Lecture 17 8 251 Spring 2007 Lecture 17 Topics Light cone elds and particles cont d Reading Sections 10 2 10 4 What are we doing now Preparing grounds to see what arises from the string How are particles described Begin with simplest particle eld the scalar eld Lagrangian density for a scalar eld x 1 1 1 2 2 2 2 L 0 M 2 2 2 The rst term represents the KE density and the second term represents the PE density Note since KE density has same units as PE density 1 1 2 2 0 2 M M 2 2 1 1 L M 2 2 2 2 S d xdtL E S Hd x 1 1 1 d x 0 2 2 M 2 2 2 2 2 d xdt M 2 d xdt M 2 2 M 2 0 2 2 M 2 0 t2 This is the equation of motion of scalard eld Next Develop notion of scalar particles How do we recognize them 1 Lecture 17 8 251 Spring 2007 Plane Waves Set scalar eld to something that could satisfy equation of motion Try p x a exp iEt i Then iE 2 i p i p M 2 0 E 2 p 2 M 2 p2 M 2 where p p p This looks sort of like a particle in quantum mechanics but a bit naive Try a exp iEt i p x a exp iEt i p x Can t anymore think of a particle with momentum p and energy E since get negative E So abandon that interpretation Quantum Field Theory The elds are dynamical variables and operations x x dP p exp ip x p 2 D dp p exp ip x p 2 D x dP p exp ip x p 2 D dp p exp ip x p 2 D 2 Lecture 17 8 251 Spring 2007 p p If know value of eld for some Ep p So geometrically the reality condition of a point Ep p in momentum space in the top hyperboloid is equal to the realty condition of the complex conjugate in the bottom hyperboloid dD p 2 M 2 exp ip x p 0 2 D dD p p2 M 2 p exp ipx 0 2 D p2 M 2 p 0 p Say p2 M 2 0 then p 0 Say p2 M 2 0 then p is arbitary This is the complete solution A little simple sounding but beautiful geometric interpretation If not on hyperboloid eld vanishes If on hyperboloid eld arbitrary subject to reality condition p determines p p 1 degree of freedom in the scalar eld 2 real numbers for two points Field Con guration 1 1 p t x a t ei p x a t e i p x v 2Ep V L1 L2 L3 Ld xi xi Li pi xi Li pi xi 2 ni pi Li 2 ni S 1 1 d xdt M 2 2 2 2 3 Lecture 17 8 251 Spring 2007 Can evaluate Can do x integral but cannot do t integral since t still arbitrary E d xH S E 1 1 dt a t a t Ep a t a t 2Ep 2 1 1 a t a t Ep a t a t 2Ep 2 1 2 a t q1 t iq2 t Thus 2 S 1 2 1 2 dt q Ep qi 2Ep i 2 i 1 This is a harmonic oscillator pi p1 ip2 S q i qi Ep 1 a t q 1 iq 2 Ep Ep Equation of motion q i Ep2 qi a t Ep2 a t a t ap e iEp t a p eiEp t No reality condition is needed E H Ep a p ap a p a p Let ap a p be destruction operations Let ap a a p p a p be creation operations ap a p 1 a p a p All other commutators 0 How do we check this is okay 4 Lecture 17 8 251 Spring 2007 qi t pj t i ij E H Ep a p ap a p a p 1 1 p t x a t ei p x a t e i p x v 2Ep 1 1 ap e iEp t ipx a p eiEp t ipx ap eiEp t i p x a p eiEp t i p x v 2Ep 1 1 iEp t i p x p t x ap e iEp t i p x a pe v 2E p p E H Ep a p ap p ap a q p q De ne a vacuum state ap 0 p E 0 Create a state a p Momentum Operator P p pa p ap Note P 0 Eq a Eq a ap q aq ap q aq a Ep q q So call ap a scalar particle of mass M momentum p and energy Ep p 2 M 2 Call a 1 particle state ap 1 ap 2 a p n n particle state of total energy Ep 1 Ep 2 Ep n and momentum p 1 p 2 p n E p1 p2 pd p p pI We have labelled the oscillators by the spatial components of the momentum which determine the energy Light cone oscillators p 2 1 pI M 2 2p 5