009 04 8 Lecturer: Bengt E W Nilsson Chapter 3: The closed quantised bosonic string. Generalised τ,σ gauges: n µ. For example n µ =,, 0,, 0).. X ±X ) =0. n x = α n p)τ n p)σ =π 0σ n P τ τ,σ )dσ σ 0, π] P τµ = πα X µ, P σµ = πα X µ Ẍ µ X µ = 0 3. Closed string periodicity X µ τ,σ) =X µ τ,σ + π). The Klein Gordon equation implies X µ τ,σ)=x L µ τ +σ)+x R µ x σ) We define u 4 τ + σ and v 4 τ σ. You can rewrite Ẍ µ X µ = 0 as u v X µ u, v) =0. So: Then u, v : we find X µ L u)= α X L u + π)+x R v π)=x L u) +X R u) n Z XL u) is π periodic X R v) is π periodic ᾱ µ n e inτ+σ), X R v) = α n Z α µ inτ σ) n e where α n µ and ᾱ n µ are independent Fourier coefficients these were identified in the open case, due to boundary conditions). Integrate: X µ L u) = X µ 0,L + α ᾱ µ 0 u + oscillators with a bar ᾱ) n 0 X µ R v) = X µ 0,R + α α µ 0 v + oscillators n0 But then the periodicity condition Xτ,σ + π) = Xτ,σ)) gives us α µ 0 = ᾱ µ 0. The momenta are the same in X L and X R. So, only one momentum only one centre of mass coordinate µ =X µ 0. X µ 0,L =X 0,R
So, closed string expansion in Minkowski space: X µ τ,σ)=x µ 0 + α α α0 τ + i n 0 ᾱ n µ n e inτ+σ) + α µ inτ n e σ)) Note: If spacetime has a compactified direction then α 0 and ᾱ 0 will not be equal for that component of X µ. Note: If on an orbifold we can even a different type of expansion. So, in the light-cone we have the following independent and canonical commutation relations: x0, p +] = i x0 I, p J ] = i δ IJ αm I, α n J ] = m δ IJ δ m+n,0 ᾱm I, ᾱ n J ] = m δ IJ δ m+n,0 all others = 0. 3.: The closed string Virasoro operators Constraint Define X ± X = X ) I + X I =4α n Z I ᾱ n p X ) I α p + ± X I ) I ᾱ p e inτ+σ) 4α n Z L n e inτ+σ) The same for X I X I ) in terms of Ln. So this means that L 0 = L 0 Level matching condition) since α 0 L 0, ᾱ 0 L 0 and α 0 = ᾱ 0! Now L 0 = α 4 pi p I + N L 0 = α 4 pi p I + N N = N Also M = p = p + p p I p I M = α N +N ) 3.3 Mass spectrum
) p +, p I no oscillators N = N = 0 and M = 4 α tachyonic! I ) α J p +, p I : N = N = and M = 0 massless states. Can also be created in field theory ᾱ by a p + p I if I = J. IJ 0 So the massless fields are metric if IJ is symmetric, Kalb Ramond field if IJ are anti-symmetric, Dilaton g µν metric Einstein B µν Kalb Ramond Kalb Ramond field theory φ dilaton dil. + couplings including the tachyon). 3) infinite set of massive fields. 3.4: String coupling and the dilaton Recall: The metric has a background value. g µν = η µν +h µν where h µν is a perturbation of the background η µν. Also written: g µν = g µν +h µν S κ d 4 x R g µν = η µν + κh µν κ is a coupling constant. d 4 x h h + κh h + ) This can also happen for a scalar: the dilaton: φx)= φ + ϕx) In string theory we put g s = e φ g s is the string coupling constant. Recall the covariant formalism S dτdσ γγ αβ α x µ β x ν η µν Induced metric g αβ = α x µ β x ν η µν ). 3
Partition function path integral): Z Dγ, x)e Sg γ,x] dτdσφx)r ) τ,σ) γ Inserting φx) = φ + ϕx): The φ dependence becomes Dγ, x)e φ dτdσr ) γ) χ = e φ χ Dγ, x) e χ: Euler number. e φ χ = g s ) χ χ = g =b 0 b +b = The g is called genus.) Figure. b 0 : number of corners. b : number of edges. b : number of sides. The sphere is topologically a cube. If we compute the g = 0 χ =) case the S ) we get L g s ) α ) 4 R = R G Newton D=0 G Newton = g s ) α ) 4 3.5: Orbifolds 4
Figure. A field theory cannot live on singular spaces, but string theory can. Consider a target space with a Z identification. x µ : x + x + x x x I x i x i, I =,, 4 x x, I = 5 Figure 3. 5
Figure 4. The spectrum of the closed string on this orbifold has two sectors: ) Untwisted. Spectrum = all states in the ordinary closed string invariant under x x. ) Twisted: A set of new states connected to the singularity the fixpoint x=0). ) Mode expansions: But we need an operator U µ 5 x as x before as before U X U = X U X µ 5 U = + X µ=5 UX 0 U = X 0, U p U = p, U α n U = α n, U ᾱ n U = ᾱ n U p +, p i, p =0 =+ p +, p i, 0, U p +, p i, p = p +, p i, p p =e ixˆ0p 0 pˆ p = pˆe ixˆ0p 0 = pˆ, e ixˆ0p] 0 =pˆ,?]e ixˆ0p 0 = p p States invariant under x x, i.e. U ) p +, p i, p + p +, p i, p i j ) α ᾱ p +, p i, p + p +, p i, p ) i α ᾱ p +, p i, p p +, p i, p ) 3.6 Twisted sector U p =U e ixˆ0p U U 0 = p =e ixˆ0 p = 0 6
Mode expansion µ 5 x as x before new! xτ,σ + π)= xτ,σ) Figure 5. X L u)=x L + i α r=half integer r ᾱ r e iru, X R v)= No momenta no mode with index = 0). No centre of mass coordinate either. X L + X R = 0) The twisted sector is fixed to x =0. 7