Phys 731: String Theory - Assignment 3

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1 Phys 73: String Theory - Assignment 3 Andrej Pokraka November 8, 27 Problem a) Evaluate the OPE T () :e ik (,) : using Wick s theorem. Expand the exponential, then sum over contractions between T and the sum (there will be ero, one, or two contractions), and then do the sum, which should simplify if you ve done the combinatorics right. Finally, Taylor expand inside the normal ordering, and get all singular terms. Show that the exponential is a tensor, of the weight given in class. b) Do the same for : e ik (, ) :: e ik2 (,) : to recover the result given in class. Carry the Taylor expansion further, up to and including terms down by 2, 2, in the OPE. Part (a) For the free scalar field, the holomorphic worldsheet energy-momentum function is Therefore, we are looking for the OPE to T () ()@ µ () :. ()@ µ () ::e ik (,) :. The product of two normal ordered operators is given by equation (2.2.) of JBBS With µ ()@ µ () and G e ik (,) we obtain : F :: G ::FG :+ cross-contractions. T () :e ik (,) : ()@ µ ()e ik (,) : cross-contractions.

2 The remaining unknown is just the sum of cross-contractions h µ ()@ µ (),e ik (,)i n n! [@ µ ()@ µ (), (ik (, )) n ] () where P [F, G] denotes the sum of all cross-contractions of the operators F and G. The last ingredient that we will need in for calculating the sum of all cross-contractions is the basic cross-contraction [@ µ (), (, )] 2 ln 2 2 µ. In equation () there will be terms with ero, one and two contractions: ) For n,therearenocontractions [@ µ ()@ µ (), ]. 2) For n, there are only terms with one contraction [@ µ ()@ µ (),ik (, )] ik µ [@ µ (), (, )] () : + ik µ [@ (), (, )] µ () : ik () :+ ik () : 2 2 ik () :. 3) For n 2, thecross-contractionsumis [@ µ ()@ µ (), (ik (, )) n ] ik...ik n [@ µ ()@ µ (), (, )... n (, )]. Note that sum of cross-contractions is completely symmetric in the indices... n.furthermore,thereareterms with one contraction and terms with two: (a) Single cross-contraction: [@ µ ()@ µ (), (ik (, )) n ] single ik...ik n [@ µ ()@ µ (), (, )... n (, )] single nik...ik n µ () 2 (, )... n (, ) : [@ µ (), (, )] + nik...ik n µ () 2 (, )... n (, ) : [@ µ (), (, )] n ik µ µ ()(ik (, )) n : The factor of n comes from the fact that can be paired up with in n different ways. 2

3 (b) Double cross-contraction: µ µ (), (ik (, )) n ] double ik...ik n [@ µ ()@ µ (), (, )... n (, )] double n (n ) ik...ik n : 3 (, )... n (, ) : [@ µ (), (, )] [@ µ (), 2 (, )] n (n ) k 2 2 :(ik (, )) n 2 : 2 Summing up the single cross-contraction terms we obtain Summing the double cross-contractions yields [@ µ ()@ µ (), (ik (, )) n ] n! single n ik n () :+ ik µ µ ()(ik (, )) n : n! n2 " # ik () :+ (n )! ik µ µ ()(ik (, )) n : n2 " # ik () :+ n! ik ()(ik (, )) n : n ik ()e ik (,) :. n n! [@ µ ()@ µ (), (ik (, )) n ] double 2 k 2 2 n2 2 k 2 2 : e ik (,) :. (n 2)! :(ik (, ))n 2 : Substituting the single and double cross-contraction sums into (), we obtain T () :e ik (,) : ()@ µ ()e ik (,) : ()@ µ ()e ik (,) : cross-contractions + ik ()e ik (,) : + k 2 2 : eik (,) :. (2) To finish calculating an the OPE we Taylor expand inside the normal ordering to obtain all singular terms. The only term that needs to be expanded () 2 () + O 2. 3

4 Then µ ()@ µ ()e ik (,) :sum of non-singular terms, and ik T () :e ik (,) : k 2 ()e ik (,) : ik 2 : eik (,) :+ ik ()e ik (,) :+non-singular terms, ()e ik (,) :+non-singular terms. In equation (2..6) of JBBS, it is shown that any tensor operator of weight (h, h) will have an OPE with T and T of the form T () :O(, ) h 2 O(, ) ) + non-singular terms with a similar expression for the OPE with T. Comparing the above to our result (2), we see that h k 2 / and a similar calculation tells us that h h. Therefore, the operator : e ik (,) : is a tensor of weight k 2 /, k 2 /. Part (b) Since there are no derivatives of fields in the product : e ik (, ) :: e ik2 (,) :,wecanuseformula(2.2.8)ofjbbs Z Z : F :: G : exp d d 2 ln 2 2 : FG : F (, ) G ( 2, 2 ) to evaluate this OPE. Thus, : e ik (, ) :: e ik2 (,) Z Z ::exp d d 2 ln 2 2 µ µ F (, ) G ( e ik (, ) e ik2 (,) : 2, 2 ) : e ik (, ) e ik2 (,) Z Z exp d 2 d 2 2 ln 2 2 µ (ik (, )) (ik 2 (, )) µ (, ) : ( 2, 2 ) : e ik (, ) e ik2 (,) Z Z exp k k 2 d 2 d 2 2 ln 2 2 (2) ( ; ) (2) ( 2 ; 2 ) : : e ik (, ) e ik2 (,) e (k k2)ln 2 : : (k k 2)/2 e ik (, ) e ik2 (,) : The factor (k k 2)/2 could be singular. We therefore expand the exponential to get a few of the less singular terms : e ik (, ) :: e ik2 (,) : (k k 2)/2 : e ik 2, 2 ) e ik 2 (,) : (k k 2)/2 : e i(k+k2) (,) n (k k 2 ) (k k 2 ) : e i(k+k2) (,) : + i (k k 2 ) + (k k 2 ) e i(k+k2) (,) : + i (k k 2 ) (k k 2 ) + : e i(k+k2) (,) : (k k 2 ) +2 (k k 2 ) 2 e i(k+k2) (,) : (k k 2 ) (k k 2 ) +2 : e i(k+k2) (,) : + + O( 2, 2 ) n : 2 (k k 2 ) + (k k 2 ) + : e i(k+k2) (,) :+O 3

5 Problem 2 (2.8 of JBBS) What is the weight of f µ e ik :?Whataretheconditionsoff µ and k µ in order for it to be a tensor? We could take the OPE of f µ e ik : with T and T to find its weight. Or we could make a rigid rotation and see how f µ e ik : transforms. Since we already know the weights µ and : e ik :, (, ) and k 2, k 2 respectively, under! and! we obtain Thus, the weight is f µ e ik :! + k 2 + k 2 f µ e ik :. + k 2, + k 2. The weight is always the coefficient of the 2 or 2 terms in the OPE expansion of T ()O(, ) or T ( )O(, ). For a general operator, the OPE expansion may have more singular terms than 2 or 2. However, the OPE of a tensor with T or T does not have terms more singular than 2 or 2. We must work out the OPE with T and T to find conditions on f µ and k µ. T () :@ µ ()e ik (,) : ()@ () ::@ µ ()e ik (,) : By symmetry we also have the OPE with T T ( ) :@ µ ()e ik (,) : ()@ ()@ µ ()e ik (,) :+ cross-contractions 2 3 ikµ ()e ik (,) :+ k 2 /+ 2 µ ()e ik (,) : + ik ()@ µ ()e ik (,) :+non-singular terms. 2 3 ik µ ()e ik (,) :+ k 2 /+ 2 µ ()e ik (,) : + ik From the above OPEs we see that f µ k µ f µ k. The intermediate results used to calculate this OPE are listed below: ( )@ µ ()e ik (,) :+non-singular terms. The sum over all contractions factories into sums with ero, one or two contractions: h ()@ (),@ µ ()e ik (,)i single- double- ()@ (),@ µ ()e ik (,)i single- ()@ (),@ µ ()e ik (,)i double- 5

6 Single cross-contraction sum: ()@ (),@ µ ()e ik (,)i single- single- 2 [@ (),@ µ ()] ()e ik (,) 2 h (),e ik (,)i ()@ µ () : 2 ()e ik (,) : 2 : n! (nik [@ (), (, )] ()@ µ () (ik (, )) n : 2 ()e ik (,) :+ ik ()@ µ ()e ik (,) : Double cross-contraction sum ()@ (),@ µ ()e ik (,)i double- double- h [@ (),@ µ (),e ik (,)i () : 2 2 n ()@ (),e ik (,)i double- µ () : [@ (),@ µ ()] h@ (),e ik (,)i () : n ()@ (), (ik (, )) n double- () : 2 ik [@ (),@ µ (), (, ) () ik ik 2 [@ (), (, (), 2 (, ) () 2 3 ikµ ()e ik (,) :+ 2 k2 µ ()e ik (,) : n n! n (ik (, ))n : n2 n! n (n ) (ik (, ))n 2 : [@ µ (), (, )] [@ µ (),@ ()] 2 ln 2 2 µ 2 µ lim! ln 2i 2 2 µ () Problem 3 (. of JBBS) Extend the old covariant quantiation, for A and general D, tothesecondexcitedleveloftheopen string. Verify the assertions made in the text about extra positive- and negative-norm states. Basically, the covariantly quantied Hilbert space is larger than the physical Hilbert space. The process of old covariant quantiation is the identification of the physical Hilbert space. 6

7 The Virasoro generators generate the left-over gauge symmetry. Therefore, the Virasoro generators must act trivially on physical states i, i h L m i. (3) The weakest condition that we can impose that satisfies the above constraint is L m i for m>. () Then, for m< the matrix element (3) vanishes by action on the left: L m L m.also,forl we need L i A i (5) where A is a possible ordering constant. Any state satisfying equations () and (5) is called physical. Note that any state of the form L n i for n> is orthogonal to all physical states h L n i hl n i. Such states are called spurious. Aspuriousstatethatisphysicaliscallednull astatethatisorthogonaltoallphysical states including itself (such states are essentially ero). We therefore need an equivalence class relation. Two physical states, i and i,areequivalent i i if i i is null. The set of equivalence classes is called the observable spectrum. Virasoro generators: L m6 2 P n µ m n nµ L P n µ n nµ Note that we have not included the ordering constant A in L. Instead we have choose to make the A dependence of explicit in the condition (5). Level : The only states at this level are ; ki. They are annihilated by L m> because there are no lower states. There are also no spurious states because there are not any lower states to raise. The physical condition becomes: (L + A),ki. Using the fact that L i k 2,wemusthaveA k 2. m 2 k 2 A/ Level : At level we have D possible states: e, ki e µ µ ; ki. The condition () yields where we have used L e, ki e 2 L e, ki p 2 k e,ki ) e µ k µ n µ m n nµ!,ki e 2 ( µ µ + µ µ),ki e µ µ,ki p 2 k µ e µ + µ,ki p 2 k e,ki. 7

8 The L condition (5) becomes (L + A) e, ki k 2 ++A e, ki ) A k 2 where we have used L e, ki 2 + n> µ n nµ! e µ µ,ki 2 + e µ µ,ki e µ µ 2,ki + µ + µ,ki e µ k 2 + µ,ki k 2 + e, ki. The mass of the e, ki sate is m 2 k 2 (A + ) /. There are three cases for invariant mass: For A> the invariant mass is positive. However, a consistent theory of interactions with A> is not known. For A< the invariant mass is negative. Clearly, this is not physical. Hence, we concentrate on the case where A. Here,thelevelinvariantmassisero,m 2. Spurious states are of the form L,ki p 2 k,ki. Spuriousstateshavee / k and are null if k 2. Since m 2, k 2 for all level states and the spurious states become physical (null). The observable spectrum has Level 2: k 2, e k, e µ e µ + k µ. Going to a reference frame for which k µ (,,,...) or equivalently k µ (,,,...) implies that e e and e, e,e 2,... e, e +,e 2,.... Choosing e we obtain e µ,,e 2,.... Therefore, there are a total of D 2 physical states at level. This is the same as in light-cone quantiation. At level 2 we have the states f,e,ki f µ µ + e µ µ 2,ki where f µ is symmetric. Since f µ is symmetric, there are (D + ) D/2+D states. The norm of a general state is hf,e,k f,e,k i 2 f µ f µ + e µe µ h,k,k i. The L condition (L ) f,e,ki k 2 + f,e,ki ) m 2 2 k 2 / where! L f,e,ki n n f µ µ + e µ µ 2 n ,ki f µ µ + e µ µ 2,ki k 2 f,e,ki + f µ µ,ki + e µ 2 2 µ 2,ki k 2 f,e,ki + f µ µ + µ + µ,ki + e µ 2 µ µ 2 2,ki k 2 f,e,ki +2f µ µ,ki +2e µ µ 2,ki k 2 +2 f,e,ki. 8

9 The L condition gives e in terms of f and removes D degrees of freedom L f,e,ki 2 e + p 2 f µ k µ,ki where we have used L f,e,ki 2 n ) e + p 2 f µ k µ (6)! n n f µ µ + e µ µ 2,ki f µ µ + e µ µ 2,ki 2 + p 2 k f µ µ + e µ µ 2,ki e µ 2 µ 2,ki + p 2 f µ k µ,ki e µ 2 µ + µ 2 2,ki + p 2 f µ k µ + µ + µ,ki 2 e + p 2 f µ k µ,ki. The L 2 condition gives a condition on the trace of f L 2 f,e,ki 2p 2 k e + f µ µ ) p 2 k e + f µ µ where we have used! L 2 f,e,ki 2 n 2 n f µ µ + e µ µ 2,ki n f µ µ + e µ µ 2,ki e µ 2 µ 2,ki + 2 f µ µ,ki,ki ) f µ µ 2 f µ k µ k (7) p 2 k e µ 2 µ + µ 2 2,ki + 2 f µ µ + µ + µ + µ + µ,ki 2 p 2 k e + f µ µ,ki. In the rest frame, k / p and k i,equations(6)and(7)implye p 2f and f ii 5f. There are D +spurious states at this level a, b; ki L a µ µ {,ki + bl 2,ki } a,ki a µ µ + 2 µ,ki + b 2 + 2,ki a µ p 2 k µ + µ 2 + µ 2,ki + b p2 k p 2 k a µ µ + a µ µ 2,ki + p 2 bk 2,ki + b 2,ki p2 k a µ + b 2 µ µ a,ki + µ + p 2 bk µ µ 2,ki f,e,ki,ki 9

10 where f µ p 2 k a µ + b 2 µ e µ a µ + p 2 bk µ. In the rest frame, e µ a µ + p 2 bk µ a µ + p 2b,µ e µ p p 2fµ 2a 2b,µ f 5 f ii b 5 2 ii b (D ) e e 2a p 2b 2 a 2 + p 2a b + p 2a b +2 b 2 In the rest frame, the states with non-negative norm are those for which a b. Problem For the µ CFT, show that the operators L m L m +(m + )v µ µ m also satisfy a Virasoro algebra, and finds its central charge (this should be relatively simple using the known result for the algebra of the L m s). This has an interpretation in terms of strings moving in a certain nontrivial spacetime. We wish to show that the operators L m satisfy a Virasoro algebra: [L m,l n](m n) L m+n + c 2 m3 m m, n for some central charge c. To show that L m satisfies the above algebra, we calculate the L m commutator using the properties of the L m and n µ [L m,l n][l m +(m + )v µ µ m,l n +(n + )v µ µ n] [L m,l n ]+(n + )v µ [L m, n]+(m µ + )v µ [ m,l µ n ]+(m + )v µ (n + )v [ m, µ n] (m n) L m+n + c 2 m3 m m, n +(n + )v µ [L m, n] µ (m + )v µ [L n, m] µ (m 3 m)v 2 m, n (m n) L m+n n(n + )v µ µ m+n + m(m + )v µ µ c m+n + m 3 m v 2 m, n 2 (m n) L m+n + m 2 + m n 2 n v µ µ c m+n + m 3 m v 2 m, n 2 (m n) L m+n +(m n)(m + n + ) v µ µ m+n + m 3 m c 2v2 2 (m n) L m+n + c 2 m3 m m, n. Thus, we see that L m+n satisfies a Virasoro algebra with central charge c c 2v 2. We have used the following facts in the derivation of [L m,l n]: m, n

11 The normal ordering of the creation and annihilation operators is defined so that all annihilation operators ( m µ with m>) aretotheleftofthecreationsoperators( m µ with m<) L m m 2 l l l ( P 2 l m l l if m P l l l if m The commutator of L m with n µ is " # [L m, n] µ m 2 l l, n µ l ( P 2 l m l l, n µ if m P l l l, n µ if m ( P 2 l m l [ l, n]+ µ m l, µ n l if m 6 2 µ [, n]+ µ 2 [, n] µ + P l l [ l, n]+ µ P l l, n µ l if m ( P 2 l l l, n µ m l +(m l) m l, n µ l if m 6 P l l l, n µ l + P l ( l) l, n µ l if m ( n µ m+n if m 6 n n µ if m n µ m+n Problem 5 Consider a CFT with the mode algebra Show that the operators L m {b m,c n } m+n,, {b m,b n } {c m,c n }. n L + (2m n) b n c m n for m 6 n (b n c n + c n b n ), n satisfy a Virasoro algebra, and determine its central charge. I suggest that, as in class, you pick out the operator (singlecommutator) terms first, and get the constant by acting with [L m,l m ] on some particular state, e.g., the one annihilated by b n and c n for n> and by c (though any state will give the same answer). Note that the mode operators b n, c n belong the the free bc CFT. A more general expression for the Virasoro generators in this theory can be found in (2.7.9) of JBBS L m ( m n) b n c m n + 2 ( ) m, n

12 (note that b n and c n anti-commute inside the normal ordering in the expression for L above). In this problem, we focus on the special case where 2. Since L m6 and L cannot be expressed in the same format., we must treat several cases of commutator [L m,l n ]: m 6,n6 and m 6 n [L m,l n ] (2m i)(2n j)[b i c m i,b j c n j ] m 6,n i,j i,j i,j i,j i,j i k (m n) (m [L m,l ] (2m i)(2n j)[b i c m i b j c n j b j c n j b i c m i ] (2m i)(2n j)[b i ( m i+j, b j c m i ) c n j b j ( n j+i, b i c n j ) c m i ] (2m i)(2n j)[( m i+j, b i c n j b i b j c m i c n j ) ( n j+i, b j c m i b j b i c n j c m i )] (2m i)(2n j)[( m i+j, b i c n j n j+i, b j c m i )] [(m i)(2n i) b m+i c n i (2m i)(n i) b n+i c m i ] [(2m k)(2n + m k) (2m + n k)(2n k)] b k c (n+m) k k n) L m+n i j i j [2 (m + n) k] b k c (n+m) k (2m i) j ([b i c m i,b j c j ]+[b i c m i,c j b j ]) (2m i) j ( m i j, b i c j i+j, b j c m i + i j, b j c m i j i+m, b i c j ) j ((m + j) b m j c j (2m + j) b j c m+j +(2m j) b j c m j (m j) b m+j c j ) j j (m + j) b m j c j + j (m + j) b m j c j + j> j< j> j (m + j) b m j c j + j (2m j) b j c m j m j k ml m k j [(m k)(2m k) b k c m k + k (2m k) b k c m k ] (2m k) b k c m k j (2m j) b j c m j + j< j (2m j) b j c m j 2

13 where we have used the intermediate results [b i c m i,b j c j ]b i c m i b j c j b j c j b i c m i b i ( m i j, b j c m i ) c j b j ( i+j, b i c j ) c m i (b j b i c j c m i b i b j c m i c j )+( m i j, b i c j i+j, b j c m i ) m i j, b i c j i+j, b j c m i, [b i c m i,c j b j ]b i c m i c j b j c j b j b i c m i b i c m i ( b j c j ) ( b j c j ) b i c m i b i c m i b i c m i b j c j b i c m i + b j c j b i c m i b i ( j+m i, b j c m i ) c j + b j ( i j, b i c j ) c m i b i b j c m i c j b j b i c j c m i + i j, b j c m i j+m i, b i c j i j, b j c m i j i+m, b i c j. The results above are only valid only up to a constant, A m, n. Therefore, the L m satisfy the algebra [L m,l n ](m n) L m+n + A m, n for an undetermined constant A. To determine A, we act with [L m,l m ] on the physical state i that is annihilated by b n> and c n L m L m i and. Taking m>, where the central charge is c A i [L m,l m ] i 2mL i L m L m i +2m i m 3m 3 +2m i m m3 i c 2 m3 m 26. We have also used the following: L i i + i (b i c i + c i b i ) i i + ib i c i i + ic i b i i i i i i 3

14 L m L m i (2m i)( 2m j) b i c m i b j c m j i { } i,j 6 if m+applejapple (2m i)( 2m j) b i c m i b j c m j i + m+applejapple m+applejapple m+applejapple m+applejapple m+applejapple m+applejapple m+applejapple m 3m3 6 i (2m i)( 2m j)( m i+j, ( i m j, c m j b i ) ( b j b i )( c m j c m i )) i i i (2m i)( 2m j) ( i,m+j ) 2 i,j+mc m j b i i,j+m b j c m i + b j c m j b i c m i i (m j)( 2m j) i i m+applejapple (2m i)( 2m j) b j c m j b i c m i i { } (m j)( 2m j) i (m j)( 2m j) i m+applejapple (m j)( 2m j)(c m j b j+m + b j c j ) i B (m j)( 2m m j b j+m i { } for m+applejapple + b j c j i { } for m+applejapple C A

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