1 Covariant quantization of the Bosonic string

Transcription

1 Covariant quantization of the Bosonic string The solution of the classical string equations of motion for the open string is X µ (σ) = x µ + α p µ σ 0 + i α n 0 where (α µ n) = α µ n.and the non-vanishing commutators are n α ne inσ0 cos nσ () [α µ m, α ν n] = mδ m+n,0 η µν, [x µ, p ν ] = iη µν () Unlike the light cone gauge, where we solved the Virasoro constraints before quantizing, here we shall quantize first by finding a representation of the entire commutator algebra () and then we shall impose the constraints as physical state conditions. To construct a representation of (), we begin with an oscillator vacuum which is annihilated by all of the positively moded oscillators, α µ n 0 >= 0, n > 0, µ = 0,..., D (3) In addition, we consider an eigenstate k > of the energy-momentum operators p µ with eigenvalues k µ p µ k >= k µ k > Then, we consider the direct product state 0, k >= 0 > k > A basis for the Fock space of quantum states is obtained by operating on this state with negatively moded oscillators, D µ=0 (α µ ) nµ n µ! D µ=0 (α µ ) nµ n µ n µ!... 0, k > This construction is manifestly covariant in that it treats all components of space-time vectors on an equal footing. The Lorentz transformations and spatial rotations are generated by the operators M µν = x µ p ν x ν p µ n n αµ nα ν n

2 which automatically have the correct commutation relations. However, this construction results in a Hilbert space with an indefinite metric. For example, the commutator algebra implies that the state α 0 0, k > has negative norm < 0, k α 0 α 0 0, k >= δ(k k ) This may not be a fatal flaw. Remember that we must still satisfy the Virasoro constraints. This means that, from the vector space of all states, we should identify a subspace of those states which satisfy the constraints as physical states. It could be that the vector space of physical states has a positive Hilbert space metric. The constraints are L n 0 where They obey the Virasoro algebra L n = α p α n p (4) p [L m, L n ] = (m n)l m+n + D (m3 m)δ m+n,0 (5) The last term on the right-hand side is an anomaly term which arises only in the quantum theory. If it were absent, the algebra would be that of the subgroup of the conformal group of the complex plane which preserves the open string boundary conditions, which is equivalent to the conformal group of the upper half of the complex plane. The last term is proportional to the identity. It commutes with the rest of the generators L n and is called a central extension. To show that it indeed appears in this algebra, it is easiest to take a matrix element of commutator where you expect it to appear in a state where it is expected to be non-zero, < 0 [L m, L n ] 0 >=< 0 L m L m 0 >, m > 0 In the equality above, we have taken into account the fact that L m 0 >= 0 since, when m > 0, all of the terms in summation (4) which defines L m contain at least one positively moded oscillator. The remaining term is L m 0 >, the norm of the state L m 0 >. If the Fock space metric were positive, we would immediately conclude that this quantity is non-zero, as it could only be zero if the operator itself annihilated 0 >. However,

3 the metric is not positive, so it is safest to simply compute the commutator. Then, L m 0 >= 0 α p α m p 0 > p= m The summation is over the range of p where both oscillators are negatively moded. These are the only terms in L m which do not annihilate 0 >. Further, L m 0 > = 4 = D m q= m 0 q=0 p= m < 0 α q α m q α p α m p 0 > q(m q) + mα p = D (m3 m) + mα p (6) The last term contributes to the term ml 0 which is expected to appear and the first term is the anomaly term. It is a simple manifestation of the fact that algebras of Poisson brackets do not always survive quantization. Note that a subalgebra of the conformal algebra, generated by {L, L 0, L } indeed does survive. They form the Lie algebra of the noncompact Lie group SL(, R). The SL(, R) symmetry will eventually be important to us. The central extension of the Virasoro algebra prevents us from requiring that all generators L n annihilate physical states, that L n phys >= 0 would be incompatible with the Virasoro algebra. We therefore impose a weaker condition L n phys >= 0, n > 0 (7) (L 0 a)phys >= 0 (8) phys > phys > if phys > phys > = null > (9) The first of the physical state conditions (7) and (8) imply that the expectation value of all of the Virasoro constraints in physical states vanish. The quantity a parameterizes the ambiguity in ordering the operators in L 0. The third condition (9) is an equivalence relation which says that two physical states are equivalent, that is describe one and the same state when they differ by a null state. A null state is a state which obeys (7) and (8) and which 3

4 has zero norm, L n null >= 0, n > 0 (0) (L 0 a)null >= 0 () < null null >= 0 () A null state itself is in the same equivalence class as the state with no string at all. Let us consider a few examples of physical states. The full space of states can be block-diagonalized by diagonalizing the level number operator N = n>0 α n α n the spectrum of which are the non-negative integers. The state with lowest level number zero is 0, k >. Since in the summation that defines L n with any positive n, all terms have at least one positively moded oscillator, L n 0, k >= 0, n > 0 Then, if k is adjusted so that (8) is obeyed, the mass of this physical state is m = k = a/α The norm of this physical state is positive, < 0.k 0, k >= δ(k k ) Now, let us consider the states at level one, ξ µ (k)α µ 0, k > Requiring that they are annihilated by L n with positive n (it is automatic for n > so there is only one condition, coming from n = ) yields the constraint ξ µ k µ = 0 These states have norm < 0, k ξ µ(k)α µ ξ µ (k)α µ 0, k >= ( ξ i ξ 0 )δ(k k ) 4

5 There are three possibilities. If k µ is spacelike, there is a reference frame where k µ = (0,..., 0, k). In that frame, the constraint requires that ξ D (k) = 0. Negative norm states (for example those with ξ i = 0, ξ 0 0) are not eliminated by the constraint. If k µ is spacelike, there is a reference frame where k µ = (m, 0,..., 0). In that frame, the constraint requires that ξ 0 (k) = 0. All of the remaining states have positive norm. There are no null states. This is a completely consistent scenario. Since k = ( a)/α, it requires that a <. Most interesting is the case where k µ is null, that is, when a =. There is a reference frame where k µ = (k, 0,..., 0, k). In that frame, the constraint requires that ξ 0 (k) = ξ D (k). Among the remaining states, there is a set which have positive norm, those states where only ξ i, with i =,..., D are nonzero. There are also null states, those where ξ i = 0 and ξ 0 (k) = ξ D (k). This means that the physical states are parameterized by polarization vectors with the equivalence relation ξ µ (k) = (ξ 0 (k), ξ i (k), ξ 0 (k)) ξ µ (k) ξ µ (k) + (η, 0,..., 0, η) Within each equivalence class there is a representative of the form ξ µ (k) = (0, ξ i (k), 0) which has positive norm. Finally, let us examine the set of states at the next level, ξ µ (k)α µ 0, k > +ζ µν (k)α µ α ν 0, k > Here, ζ µν is a symmetric tensor. These states have norm η µη µ + ζ µνζ µν which can be negative. The physical state condition (8) requires that m = k = ( a)/α. Since, in the above, we found that physical states had non-negative norm only when a, this state must be massive, that is k µ In this case the level zero state constructed above is a tachyon with mass m = /α and the level one state that we are discussing is a massless vector, matching the spectrum of the open bosonic string that if found in the light-cone gauge. 5

6 must be time-like and we can find a reference frame where k µ = (m, 0,..., 0). The nontrivial physical state conditions, that they are annihilated by L and L yield the conditions ( α p µ α µ + α µ α µ )(ξ µ (k)α µ + ζ µν (k)α µ α ν ) 0, k >= 0 (3) ( α p µ α µ + α µα µ )(ξ µ (k)α µ + ζ µν (k)α µ α ν ) 0, k >= 0 (4) These result in In the rest frame k µ = (m, 0,..., 0), these give ξ µ + α k ν ζ µν = 0 (5) α k µ ξ µ + ζ µ µ = 0 (6) ξ µ + α mζ µ0 = 0 ζ µ0 = α m ξµ (7) ( α mξ 0 + ζ µ µ = 0 ζi i = ) α m + ξ 0 (8) α m Then, when we calculate the norm of the physical states, it is convenient to separate ζµνξ µν = ζ 00 ζ 0i + ζij δij D ζkk + D ζkk Then, using our solution of the constraints in the rest frame, the norm is ξµξ µ + ζµνξ µν = ξ i ξ 0 + α m ξ0 α m ξi + ζ ij δij D ζkk + ( D ) α m + ξ 0 α m ( ( ξµξ µ +ζµνξ µν = ξ i )+ ξ 0 α m D + ζij δij D ζkk 6 ( α m + ) ) + α m α m

7 Remembering that α m = a, we see that the coefficient of ξ i is zero when a = (so there are null vectors) and is positive when a <. The coefficient of ξ 0 is non-negative when D + [4( a) + ] [( a) ] When a = we find D 6. At the critical dimension and intercept D = 6 and a =, there are null states. Any state for with ζ ij δij D ζkk = 0 is a null state. The multiplet of physical states there are parameterized by the symmetric traceless tensor ζ ij δij D ζkk = 0 which are just enough to form a massive tensor field. In four dimensions it would be a massive spin field. 7