of Hˆ , i.e. a form with the annihilation
operator aˆP on the right. N.B. After normal-ordering, Hˆ is still hermitian. Clearly, a normal-ordered operator has the vacuum as an eigenstate with eigenvalue zero. ● However, we should not think that this means vacuum fluctuations are not there. They give rise to real effects, such as spontaneous transitions and . . .
25 The Casimir Effect ● Suppose a region of vacuum is bounded by the plane surfaces of two semi-infinite conductors. Then some (low-frequency) vacuum fluctuations are forbidden by the boundary conditions (E = B = 0). Thus the vacuum & ⊥ energy is reduced, corresponding to an attractive force between the conductors.
z unit area
a
26 ● Since the vacuum energy density is proportional to !c, by dimensional analysis this Casimir force per unit area must be !c F C ∝ a4
● To find the constant of proportionality, note that kz = nπ/a where n = 0, 1, 2, . . ., so ω = c k = c k2 + (nπ/a)2 | | & 3
● For n = 1, 2, 3 . . . there are 2 polarizations; for n = 0 only one polarization is allowed, since E = 0. Hence the energy per unit area is & 2 d k ∞ 2 1 & 2 E = 2 !c k + 2 k + (nπ/a) (2π)2 | &| & & n=1 ' ! 2 3
27 ● This is assumed to be cut off by new physics at wave numbers k > Λ. Thus | | !c ∞ E = 1 F (0) + F (n) 2π 2 & n=1 ' 2 where F (n) = k dk k2 + (nπ/a)2f k2 + (nπ/a)2 & & & & ! 3 "3 # with f(k) = 1 for k Λ and f(k) = 0 for k Λ. - . ● Changing variable from k to k = k2 + (nπ/a)2, & & 3 ∞ F (n) = k2 dk f(k) !nπ/a Removing the boundary conditions would allow n to be a continuous variable:
!c ∞ E = dn F (n) 0 2π !0
28 Hence the change in the vacuum energy per unit area is
!c ∞ ∞ δE = E E = 1 F (0) + F (n) dn F (n) − 0 2π 2 − & n=1 0 ' 2 !
● The Euler-McLaurin formula tells us that
∞ ∞ 1 1 (2n 1) dn F (n) = F (0) + F (n) + B F − (0) 2 (2n)! 2n 0 n=1 ! 2 * + where B2n are Bernoulli numbers: 1 1 B = , B = , . . . 2 6 4 −30 Hence !c 1 1 δE = F !(0) + F !!!(0) + 2π −12 720 · · · * + But 2 ∞ 2 π nπ nπ F (n) = k dk f(k) F !(n) = f nπ/a ⇒ − a a a ! ( ) ( ) 29 3 and so F !(0) = 0, F !!!(0) = 2(π/a) and − π2 !c δE = −720 a3
● Thus the Casimir force is d π2 !c F = δE = C da 240 a4 This is very small, 27 1.3 10− F = × Pa m4 C a4 but it has been measured (Sparnaay, 1957).
30 Complex Fields
● Suppose φˆ is the second-quantized version of a complex field, i.e. φˆ† = φˆ. We / can always decompose it into 1 φˆ = φˆ1 + iφˆ2 √2 ( ) where φˆ1 and φˆ2 are hermitian. Then
ikx iωt ikx+iωt φˆ(x, t) = dk N(k) aˆ(k)e − + ˆb†(k)e− ! 0 1 where 1 1 aˆ = (aˆ1 + iaˆ2) , ˆb† = (aˆ† + iaˆ†) = aˆ† √2 √2 1 2 /
● From the canonical commutation relations
φˆ (x, t), πˆ (x!, t) = iδ δ(x x!) (j, l = 1, 2) j l jl − we can deduce 0 1 N(k) aˆ (k), aˆ†(k!) = δ δ(k k!) j l jl − 0 1 31 and hence that
N(k)[aˆ(k), aˆ†(k!)] = N(k)[ˆb(k), ˆb†(k!)] = δ(k k!) , − [aˆ(k), ˆb†(k!)] = [aˆ(k), ˆb(k!)] = [aˆ†(k), ˆb†(k!)] = 0 .
Hence aˆ† and ˆb† are creation operators for different particles. ● The Lagrangian density = [φˆ ] + [φˆ ] L L 1 L 2 can be written as ˆ ˆ ˆ ˆ ∂φ† ∂φ ∂φ† ∂φ 2 = m φˆ†φˆ L ∂t ∂t − ∂x ∂x − The canonical momentum density is thus
∂ ∂φˆ† πˆ = L = ˆ ∂φ˙ ∂t and the Hamiltonian density is ˆ ˆ ˆ ˆ† ∂φ† ∂φ 2 ˆ = πˆφ˙ + πˆ†φ˙ ˆ = πˆ†πˆ + + m φˆ†φˆ H − L ∂x ∂x
32 ● Using the Fourier expansion of φˆ and integrating over all space, we find
Hˆ = dx ˆ = 1 dk N(k) ω(k) H 2 × ! ! [aˆ(k)aˆ†(k) + aˆ†(k)aˆ(k) + ˆb(k)ˆb†(k) + ˆb†(k)ˆb(k)]
or after normal ordering
Hˆ = dk N(k) ω(k) [aˆ†(k)aˆ(k) + ˆb†(k)ˆb(k)] ! ˆ Thus both aˆ† and b† create particles with positive energy !ω(k).
33 Symmetries and Conservation Laws ● We want to find a current and a density that satisfy the continuity equation for the complex Klein-Gordon field. We use an important general result called Noether’s theorem (Emmy Noether, 1918), which tells us that there is a conserved current associated with every continuous symmetry of the Lagrangian, i.e. with symmetry under a transformation of the form φ φ + δφ → where δφ is infinitesimal. Symmetry means ∂ ∂ ∂ δ = Lδφ + L δφ! + Lδφ˙ = 0 L ∂φ ∂φ! ∂φ˙ where ∂φ ∂ δφ! = δ = δφ ∂x ∂x " # ∂φ ∂ δφ˙ = δ = δφ ∂t ∂t " # (easily generalized to 3 spatial dimensions).
34 ● The Euler-Lagrange equation of motion ∂ ∂ ∂ ∂ ∂ L L L = 0 ∂φ − ∂x ∂φ − ∂t ˙ " ! # " ∂φ # then implies that ∂ ∂ ∂ ∂ δ = L δφ + L (δφ) L ∂x ∂φ ∂φ ∂x " ! # ! ∂ ∂ ∂ ∂ + L δφ + L (δφ) = 0 ∂t ˙ ˙ ∂t " ∂φ # ∂φ ∂ ∂ ∂ ∂ L δφ + Lδφ = 0 ⇒ ∂x ∂φ ∂t ˙ " ! # " ∂φ # ● Comparing with the conservation/continuity equation (in 1 dimension) ∂ ∂ρ (J ) + = 0 ∂x x ∂t shows that the conserved density and current are (proportional to) ∂ ∂ ρ = Lδφ , Jx = L δφ ∂φ˙ ∂φ!
35 ● In 3 spatial dimensions ∂ ∂ J = L δφ , J = L δφ , . . . x ∂(∂φ/∂x) y ∂(∂φ/∂y) and hence in covariant notation ∂ J µ = L δφ ∂(∂µφ)
● If the Lagrangian involves several fields φ1, φ2, . . ., the symmetry may involve changing them all: invariance w.r.t. φ φ + δφ conserved Noether j → j j ⇒ current ∂ J µ = L δφ ∂(∂ φ ) j j µ j 2
● After second quantization, the same procedure (being careful about the order of operators) can be used to define conserved current and density operators.
36 Phase (Gauge) Invariance ● The Klein-Gordon Lagrangian density ˆ ˆ ˆ ˆ ∂φ† ∂φ ∂φ† ∂φ 2 = m φˆ†φˆ L ∂t ∂t − ∂x ∂x − or in 3 spatial dimensions ˆ ˆ ∂φ† ∂φ 2 = φˆ† φˆ m φˆ†φˆ L ∂t ∂t − ∇ · ∇ − µ 2 = ∂ φˆ†∂ φˆ m φˆ†φˆ µ − is invariant under a global phase change in φˆ:
iε φˆ e− φˆ φˆ iεφˆ → 0 − +iε φˆ† e φˆ† φˆ† + iεφˆ† → 0 i.e. δφˆ iφˆ, δφˆ† +iφˆ†. ∝ − ∝ ● The corresponding conserved Noether current is just that derived earlier:
µ ∂ ∂ µ µ Jˆ = L δφˆ + δφˆ† L = i(∂ φˆ†)φˆ + iφˆ†(∂ φˆ) ˆ ˆ ∂(∂µφ) ∂(∂µφ†) − 37 ● We can define an associated conserved charge, which is the integral of ρˆ over all space: Qˆ = ρˆd3r ! dQˆ ∂ρˆ = d3r = Jˆ d3r = Jˆ dS = 0 dt ∂t − ∇ · − sphere · ! ! !∞
● In this case ˆ ˆ ∂φ† ∂φ 3 Qˆ = i φˆ φˆ† d r − ∂t − ∂t ! 4 5 Inserting the Fourier decomposition of the field,
3 ik x ik x φˆ = d k N(k) aˆ(k)e− · + ˆb†(k)e · ! 0 1 3 1 where N(k) = [(2π) 2ω(k)]− , we find
3 Qˆ = d k N(k) aˆ†(k)aˆ(k)–ˆb†(k)ˆb(k) ! 0 1 38 ● Comparing with the energy
3 Hˆ = d k N(k)ω(k) aˆ†(k)aˆ(k)+ˆb†(k)ˆb(k) ! 0 1 we see that the particles created by aˆ† and ˆb† have opposite charge. ● Thus we see that in quantum field theory all the ‘problems’ with the negative-energy solutions of the Klein-Gordon equation are resolved, as follows ❖ The object φ that satisifies the KG equation is in fact the field operator φˆ. ❖ The Fourier decomposition of φ has a positive frequency part that annihilates a particle (with energy !ω and charge +1) AND a negative frequency part that creates an antiparticle (with energy !ω and charge 1). − ❖ Similarly, φˆ† creates a particle or annihilates an antiparticle.
39 The Dirac Field ● The Lagrangian density that gives the Dirac equation of motion,
iγµ∂ ψ mψ = 0 µ − is = ψ¯iγµ∂ ψ mψ¯ψ LD µ −
0 ● As in the Klein-Gordon case, we should treat ψ and ψ¯ = ψ†γ as independent fields. The Euler-Lagrange equation for ψ¯ gives the Dirac equation for ψ: ∂ ∂ LD ∂ LD = iγµ∂ ψ mψ = 0 ∂ψ¯ − µ ∂(∂ ψ¯) µ − " µ # while that for ψ gives the Dirac equation for ψ¯: ∂ ∂ LD = ∂ LD = mψ¯ i∂ ψ¯γµ = 0 ∂ψ µ ∂(∂ ψ) − − µ " µ # . / µ 0 µ 0 0 µ µ Check: i∂ ψ†γ † mψ† = 0, ψ¯ = ψ†γ , γ †γ = γ γ i∂ ψ¯γ = mψ¯. − µ − ⇒ − µ 40 ● The generalized momentum densities are
∂ D 0 π = L = ψ¯iγ = iψ† ∂ψ˙ ∂ π¯ = LD = 0 ∂ψ¯˙
● Hence the Hamiltonian density is ∂ψ = πψ˙ = ψ¯iγ0 HD − LD ∂t − LD = ψ¯iγ ψ + mψ¯ψ − · ∇ and the total energy is given by
H = d3r = d3r ψ¯( iγ + m)ψ HD − · ∇ ! ! 3 = d r ψ†( iα + βm)ψ − · ∇ ! which is indeed the expectation value of the original Dirac Hamiltonian.
41 ● Now we second quantize by expressing the field operator ψˆ as a Fourier integral over plane-wave spinors with operator coefficients:
ˆ 3 ik x ˆ +ik x ψ(r, t) = d k N(k) [cˆs(k)us(k)e− · + ds†(k)vs(k)e · ] s ! 2 where s = 1, 2 label spin up/down, i.e. u1 = u↑ free particle spinor, etc.
● We expect that the operator cˆs(k) annihilates a particle (e.g. an electron) of ˆ momentum !k, spin orientation s. while ds†(k) creates an antiparticle (positron) of the same momentum and spin. But the Hamiltonian operator is
3 Hˆ = d r ψˆ†( iα + βm)ψˆ − · ∇ ! and using the Dirac equation
( iα + βm)u = Eu = ωu − · ∇ s s s ( iα + βm)v = Ev = ωv − · ∇ s − s − s
42 we find that
ˆ 3 ˆ ˆ H = d k N(k)ω(k) cˆs†(k)cˆs(k)–ds(k)ds†(k) s ! 2 0 1
● Thus there appears to be a problem: unlike the Klein-Gordon case, we seem to get a negative contribution to the energy from the antiparticles! ● The Dirac equation also has phase (gauge) symmetry:
iε ψˆ e− ψˆ ψˆ iεψˆ → 0 − +iε ψˆ† e ψˆ† ψˆ† + iεψˆ† → 0 with corresponding Noether current
µ ∂ ∂ Jˆ = L δψˆ + δψˆ† L ˆ ˆ ∂(∂µψ) ∂(∂µψ†) = ψˆ¯ iγµ( iψˆ) = ψˆ¯γµψˆ −
43 and conserved charge
ˆ 3 ˆ0 3 ˆ ˆ 3 ˆ ˆ Q = d r J = d r ψ†ψ = d k N(k) cˆs†(k)cˆs(k)+ds(k)ds†(k) s ! ! ! 2 0 1 which also seems to have the wrong sign for the antiparticle contribution. ˆ ˆ ● However, we actually need the normal-ordered operators, which involve ds†ds, ˆ ˆ not dsds†. Hence if ˆ ˆ ˆ ds†ds = –dsds† + const. then all signs are correct. ● Thus we are forced to give the creation and annihilation operators for spin one-half particles anticommutation relations:
cˆs(k), cˆ† (k!) cˆs(k)cˆ† (k!)+cˆ† (k!)cˆs(k) s! ≡ s! s! 6 3 7 3 = (2π) 2ω(k) δss δ (k k!) = dˆs(k), dˆ† (k!) ! − s! while 6 7 cˆ (k), cˆ (k!) = dˆ (k), dˆ (k!) = 0 { s s! } { s s! }
44 ● This means that two-particle states are antisymmetric:
3 3 φ = d k d k φ˜ (k , k )cˆ† (k )cˆ† (k ) 0 | 12' 1 2 s1s2 1 2 s1 1 s2 2 | ' s ,s ! 21 2 3 3 = d k d k φ˜ (k , k )cˆ† (k )cˆ† (k ) 0 − 1 2 s1s2 1 2 s2 2 s1 1 | ' s ,s ! 21 2 3 3 = d k d k φ˜ (k , k )cˆ† (k )cˆ† (k ) 0 − 1 2 s2s1 2 1 s1 1 s2 2 | ' s ,s ! 21 2 = φ −| 21' ● Thus spin one-half particles must be fermions. This is the spin-statistics theorem.
45 Interacting Fields ● We introduce e.m. interactions into the Dirac Lagrangian by the usual minimal substitution, ∂µ ∂µ + ieAµ: → = ψ¯iγµ (∂ + ieA ) ψ mψ¯ψ LD µ µ − = eA ψ¯γµψ L0 − µ where is the free-particle Lagrangian density. L0 ● Notice that the canonical momentum π = ∂ /∂ψ˙ is unchanged, and so the LD Hamiltonial density is = πψ˙ = + HD − LD H0 HI where the interaction Hamiltonian density is = eA ψ¯γµψ HI µ
● In the second-quantized theory becomes an operator, HI ˆ = eAˆ ψˆ¯γµψˆ HI µ 46 where all the field operators are capable of creating or annihilating particles:
ˆ 3 ik x +ik x Aµ = d k N(k) aˆP (k)εP µe− · + aˆP† (k)εP∗ µe · ! P 2, - ˆ 3 +ip! x ip! x ψ¯ = d p! N(p!) cˆ† (p!)u¯ (p!)e · + dˆ (p!)v¯ (p!)e− · s! s! s! s! ! s! 2, - ˆ 3 ip x ˆ +ip x ψ = d p N(p) cˆs(p)us(p)e− · + ds†(p)vs(p)e · ! s 2, - ● In first-order perturbation theory, the transition matrix element is
= i d4x f i Afi − ( |HI| ' !
● Suppose for example that the initial state i contains an electron of | ' momentum pi, spin orientation si, and a photon of momentum q, polarization R:
i = cˆ† (p )aˆ† (q) 0 | ' si i R | '
47 ● Using the (anti)commutation relations for the cˆ’s and aˆ’s, the positive-frequency parts of Aˆ and ψˆ give
(+) (+) i(p +q) x Aˆ ψˆ i = ε (q)u (p )e− i · 0 µ | ' Rµ si i | '
● Similarly, if f contains an electron of momentum p , spin s , | ' f f
f = cˆ† (p ) 0 | ' sf f | ' f = 0 cˆ (p ) ( | ( | sf f and hence the negative-frequency part of ψˆ¯ gives
( ) ˆ − +ipf x f ψ¯ = u¯ (p )e · 0 ( | sf f ( |
● Putting everything together, we have
( ) (+) ˆ − µ (+) µ i(pf pi q) x f Aˆ ψ¯ γ ψˆ i = ε u¯ (p )γ u (p )e − − · ( | µ | ' Rµ sf f si i 48 which gives
= ie(2π)4 δ4(p p q) ε u¯ (p )γµu (p ) Afi − f − i − Rµ sf f si i corresponding to the Feynman rule for the vertex p i
p q f
● Similarly, if the initial and final fermions are positrons, then the term
(+) (+) ˆ µ ( ) f Aˆ ψ¯ γ ψˆ − i ( | µ | ' gives the expected result
= ie(2π)4 δ4(p p q) ε v¯ (p )γµv (p ) Afi − f − i − Rµ si i sf f
49