Solutions for String Theory 101

Solutions for String Theory 0 Lectures at the International School of Strings and Fundamental Physics Munich July 6 - August 6 00 Email: Neil Lambert Theory Division CERN Geneva 3 Switzerland neil.lambert@cern.ch

Problem: Show that if is satisfied then so is m d dτ ( v i ) = qf v i0 + qf ij v j (0.) m d dτ ( ) = qf v 0i v i (0.) Solution: We simply multiply (0.) by v i and use the anti-symmetry of F ij to deduce m d dτ ( v i ) v i = qf v 0i v i (0.3) Now the left hand side is m d ( v i ) v i = mv d dτ v dτ ( ) m dvi vi v v dτ = mv d ( ) m dv dτ v v dτ = mv d ( ) m( v ) d ( dτ v dτ = m d ( ) dτ v v ) (0.4) This agrees with the left hand side of (0.) and since the right hand sides already agree we are done. Problem: Show that, in static gauge X 0 = τ, the Hamiltonian for a charged particle is H = m + (p i qa i )(p i qa i ) qa 0 (0.5) Solution: In static gauge the Lagrangian is so the momentum conjugate to X i is L = m ẊiẊi + qa 0 + qa i Ẋ i (0.6) p i = L Ẋi = m Ẋi v + qa i (0.7)

Inverting this gives Ẋ i v = (pi qa i )/m (0.8) We square to find v v and hence v = (p qa) /m v = Ẋ i = (p qa) m + (p qa) (0.9) p i qa i m + (p qa) (0.0) Finally we calculate H = p i Ẋ i L = (p i qa i )p i m + (p qa) + m m + (p qa) qa (p i qa i )A i 0 q m + (p qa) = m + (p qa) qa 0 (0.) Problem: Find the Schödinger equation, contraint and effective action for a quantized particle in the backgroud of a classical electromagnetic field using the action S pp = ( e ) e Ẋµ Ẋ ν η µν + m A µ Ẋ µ (0.) Solution: Proceeding as before we first calculate p µ = L Ẋµ = eẋν η µν + A µ Inverting this gives (0.3) Ẋ µ = eη µν (p ν A ν ) (0.4) Thus the main effect is merely to shift p µ p µ A µ. The constraint is unchanged as the new term is independent of e: however in terms of the momentum it becomes e Ẋµ Ẋ ν η µν + m = 0 (0.5) (p µ A µ )(p ν A ν )η µν + m = 0 (0.6) 3

In the calculation of the Hamiltonian we have two effects. The first is that we which find from the replacement p µ p µ A µ in the old Hamiltonian. The second is the addition of the A µ Ẋ µ term which leads to an addition term A µ Ẋ µ = eη µν (p ν A ν )A µ (0.7) The factors of A µ from these two effects combine and we find H = e ( η µν (p µ A µ )(p ν A ν ) + m ) (0.8) Next consider the quantum theory where we consider wavefunctions Ψ(X µ, tau) and promote ˆp µ Ψ = i Ψ X, µ ˆXµ Ψ = X µ Ψ (0.9) Thus the Schrodinger equation is i Ψ τ = e ( ( η µν and the constraint is ( η µν ( X ia µ µ X ia µ µ ) ( ) ) X ia ν ν + m Ψ (0.0) ) ( ) ) X ia ν ν + m Ψ = 0 (0.) Thus we again find that Ψ is independent of τ. The effective action is just found by replacing µ ip µ + A µ and hence we have S = d D x( µ Ψ ia µ Ψ)( ν Ψ ia ν Ψ)η µν + m Ψ (0.) You should recognize this as a Klein-Gordon scalar field coupled to a background Electromagnetic field. Problem: Show that by solving the equation of motion for the metric γ β on a d- dimensional worldsheet the action S HT = d d σ det(γ) ( γ β X µ β X ν η µν m (d ) ) (0.3) one finds the action S NG = m d d d σ det ( X µ β X ν η µν ) (0.4) for the remaining fields X µ, i.e. calculate and solve the γ β equation of motion and then substitute the solution back into S HT to obtain S NG. Note that the action S HT is often referred to as the Howe-Tucker form for the action whereas S NG is the Nambu-Goto form. (Hint: You will need to use the fact that δ det(γ)/δγ β = γ β det(γ) 4

Solution: From S HT we calculate the γ β equation of motion X µ β X ν η µν + 4 γ ( β γ γδ γ X µ δ X ν η µν m (d ) ) = 0 (0.5) This implies that γ β = b X µ β X ν η µν (0.6) for some b. To determine b we substitute back into the equation of motion to find + b 4 (d/b m (d )) = 0 (0.7) where we have used the fact that if g β = X µ β X ν η µν then γ β g β = d/b (0.8) This tells us that b = m. Substituting back into S HT gives S HT = m d d d σ det g ( dm m (d ) ) = m d d d σ det g (0.9) where again g β = X µ β X ν η µν. This is precisely S NG. Problem: What transformation law must γ β have to ensure that S HT is reparameterization invariant? (Hint: Use the fact that why?) σ γ σ σ β σ γ = δβ (0.30) Solution: Under a reparameterization σ = σ (σ ) we have that X µ σ Since the m term is invariant it must be that = Xµ σ β σ β σ (0.3) γ β X µ β X ν η µν (0.3) is invariant in order for the expression to make sense. Thus we are lead to postulate that γ β = σγ σ δ σ σ γ γδ γ β = σ σ β σ γ δ γ γδ (0.33) 5

since σ γ σ β σ σ = γ δβ It remains to check that and σ γ σ σ β σ γ = δβ (0.34) d d σ det(γ) (0.35) is invariant. However this follows from the above formula and the Jacbobian transformation rule for integration ( ) σ d d σ = det d d σ (0.36) σ β Problem: Show that if x µ, p µ 0 then we also have with the other commutators vanishing. Solution: Recall that we have and we require ˆX µ = x µ + p µ τ + ˆP µ = p µ π [x µ, p ν ] = iη µν (0.37) ( a µ ) n n ein(τ+σ) + ãµ n n ein(τ σ) i n 0 a µ ne in(τ+σ) + ã µ n 0 ne in(τ σ) (0.38) n 0 [ ˆX µ (τ, σ), ˆP ν (τ, σ )] = iδ(σ σ )δ µ ν (0.39) In the lectures we considered terms that come from two oscillators, i.e. terms with a factor of e inτ+imτ. It should be clear that any term in the commutator with a single exponential must also vanish. Thus we see that the commutator of x µ, w µ, p µ with any oscillators a µ n, ã µ n must vanish. Thus the remaining terms are π [xµ + w µ σ, p ν ] = i π δµ ν (0.40) where on the right hand side we have included a left over term from the calculation of the oscilators (i.e. the n = 0 term from the Fourier decomposition of δ(σ σ )). Okay, it is clear that the term linear in σ on the left hand side must vanish. Thus we find Problem: Show that in these coordinates [x µ, p ν ] = iδ µ ν, [w µ, p ν ] = 0 (0.4) ˆT ++ = + ˆXµ + ˆXν η µν ˆT = ˆXµ ˆXν η µν ˆT + = T + = 0 (0.4) 6

ˆT = ˆXµ ˆXν η µν (0.45) Solution: We have The new coordinates are ˆT β = X µ β X ν η µν η βη γδ γ X µ δ X ν η µν (0.43) σ + = τ + σ σ = τ σ σ++σ τ = σ = σ+ σ (0.44) and hence it follows that ds = dτ + dσ = dσ + dσ. From this we read off that η ++ = η = 0 and η + = η + =. Hence η++ = η = 0 and η + = η + =. Thus we see that ˆT ++ = + ˆXµ + ˆXν η µν For the ˆT + components we note that η γδ γ X µ δ X ν η µν = 4 X µ + X ν η µν (0.46) so that ˆT + = X µ + X ν η µν 4 X µ + X ν η µν = 0 (0.47) Problem: Show that < 0, 0; 0 : L :: L : 0; 0, 0 >= D (0.48) Solution: We have so > = L 0; 0, 0 > = η µν µ nn 0; ν 0, 0 > n = η µν µ ν 0; 0, 0 > (0.49) < > = 4 η µνη λρ < 0, 0 : 0 λ ρ µ 0; ν 0, 0 > = 4 η µνη λρ < 0, 0 : 0 λ µ ρ 0; ν 0, 0 > + 4 η µνη λρ η ρµ < 0, 0 : 0 λ 0; ν 0, 0 > = 4 η µνη λρ η ρν < 0, 0 : 0 λ 0; µ 0, 0 > + 4 η µνη λρ η ρµ η λν < 0, 0 : 0 0; 0, 0 > 7

= 4 η µνη λρ η ρν η λµ < 0, 0 : 0 0; 0, 0 > + 4 η µνη λρ η ρµ η λν < 0, 0 : 0 0; 0, 0 > = η µνη λρ η ρν η λµ = D (0.50) Problem: Show that the state (a 0 + a ) 0 > has zero norm. Solution: < 0 (a 0 + a )(a 0 + a ) 0 > = < 0 a 0 a 0 + a a 0 > = η 00 + η = 0 (0.5) Problem: Show that the boundary conditions on an open string are at σ = 0, π. η µν δx µ σ X ν = 0 (0.5) Solution: In calculating the Euler-Lagrange equations for the action one integrates by parts: d ση µν η β X µ β δx ν = d ση µν η β β X µ δx ν + d ση µν η β β ( X µ δx ν ) (0.53) Thus we need to discard the second term d σ β (η µν η β X µ δx ν ) = η µν σ X µ δx ν (0.54) where we have used the fact that the normal to the boundary is σ = σ. Locally implies that this term should vanish at each end point separately. Problem: Show that the constraints imply that p µ G µν = p ν G µν = 0 for the level one closed string states G µν >= G µν µ ν 0; p > Solution: Consider L first. We find L G µν µ 0; ν p > = η λρ n λ n ρ µ 0; ν p > n (0.55) 8

Now if n > then ρ n can be commuted through until is annhilates 0; p >. Similarly if n < 0 λ n can be commuted through until is annhilates 0; p >. Thus we have L G µν µ ν 0; p > = η λρ( λ 0 ρ + λ ρ 0)G µν µ ν G µν 0; p > = η λρ 0 λ ρ µ G ν µν 0; p > = η λρ 0 λ ν ρ G µ µν 0; p > = η λρ 0 λ [ ν, ρ ]G µ µν 0; p > = η λρ 0 λ η ν ρµ G µν 0; p > = pµ G ν µν 0; p > (0.56) Since this must vanish we find p µ G µν = 0. Similarly evaluating L G µν µ ν 0; p >= 0 will lead to p ν G µν = 0. Problem: Show that g µν = G (µν) D ηλρ G λρ η µν b µν = G [µν] φ = η λρ G λρ (0.57) will transform into themselves under spacetime Lorentz transformations. Solution: Let us adopt a matrix notation. Under a Lorentz transformation a tensor G transforms as G = ΛGΛ T (0.58) and Lorentz transformations satisfy η = ΛηΛ T. Now b = (G G T ) = (ΛGΛT (ΛGΛ T ) T ) = (ΛGΛT ΛG T Λ T ) = ΛbΛ T (0.59) so indeed b transforms into itself. It also follows that the symmetric part of G transforms into itself so we need only show that φ = Tr(η G) is invariant. To do this we note that η = (Λ ) T η Λ (0.60) 9

so that φ = Tr(η G ) = Tr(η ΛGΛ T ) = Tr((Λ ) T η Λ ΛGΛ T ) = Tr((Λ ) T η GΛ T ) = Tr(η G) = φ (0.6) Problem: Show that in light cone gauge ( X = x + p τ + i ) n n n e inσ+ + n n e inσ where and similarly for n. Solution: We need to solve (0.6) n = p n m i mδ j + ij (0.63) m p + Ẋ + Ẋi Ẋ j δ ij + X i X j δ ij = 0 p + X + Ẋi X j δ ij = 0 (0.64) We have that where i 0 = i 0 = Ẋ i = X i = /p i. Thus ( i n e inσ + + ) ne i inσ n ( i n e inσ + ) ne i inσ n (0.65) Ẋ i X j δ ij = δ ij From the second equation we find that X = F (τ) + i 4p + δ ij m+n 0 n i me j i(n+m)σ + n i me j i(n+m)σ (0.66) nm i n j m n + m e i(n+m)σ + + i n j m n + m e i(n+m)σ (0.67) 0

where F (τ) is an integration constant. Let us now consider the first equation so we calculate Ẋ i Ẋ j δ ij = δ ij X i X j δ ij = δ ij and hence n i me j i(n+m)σ + + n i me j i(n+m)σ + (n i m j + n i m)e j inσ + imσ nm n i me j i(n+m)σ + + n i me j i(n+m)σ (n i m j + n i m)e j inσ + imσ nm (0.68) Ẋ i Ẋ j δ ij + X i X j δ ij = δ ij n i me j i(n+m)σ + + n i me j i(n+m)σ (0.69) Substituting our solution into the first equation leads to This implies that 0 = p + F δ ij + δ ij nm n i me j i(n+m)σ + + n i me j i(n+m)σ n+m=0 n i me j i(n+m)σ + + n i me j i(n+m)σ (0.70) nm p + F = δ ij Thus F = p τ + x is linear with p i p j + p i p j (0.7) p p + p + δ ij p i p j + p i p j = 0 (0.7) Seperating out the zero-mode piece we can rewrite this as where p 4 p + p + p i p j δ ij + (N + Ñ = 0 (0.73) N + Ñ = δ ij n i n j + n i n j (0.74) can be identified with the total oscillator number of the transverse coordinates. Lastly we summarise our expression by writing n 0 X = x + p τ + i n 0 n n e inσ + + n n e inσ (0.75) where n = 4p n m i mδ j + ij n = i 4p + n m mδ j ij (0.76) m m

Problem: Show that for a periodic Fermion, where L 0 = l ld l d l + 4 and {d n, d m } = δ n, m, one has Z = q 4 ( + q l ) (0.77) l= and for an anti-periodic Fermion, where L 0 = r rb r b r 48 r, b s } = δ r, s and r, s Z +, one has Z = q 48 ( + q l ) (0.78) l= Solution: Everything follows as it did for the Boson. However there are only two possible cases for each oscillator d l, either it isn t present or it is present once. In other words because of the anti-commutivity there are just two states 0 > and d n 0 > so one has q ld ld l = + q l and hence Z = q 4 ( + q l ) (0.79) l= Similarly for the anti-periodic Fermion only now we simply write r = l with l =,, 3,... to find Z = q 48 ( + q l ) (0.80) l= Problem: Obtain the equations of motion of S effective = and show that they agree with d 6 x ge φ ( R 4( φ) + H µνλh µνλ ) (0.8) R µν = 4 H µλρh λρ ν D λ H λµν = D λ φh λµν + D µ D ν φ 4D φ 4(Dφ) = R + H (0.8) You may need to recall that δ g = ggµν δg µν and g µν δr µν = D µ D ν δg µν g µν D δg µν. Solution: The equations of motion for φ follows pretty much as normal and one finds 8D µ (e φ D µ φ) e φ ( R 4 µ φ µ φ + H µνλh µνλ ) = 0 (0.83)

or 8D φ 8D µ φd µ φ R 6 H µνλh µνλ = 0 (0.84) The equation for b µν is also fairly standard and leads to D µ (e φ D [µ b νλ] ) = 0 (0.85) or D µ H µνλ D µ φh µνλ = 0 (0.86) The important point here is that when we vary the metric we find a term like ge φ g µν δr µν (0.87) appearing. We have that g µν δr µν = D µ D ν δg µν g µν D δg µν (0.88) is a total derivative. But now this won t be the case. Integrating the above term by parts gives ge ge φ g µν φ δr µν = ( D µ D ν δg µν g µν D δg µν) = ge φ ( 4D ν φd µ φ 4D λ φd λ φg µν D µ D ν φ + D φg µν ) δg µν (0.89) Thus one finds, after including all the usual terms, 0 = R µν g µνr + 4D µ φd ν φ 4D λ φd λ φg µν D µ D ν φ + D φg µν 4D µ φd ν φ + D λ φd λ φg µν + 4 H µλρh λρ ν = R µν g µνr D λ φd λ φg µν D µ D ν φ + D φg µν 4 g µνh λρσ H λρσ + 4 H µλρhν λρ 4 g µνh λρσ H λρσ (0.90) Next we substitute in the scalar equation, writen as R = 4D φ 4D µ φd µ φ H µνλh µνλ (0.9) and find that 0 = R µν D µ D ν φ + 4 H µλρh λρ ν (0.9) 3

Problem: Show that S = 4π d σ X µ β X ν η µν η β + i ψ µ γ ψ ν η µν (0.93) is invariant under δx µ = i ɛψ µ, δψ µ = γ X µ ɛ (0.94) for any constant ɛ. Here ψ = ψ T γ 0 and γ are real matrices that satisfy {γ, γ β } = η β. A convenient choice is γ 0 = iσ and γ = σ. Solution: First we note that δs = d σ 4π X µ β δx ν η µν η β + iδ ψ µ γ ψ ν η µν + i ψ µ γ δψ ν η µν (0.95) Looking at the final term we can write it as i ψ µ γ δψ ν η µν = (i ψ µ γ δψ ν η µν ) i ψµ γ δψ ν η µν i ψµ γ δψ ν η µν (0.96) where we dropped a total derivative. Next we note that (using a, b for spinor indices) ψµ γ δψ ν η µν = ψ a µ (γ 0 γ ) ab δψb ν η µν = δψb ν (γ 0 γ ) ab ψ a µ η µν = δ ψ µ γ ψ ν η µν (0.97) In the second line we used the fact that spinors are anti-commuting and in the third line we used that γ 0 γ is a symmetric matrix (convince yourself of this!) and swapped µ ν. Thus putting these together we find δs = π d σ X µ β δx ν η µν η β + iδ ψ µ γ ψ ν η µν (0.98) To continue we observe that δ ψ µ = (γ X µ ɛ) T γ 0 = ɛ T (γ ) T γ 0 X µ = ɛ T γ 0 γ γ 0 γ 0 X µ = ɛγ X µ (0.99) 4

where in the third line we have used (γ ) T = γ 0 γ γ 0 (convince yourself of this too!). We now have δs = d σi π X µ β ɛψ ν η µν η β i ɛγ β β X µ γ ψ ν η µν = d σi π X µ β ɛψ ν η µν η β i ɛ β X µ (η β + γ β ) ψ ν η µν = d σ i ɛ π β X µ γ β ψ ν η µν = ) d σ i π ( ɛ β X µ γ β ψ ν η µν 0 (0.00) Problem: Show that S = 4π is invariant under d σ X µ β X ν η µν + i ψ µ γ ψ ν η µν + i λ A +γ λ B +δ AB (0.0) δx µ = i ɛ + ψ µ δψ µ = γ X µ ɛ + δλ A + = 0 (0.0) provided that γ 0 ɛ + = ɛ +. Solution: This calculation is an exact copy of the previous problem. The important point is that the ɛ + generator does not involve the wrong chiral component ψ µ + of ψ µ : ( γ 0)δψ µ = ( γ 0)γ X µ ɛ + = γ X µ ( + γ 0 )ɛ + = γ X µ ɛ + = δψ µ (0.03) Problem: Show that the action S = d σ 4π X µ β X ν η µν + i ψ γ µ ψ η ν µν + i λ A +γ λ B +δ AB (0.04) can be written as S = d σ 4π X µ β X ν η µν +i(ψ ) µ T ( τ σ )ψ η ν µν +i(λ A +) T ( τ + σ )λ B +δ AB (0.05) So that ψ µ and λ A + are indeed left and right-moving respectively. 5

Solution: Simply write and ψ µ γ ψ ν η µν = (ψ µ ) T γ 0 (γ 0 τ + γ σ )ψ ν η µν = (ψ ) µ T ( τ + γ 0 σ )ψ η ν µν = (ψ ) µ T ( τ σ )ψ η ν µν (0.06) λ A +γ λ B +δ AB = (λ A +) T γ 0 (γ 0 τ + γ σ )λ B +δ AB = (λ A +) T ( τ + γ 0 σ )λ B +δ AB = (λ A +) T ( τ + σ )λ B +δ AB (0.07) 6