String Theory- Light-Cone Fields II

� � � � � � � � � � � Lecture 17 8.251 Spring 2007 Lecture 17 -Topics Light-cone fields 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/field: the scalar field. Lagrangian density for a scalar field φ(x): 1 11 L = 2(∂0φ)2 − 2(�φ)2 +2 M2φ2 The first term represents the KE density and the second term represents the PE density. Note since KE density has same units as PE density: 11(∂0φ)2 = M2φ2 [M]22 ⇒ L = − 21 ηµν ∂µφ∂ν φ − 21 M 2φ2 S = d�xdtL 1 11 E = Hd�x = d�x((∂0φ)2 +(�φ)2 + M2φ2)222 δS = d�xdt(−ηµν ∂µ(δφ)ν φ − M2φδφ) = d�xdtδφ(ηµν ∂µ∂ν φ − M2φ) (∂2 − M2)φ =0 ∂2φ − ∂t2 + �2φ − M2φ =0 This is the equation of motion of scalard field.Next: Develop notion of scalar particles. How do we recognize them?1 Lecture 17 8.251 Spring 2007 Plane Waves Set scalar field to something that could satisfy equation of motion. Try: φ = a exp(−iEt + i�p �x)· Then: −(−iE)2 +(i�p)(i�p) − M2 =0 · E2 − p�2 = M2 ⇒−p 2 = M2 ( where p = pµpµ) This looks sort of like a particle in quantum mechanics, but a bit naive. Try: φ = 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 fields are dynamical variables and operations. � dP p (2π)D exp (ip · x)φ(p)φ(x)= (φ(x))∗ = exp (ip x)(φ(p)) (φ(x))∗ = � (2dπp)p D exp (−ip · x)(φ(p))∗ = � (2dπP )p�D exp (ip · x)(φ(−p))∗ � dpp�(2π)D · 2 � � Lecture 17 8.251 Spring 2007 [φ(p)]∗ = φ(−p) If know value of field 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. (∂2 − M2) dDp (2π)D exp (ip · x)φ(p) = 0 � dDp (−p 2 − M2)φ(p)exp(ipx)=0 (2π)D (p 2 + M2)φ(p)=0 ∀p Say p2 + M2 = 0 then �φ(p)=0 Say p2 + M2 = 0 then φ(p) is arbitary. This is the complete solution. A little simple sounding, but beautiful geometric interpretation. If not on hyperboloid, field vanishes. If on hyperboloid, field arbitrary (subject to reality condition). φ(p) determines φ(−p)=(φ(p))∗ 1 degree of freedom in the scalar field. (2 real numbers for two points). Field Configuration φp(t,�x)= √1 v � 21 Ep (a(t)e i�p·�x + a∗(t)e−i�p·�x)V = L1L2L3 ...Ldii + Lix ≈ x pi(xi + Li)= pixi +2πni piLi =2πni 11 S = d�xdt(− (∂µφ)(∂µφ) − M2φ2)22 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 11 S = dt a˙∗(t)a(t) − Epa∗(t)a(t) (1)2Ep 2 11 E = a˙∗(t)˙a(t)+ Epa∗(t)a(t) (2)2Ep 2 a(t)= q1(t)+ iq2(t) Thus: 2S = �� dt � 2E1 p q˙i 2 − 21 Epqi 2 � i=1 This is a harmonic oscillator. ∂S q˙i pi == ∂qi Ep 1 a˙(t) p1 + ip2 = (˙q1 + iq˙2)= Ep Ep Equation of motion: q¨i = −Ep 2 qi a¨(t)= −E2 a(t)p a(t)= ape−iEpt + a∗ e iEpt −pNo reality condition is needed. E = H = Ep(a∗ap + a∗ a)p−p−pLet ap�, ap be destruction operations. Let a∗ a + ,a∗ a + be creation −�p�→ p�−p�→ −p�operations. [ap,a +]=1=[a−p,a + ]p −pAll other commutators = 0. How do we check this is okay? 4 � � � � �� | � Lecture 17 8.251 Spring 2007 [qi(t),pj (t)] = iδij ++E = H = Ep(ap ap + a−pa−p) 11 φp(t,�x)= √v � 2Ep (a(t)e i�p·�x + a∗(t)e−i�p·�x) =1 � 1(ape−iEpt+ipx + aiEpt+ipx + ap+e iEpt−i�p·�x + aiEpt−i�p�x)√v 2Ep −pe −pe 1 � 1 ape−iEpt+i(p��x) + a + iEpt−i(p��x)φp(t,�x)= √v � 2Ep ·p e ·p�+E = H = Epa p�ap�p�[ap�,a +q�]= δ�p,�q Define a vacuum state |Ω�: ap�|Ω� =0∀p�E |Ω� =0 Create a state a + p�|Ω� Momentum Operator: P p��p ap. P |�= pa+ Note �Ω� =0 + +++= Eq aq aqap�|Ω� = Eqaq [aq ,a ] |Ω� = Ep�(ap |Ω�) �qq So call ap�Ω� a scalar particle of mass M, momentum �p, and energy Ep�= p2 + M2 Call a 1-particle state a + p�1 ,a + p�2 ,...,a + p�n |Ω� = n − particle state of total energy Ep�1 + Ep�2 + ... + Ep�n and momentum p�1 + p�2 + ... + p�n (E,p1 ,p 2 ,...,p d)(p + ,p−,p I )↔ We have labelled the oscillators by the spatial components of the momentum which determine the energy. Light-cone oscillators: 1 I2 p− =(p + M2)2p+ 5