Recursions in Calogero-Sutherland Model Based on Virasoro Singular Vectors
aa r X i v : . [ h e p - t h ] A ug Recursions in Calogero-Sutherland Model Based onVirasoro Singular Vectors
Jian-Feng Wu ∗ , Ying-Ying Xu † , and Ming Yu ‡ Institute of Theoretical Physics,Chinese Academy of Sciences, Beijing, 100190, China
Abstract
The present work is much motivated by finding an explicit way in the construction of the Jacksymmetric function, which is the spectrum generating function for the Calogero-Sutherland(CS)model. To accomplish this work, the hidden Virasoro structure in the CS model is much explored.In particular, we found that the Virasoro singular vectors form a skew hierarchy in the CS model.Literally, skew is analogous to coset, but here specifically refer to the operation on the Youngtableaux. In fact, based on the construction of the Virasoro singular vectors, this hierarchical struc-ture can be used to give a complete construction of the CS states, i.e. the Jack symmetric functions,recursively. The construction is given both in operator formalism as well as in integral representa-tion. This new integral representation for the Jack symmetric functions may shed some insights onthe spectrum constructions for the other integrable systems.PACS: 11.25.Hf,02.30.Ik,02.30.Tb
Keywords:
Calogero-Sutherland Model, Jack Symmetric Function, Recursion Relations, IntegralRepresentation ∗ [email protected] † [email protected] ‡ [email protected] Introduction and Conclusion
The Calogero-Sutherland (CS) Model, which is an integrable 1d many-body system, plays importantroles in many different research areas in physics and mathematics. Among them are the 2D confor-mal field theories(CFTs)[1, 16], the generalized matrix models[2, 4], the fractional quantum hall ef-fects(FQHE) [6, 7, 11, 12], and there is an even more surprising correspondence related to the N = ∗
4D supersymmetric gauge systems[23–29]. The spectrum of the CS model can be generated from theJack polynomials[5, 17–19, 32]. From the CFT point of view, Jack symmetric functions are naturallythe building blocks for the conformal towers, the characters of which encode the (extended) conformalsymmetries. For instance, the Jack functions related to a given Young tableaux are believed to be in oneto one correspondence with the singular vectors of the W -algebra[1], which reflects the hidden W + ∞ symmetry of the CS model. Singular vectors in 2d CFT are the keys to the calculation of the correlationfunctions in CFT and may also reveal important physical properties of the CS model. Unfortunately,on the CFT side, it is not clear how to relate the construction of the secondary states in the conformaltower to that of the Jack functions, except for some simple cases, i.e. the Virasoro singular vectors [10].On the 2d CFT side, the calculation of conformal blocks is based on the conformal Ward-identities, [ L n , V h ( z )] = ( z n + ∂ z + ( n + hz n ) V h ( z ) . And the calculation is carried out perturbatively level by level [8, 9]. In some special cases, the decou-pling of the Virasoro null vectors can be implemented as differential equations for the conformal blocks.For the general case, recursion relations have been proposed by Zamolodchikov on the meromorphicstructures of the conformal blocks either in complex c -plane or h -plane. However, it remains unclear(to us) how to construct the basis vectors in the given conformal tower by making use of the Zamolod-chikov’s recursion formulae explicitly. In contrast, there are various ways in the explicit constructionsof the Jack polynomials, each follows different strategies. For example, there are two integral represen-tations. The one given in [3] is based on the W + ∞ symmetry hidden in the CS model, the other, by theauthors of [13], starts from the so called shift Jack polynomials. There is also an operator formalism[14, 15] based on the Dunkel (exchange) operators. However, here we follow a different strategy inconstructing the Jack polynomials which are intrinsically related to the singular vectors in the Virasoroalgebra, and present a new recursion relation derived from the construction. We also do not need toinvoke the hidden W -algebra. In fact there is a hidden Virasoro structure in the Hilbert space of theCS model which can be used recursively in our construction. To be more specific, the ket states in theFock space realization of the CS model is mapped a la Feigin-Fuks-Dotsenko-Fateev Coulomb gas for-malism [20][21] to the singular vectors of the Virasoro algebra and its skew hierarchical descendants.The construction of the singular vectors defines a new recursion relation for the Jack functions andfinally leads to a new integral representation which differs from the one appeared in [3, 13]. Hence ourapproach may supply new insights in dealing with CS model or more general integrable systems.The structure of this article is organized as following. In section 2, we review some useful propertiesof the Jack polynomials and the CS model. In section 3, we review the construction of the Virasorosingular vectors which are related to the Jack polynomials with rectangular Young tableaux. It shouldbe stressed again that the Virasoro symmetry is hidden in the ket (or bra) Hilbert space only, and is notthe true symmetry in the CS model. The reason is that the prescribed Hermiticity in the CS model is notrespected by the conjugation of the hidden Virasoro algebra. I.E., the Virasoro structure in the bra andket Hilbert spaces, respectively, are not related by the Hermitian conjugation in the CS model. Our mainresults are given in sections 4 and 5. In section 4, we propose a skew (recursion) formula for generatingnew Jack functions stating from the simple ones which inevitably involve the Virasoro singular vectors,1r equivalently, the Jack functions of the rectangular graph. Our proof of the skew (recursion) formulacan be considered as an operator formalism generalization of its counterpart for the Jack symmetricpolynomials found by Kadell in [18]. The basic skew relation is further developed recursively in section5. This can be made explicit first in the operator formalism, based on which we develop a new integralrepresentation for the Jack symmetric functions associated with any generic Young tableaux. Oneimmediately sees the advantages of our operator formalism of the skew (recursion) formula over theone proposed by Kadell. Since our formalism does not depend on explicitly the number of argumentvariables { z i } for i = , · · · , N , so the recursion is done without worrying the change of the totalnumber of arguments. Finally, the integral representation of the Jack symmetric polynomials whichdepend explicitly on finite number of variables { z i } , for i = , · · · , N , are presented as a by-product. Now we review the Jack polynomials and the Calogero-Sutherland model. The Jack polynomials can beviewed as a special one parameter generalization of the Schur polynomials[32] and the Jack symmetricfunction is the large N limit of the Jack polynomials. For physicists, the most familiar integrablesystem which involves Jack symmetric functions as its spectrum functions is the Calogero-Sutherlandmodel. This model is an integrable system and shows a great deal of interesting aspects, such asduality, conformal invariance, and even the combinatorial property of the spectrum etc. An elementaryintroduction can be found, for instance, in [11] and further studies in [19] and [32]. Here we onlyreview some basics of the Jack polynomials.
Given a partition: λ = ( λ , λ , . . . , λ n ) , λ ≥ λ ≥ · · · ≥ λ n , n ≡ l ( λ ) , one defines the relatedJack polynomial as the basis function for the symmetric homogeneous polynomials in N variables { z i } , i = , , ..., N , of degree | λ | = P λ i J λ ( { z i } ) = X λ ′ ≤ λ, l ( λ ′ ) ≤ N C λ ′ λ z λ ′ , (1)which should satisfy the second order differential equation: H J J λ = E λ J λ , H J = H J + β H IJ (2) H J = N X i (cid:0) z i ∂ z i (cid:1) , H IJ = X i < j z i + z j z i − z j ( z i ∂ z i − z j ∂ z j ) , (3)here z λ : = X P z λ P (1) · · · z λ N P ( N ) , (4) λ ′ ≤ λ ⇒ j X i = λ ′ i ≤ j X i = λ i , f or j = , , · · · l ( λ ′ ) P means the permutations of N objects. We have also defined λ i = for i > l ( λ ) and λ + λ + · · · + λ l ( λ ) = | λ | , | λ | is the level of the partition. λ can be represented graphically as a Young tableau2 = { ( i , j ) | ≤ i ≤ l ( λ ) } , ≤ j ≤ λ i . And the corresponding transposed Young tableau is representedas λ t = ( λ t , λ t , . . . , λ t λ ) ⇒ λ = ( λ , λ , . . . , λ λ t ) . It is clear that l ( λ ) ≡ λ t . We shall see later that the defining differential equation can be derived fromthe CS Hamiltonian.The Jack polynomials can be generated by the power sum symmetric polynomials as well: p l = P Ni = z li J λ ( p ) = X | λ ′ | = | λ | d λ ′ λ p λ ′ , p λ ′ = p λ ′ p λ ′ · · · p λ ′ m , d | λ | λ = when N large, J λ spans the Hilbert space of free oscillators, and each power sum symmetric polynomialbehaves as a single oscillator. By using the conjugacy class representation of the Young tableau λ = { i m i } , i = , , . . . , where m i means the multiplicity of the rows of i squares in Young tableau λ , thenormalization of the power sum symmetric polynomial is derived from that of the oscillators, h p λ , p λ ′ i = h a λ a − λ ′ β l ( λ ) β l ( λ ′ ) i (5) = δ λλ ′ i m i m i ! β − l ( λ ) h a n a − m i = δ n , m n , β = k , In fact, the above normalization is consistent with that of Jack symmetric functions[11, 19]. Thenormalization of the Jack polynomials is derived from that of the wave function in the CS model: N Y i ˆ π dx i J λ ′ ( p ∗ ) J λ ( p ) Y i < j | z i − z j | β = Γ N δ λ,λ ′ j λ ¯ A λ, N ¯ B λ, N , (6)here j λ = A /βλ B /βλ , z i = e ix i (7) A /βλ = Y s ∈ λ (cid:16) a λ ( s ) β − + l λ ( s ) + (cid:17) , B /βλ = Y s ∈ λ (cid:16) ( a λ ( s ) + β − + l λ ( s ) (cid:17) a λ ( s ) and l λ ( s ) are called arm-length and leg-length of the box s in the Young tableau λ : a λ ( s ) = λ i − j , l λ ( s ) = λ tj − i , λ tj is the j -th part of the partition related to the transposed Young tableau λ . ¯ A λ, N = Y s ∈ λ (cid:0) N + a ′ λ ( s ) /β − l ′ λ ( s ) (cid:1) , ¯ B λ, N = Y s ∈ λ (cid:0) N + ( a ′ λ ( s ) + /β − ( l ′ λ ( s ) + (cid:1) , a ′ λ ( i , j ) = j − , l ′ λ ( i , j ) = i − denote the co-arm-length and co-leg-length for the box s = ( i , j ) inYoung tableau λ . Γ N ≡ π N Γ (1 + N β ) Γ N (1 + β ) is the normalization of the ground state. When N → ∞ , andafter dividing out the ground state normalization Γ N , we obtain the normalization for the Jack functions: h J λ , J λ ′ i = ˆ J λ ′ ( p ∗ ) J λ ( p ) Q i < j | z i − z j | β Γ N N Y i = dx i | N →∞ = δ λ,λ ′ j λ (8)3e shall see in the next section that eq.(6) implies the following integral formula J λ ( a − k ) = ˆ e k P n > a − nn p n J λ ( p ∗ ) Q i < j | z i − z j | β Γ N N Y i = dx i . (9)Now we shall clarify some notations used in this work. J λ ( p ) means Jack polynomials in the powersum polynomial basis, J λ ( ak ) the annihilation operator valued Jack symmetric function, i.e. with thesubstitution p n → a n k , and J − λ ≡ J λ ( a − k ) the creation operator valued Jack symmetric function, i.e.with the substitution p ∗ n → a − n k . The CS model is introduced for studying Coulomb interacting electrons distributed on a circle. TheHamiltonian for this system can be written as : H CS = − N X i = ∂ i + X i < j β ( β − ( x i j ) (10)here ∂ i = ∂ x i , ~ / m = , β = k , k is the charge of the N identical electrons. This is an exactsolvable system. However, let’s consider another auxiliary Hamiltonian which is positive definite anddiffers from the original one only by a shift of the constant ground state energy H = − N X i = ( ∂ i + ∂ i ln Y l < j sin β x l j )( ∂ i − ∂ i ln Y l < j sin β x l j ) (11) = − N X i = ∂ i + β ( β − X i < j x i j − β N ( N + N − = H CS − E , E = β N ( N + N − . In going from the first line to the second line of eq. (11), we have used the identity X i , j , k cot x i j cot x ik + X j , i , k cot x ji cot x jk + X k , i , j cot x ki cot x k j (12) = X i , j , k − cos x i j − cos x ik cos x jk sin x ik sin x jk = X i , j , k ( − = − N ( N − N − . where x i j ≡ x i − x j . It is also convenient to define z i = e ix i for later use. The ground state should bea solution to the eigen-equation: H CS ψ = E ψ , For convenience, we set the circumference of the circle as L = π ψ = Y i < j (2 sin β x i j ) , where the factor of 2 is included for normalization reason. If one defines the excited state as ψ λ = J λ ψ , then it can be shown that this state actually satisfies the energy eigen-equation: H ψ λ = H ψ J λ ( p ) = ψ ( ψ − H ψ ) J λ ( p ) (13) = ψ H J J λ ( p ) = ψ E λ J λ ( p ) , H J = ψ − H ψ = − N X i = ( ∂ i + β X j , i cot x i j ) ∂ i , (14)thus the eigen-equation can be rewritten as H J J λ = − N X i = ∂ i − β X i < j cot x i j ( ∂ i − ∂ j ) J λ = E λ J λ . (15)We see that this coincides with the defining differential equation eq.(2) for the Jack polynomials. Theeigenstates of H J in the form of eq.(14) and(15), means that H J is triangular with respect to the sym-metric monomials. J λ ∼ l ( λ ) Y i = z λ i i + symmetrization + daughter terms . Here the daughter terms are the symmetrized monomials associated with Young tableau λ ′ < λ . That isto say, given a Young tableau λ , one can squeeze the partition { λ } to other partitions by moving squaresin λ downwards to get new Young tableaux. These terms actually reflect the triangular property of theinteraction H IJ as in eq.(2). fig.1 gives an example of squeezing. It is easy to read out the eigenvalueof H J which can be read off from the diagonal value of the leading term of the eigenstate E λ = − N X i ( − λ i ) − constant term part X i < j i β ( z i + z j )( z i − z j ) ( z λ i i z λ j j − z λ i j z λ j i )( z λ i i z λ j j + z λ i j z λ j i ) (2 i λ i − i λ j ) = N X i λ i + β X i < j ( λ i − λ j ) = N X i λ i + β N X i ( N − i + λ i . (16)Here in the last step of eq.(16), we have used the fact that N X i < j ( λ i − λ j ) = ( N − λ + ( N − − λ + · · · + ( N − i + λ i · · · = N X i ( N − i + λ i .
21 2 33
Figure 1: An example for squeezing Young tableau, where the square 3 has been squeezed downwardto form a different Young tableau.Here, since we are concerning ourselves to all the Young tableaux λ ′ ≤ λ with the restriction l ( λ ′ ) ≤ N ,so we redefine λ as well as λ ′ to include the trailing null parts such that l ( λ ′ ) = l ( λ ) = N . Actually,eq.(16) can be written as the following more compact formula: E λ = k (cid:16) k − || λ || − k || λ t || + kN | λ | (cid:17) (17) || λ || ≡ N X i λ i , || λ t || ≡ N X i ( λ ti ) , | λ | ≡ N X i λ i . In fact, the second quantized form of the CS model can be realized as a theory of 2D scalar field ϕ ( z ) [4].In the corresponding CFT, ϕ ( z ) is a scalar defined on the unit circle but can be analytically continuedto complex plane. The vertex operator for CS model reads V k ( z ) = : e k ϕ ( z ) : (18) ϕ ( z ) = q + plnz + X n ∈ z , n , a − m m z m h ϕ ( z ) ϕ ( w ) i = log( z − w ) , here [ a n , a m ] = n δ n + m , , [ p , q ] = , ϕ ( z ) † = − ϕ ( z ) . It is easy to show that the ground state of the CS model can be written as the holomorphic part of thecorrelation function in conformal field theory h k f | V k ( z ) · · · V k ( z n ) | k i i = N Y i < j ( z i − z j ) k N Y j = z k i · kj . (19)6f one choose a | k i i = k i | k i i , k i = k (1 − N ) , the correlation function reproduces the ground state ofCS model up to a phase factor : N Y i < j ( z i − z j ) k N Y j = z k i · kj = ( i ) k N ( N − Y i < j (2 sin β x i j ) . (20)Noticing that the ground state actually comes from the contraction of the vertex operators, then we candefine the state | ψ i as | ψ i = n Y j = V k ( z j ) | k i i ∼ Y i < j (2 sin β x i j ) : N Y j = V k ( z j ) : | k i i (21) = ψ ( x i ) e k P n > a − n p n / n | N + k i ≡ ψ ( x i ) V ( − ) k ( p ) | N + k i , with the action of the CS Hamiltonian, one obtains: H | ψ i ∼ − ψ X i ( ∂ i + β X j , i cot x i j ) ∂ i V ( − ) k ( p ) | k i i = ψ X n > a − n a n ( β N + n (1 − β )) + k X n , m > ( a − n − m a n a m + a − n a − m a n + m ) V ( − ) k ( p ) | k i i . Then we get the second quantized form of the Hamiltonian, ˆ H : = X n > a − n a n ( β N + n (1 − β )) + k X n , m > ( a − n − m a n a m + a − n a − m a n + m ) , (22)and the wave function in coordinate space: ψ λ ( { z i } ) = h k f | J λ ( a / k ) ψ V ( − ) k ( p ) | k i i (23)does satisfy the eigen-equation H ψ λ = h k f | J λ ( a / k ) H ψ V ( − ) k ( p ) | k i i (24) = h k f | J λ ( a / k ) ψ ˆ HV ( − ) k ( p ) | k i i = E λ ψ λ , provided the following defining operator equation for the Jack functions is satisfied. h | J λ ( a / k ) ˆ H = h | J λ ( a / k ) E λ . (25) For simplicity, we drop this phase factor in the following context. .4 Duality Relation Since for CS system a =
12 ( N + k , the Hamiltonian, eq.(22) can be written in a more compact formas ˆ H = X n , m > k ( a − n a − m a n + m + a − n − m a n a m ) + X n > (2 a a − n a n + (1 − β ) na − n a n − β a − n a n ) (26) = k ˛ ( z ∂ z ϕ ( z ) − a ) dz π iz + X n > a a − n a n + X n > (1 − β ) na − n a n − β a − n a n ≡ k (cid:16) ˆ H ′ ( k ) + (2 a − k ) a − n a n (cid:17) . There exists an explicit duality relation which can be read off as follows. First, we have the non-zeromode part of ˆ H as ˆ H ′ ( k ) = X n , m > ( a − n a − m a n + m + a − n − m a n a m ) + X n > ( k − − k ) na − n a n , (27)it has an apparent symmetry k − ↔ − k , (28)namely, let ˜ k = − k − ˆ H ′ (˜ k ) = ˆ H ′ ( k ) . (29)Now we shall show that k → ˜ k sends Young tableau λ to its dual diagram (transposed diagram) λ t = { λ t , λ t , . . . , λ tN } . Since ˆ H ′ ( k ) acts on Jack function gives ˆ H ′ ( k ) | J λ i = E ( k ) Y | J λ i (30) E ( k ) λ = X i (cid:16) λ i k − − (2 i − λ i k (cid:17) (31) = X i (cid:16) λ t i ˜ k − − (2 i − λ ti ˜ k (cid:17) = E (˜ k ) λ t Here we have used the following identity l ( λ ) X i = (2 i − λ i = l ( λ t ) X j = ( λ tj ) . We now conclude that J /βλ ( a − k ) = ( − β ) | λ | J βλ t ( − ka − ) . (32)Finally, notice that the inclusion of of the a part of ˆ H , (2 a − k ) a n a n = Nk | λ | , will not change eq.(32),only the eigenvalue of ˆ H is different on the two sides of eq.(32). For convenience, we neglect the summation symbols, one can recover them whenever one needs. .5 Generating Function and Skew-Fusion Coefficient The bra state | ψ i defined in eq.(21), is actually the second quantized form of the wave packet, which is alinear superposition of the incoming energy eigenfunctions defined on the unit circle. The superpositioncoefficients are understood as the creation operators creating incoming energy eigenstates. To see this,from eq.(32) and orthogonality condition, eq.(8), we can expand , exp( k X n > a − n n p n ) = X λ J /βλ ( a − k ) j /βλ J /βλ ( p ) ⇒ (33) exp( − k X n > a − n n p n ) = X λ J /βλ ( a − k )( − ) | λ | P βλ t ( p ) A /βλ . Here, the last equality in eq.(33) comes from the duality relation, eq.(32), and we have also defined P βλ ( p ) = J βλ ( p ) Q s ∈ λ ( a λ ( s ) β + l λ ( s ) + = J βλ ( p ) A βλ , which is proportional to the Jack polynomials but normalized differently, P /βλ ( p ) = z λ + X λ ′ <λ m λ ′ λ z λ ′ , (34)Similarly ψ † creates outgoing states, exp( 1 k X n > a n − n p − n ) = X λ J /βλ ( ak )( − ) | λ | P βλ t ( p ∗ ) A /βλ ; (35) exp( − k X n > a n − n p − n ) = X λ J /βλ ( ak ) J /βλ ( p ∗ ) j /βλ . (36)Besides being a wave packet, the state | ψ i is also a coherent state which is the eigenstate for all theannihilation operators J λ ( k − a ) with eigenvalue J λ ( p ) for each Young tableau λ . To show that ther.h.s of eq.(33) is actually a coherent state, we need define a 3-point function[19] h J µ J ν J − λ i ≡ g λµν ⇒ J µ ( p ) J ν ( p ) = P λ g λµν j − λ J λ ( p ) , and J µ J − ν |i = P λ g νµλ j − λ J − λ |i : = J − ν/µ |i is called the skew Jacksymmetric function. Hence we have the following equation. J µ | ψ i = X ν J µ J − ν J ν ( p ) j ν | k i i = X λ,ν g νλ,µ J − λ j λ J ν ( p ) j ν | k i i = X λ J − λ J λ ( p ) j λ J µ ( p ) | k i i (37) = J µ ( p ) | ψ i In general, the fusion coefficient g λµν is not a simple expression. However, if a rectangular Young tableauis involved, then g s r λ, s r /λ can be derived from the generating function, eq.(33) and the normalization9ondition, eq.(6). One gets J λ J − s r | i = A λ ‹ V ( − ) k ( p ) P λ ( p ) z − s − i dz i Q i < j | z i − z j | β Γ r B s r | i (38) = A λ J /β s r /λ ( a − k ) ¯ A s r /λ , rA s r /λ ¯ B s r /λ, r B s r | i , In deriving this , use has been made of the following identity [18], P λ ( p ) z − si = P s r /λ ( p ∗ ) . (39)Thus g s r λ, s r /λ j − s r /λ = A λ ¯ A s r /λ, r A s r /λ ¯ B s r /λ, r B s r , r (40) = B s r , r ¯ A λ, r ¯ B s r /λ, r . In reaching the last line in the above equation, we have used the following interesting identity: A λ ¯ A s r /λ, r = A s r /λ ¯ A λ, r . (41)This identity can be proven diagrammatically by moving squares in the Young tableaux. The detailedpresentation on this diagrammatic proof will appear elsewhere [31].Another example which involves the skew Jack function is of two sets of oscillators. Let’s considerthe following expansion: exp( X n > k ( a − n + ˜ a − n ) p n n ) = X l ( λ ) ≤ N J λ ( a − + ˜ a − k ) J λ ( p ) j λ . (42)One can also expand exp( P n > k ( a − n + ˜ a − n ) p n n ) in another way, exp( X n > k a − n p n n ) exp( X n > k ˜ a − n p n n ) (43) = X µ,ν J µ ( a − k ) j µ J µ ( p ) J ν ( ˜ a − k ) j ν J ν ( p ) = X | µ | + | ν | = | λ | J µ ( a − k ) j µ J ν ( ˜ a − k ) j ν g λµν J λ ( p ) j λ = X µ,λ J µ ( a − k ) j µ J λ/µ ( ˜ a − k ) J λ ( p ) j λ . In this particular case, s r /λ is taken to represent the Young tableau as in fig.3 and J s r /λ the corresponding Jack functionassociated with it. We drop the superscript 1 /β of A /βλ etc and add it explicitly when necessary. J λ ( a − + ˜ a − k ) = X µ J µ ( a − k ) j µ J λ/µ ( ˜ a − k ) . (44)Such that the skew Jack function can be obtained from the inner product J λ/µ ( ˜ a − k ) = h | J µ ( ak ) J λ ( a − + ˜ a − k ) | i . (45)Here, h | and | i are the bra and ket vacuum states for the a n ’s only. Eq.(45) turns out to be very usefulwhen we develop a skew-recursive integral for the construction of Jack states in section 5. From the discussions in the previous sections, we can see that there exist apparent similarities betweenthe CS model and the Coulomb gas picture. The Coulomb gas picture endowed with screening chargesoriginated in [21] and [20]. This method plays an important role in the calculations of the correlationfunctions in 2D conformal field theories. The conformal blocks are calculated with the insertions ofthe primary vertex operators which usually ends up with a charge deficit. In Coulomb gas picture, suchkind of charge deficit can be compensated by sandwiching a number of conformally invariant screeningcharges to make charge balanced while keeping conformal invariance intact. To see the similaritiesbetween the CS model and the Coulomb gas picture, notice the following:1) For one scalar theory, we have two kinds of vertex operators which may be interpreted as screen-ing vertex operators with the charges α ± in CFT and ± k ∓ in CS model.2) In both cases there are zero norm states.3) In CFT the descendant states are generated by the Virasoro algebra, while in the CS model theJack symmetric functions. Both expands a complete set of basis.However, despite all of those similarities, we have to address some apparent dis-similarities:1) α + α − = − while k ± ( − k ) ∓ = −
2) In CFT, zero norm state exists for generic α ± , while in CS model only for k ≤ , see eq.(6)3) In CFT the conjugate state is defined by L †− n = L n , while in CS model a †− n = a n . The twoconjugations coincide only in the case when k ≤ ⇒ c ≥ .Combining the above comparison 2) and 3) we see that there is an chance to map the two systemsinto each other in the case of Liouville type CFT, provided we can solve the problem 1), i.e. mappingbetween α ± and ± k ∓ . It turns out that it can be solved by introducing an additional scalar field. Forexample, in AGT conjecture, an additional U (1) scalar is needed to make the comparison betweenNekrasov instanton counting and the conformal blocks of the Liouville type, where the Virasoro struc-ture is explicitly shown ([23, 24]). In that case, Jack functions are the essential ingredients in buildingup the desired conformal blocks. We shall postpone our discussion on this point until our next paper[30]which is finishing soon. However, in the present paper, we shall restrict ourselves to the case of one setof oscillators in the operator formalism and to the case of generic k . In this case, we shall see that theVirasoro structure is implicit. 11 z z r Figure 2: Felder’s integration contour
The existence of the Virasoro structure in the Jack symmetric function has been investigated by theauthors of [1, 2]. In particular, it has been found there is a direct map between Virasoro singularvectors and the Jack functions of the rectangular Young tableau. Although it was suggested in [1] thatsuch relationship should lead to an integral representations for the Jack functions, only in some simplecases, the explicit construction was found. Starting from the next section, we shall present a completeconstruction for the Jack functions based on the Virasoro null vectors and their skew hierarchies. Here,to see how it works, we shall make some preparations. Let’s rewrite the Hamiltonian ˆ H as ˆ H = X n > (cid:16) α + ˜ a − n ˜ L n + ( N β + β − − α + ˜ a ) ˜ a − n ˜ a n (cid:17) , (46)here we have redefined ˜ a n = √ a n , ˜ a − n = a − n √ , n > , ˜ a = √ a , α ± = ± √ k ± , α + + α − = α ,and the Virasoro generator ˜ L n = X m ∈ Z : ˜ a m ˜ a n − m : − ( n + α ˜ a n . (47)Notice that in this convention, the Hamiltonian separates into two parts, one for the ”Virasoro part”which is proportional to ˜ L n and the other part is in fact the conserved charge and is always diagonalon Jack functions and its eigenvalue proportional to the norm of the Young tableau. It is clear thatany “Virasoro” singular vector | χ r , s i is an eigenstate of ˆ H whose eigenvalue suggests that | χ r , s i isproportional to the Jack state J { s r } . Of course, The singular vector in the “Virasoro” sector is notsingular on the CS model side, since for generic k , Jack functions has non-zero norm. This is becausethe redefinition, eq.(47),is not unitary and the conjugation in the “Virasoro” sector is not hermitian.While in the CS model, the conjugation is always Hermitian for real k .To make the comparison more clear, we shall assume that the a eigenvalues differ from k i definedin eq.(19). Consider a general vacuum state | p i in the CS model, which is mapped to a highest weightstate with conformal dimension h p = p ( p − α ) in the Virasoro sector. The singular vectors appears This redefinition is not unitary but it makes the following computation simpler. p p = p r , s ≡
12 (1 − r ) α + +
12 (1 − s ) α − . and at the level rs . And this null vector can be constructed explicitly by making use of the fact that h p r , s = h p − r , − s , | χ r , s i = S r | p r , − s i which satisfies ˜ L n | χ r , s i = δ n , ( h p r , s + rs ) | χ r , s i , n ≥ (48) ˜ a | χ r , s i = p − r , − s | χ r , s i . Here S ≡ S + = ¸ V + ( z ) dz , V ± ( z ) = : exp( α ± ˜ ϕ ( z )) : , α ± = ± √ k ± are called the screening chargesin the Virasoro sector. When multiple S ’ act together, we take Felder’s contour [22] for S r (see fig.2).to get | χ r , s i = S r | p r , − s i = ‹ r Y i < j | z i − z j | β e k P n > a − n p n r Y i = z − s − i dz i | p − r , − s i ∝ J − s r | p − r , − s i . (49)Notice that in the equation above we have used a − n instead of ˜ a − n to make the comparison with eq.(21). The construction of the Jack states for the rectangular diagrams, as well as the null vectors of theVirasoro algebra hidden in the CS model, thus reduces to the evaluation of the multi-integrals of theSelberg type in eq.(49). Since there is no closed formula for such type of operator valued multi-integrals, we choose to discuss some simple cases here. The simplest one is the case of one screeningcharge for the Young tableau { n } . From eq.(49) and duality relation eq.(32), one can verify that thestate | J n i = ˛ e − k P m > a − mzmm ( − n n ! z − n − dz | p − n , − i (50) = ˛ e α − ˜ ϕ ( z ) ( − n n ! dz | p − n , i = ˛ e p − n , ˜ q e z ˜ L − ( − n n ! z − n − dz | p , − i = e p − n , ˜ q ( − ˜ L − ) n | p , − i reproduces the Jack polynomial J { n } . To take its conjugate state we have to be careful to take itsHermitian conjugation. Now let’s workout the Hermitian conjugate of ˜ L − . ˜ L − = P n ≥ ˜ a − n − ˜ a n = P n ≥ a − n − a n ≡ L − . Here we have defined L n = P m ∈ Z : a m a n − m : . It can be checked that L †− n = L n and L | p , − i = k | p , − i Thus the normalization of J n reads h p , − | ( L ) n ( L − ) n | p , − i = (2 h + n − n )(2 h + n − n − · · · (2 h ) · (51) = Y s ∈ n ( l ( s ) + + a ( s ) 1 β )( l ( s ) + ( a ( s ) +
1) 1 β ) We have dropped the factor π i for convenience. We also use the label J s r instead of J { s r } for the same reason. k → − / k and meanwhile transpose the parti-tion(Young tableau), the theory doesn’t change. This implies one can define the Jack polynomial withYoung tableau { n } as: h k | ( L ) n up to a normalization factor k − n , J n = k − n h k | ( L ) n = n ! k − n h | ˛ e − k ϕ ( w ) w n − dw . In the previous section we have shown that any “Virasoro” null vector, represented by a multiple integralof the Selberg type, is a Jack state of the rectangular graph up to normalization. One may naturally askhow the other Jack states be represented. Our answer to this question is positive. In this and thefollowing sections we shall show that any “Virasoro” null vector, or equivalently, the Jack state ofthe rectangular graph, skewed by another Jack state is again a Jack state. In this way we can buildany desired Jack state recursively either in operator or multiple integral formalism. There are alreadytwo kinds of integral representations of the Jack symmetric polynomials[1][13]. Both are based onthe method that the number of arguments N in J λ ( p ) are increased recursively. The method we havedeveloped is, however, in a different manner. While other methods are based on adding blocks ofsquares to the Young tableau, we are trying to subtract a block of squares from a given rectangular one.And the other difference is that we first build an operator formalism, and later an integral formalismbased on it (in contrast to the pure operator formalism, [14]. The way to subtract a block of squaresfrom a given Young tableau is described in mathematical language as ”skewing”. We have already seenthis method in section 2.5. The skewing of λ by µ when λ is a rectangular one is, however, simpler. Inthis case, the summation only contains one term. This fact is proven by Kadell in [18] and is presentedas P λ ( p ) Q Ni = z − ni = P n N /λ ( p ∗ ) with the Young tableau n N /λ : = { n , · · · , n , n − λ l , n − λ l − , · · · , n − λ } ⇒ λ ≤ n and l ≤ N . In fact, in eq.(38), we have made use of this identity in the calculation ofthe fusion coefficients. Here, however, we shall show that this particular skew relation has profoundmeaning related to the Virasoro singular vectors. One can also view our method as an alternative proofon Kadell’s formula, eq.(39). Proposition 1
Given a Jack bra state of the rectangular graph, | p − N , − n i { n N } = J − n N | p − N , − n i , if it is acted from the left by a Jack annihilation operator J λ , λ ≺ n N , J n N /λ | p − N , − n i : = J λ ( ak ) | p − N , − n i { n N } ,then J n N /λ | p − N , − n i is again a Jack bra state up to a normalization constant. J λ ( ak ) | p − N , − n i { n N } ∝ | p − N , − n i { n N /λ } . Here, the introducing of p − N , − n for the oscillator vacuum state is artificial. It just make the comparisonwith the “Virasoro” null vector easier. The Young tableau { n N } /λ is shown in fig.3. Before rushing to14 n λ λ R Figure 3: The Young tableau for n N /λ , the shadowed part labeled as λ R has been cut out from n N .the proof of the proposition, we start from some simple examples according to the level of the graphsbeing cut. In order to show that | p − N , − n i { n N } ∝ | χ N , n i , (52)we just have to calculate ˆ H | χ N , n i = ( N β + β − − α + ˜ a ) ˜ a − n ˜ a n | χ N , n i (53) = Nn | χ N , n i Notice that E n N = Nn , eq.(52) is proved. Level one graph is just a single square. If we cut a square in the SE corner of the rectangular graph, theresulting state is proportional to ˜ a J n N | p − N , − n i . It is easy to show that the ”Virasoro part” of the Hamiltonian have eigenvalue on the resulting state α + ˜ a − n ˜ L n ˜ a | p − N , − n i { n N } = (( N − β − ( n − a | p − N , − n i { n N } , And the diagonal part of ˆ H has the eigenvalue ( β N + β − − α + ˜ a )( nN −
1) ˜ a | p − N , − n i { n N } = n ( nN −
1) ˜ a | p − N , − n i { n N } . Combining the two parts together, we have ˆ H ˜ a | p − N , − n i { n N } = E n N / (cid:3) ˜ a | p − N , − n i { n N } , Again this state has the correct property corresponding to the skew Young tableau { n N / (cid:3) } .15 .1.3 Example 2: level 2 There are two different Young tableaux λ (1) and λ (2) at level 2. If we cut these Young tableaux from arectangular one s r , the resulting states will span a two dimensional Hilbert space. Let us denote themas | χ i = ( ˜ a k + A ˜ a ) | ψ i { n N } , here A is an undetermined parameter. Note that the “diagonal part” of the Hamiltonian only shift theeigenvalue by a global constant. So for the eigen-equation ˆ H | χ i = E χ | χ i . we can drop this diagonal term and consider only the ”Virasoro part” of the Hamiltonian. After a simplecomputation, one finds A = √ k or − √ k corresponding to λ (1) = { } and λ (2) = { } respectively. It is straightforward to continue on to level 3 graphs being cut. The resulting state is denoted as | χ i = ( ˜ a √ k ) + A ˜ a ˜ a + B ˜ a | ψ i { n N } , here A , B are undetermined parameters. There are three independent solutions for the eigen-equationcorresponds to the three Young tableau at level 3.For the horizontal Young tableau { , } , one gets : A = / k , B = √ / k . For the vertical Young tableau { , , } , A = − / k , B = √ / k . For the symmetric Young tableau { , } , A = − k (1 − k ) , B = − √ k . These results reproduce the level 3 Jack polynomials.
Having checked for the low level skew Jack states, we are encouraged to find a more general proof forthe proposition 1. In this section, we shall show that if h J λ | is a Jack symmetric function related toYoung tableau λ , then J λ | χ r , s i is proportional to a Jack symmetric function related to a Young tableau s r /λ , with λ ≺ s r . Here, | χ r , s i is a Virasoro singular vector descendant from | p − r , − s i , see eq.(49). Wecan prove this in operator formalism first by “brute force”. Later in the next section we shall present itin a more compact manner. To proceed, we need to write the operator valued Jack function as follows: J λ = X λ ′ , | λ ′ | = | λ | C λ ′ λ a λ ′ · · · a λ ′ s . (54)16hen consider the commutator of J λ and ˆ H defined in eq.(22), [ J λ , ˆ H ] = X λ ′ , l C λ ′ λ a λ ′ · · · [ a λ ′ l , ˆ H ] · · · a λ ′ s = X λ ′ , l C λ ′ λ a λ ′ · · · [(1 − β )( λ ′ l ) a λ ′ l + k λ ′ l L ′ λ ′ l + β Nla λ ′ l ] · · · a λ ′ s , here L ′ l = X m ∈ Z (: a m a l − m :) − a a l . In deriving this, we have used the commutation between a l and ˆ H : [ a l , ˆ H ] = (1 − β ) l a l + X m > kla − m a l + m + X l > m > kla l − m a m + β Nla l (55) = (1 − β ) l a l + klL ′ l + β Nla l . In moving L ′ λ ′ l to the most left by the commutation relation [ L ′ n , a m ] = − ma m + n for n , m > , moreterms are generated, [ J λ , ˆ H ] = X λ ′ , l C λ ′ λ a λ ′ · · · a λ ′ l − [(1 − β )( λ ′ l ) a λ ′ l + β N λ ′ l a λ ′ l ] · · · a λ ′ s (56) + X λ ′ , n < l C λ ′ λ (2 k λ ′ l λ ′ n ) a λ ′ · · · a λ ′ n + λ ′ l · · · a λ ′ l − a λ ′ l + · · · a λ ′ s + X λ ′ , l C λ ′ λ (2 k λ ′ l ) L ′ λ ′ l a λ ′ · · · a λ ′ l − a λ ′ l + · · · a λ ′ s . Let us define some notations to simplify our calculation. We denote the first line on the r.h.s. of eq.(56)as A since this term retain the same number of a n ’s comparing to the original terms in J λ , the secondline is named as A − since it contains one less a n comparing to the original term in J λ , the third lineseparates into two terms A + + / A , which are defined as A + = X λ ′ , l ,λ ′ l > m > C λ ′ λ k λ ′ l a λ ′ · · · a λ ′ l − ( a λ ′ l − m a m ) a λ ′ l + · · · a λ ′ s / A = X λ ′ , l , m > C λ ′ λ (2 k λ ′ l )( a − m a λ ′ l + m ) a λ ′ · · · a λ ′ l − a λ ′ l + · · · a λ ′ s . If we apply eq.(56) to a bra vacuum sate h | , the contribution of / A vanishes. Since h J λ | is an eigenstateof ˆ H , we conclude h | J λ ˆ H = h | J λ E λ ⇒ h J λ , ˆ H i = E λ J λ + / A (57) E λ J λ = A + + A − + A (58)17ow we calculate the action of ˆ H on the ket state J λ | χ r , s i ˆ H J λ | χ r , s i = [ ˆ H , J λ ] | χ r , s i + J λ ˆ H | χ r , s i = [ ˆ H , J λ ] | χ r , s i + E rs J λ | χ r , s i . (59)By moving L ′ λ l in eq.(55) to the most right, we get [ ˆ H , J λ ] = E λ J λ − A + − A (60) − X λ ′ , l (2 k λ ′ l ) C λ ′ λ a λ ′ · · · a λ ′ l − a λ ′ l + · · · a λ ′ s L ′ λ ′ l − X λ ′ l > m > a λ ′ l − m a m . and A + + A + X λ ′ , l (2 k λ ′ l ) C λ ′ λ a λ ′ · · · a λ ′ l − a λ ′ l + · · · a λ ′ s L ′ λ ′ l − X λ ′ l > m > a λ ′ l − m a m (61) = X λ ′ , l k λ ′ l C λ ′ λ a λ ′ · · · a λ ′ l − a λ ′ l + · · · a λ ′ s × [ X m > ( a − m a λ ′ l + m ) + X λ ′ l > m > ( a λ ′ l − m a m ) + [ − ( k − k ) λ ′ l + Nk ] a λ ′ l ] = k ( √ α − √ a ) + Nk ) | λ | J λ + X λ ′ , l k λ ′ l C λ ′ λ a λ ′ · · · a λ ′ l − a λ ′ l + · · · a λ ′ s ˜ L λ ′ l , In deriving these, use has been made of eq.(58) and eq.(47). Substituting the results in eqs.(60-61) toeq.(59) and using the property of the Virasoro singular vector, ˜ L l | χ r , s i = , l > , we conclude ˆ H J λ | χ r , s i = h E λ + E r , s − | λ | ˆ M i J λ | χ r , s i , (62)here ˆ M = k ( √ α − √ a + Nk ) , on | χ r , s i , ˜ a gives ˜ a | χ r , s i = + r α + + + s α − ! | χ r , s i . (63)The establishment of eq.(62) finishes the proof of the proposition 1 we proposed before, that is, Jackpolynomials for rectangular Young tableaux, skewed by an Jack state is again a Jack state. The proposition 1 is proven in the previous subsection by making use of the eigen-equation for ˆ H .However, we know that the eigenstate of ˆ H can always been written as an integral transformation, h J λ | ∝ h | ˛ e k P n > an − n p − n Y i < j | z i − z j | β J λ ( { z i } ) Y i dz i z i (64) ≡ h | ˛ F λ ( a , z ) Y i dz i z i . a ≡ { a n } , z ≡ { z i } , and in the following the integration measure Q i dz i z i will be implied withoutwritten out explicitly. With J λ realized in this way, we found that the brute force proof can be rewrittenin a more compact form with less indices involved. Using: a − m e k P n > an − n p − n = e k P n > an − n p − n ( a − m + k p − m ) (65) e k P n > an − n p − n a − m = ( a − m − k p − m ) e k P n > an − n p − n , (66)we have ˆ F λ ( a , z ) ˆ H = ˆ ∞ X n , m = k (( a − n − k p − n )( a − m − k p − m ) a n + m + ( a − n − m − k p − n − m ) a n a m ) (67) + ∞ X n = ( a − n − k p − n ) a n ( β N + n (1 − β )) F λ ( a , z ) = ˆ ˆ HF λ ( a , z ) + ˆ ∞ X n , m = ( k p − n p − m a n + m − k p − m a − n a n + m − k p − n − m a n a m ) − ∞ X n = k p − n a n ( β N + n (1 − β )) F λ ( a , z ) . Since the terms containing a − n ’s on the most left will annihilate the bra vacuum h | , we conclude thefollowing identity h J λ , ˆ H i = E λ J λ − ˆ k ∞ X n , m = p − m a − n a n + m F λ ( a , z ) (68)will be true. Comparing eq.(67) and eq.(68), we have ˆ ∞ X n , m = ( k p − n p − m a n + m − k p − n − m a n a m ) (69) − ∞ X n = k p − n a n ( β N + n (1 − β )) F λ ( a , z ) = E λ J λ . Now we move a − n ’s in the last term in eq. (68) to the most right to get: h J λ , ˆ H i = E λ J λ − ˆ k ∞ X n , m = F λ ( a , z ) p − m ( a − n + k p − n ) a n + m . (70)Using eq.(69), the last term in the above eqation can be rewritten as − E λ J λ − ˆ F λ ( a , z ) ∞ X n , m = k ( p − m a − n a n + m + p − n − m a n a m ) (71) + ∞ X n = k p − n a n ( β N + n (1 − β )) , h J λ , ˆ H i = − E λ J λ − k ˆ F λ ( a , z ) ∞ X n , m = p − m ∞ X n = ( a − n a n + m ) + m − X n = a n a m − n (72) + ∞ X m = p − m a m ( kN + m ( k − − k )) = − E λ J λ − k ˆ F λ ( a , z ) ∞ X m = p − m ( ˜ L m + a m ( kN − ( k − − k )) − a ) , where ˜ L m is the same as what we defined in eq.(47). When we apply eq.(72) to a Virasoro singularvector | χ rs i , ˜ L n | χ rs i = implies: h ˆ H , J λ i | χ rs i = E λ J λ + k ˆ F λ ( a , z ) X m > p − m a m ( kN + ( k − k − ) − a ) | χ rs i . (73)Now we can check, using eqs.(64-65), − k ˆ F λ ( a , z ) X m > p − m a m = [ J λ , X m > a − m a m ] = | λ | J λ , which leads to ˆ H J λ | χ rs i = (cid:16) E λ + E χ rs − k | λ | ( kN + k − k − − a ) (cid:17) J λ | χ rs i . (74)Here and before we have assumed that Virasoro ˜ L n singular state | χ rs i is an eigenstate for CS Hamilto-nian ˆ H with eigenvalue E χ rs . This can be checked as follows. From the formula, eq.(46) H = k ∞ X n = a − n ˜ L n + ∞ X n = ( β N + β − − ka ) a − n a n , we arrive at: E χ rs = ( β N + β − − √ k p − r , − s ) l , here l is the level of the decendant states. By theconstruction of Virasoro singular vectors, we know l = rs , √ p − r , − s = (1 + r ) k − (1 + s ) k − , hence E χ rs = " β N + β − − k ((1 + r ) k − (1 + s ) 1 k ) | λ | (75) = rs + β ( N − r ) rs = E { s r } . Thus eq.(73) implies that J λ | χ i is an eigenstate of ˆ H with the eigenvalue E λ + E s r − k | λ | ( kN + k − k − − (1 + r ) k + (1 + s ) k − ) (76) = E λ + E s r − | λ | (( N − r ) β + s )) = E s r /λ . This concludes our proof of proposition 1. 20
Skew-Recursive Construction of Jack States
In the previous sections we have shown that if we cut, inside a rectangular Young tableau of size r × s , any sub-Young tableau in a skew way, the resulting Young tableau is unique and hence thecorresponding Jack function, which is named as J s r /λ . This Jack function, J s r /λ , can be used againto cut another bigger rectangular Young tableau of size r × s to get J s r / ( s r /λ ) and so forth. If weknow the construction of the Jack function for a definite Young tableau, we can build a tower of Jackfunctions upon it in such a skew way. Of course, if we start with a trivial Young tableau (empty), thenthe tower of Jack functions is built upon the constructions of the Jack function for rectangular Youngtableau only, which are in fact Virasoro singular vectors. Following is the precise procedure whichleads to the recursive construction of the Jack functions. First, J − λ acts on the left vacuum to create a bra state λ h | ≡ h | J λ , J λ acts to the right will produce a skew ket state J λ | i { s r } ≡ J λ J − s r | i ≡ J − s r /λ | i (77) = g s r λ, s r /λ j − λ, s r ] J − [ λ, s r ] | i = g s r λ, s r /λ j − λ, s r ] | i { s r /λ } . Here we use the symbol [ λ, s r ] to represent the unique Young tableau s r /λ , see fig.3, where λ R means λ rotated by π angle. Such type of Young tableau, i.e., a rectangular one cut in the SE corner by a rotated λ , will be frequently used recursively. For example, [[ λ, s r ] , s r ] will define another Jack functionassociated with the Young tableau s r cut in the SE corner by [ λ, s r ] rotated.To facilitate such recursive procedure, we shall define the following abbreviation [ r , s ] λ, n ≡ h · · · [[ λ, s r ] , s r ] , · · · , s r n n i (78) h J ( r , s ) λ, n + ≡ h J s rn + n + J − ( r , s ) λ, n , J ( r , s ) λ, = J λ . Here and after, however, we shall take λ to be the empty Young tableau, so we shall use the abbreviation [ r , s ] n ≡ h · · · [[ s r , s r ] , s r ] , · · · , s r n n i (79) h J ( r , s ) n + | ≡ h J s rn + n + J − ( r , s ) n | J ( r , s ) = . It is clear that any regular Young tableau can be represented uniquely by two integer vectors of di-mension n each, [ r , s ] n , where n − is the total number of skews for the Young tableau considered21ccording to our convention. From the definition eq. (77), we know that J ( r , s ) n differ from the standardJack symmetric function J [ r , s ] n only by a normalization constant. For example, | J − ( r , s ) i = J − s r | i (80) h J ( r , s ) | = h | J s r J − s r ≡ h | J s r / s r (81) = g s r s r , s r / s r j − s r / s r h | J [ s r , s r ] J − ( r , s ) | i = J ( r , s ) J − s r | i (82) = g s r s r , s r / s r j − s r , s r ] J [ s r , s r ] J − s r | i = g s r s r , s r / s r j − s r , s r ] g s r [ s r , s r ] , [[ s r , s r ] , s r ] × j − s r , s r ] , s r ] J − [[ s r , s r ] , s r ] | i In general, the normalization constant can be determined as following. Suppose J ( r , s ) n = C [ r , s ] n J [ r , s ] n , then h J ( r , s ) n + | = h | J s rn + n + J − ( r , s ) n = h | J s rn + n + J − [ r , s ] n C [ r , s ] n (83) = g s rn + n + [ r , s ] n , [[ r , s ] n , s rn + n + ] j − r , s ] n , s rn + n + ] h J [ r , s ] n + | C [ r , s ] n + , so C λ can be defined recursively: C [ r , s ] n + = C [ r , s ] n g s rn + n + [ r , s ] n , [ r , s ] n + j − r , s ] n + , (84)where the fusion coefficient g s rn + n + [ r , s ] n , [ r , s ] n + is calculated in eq.(40). In practice, an integral formalism is more useful in analysis. Based on the operator formalism, wederive the following integrals for building the Jack symmetric functions.
Since J s r ’s are essentially the building blocks for any generic Jack function J [ r , s ] n , we come back to theconstruction of J s r by the following integral, J − s r | p i = ˆ [ dz ] + r r Y i = z − s − i e P n > a − npnn | p i , (85)22ere we have defined [ dz ] + r ≡ B s r Γ r Y i < j | z i − z j | β r Y i = dz i [ dz ] − s ≡ ( − sr A s r Γ r Y i < j | z i − z j | /β s Y i = dz i . To relate J − s r | p i to a Virasoro singular vector, we introduce two scalar field, ϕ (0) and ϕ (1) to providethe right integration measure [ dz ] , h ϕ ( i ) ( z ) ϕ ( j ) ( z ′ ) i = δ i j log( z − z ′ ) , and define the vertex operator integral V ± = ˛ : e k ± ( ϕ (0) + ϕ (1) )( z ) : . Clearly, V ± is the screening charge for the Virasoro algebra L ± n respectively. Here T , ± ( z ) ≡ X n ∈ Z L ± n z − n − =
14 ( ∂ z ( ϕ (0) + ϕ (1) )) ±
12 ( k − k ) ∂ z ( ϕ (0) + ϕ (1) ) . Define | χ + rs i = ( V + ) r | p r , − s ih χ − rs | = h p − r , s | ( V − ) s , clearly we have L + n | χ + rs i = , n > h χ − rs | L −− n = , n > . However, to get J s r , we have to project out one of the two scalar fields, say, ϕ (0) and from eq.(45) weget, J s r ( a (1) − k ) | p i ∝ h p | χ + rs i (86) h p | J s r ( a (1) k ) ∝ h χ − rs | p i , (87)so that now J ± s r contains only a (1) ± n ’s.Now the Jack states read | J − s r i = ˆ r Y i = z − s − i [ dz ] + r e k P n > a (1) − n pnn | p i ≡ J − ( r , s ) i (88) h J s r | = h p | ˆ e k P n > a (1) n p − n − n s Y i = z r − i [ dz ] − s ≡ h J ( r , s ) , (89)23ere | p i i is the vacuum state (no oscillator excitations) for the ϕ ( i ) scalar a ( i ) n | p i i = δ n , p ( i ) | p i i n ≥ (90)Notice that since a (0) − n has been projected out, J − s r is no longer a null vector for L + n . However, J − s r isstill a null vector for the modified Virasoro generator ˜ L n constructed with ϕ (1) only, see, eq. (47). Now we shall specify how the bra state h p + p r , s | and the ket state | p − p r , s i are labeled.Since we have L ± n acts on ket-state and bra-state respectively, so we have different screening chargesfor L ± n respectively. α ++ = √ k , α + − = − √ k − for L + n , and α − + = − √ k , α −− = √ k − for L − n . If we combine ϕ ( i ) + ϕ ( i + into a single scalar, ϕ = √ ϕ ( i ) + ϕ ( i + ) , and a | p r , s i = p + r , s | p r , s ih p r , s | a = h p r , s | p − rs , then we define p + rs =
12 (1 − r ) α ++ +
12 (1 − s ) α + − (91) =
12 (1 − r ) √ k −
12 (1 − s ) √ k p − rs = (1 + r ) α − + + (1 + s ) α −− (92) = − (1 + r ) √ k + (1 + s ) √ k Now consider a | p + r , s i i , i + = p + r , s | p r , s i i , i + . However, when, say ϕ ( i ) is projected out, then a ( i + | p + r , s i i + = √ p + r , s | p + r , s i i + . h p − r , s | , the projection is similar. To see this notation will provide the correct integration measure,one could check: h p −− r , s | ( V − ) s / Γ s | p + − r , − s i i (93) = h p −− r , − s | ˆ Q i < j ( z i − z j ) k Q si = z √ k a i Γ s e √ k P n > an − n p − n Y dz i | p + − r , − s i i = i + h p −− r , − s | ˆ Q i < j ( z i − z j ) k Γ s s Y i = z √ k ( (1 − r )( − √ k ) + (1 − s ) √ k ) i e k P n > a ( i + n − n p − n Y i dz i = i + h p −− r , − s | ˆ Y i < j " ( z i − z j ) z i z j k e k P n > a ( i + n − n p − n s Y i = z r − i dz i / Γ s = i + h χ rs | ∝ h p −− r , − s | ˆ e k P n > a ( i + n − n p − n s Y i = ( z i ) r − [ dz ] − s = i + h p −− r , − s | J s r ( a ( i + k ) , produces the Jack states of rectangular graph. Now we have | J − s r i = | J − ( r , s ) i = h p | χ r , s i (94) = ˆ e P n > a (1) − n pnn k r Y i = ( z , i ) − s − [ dz ] + r | p + − r , − s i p = √ p + − r , − s =
12 (1 + r ) k −
12 (1 + s ) 1 k . For one skew Young tableau of the type as in fig.4.a, we have to introduce ϕ (2) scalar and project out ϕ (1) scalar. The resulting state is actually the skew Jack state, as what has been shown in eq.(45); Weproceed to construct h J ( r , s ) | = h χ r , s | e δ k q J − ( r , s ) | p + − r , − s i (95) = h p −− r , − s | ( V − ) s e δ k q (1) J − ( r , s ) | p + − r , − s i = h p −− r , − s | ¨ e P n > k a (2) n P z − n , i − n s Y i = ( z , i ) r − Y s , r (1 − z z ) r Y i = ( z , i ) − s − [ dz ] − s [ dz ] + r . Here we have defined s m , r n Y (1 − z n z m ) ≡ s m Y i = r n Y j = (1 − z n , j z m , i ) . r ss a br s r s r s Figure 4: a. Young tableau { s r } / { s r } b. Young tableau { s r } / ( { s r } / { s r } ) , this is a three-ladderYoung tableau.and e δ k q is introduced to eliminate the charge deficit in ϕ (1) sector, that is h p −− r , − s | e δ k q (1) | p + − r , − s i , . (96)will give the following equation, √ p −− r , − s = δ k + √ p + − r , − s (97) δ k = ( p −− r , − s − p + − r , − s ) 1 √ (98) =
12 ( 1 k − k ) − k + r − r ) + k (1 + s − s ) = √ n α − + p + r − r , s − s o α ± = α ± + + α ±− . For two skew Young tableau,fig.4.b, ϕ (3) is introduced and ϕ (2) eliminated. | J − ( r , s ) i = h p −− r , − s | J ( r , s ) e δ k q (2) | χ r , s i (99) = h p −− r , − s | J ( r , s ) | e δ k q (2) ( V + ) r | p + − r , − s i = ˆ r Y i = ( z , i ) − s − [ dz ] + r s , r Y − z z ! s Y i = ( z , i ) r − [ dz ] − s × s , r Y − z z ! r Y i = ( z , i ) − s − [ dz ] + r exp k X n > a (3) − n n X i z n , i | p + − r , − s i δ k + √ p + − r , − s = √ p −− r , − s (100) δ k = √ − α + − p + r − r , s − s ) = √ α − + p − r − r , s − s ) =
12 ( 1 k − k ) − k + r − r ) + k (1 + s − s ) . In general, proceed recursively, we have, for n odd J − ( r , s ) n | p + − r n , − s n i n = n − h p −− r n − , − s n − | J ( r , s ) n − e δ k n − , n q ( n − | χ r n , s n i n − , n (101) = ˚ exp k X m > a ( n ) − m P r n i = z mn , i m r n Y i = ( z n , i ) − s n − s n − , r n Y (1 − z n z n − ) × s n − Y i = ( z n − , i ) r n − − s n − , r n − Y (1 − z n − z n − ) · · · s Y i = ( z , i ) r − × s , r Y (1 − z z ) r Y i = ( z , i ) − s − [ dz ] o [ n ]! | p + − r n , − s n i n . Here δ k n − , n = √ (cid:16) − α + − p + r n − − r n , s n − − s n (cid:17) (102) = √ (cid:16) α − + p − r n − r n − , s n − s n − (cid:17) =
12 ( 1 k − k ) − k + r n − r n − ) + k (1 + s n − s n − ) . For n even, n h p −− r n , − s n | J ( r , s ) n = h χ r n , s n | e δ k n , n − q ( n − J − ( r , s ) n − | p + − r n − , − s n − i n − (103) = n h p −− r n , − s n | ˚ exp k X m > a ( n ) m P s n i = z − mn , i − m s n Y i = ( z n , i ) r n − s n , r n − Y (1 − z n − z n ) × r n − Y i = ( z n − , i ) − s n − − s n − , r n − Y (1 − z n − z n − ) · · · s Y i = ( z , i ) r − × s , r Y (1 − z z ) r Y i = ( z , i ) − s − [ dz ] e [ n ]! . δ k n , n − = √ (cid:16) − α + − p + r n − r n − , s n − s n − (cid:17) (104) = √ α − + p − r n − − r n , s n − − s n ) =
12 ( 1 k − k ) − k + r n − − r n ) + k (1 + s n − − s n ) . The integration measures are defined as following: for n odd, [ dz ] o [ n ]! ≡ [ dz ] + r [ dz ] − s · · · [ dz n ] + r n . For n even, [ dz ] e [ n ]! ≡ [ dz ] + r [ dz ] − s · · · [ dz n ] − s n . Eq.(101) and eq.(103) are the main results of our present work. It provides an integral represen-tation for any Jack symmetric function which, in our formalism, is labeled by two integer vectors ofdimension n each, ( r , s ) n .The integral representation not only provide a useful tool in analyzing problems involving Jacksymmetric functions, but also give an explicit construction of the Jack symmetric functions in termsof free bosons. It is also desirable to work out explicitly the Selberg type multi-integrals appearing ineq.(101) and eq.(103). Having got the integral representation for a general Jack symmetric function, it is then straightforwardto get the Jack symmetric polynomials in any number N of arguments z i . Notice that in the followingwe shall present the unnormalized Jack polynomials. However, the normalization constants can beeasily worked out.First, let us consider n even, thus J / k ( r , s ) n ( { z i } ) ≡ h J ( r , s ) n exp k X m > a ( n ) − m m N X i = z mi | p + n i n (105) = ˚ s n , N Y (1 − zz n ) s n Y i = ( z n , i ) r n − [ dz n ] − s n s n , r n − Y (1 − z n − z n ) r n − Y i = ( z n − , i ) − s n − − [ dz n − ] + r n − × s n − , r n − Y (1 − z n − z n − ) s n − Y i = ( z n − , i ) r n − − [ dz n − ] − s n − · · ·× s , r Y (1 − z z ) r Y i = ( z , i ) − s − [ dz i ] + r . In fact, one can easily see that the distinguishment between even and odd skews is artificial. n odd, J / k ( r , s ) n ( { z − i } ) ≡ n h p − n | exp k X m > a ( n ) m − m N X i = z − mi J − ( r , s ) n i (106) = ˚ N , r n Y (1 − z n z ) r n Y i = ( z n , i ) − s n − [ dz n ] + r n s n − , r n Y (1 − z n z n − ) s n − Y i = ( z n − , i ) r n − − [ dz n − ] − s n − × s n − , r n − Y (1 − z n − z n − ) r n − Y i = ( z n − , i ) − s n − − [ dz n − ] + s n − · · ·× s , r Y (1 − z z ) r Y i = ( z , i ) − s − [ dz ] + r . Now p ± n can be easily worked out, p + n = √ p −− r n , − s n = √
12 (1 − r n ) α − + +
12 (1 − s n ) α −− ! (107) = − k − r n ) + k (1 − s n ) p − n = √ p + − r n , − s n = k + r n ) − k (1 + s n ) . (108) This work is part of the CAS program ”Frontier Topics in Mathematical Physics” (KJCX3-SYW-S03)and is supported partially by a national grant NSFC(11035008).
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