MA4445: Quantum Field Theory 1

MA4445: Quantum Field Theory 1 Dr. Samson Shatashvilli September 7, 01 1 Note These notes are absolutely awful, there was no way to definitely take down all of the equations, so... sorry? I will try to add as I go along too. Introduction.1 Books Peskin & Schroeder Schwever - Introduction to QFT P.A.M. Dirac lectures of QM. - only use 0 pages but is much better. Why; STR QM Particle Physics TP Quantum Many Body Sytem Phys End QED; (abc ABC...) M(e + e µ + µ ) = i=0 αi M i () α = e 4π c = 1 137.3 Alternate QFT s 1. Quantum Yang Mill s (classical Yang Mills has i = 0, Weak, Strong. Quantum Gravity 3. Cosmology e + e µ + µ Where the total momentum p tot = 0. p + m e = k + m µ k = k = p = p Θ; m e, m µ, m µ >> m e E >> m e, m µ E = p E = p + m ; E = p + m 1.E = p.θ Number Of Events σ = ρ A ρ B e A e B A Where σ is the Cross section. Could also compute the differential cross section dσ dω dσ dω (E, Θ) = 1 64π E M(Θ) 1

.4 Dimensional Analysis (units) = c = 1 [Length]=[Time]=[Energy] 1 =[mass] 1 [σ=[length] =[Energy].5 M-Scattering Amplitude Initial state is e + e, and final state is µ + µ [ ] dσ 1 = dω [Energy] M Final State H I Initial State e + e H I µ + µ = 0 } e + e H I γ µ µ + µ 0 H I γ µ s = t x = µν X µ η µν X ν M µ + µ H I γ µ η µν γ H I e + e 1 A µ B µ = A µ η µν B ν Spin Initial, Spin Final Total spin is one, from the polarisation of 1/ spin electrons. (LR) (LR) dσ dω (0, x, y, z) e (0, 1, i, 0) e (0, cos θ, i, sin θ) = {ε µ } 1.. 3. 4. (LR) (LR) e (1 cos θ) (RR) (LR) = 0 (RR) (LL) = 0 M(RL LR) = e (1 cos θ) M(LR RL) = e (1 cos θ) M(LR LR) = e (1 + cos θ) dσ dω = 1 σ = π 1 64φ E M spin = α dσ α dω = π dω 4E π 0 σ tot = 4πα 3E 4E (1 + cos θ) (1 + cos θ)( d cos θ) α = e 4π This is nd order fine string constant, eventually we will find that the constant is 1. 1. E >> M µ, relax. E >> M e, relaxed

3. O(α ) + O(α 4 ), compute 4. e + e µ + µ infinitely many more processes e + e anything, implies nothing anything. M s s r r (p, p, k, k ) = V s (p )( ieγ µ )U s (p) i ν µν q Ū r (k)( ieγ ν V ν )(k ) 1. Relativistic Waves; E + p m E = p + m. Causality; s = t (x 1 x ) < 0 U(t) = x e iht d 3 p x 0 = (π) 3 x e iht p p x o = x p = e ipx ˆp = i x ˆ pe i p v = pe i p x (π) 3 p p p i p i = 1 U(t) = e iht p i p = e π t p p (π) 3 e i = p sin θ dp dθ dφ = π t ip(x x0) = e i (π) 3 t x p p x 0 1 1 p ip(x x0) cos θ (d cos θ)e 1 e ip(x x0) e ip(x x0) = < 0 ip(x x 0 ) = sin p(x x 0) p(x x 0 ) U(t) = x iht x 0 = x e iˆp +m t x 0 1 π x x 0 s = t (x x 0 ) << 0 implies that U(t) = 0 p dp e f(p) = e f(x a) = 0 p dp sin(p x x 0 (e it p +m e f0(a) sqrtf( (a) by doing a taylor expansion [f(x) = f 0 + (x a)f (a) + (x a) f (a) +...] p = i m(x x 0) (x x0 ) t but as x x 0, then this does not equal to zero, which is bad, and shows incompatability. Antiparticles Multiparticle states spin and statistics (Pauli) Probability any commuon (event) space like regions forbidden. H = ˆp m = ˆp + m x e ih(t t ) x Which does not go to zero. Thus need a Quantum Field Theory. 3

3 Field Theory (classical) q 1,..., q N, q 1,..., q N φ x (t) = φ( x, t) φ x (t) φ( x, t) x µ x µ R 1,3 x 0 = t, x S = t x = x 0 x = X µ η µν X ν 1 0 0 0 η = 0 1 0 0 0 0 1 0 0 0 0 1 S = t t 1 dt L(q, q) The equation of Motion is such that δs = 0 S = dt d 3 xl(φ(x), µ φ(x)) δs = 0 = d 4 x Assume φ(x), µ φ) = 0 on boundary of R 1,3 δs = d 4 x ( δφ(x) δφ(x) + δ( µ φ) = µ (δφ) ([ ( )] ) L δφ(x) µ δφ + δ µ φ ) δ µ φ µ(δφ) [ ] d 4 x µ δ µ φ δφ from the E.o.m There are N-variables δφ(x) δ(l µ δ µ φ = 0 H = N i=1 p i q i L(q, q) L p i = q i (q, q) H(p, q, q) H(p, q) p i A ( ij q) q j j q = j (A 1 (q)) ij p j dh = i=1 ( ) L det 0 deta(q) 0 q i q i S = t t 1 dt L(q, q S = dt d 3 x L(φ(x), µ φ(x)) ( H dp i + H dq i + H ) d q i = q i do i L ( dq i + p i L ) d q i p i q i q i q i q i p(x) = H(p, q) L φ( x, t) = Π( x, t)d3 x 4

H = Π(x, t) = δφ(, tx) [ d 3 x Π( x, t) φ( x, ] t) L(φ, µ φ) Example example E.O.M. µ φ µ φ = H = Π( x, t) φ( x, t) L(φ, µ φ) 3 µ=0 µ φ µ φ = 0 φ 0 φ i = ( t φ) ( φ) i φ i φ L = 1 ( µφ) m φ = 1 φ 1 ( φ) 1 m φ ( + m )φ = 0 ( ( ) ) t + m = 0 H = Π( φ L = 1 Π + 1 ( φ) + m φ 0 *** I APOLOGISE FOR THE ILLEGABILITY OF THIS, SHATASHVILLI WRITES IN A VERY INCO- HERENT FASHION *** 4 Noethers Theorem φ(x) φ (x) = φ(x) + α φ(x) Suppose φ is chose such that the equations of motion do not change. L L = L + α L α L = ( δφ (αδφ) + δ µ φ ] = φ ] α φ) + µ ( δ µ φ αδφ j µ Q = α L = dµj µ δ µ φ φ J µ µ j µ = 0 j 0 d 3 x = j µ σ µ dotq = j 0 d 3 x = ) µ (α φ) [ µ ( δ µ φ jd 3 x = 0 )] α φ m = 0; KG L = 1 ( µφ) φ = 0 ( ) = 6 t = µ µ φ φ + α( φ) φ = 0 0 = L = µ J µ 5

Ex.: Complex KG φ complex C µ j µ 1µ µ φ φ = 0 L = 1 µφ 1 m φ φ = φ 1 + iφ ( + m )φ = 0 ( + m ) φ = 0 φ φ = e iα φ = φ + iαφ +... φ φ = e iαφ = φ iα φ +... j µ = φ = iφ; φ = i φ j µ = δ µ φ φ J µ δ µ φ iφ + φ) δ µ φ( i = 1 ( µ φ iφ + µ φ( i φ)) = 1 µ φ φ µ φphi) φ(x) φ = φ(x + a) = φ(x) + a µ µ φ +... d 4 x L d 4 xl + d 4 x a µ µ L L L = L + a ν µ (δ µ ν L) J µ = δ µ ν L T ν µ = j ν µ = δ µ φ νφ J ν µ Q v = Tν 0 d 3 x Q 0 = H = T0 0 d 3 x = Q i = P = Ti 0 d 3 x ( + m )φ = 0 Π( x, t) = φ( x, t) φ φ Ld3 x H = Π + ( φ) + m φ {p i, q j } = δ ij [p i, q j ] = i δ ij p i = i q i {Π( x, t), φ( y, t)} = δ (3) ( x y) ( ( ) 3 ( ) ) + m t x i=1 i (π) 3 ei p x φ( p, t) = 0 [ ] φ( p, (π) 3 ei p x t)) + p φ( p, t) + m φ( p, t) = 0 [ ] t + ( p + m ) φ( p, t) = 0 6

ω p = p + m [ ] t + ω p φ(p, t) = 0 H SHO = 1 Π + 1 ω φ φ = 1 aω (â + â ) SHO ω Π = i (â â ) [Π, φ] = 1 [â, â ] = 1 H SHO = ω(â â + 1 ) [H, â ] = ωâ [H, â] = ωâ â 0 = 0 (thegroundstate) H 0 = ω(â + â + 1 ) 0 = 1 ω 0 = E 0 0 H(â 0 ) = ( [H, â ] + â H ) 0 = (ωâ + 1 ) ωâ 0 = (ω + 1 ω)(â 0 ) H(â n 0 ) = (n + 1 )ω(â n 0 ) E n = (n + 1 ω φ( p, t) = 1 ωp (a( p, t) + a ( p, t)) φ( x, t) = Π( x, t) = ωp Π( p, t) = i d 3 p (π) 3 ei p x φ( p, t) = (a( p, t) a ( p, t)) 1 (π) 3 (e i p x a( p, t) + e i p x a ( p, t)) ωp ωp d 3 p d (π) 3 ei p x Π( x, 3 p t) = (π) 3 ( 1) (ei p x a( p, t) e i p x a ( p, t) [ ] ˆΠ( x, t), ˆφ( y, t) = iδ (3) ( x y) [â(p, t), â (p, t) ] = δ (3) ( p p ) H = d 3 xh = d 3 x (π) 3 (π) 3 ( 1) ωp ω p (e i p x a(p, t) e i p x a (p, t))(e i p x a(p, t) e i p x a (p, t)) = d 3 x p 1 ω pa (p, t)a(p, t) + a(p, t)a (p, t)) E 0 = δ (3) (0) ω p H KG = ω p â (p, t)â(p, t) Hamiltonians H o = 0 0 [H, â (p, t)] = ω p â (p, t) H, â(p, t)] = ω p â(p, t) 7

NO HATS FROM NOW ON H KG = ω p â (p, t)â(p, t) ω p = E p, a (p, t) 0 = p > such that = [a (p )a(p ), a (p)] = a (p )[a(p ), a (p)] â(p, t) 0 = 0 H(a (p, t) 0 ) = ω p (a (p) 0 p 1, p = a (p 1 )a (p ) 0 H p 1, p = (ω p1 + ω p ) p 1, p p 1,..., p N = a (p 1 )...a (p N ) 0 > N H p 1,..., p N = ω pi p 1, p i=1 H = T0 0 d 3 x p i = Ti 0 d 3 x = p i a (p)a(p) H = ω p a (p)a(p) ω p = p + m p = pa (p)a(p) 0 0 = 1 a (p) 0 = p (ω p, p) a (p 1 )a (p ) 0 = p 1, p (ω p1 + ω p, p 1 + p ) p 1 p = 0 a(p 1 )a (p ) 0 = 0 [a(p 1 ), a (p )] + a (p )a(p 1 ) 0 But the commutator bracketrs are equal to δ (3) ( p 1 p ) p 3 = γ(p 3 + βe) E = γ(e + βp 3 ) δ (3) ( p 1 p ) δ (3) ( p 1 p ) dp 3 dp 3 δ (f(x) f(x 0 )) δ(x x 0 ) δ (f(x) f(x 0 )) 1 f (x 0 ) δ(x x 0) df(y) dxδ(x) = 1 dy dyδ(f(y) = 1 δ(f(y))f (y) = δ(y) p 1 p = δ (3) ( p 1 p ) vecp = E p a (p) 0 p 1 p = E p1 δ (3) ( p 1 p ) = E p 1 δ (3) (p 1 p ) E p1 δ (3) (p 1 p ) = E p 1 δ(p 1 p ) φ( x, t = 0) φ( x, t) = e iht φ( x, t = 0)e iht i φ t = [H, φ] 8

i t π( x, t) = [π(x, t), ] i [φ( x, t φ( x, t) = t), d 3 x { 1 π (x ) + 1 ( φ) + 1 m φ ] [ d 3 x x φ( x, t) ] x ( iδ( x x ) + m ( o)φ( x, t)δ( x x ) = im φ + i φ( x, t) t π( x, t) = m φ + φ( x, t) π( x, t) + φ(x, t) π( x, t) = (3) φ( x.t) m φ( x, t) φ( x, t) (3) φ( x, t) + m φ( x, t) = 0 ( + m )φ = 0 H = E p a (p )a(p ) n Ha(p) = a(p)(h E p ) [H, a(p)] = E p a)p) t n i n Hn n! a(p) = n a(p) (H E p) n t n n! e iht a(p) = a(p)e i(h Ep)t e iht a(p)e iht = a(p)e iept a(p, t) = e iept a(p, t = 0) a (p, t) = e iept a (p, t = 0) 1 ( φ( x, 0) = a(p)e i px (π) 3 + a (p)e e p x) Ep 1 ( φ( x, t) = a(p)e ip x (π) 3 + a (p)e ip x) Ep p x = Et p x = p µ x µ = p 0 x 0 pcdot x 1 0 φ(x)φ(y) 0 = D(x y) = (π) 3 e ip (x y) E p a(p) 0 = 0 0 a = 0 [a(p), a (p )] = δ (3) ( p p ) δ(f(x)) = p = m p 0 p = m N i=1 1 f (x i ) δ(x x i) f(x i ) = 0 i = 1,..N δ(p 0 ( p + m ) = 1 δ(p 0 E p ) 1 δ(p 0 + E p ) E p E p D(x y) = (π) 3 πδ(p m ) p0>0e ip(x y) p o t p( x y) d 4 p (π) 4 πδ(p m )e i p +m t 4π 1 p i dp p +m t (π) 3 = 1 de(e m 0 p + m (π) m E e iet 9

1.. D(x y) = x 0 y 0 = t x y = 0 s = (x 0 y 0 ) ( x y) > 0 e i p r t >> 0 ψ ψ = e imt x y = r p + m (π) 3 = 1 (π) 1 dp (π) p eipr e ipr 0 E p ipr 0 = i (π) e ip rη p + m p dp sin θdθdη e ipr dp p p + m e ipr p + m e mr im Need [φ(x), φ(y)] to be zero. 0 [φ(x), φ(y)] 0 = 0 1 [φ(x), φ(y)] = D(x y) D(y x) = (π) 3 (e ip(x y) e ip(x y) ) E p Somethingsomethingsomething φ-complex Noether current Q = d 3 xi(φ φ φ φ) (x y) (x y)? 0 [φ(x), φ (y)] 0 [Q, φ] = φ positive energy [Q, φ ] = φ negative energy Positively charged particles with positive energy cancel negatively charged particles with negative energy. Something something something antiparticle, something something something going backwards and forwards in time something something. 0 [φ(x), φ(y)] 0 = (π) 3 ( e ip(x y) E p eip(x y) E p p(x y) = p 0 (x 0 y 0 ) + p( x y) = p(x y) p 0 : p 0 = E p = p(x y) 0 [φ(x), φ d 4 p e ip(x y) (y)] 0 = (π) 4 p = G(x y) m Θ(x 0 y 0 ) 0 [φ(x), φ(y)] 0 = G R (x y) ( + m )G R (x y) = iδ (4) (x y) e ( + m ip(x y) )G R = (π) 4 d 4 p = δ (4) (x y) Θ(y 0 x 0 ) 0 [φ(x), φ(y)] 0 = G Adv (x y) Θ(x 0 y 0 ) 0 φ(x)φ(y) 0 + Θ(y 0 x 0 ) 0 φ(y)φ(x) 0 = G F (x y) = 0 T (φ(x)φ(y)) 0 d 4 p i G F (x y) = (π) 4 p m + iɛ e ip(x y) p 0 = ± p + m ± iɛ ( + m )φ(x) = j(x) L = 1 ( µφ) 1 m φ + j(x)φ(x) ) 10

Imagine φ 0 is a solution to ( + m )φ 0 = 0 φφ 0 + G(x y)j(y)d 4 y G : ( + m )G(x y) = δ (4) (x y) d 3 p 1 (π) 3 (a(p)e ipx + a (p)e ipx 1 ) + Ep (π) 3 Θ(x 0 y 0 )(e ip(x y) e ip(x y) )j(y)d 4 y E p = H = [( ) ] 1 (π) 3 a(p) + i j(p) e ip x + c 1 c 1 Ep Ep a(p) a(p) + ( (π) 3 E p a (p) 0 h 0 = 1 0 N 0 = i Ep j(p) ) ( i j (p) a(p) + Ep (π) 3 j(p) 0 a 1 Ep jp < ) i j(p) Ep 11