(a p (t)e i p x +a (t)e ip x p

Transcription

1 5/29/3 Lecture outline Reading: Zwiebach chapters and. Last time: quantize KG field, φ(t, x) = (a (t)e i x +a (t)e ip x V ). 2Ep H = ( ȧ ȧ(t)+ 2E 2 E pa a) = p > E p a a. P = a a. [a p,a k ] = δ p,k, [a p,a k ] = [a p,a k ] =. Likewise, for the Maxwell field A µ ; gauge invariance δa µ (p) = ip µ ǫ(p). In light cone gauge, since p +, can set A + (p) =. Then get A = (p I A I )/p +, i.e. A is not an independent d.o.f., but rather constrained, and the Maxewell EOM gives p 2 A µ (p) =. For p 2, require A µ (p) =, and for p 2 = get that there are D 2 physical transverse d.o.f., the A I (p). The one-photon states are D I=2 ξ I a I p +,p T ω. Gravitational light cone gauge conditions: h ++ = h + = h +I =. Other light cone components are constrained. So physical d.o.f. are specified by a traceless symmetric matrix h IJ in the D 2 transverse directions. So 2D(D 3) d.o.f.. Recalltherelativisticpointparticle, withs = Ldτ andl = m ẋ 2, where d dτ. (τ is taken to be dimensionless.) The momentum is p µ = L/ ẋ µ = mẋ µ / ẋ 2 and the EOM is ṗ µ =. In light cone gauge we take x + = p + τ/m 2. Then p + = mẋ + / ẋ 2 and the light cone gauge condition implies ẋ 2 = /m 2, so p µ = m 2 ẋ µ. Also, p 2 + m 2 = yields p = (p I p I +m 2 )/2p +, which is solved for p and then ẋ = p /m 2 is integrated to x = p τ/m 2 +x. Also, xi = x I +p I τ/m 2. The dynamical variables are (x I,x,pI,p + ). Heisenberg picture: put time dependence in the operators rather than the states, with [q(t),p(t)] = i and i d O(t) = i O dt t +[O,H]. For time independent Hamiltonian, we have ψ(t) S = e iht ψ H and O H = e iht O S e iht.

2 Quantize the point particle in light cone gauge by taking the independent operators (x I,x,pI,p + ), with [x I,p J ] = iη IJ and [x,p+ ] = iη + = i. These commutators are for either S or H picture, with the operators being functions of τ in the H picture. The remaining variables are defined by x + (τ) = p + τ/m 2, x (τ) = x + p τ/m 2, p = (p I p I +m 2 )/2p + (the first two are explicitly τ dependent even in the S picture). The Hamiltonian is p, which generates x + translations. Since τ = p+ m 2 p + m 2 p the Hamiltonian is Verify e.g. H = p+ p m 2 = 2m 2(pI p I +m 2 ). i d dτ pµ = [p µ,h] =, i dxi dτ = [xi,h] = i pi m 2, x + reproducing the correct EOM. Likewise, verify ẋ = and ẋ + = τ x + = p + /m 2. The momentum eigenstates are labeled by p +,p I and these are also energy eigenstates, H p +,p I = 2m 2 (p I p I +m 2 ) p +,p I. Connect the quantized point particle with the excitations of scalar field theory via p +,p I a p +,p I Ω. The S.E. of the quantum point particle wavefunction maps to the classical scalar field equations, e.g. in light cone gauge: (i τ 2m 2(pI p I +m 2 ))φ(τ,p +,p I ) = is either the quantum S.E. of the point particle or the classical field equations of a scalar field. (Aside: the light cone is here used as a trick to get to second quantization. First quantization is what you learn the first time you study (non-relativisitic) QM: replace coordinates and momenta with operators, and Poisson brackets with commutators. Second quantization is for field theory, replacing the fields and their conjugate momenta with operators, and their PBs with commutators, leading to multi-particle states. Here lightcone first quantization of the point particle leads to a Schrodinger equation that agrees with the classical EOM of a light-cone field theory, which we then need to quantize again to get second quantization. 2

3 Lorentz transformations correspond to the inf. transformations δx µ = ǫ µν x ν, and the corresponding conserved Noether charges are the generalized angular momenta M µν = x µ p ν x ν p µ. These generate rotations and boosts. Have e.g. [M µν,x ρ ] = iη µρ x ν iη νρ x µ and [M µν,m ρσ ] = iη µρ M νσ ±(perms). In light cone coordinates, have M IJ, M ±I, M +, with e.g. [M +,M +I ] = im +I and [M I,M J ] =. There are some ordering issues for some of these, where operators which don t commute need to be replaced with symmetrized averages, e.g. M + = 2 (x p+ + p + x ) and M I = x pi 2 (xi p +p x I ). Open string. Imposed constraints (Ẋ ±X ) 2 = to get P σµ = 2πα Xµ, P τµ = 2πα Ẋµ. In light cone gauge, much as with the point particle, the independent variables are (X I,(σ)x,PτI (σ),p + ). In the H picture the capitalized ones depend (implicitly) on τ too. The commutation relations are [X I (σ),p τj (σ )] = iη IJ δ(σ σ ), [x,p+ ] = i. The Hamiltonian is taken to be π π H = 2α p + p = 2α p + dσp τ = πα dσ(p τi P τi +X I X I (2πα ) 2 ) Can write H = L since L = 2α p + p. This H properly yields the expected time derivatives, e.g. Ẋ I = 2πα P τi. Recall the solution with N BCs: X I (τ,σ) = x I + 2α α I τ +i 2α n The needed commutators are ensured by n αi n cosnσe inτ. () [α I m,αj n ] = mηij δ n+m,. Also, as before, we define α I 2α p I. Now define α µ n> = na µ n and αµ n = aµ n n to rewrite the above as [a I m,aj ] = δ m,n η IJ. (2) 3

4 The transverse light cone coordinates can be described by S l.c. = dτdσ Ẋ I X I X I ). 4πα (ẊI Gives correct P τi = L/ ẊI and correct H = dσ(p τi Ẋ I L). WritingX I (τ,σ) = q I (τ)+2 α qi N (τ)n /2 cosnσ andplugginginto theaction above gives S = dτ[ 4α qi q I + ( 2n qi n qi n n 2 qi n qi n )] and H = α p I p I + n 2 (pi np I n +q I nq I m). A bunch of harmonic oscillators. Relate to () and (2), showing that the a m can be interpreted as the usual harmonic oscillator annihilation operators. Summary: we fix X + to be simply related to τ, find that the X I are given by simple harmonic oscillators, and X is a complicated expression, fully determined in terms of the transverse direction quantities: X + (τσ) = 2α p + τ = 2α α + τ. For X recall expansion, with 2α α n = p L + n, where L n 2 p αi n p αi p is the transverse Virasoro operator. Recall [αi m,αj n ] = mδ IJ δ m+n,. There is an ordering ambiguity here, only for L : L = 2 α α + 2 α I p αi p + 2 p= α I p αi p. The ordering in the last terms need to be fixed, so the annihilation operator α p is on the right, using α I p αi p = αi p αi p +[αi p,αi p ], which gives L = α p I p I + where the normal ordering constant has been put into p= na I n ai n, 2α p = p +(L +a), a = 2 (D 2) p. p= This leads to M 2 = α (a+ na I n a I n). 4

5 The divergent sum for a is regulatedby using ζ(s) = n s and analyticallycontinuing to get ζ( ) = /2. So a = 24 (D 2). Virasoro generators and algebra (corresponds to worldsheet energy-momentum tensor). Since (α I n ) = α I n, get L n = L n. Also, [L m,αi n ] = nαi n+m. [L m,l n ] = (m n)l n+m + D 2 2 (m3 m)δ m+n,. Spacetime Lorentz symmetry corresponds to conserved currents on worldsheet, with conserved charges M µν = π (X µ P τ ν (µ ν))dσ. Plug in P τ ν = 2α Ẋ µ and plug in oscillator expansion of X µ to get M µν = x µ pν x ν p µ i n (αµ nα ν n α ν nα µ n). In light cone gauge, have to be careful with M I, since X is constrained to something complicated, X (τ,σ) = x + 2α α τ +i 2α n n α n e inτ cosnσ, 2α α n = p +L n. and also careful to ensure that [M I,M J ] =. Find, after appropriately ordering terms, M I = x pi ( x I 4α p + (L ) i +a)+(l +a)xi 2α p + n (L n αi n αi n L n ). Then get [M I,M J ] = α p +2 (α I m αj m (I J))[m( ((D 2)/24))+m (((D 2)/24)+a)]. m= Since this must be zero, get D = 26 and a =. The worldsheet Hamiltonian is thus H = 2α p + p = L. 5