Unformatted text preview: Lecture 1 Quantum Field Theories: An introduction
The string theory is a special case of a quantum field theory (QFT). Any QFT deals of Riemannian manifolds, the dimension of is with smooth maps the dimension of the theory. We also have an action function defined on the set Map of smooth maps. A QFT studies integrals (1.1) Map Here stands for some measure on the space of paths, is a parameter (usually very small, Planck constant) and Map is an insertion function. The number should be interpreted as the probability amplitude of the contribution of the map to the integral. The integral (1.2) Map is called the partition function of the theory. In a relativistic QFT, the space has a Lorentzian metric of signature . The first coordinate is reserved for time, the rest are for space. In this case, the integral (1.1) is replaced with (1.3) Map Let us start with a dimensional theory. In this case is a point, so is and is a scalar function. The Minkowski partition function a point of the theory is an integral (1.4) Following the Harvard lectures of C. Vafa in 1999, let us consider the following example: 1 % 8 Q Uv$ ! t h GXWVf" b d I `y Q Uv$ #! R " t Dx8 [email protected] w8 7 GXWVU"TS d !b GrR1tututs5r qp " 8 ig2f06 ( & $ #! R eHb h d c I F PHG 7 ( &$" DB E8 [email protected] 98 32547 610)'%#! aWVGX`U"Y ST QR& $ DB E8 [email protected] 2 LECTURE 1. QUANTUM FIELD THEORIES: AN INTRODUCTION Example 1.1. Recall the integral expression for the function: but can be meromorphically extended to the This integral is convergent for Re whole plane with poles at . We have By substituting Although in the substitution above is a positive real number, one can show that formula (1.6) make sense, as a Riemann integral, for any complex with Re . When Re this is easy to see using the Hankel representation of as a contour integral in the complex plane. When is a pure imaginary, it is more delicate and we refer to [KratzerFranz], 1.6.1.2. Taking , we can use to define a probability measure on . It is called the Gaussian measure. Let us compute the integral Here Obviously, Also & h h & $ d fo p 1u ufo d u o r W d h g & h & $ & ty d h g & p t h i g u Ap d W b F Uo d & & Tv Ah & $ Uv$ t h f d 'h f d W ! b } I & h ih~$ d ~h } u d v v 9 y i v j y H { v v & v g ihGR& $ x t vu zd gV v d h y qw v s d o t u std o r q'o d fo p wf d fdr 9 nm l y k w j hi&'u$R& & 'u$R& t h Gqgfe dd h Aa d 9 in (1.5), we obtain the Gauss integral: We have (1.5) (1.6) 3 where is equal to the number of ways to arrange objects in pairs is the same as to make a labelled 3valent Observe that to arrange graph with vertices by connecting 1valent vertices of the following disconnected graph: b1 c1 a1 a2 Fig. 1 This graph comes with labeling of each vertex and an ordering of the three edges emanating from the vertex. Let be such a graph, be the number of its vertices and be the number of its edges. We have , so that for some . Let Then where the sum is taken over the set of labeled trivalent graphs. Let be the number of labelled trivalent graphs which define the same unlabelled graph when we forget , where is the number of about the labelling. We can write labelling of the same unlabelled 3valent graph . Thus where the sum is taken with respect to the set of all unlabelled 3valent graphs. It is easy to see that Aut d W 7 We We t u 5 w g Uo qp ro r U p Y % u1 o d 5 g d 5 u u b2 b 2n c2 a 2n c
2n w d u " t "'3 c o '35 w 7 Ap d W WW 7 7 5 W p o d W b r W we d w p w ro d w b r w We g E p d d w b ~ d ww d objects in pairs. This gives us (1.7) 4 so that LECTURE 1. QUANTUM FIELD THEORIES: AN INTRODUCTION Aut Aut Given an unlabelled 3valent graph with vertices, we assign to each verte Let be functions on The righthand side is called the path integral with insertion functions lefthandside is called the correlation function. In the example above Exercises 1.1 Find the Feynman rules to compute Compute the coefficient at 1.2 Show that the distribution is a generalized solution of the equation
V g V (defined to be zero for (you have to give the meaning of the righthandside). 1.3 Show that, for any , the function is a generalized eigenfunction of the operator in and any generalized eigenfunction coincides with one of these functions. 1.4 Find the Fourier transform and the derivative of the Dirac function . y p v W v$ g y { I 5E i d 1 o p p wye d E g & v Ah & $ t h d W b . For any function set Map defined by and a point one can consider a function on the and , one can consider the integral Map ptp w
. The ) 5t w w e d D w B f D B 7 R1ttuut 7 $vQ " #! RT t 8 @ U"TS 98D B 7 1tut 98D B 7 d ! D B 7 utut1t D B 7 t 7 R1tutut 7 utut1t 5t w8 7 d 98D B 7 ' 7 & v & $ 5t w w e d " w d G d x v x v e d A v w eW˩ d u w a U d " & v & $ e f e f 16 LECTURE 1. QUANTUM FIELD THEORIES: AN INTRODUCTION Lecture 2 Partition function as the trace of an operator
of an operator in a finite dimensional Hilbert space is Recall that the trace Tr equal to the sum of the diagonal entries of a matrix of with respect to any basis. If we choose an orthonormal basis , then Tr (2.1) If is a normal operator (e.g. Hermitian or unitary), then one can choose an orthonormal basis of consisting of eigenvectors of . In this case h Tr where Sp is the spectrum of (the set of eigenvalues) and is equal to the dimension of the eigensubspace corresponding to the eigenvalue . Notice that i l k j This gives There are several approaches to generalize the notion of the trace to operators in infinitedimensional Hilbert spaces. We shall briefly discuss them. First assume that is a bounded operator. First we try to generalize the definition of a trace by using (2.1). One chooses a basis and sets Tr 17 v $ v $ v p d $ $ 1tutut Rutut1t ' Tr k ' ( Sp h ' ( Sp $ ptA d " "t d " h d v $v v $ tp d $ $ Rut1tut j $ m ( $ 7 (2.2) (2.3) 18 LECTURE 2. PARTITION FUNCTION AS THE TRACE OF AN OPERATOR if the series convergent. If the convergence is absolute, then this definition does not depend on the choice of a basis. In this case is called a traceclass operator. For example, one can show that Tr Tr if both and are traceclass. An example of a traceclass operator is a HilbertSchmidt operator in the space . If is its kernel, then Tr When is a selfadjoint HilbertSchmidt operator, the two definitions coincide. This follows from the HilbertSchmidt Theorem.
n q Example 2.1. Let . Then
n be a finite set equipped with the measure with inner product
G q G so that is a linear operator defined by the matrix
s Rs r Tr This agrees with definition (2.1) when we take the standard orthonormal basis of As we have already mentioned, in physics one deals with unbounded linear operalike a differential operator. One tries to generalize definition (2.3). tors in Notice that u h m ( k j m t Now for any such that has a basis of eigenvectors of zetafunction of as follows. Let positive eigenvalues of . One sets
0 0 0 0 v one can define the be the sequence of where is the multiplicity of , i.e. the dimension of the eigensubspace of eigenvectors with eigenvalue . When , where is a compact manifold of dimension and is a positive elliptic differential operator of order , one can show
w I s s r s s ' s r h where with a matrix . It is clear that can be identified and . Then its trace is equal to 'h g } } d 'h ~h g d h } ~ d w g y 1tutt 1tutt v v v v t & h p h d d ~h g } t d v v d W8 v d 8 E We v v v GRut!ut1t d 5 v R1tutut v d v d ~h 8 d 8 } o n ! t 'h We d } 3 d n o ph utut1t~o ! o n @q } I d 'h g G We . 19 is an analytic function for Re and it can be analytically extended that to an open subset containing . In this case we define
w y x k j ' This obviously agrees with (2.3) when that for any positive number h k j Example 2.2. Consider the operator which acts on the space , where with the usual measure descended to the factor. Note that in this measure the length of is equal to , i.e. is the circle of radius . The measure corresponds to the choice of metric on the circle determined by its radius. The normalized eigenvectors of are . The positive part of the spectrum consists of numbers with . Thus
w v w z w w z  { g w v where is the Riemann zeta function. It is known to be an analytic function for Re . This agrees with the above since is onedimensional and is an elliptic operator of second order. We have
m ( v v Thus and extends to a map of the universal coverings . It satisfies for some integer (equal to the degree of the map of oriented manifolds). Let Map be the set of maps corresponding to the same . It is clear that each Map can be uniquely written in the form w } w } Q A map w Example 2.3. Let us consider the path integral when We use the action
~ 5 d r d r GG r r u r %Y I I ~i u t h g w g o d Um u I d m u I d g t g d g h h qp w u 5 p d Wy w u } Q w det ' This of course agrees with the finitedimensional case because t o p d wy w o p d wy u & p g td g dd 9 g m g m d $v A A fV h u h p h d 9 g 7 E d Wy 5t " d &$ 7 "t d y h ) w j
and
j v k w j m t y x ' v ' h m u I d j Uo w w w w v d v is finitedimensional. Also it is easy to see (2.4) . (2.5) . z {$ {$ ~ z z { G ~ G d u~ X~ d u~ z {$ ~ {$ {$ h h v & $ h h v $ gv & $ t V d $v d g g hh v $ VU g v & $ d d h h v & $ h & $ h & $ t V d V d s d b g d h & $ d u h h v $ db d G {$ {$ {$ h & $ t V o s v $ ! th h s d h } w } Q w ~ & h & $v p w 5 u " g V 1" o d w w e ! Q Uv "$ RT h wye d @ U" u v $ Q Uv "$ Q Uv$ #! " " t " @ 3 h h d @ 3 d !b t h g w g o r g g u d t y d Wy p d h w g d h w w g u } } Q w } ~ w } ~
Map t h g w r w w r g w g o d R d r u k g
} Q Let us compute the trace of the operator eigenfunctions in are the functions
} w Y } } {$ os {$ ~ Taking to be consistent with the previous computation of the path integral. This gives where Observe that we must have The Minkowski partition function is Thus We have where on such 20 Now we apply the Poisson summation formula LECTURE 2. PARTITION FUNCTION AS THE TRACE OF AN OPERATOR satisfies is equal to Tr , we get Map , hence belongs to Map . Its normalized . By (2.1), we have . The value of (2.6) w 21 Comparing this with (2.6), we see that Remark 2.1. If we repeat the computations for the Euclidean partition function (replacing with ) we get This shows that This is the first glimpse of the Tduality. Let be the modular form associated to the quadratic form ( equal to the value at zero of the Riemann theta function in one variable). It satisfies the functional equation (the proof uses the Poisson summation formula). Observe that
w } w } This is our first encounter with the theory of modular forms. There is another way to compute the partition function for the action Map where is the circle of radius
~ . Notice that
~ ~ Thus Map & "$ " ~ " ~ Qh t " 3ww p p d h ww p @ h g w g 5t g W d c b h pt i Ap d o qp g i d h v $ t y r d d W j 3 w If we modify the partition function by inserting the factor
w } } w } h h & $ t V d
, we get
z ~ { # q {$ w d } d } ~ pt c b d c b s o pt ~ ! " z } d c )b g p g w d h g w p " R d o p db S c w {$ w } } w } {$ w } g F where . v&$ p g db d
Tr
d u~ cb s d cb h h & $ V s d c)b h h g q p d 22 LECTURE 2. PARTITION FUNCTION AS THE TRACE OF AN OPERATOR can be thought as a generalization of the Gaussian integral since is a quadratic form in . If and is a positivedefinite selfadjoint operator, we could use the orthogonal change of variables to diagonalize and write ' Here we assumed that all eigenvalues are positive, or equivalently, that the quadratic form is positive definite. To get rid of let us change the measure on replacing with so that
k j ' d Now, for any normal positive definite operator in a Hilbert space , we can write any element as a sum , where is an orthonormal basis of eigenvectors of . The coordinate is an analog of the coordinate from above. This motivates the following definition det d W u ( Here the measure is defined up to some multiplicative constant be defining the correlation functions by the formula so the choice of the constant will not matter. We would like to appy this to the operator in . However, not all of its eigenvalues are positive. Constant functions form the nullspace of this operator. If we decompose each vector as a sum of normalized eigenvectors, then coefficients will be analogs of the coordinates in . So, we can write our space as the product of the space of constant functions and functions with . The coefficient of the constant function at is equal to . Thus the integral over the space of constant functions is equal to So, using (2.5), we obtain
} w } w } w } Q } Q } Q } Q z { } Q G u F G R& t s d s db u u d t s td s h s g u u u u u s o d o y d v s I v hh v p d I g u vQ $ B D [email protected] U" D vQ $ C @ " 3 7 u1 7 d 7 1tut1t5 7 B D [email protected] B R& &$ t g V ou w d @ v e8 8 v 98 v h & &$ & t d h ut1tt h u s g h h v I u &d v v v v v v h & h &$ t ou d g V u d g s V u d u y d h h & & h & $ & $ d h 1tutt h d h ututt h I d 7 9 g d 7 5 & $ 5 @ " ~ " ~ k j k j q ' l W l F where h h g p d . The integral (2.7) . In fact we will 23 This agrees with the computations in example 2.3 if we switch from the Minkowski partition function to the Euclidean one. Here is another application of the Gaussian integral for quadratic functionals. Consider the action functional defined by some Lagrangian . We know that its stationary points are classical solutions. Write , where is a classical solution. Then This gives a semiclassical approximation: This approximation is exact when the action is quadratic in . Example 2.4. We take and and the Minkowski partition function is It is called the path integral of the harmonic oscillator. The kernel of the operator is given by Choose a critical path cl for the action defined by our Lagrangian and decompose the action in the Taylor expansion at cl .
cl The classical path is a solution of the Lagrangian equation: Its solution satisfying the initial condition
m ( cl g g p g o d I d I d u p p h )& g t D g ou Bp F qi B D [email protected] F R qi g r g g o d ) r classical solutions . The Lagrangian is cl cl is g 5t p g o d A r p " % t h g g p g D [email protected] " ' d a B terms of higher order in t d I rڧ k j p t h g g p g D [email protected] d ! b B t p p H H d A t y d g m t m t r d r ww q p H p H d d p h h q Q m t )
e d e & v & $ " (2.8) l Of course here we use a "physicists's argument" since we don't have the right to write the product as the product of two infinite products , one of which is divergent (see the next remark for an attempt to justify the argument). Now we use that the first product corresponds to the action with . So to be consistent with our previous computation we must have l l l k j k j Q e e ) h u t ou s d y Ey e d & g U u o y d g g u ~ u 5t g u g p o g U u o d g p g U u A d @ u g op g o m u d U u y d d pg d t @ t g p g hh p d @ " h h & x p B " @ u o d @ o Ap D [email protected] g p " % d h g r g h p D B h [email protected] " % y d d y p d g p d d `u " % B t h A p g r g h h A p p D [email protected] " % p A qi d e g g t h g r g h h d g g g g w p H t p p H g r g d h p o d  8
cl cl cl m ( Thus we can rewrite (2.8) in the form The second variation of the action functional is The value of the action functional on the classical solution is 24 LECTURE 2. PARTITION FUNCTION AS THE TRACE OF AN OPERATOR cl e The eigenfunctions of , where equal to
m t e Q e e where Now let us make the variable change replacing with . The limits in the cl path integral change to . The paths we integrate over are periodic with the period satisfying . Using the generalization of the Gaussian integral to functional integrals we have Map satisfying the condition are the functions and . The corresponding eigenvalues are . We know that
cl cl ' ) n ' q `u ' ) ' ) ' ) v&$ h v & $ h Ah & $ V t t Rt1utdr g WV" u g x d E i i"" v &$ v&$ p g d g r g r g r g ip p vg &$ p h g E v & $ o s u d a V ' ' ' ) ' ) ' ) ' ) ' ' ) i ' m t k j v &$ v&$ p g r g r g g ip t g r v & $ v & $ p v & $ g r o g r g i d g r o gv & $ p gv & $ p d v & $ o p v g & $ r oo g r g i d p p v g & $ p h u tpA qW g E v & $ o s u d g V A qi V & u p d g V @ o qi d e g g pt g u g p o d h i l Y m t The two computations disagree. The way out of this contradiction is the choice of the normalizing constant which we used to define the Gaussian integral. It shows that we have to choose . Now we use the Euler infinite product expansion for the sine function: u s d hh v g p Note that if we compute the product using the zeta function of the operator on the sapce of functions on satisfying we get u g u t d U u o y d 8 d 8 D B ~ 8
l From this we deduce that cl ' ) Let us rewrite cl ' ) ' ) Using this we obtain
cl in the following form cl cl 25 26 LECTURE 2. PARTITION FUNCTION AS THE TRACE OF AN OPERATOR Now recall that the kernel of a HilbertSchmidt unitary operator can be written in the form where is the normalized eigenfunction with the eigenvalue . In our case, the eigenvalues of must be equal to . Thus the eigenvalues of the Hamiltonian are . We shall see in the lecture that the eigenvectors of are where are the Hermite polynomials. , we get This checks the When first term. Exercises
be a nondecreasing sequence of positive real numbers. 2.1 Let Define provided that t the observable . Any can be written as G G d h G G G Of course this has to be understood as the equality of distributions. For any test function we have The expectation value of is equal to d Consider the deltafunction as a state (although it does not belong to . Then the probability of to take a value in the state is equal to . The inner product is of course not defined but we can give it the following meaning. We know when tends to that is equal to the limit of tempered distributions zero. Thus we can set When is a variable, we get ~ ~ G ~ ~ n Example 3.2. Let Its eigenvectors are the functions G and equal to the momentum operator with eigenvalue . We have h I g v&$ u v d h fV W g s d p i v & $ g o fV d g d u x h R& '$ I d x x & p pt v W d d g V " ou s TT d x d h & & $ u h & & $ x x u t g V " g o s TT d V " o s T d k x g h & & $ g x x V " x 8 d h h i% e f G e t h d i h I g pt d h i d 5 i
d f 5t w8 d h & & x t h g d & 8 & & d h i8 h G i h G G h h } Q G } Q e d d h G e i 8 G . 29 is the th Fourier coefficient of . So, the wave function of is the function on Sp . The probability that takes value at the state is equal to The expectation value is equal to
G The dynamics of a quantum system is defined by a choice of a selfadjoint operator , called the Hamiltonian operator. In Schr dinger's picture the operators do not o change with time, but the states evolve according to the law Here is a fixed constant, the Planck constant. Equivalently, Schr dinger equation: o is a solution of the In Heisenberg's picture, the states do not change with time but the observables evolve according to the law
d n d n We have the Hamiltonian equation:
n where If is an eigenvector of , then and hence the corresponding state (equal to the line spanned by ) does not change with time, i.e. describes a stationary state. Usually one measures observables at the stationary states of . There are two ways to define a quantum mechanical system. One (due to Feynman) uses the path integral approach. Here we take as in Lecture 1 and define the Hamiltonian by means of the path integral. The choice here is the action functional. It is defined in such a way that its stationary paths describe the motions of a classical mechanical system. Another approach is via quantization of a classical mechanical system. Recall that the latter is defined by a Lagrangian which, in its turn, defines an action functional on the space Map
F A critical point of this functional defines a motion of the mechanical system. The equations for a critical point are called the EulerLagrange equations. If one chooses g v v C 5t D G B I š v i G ~h g d } G d G h &$ 6 d D t5 p F d X l B D X 5 B d h h $ &$ t 6 6 d s 5t d h h qF x t h A p d n ` t g v d o &$ t H 6 d F G G G f o d {$ ` F n n q e G G G G G o n F d m F F . (3.2) G 30 local coordinates in LECTURE 3. QUANTUM MECHANICS
and the corresponding local coordinates (so that , the equations look as
s W i v Q W d W Here the lefthand side is evaluated at a path in given by . For example, if we assume that the restriction of to each tangent space is a positivedefinite quadratic form, we can use to define a Riemannian metric on . A critical path becomes a geodesic. Another way to define the classical mechanics is via a Hamiltonian function which is a function on the cotangent bundle . Recall that any nondegenerate quadratic form on a vector space defines a on the dual space . If we view a quadratic form as a symmetric quadratic form bilinear form, and hence as a linear map , then is the inverse map. Let us see that, for any , is equal to the maximum (if ) or thof the Hamiltonian vector field . The analog of the dynamical system (3.5) is the Hamiltonian equation in quantum mechanics (3.2). For example, when a mechanical system is given on the configuration space with coordinate functions we need to assign some operators to the coordinate functions: n F n n d F h n By analogy with (3.4), we should have
d d s Recall that these are unbounded selfadjoint operators. (resp. ) is called the position (resp. momentum) operator . The operator
d Let us give an example of quantization of a classical mechanical system given by a harmonic oscillator. It is given by the Lagrangian So the motion does not depend on the mass but the total energy does. In the sequel we shall assume for simplicity that . The Hamiltonian operator can be written in the form
F ) ) d F F r Gv v d Fr p `v d g g r g o d v o d g t g p d g hh where we used that to express viewed as the total energy of the system. The corresponding Newton equation is ҷ a via . The function e 5 g g r g o X d g g ip g i o X d d p d e where is is the mass and is the frequency. The corresponding Hamiltonian function # So we have to find an appropriate Hilbert space fying (3.7). We take and define
d pA g g p g io d v v v v v v t 8 F 8 8 8 v v I g 7i v v 7 v dv 7 d X D B y d X D B d X D B v v t ARut1tut'o d d d v v I # B X D t X D B T X d X D B TT X d d " and operators
d F s n s d d n n s d n d F F n F (3.7) satis can be (3.8) 35 where are the annihilation and the creation operators. We shall see shortly the reason for these names. They are obviously adjoint to each other. Using the commutator relation , we obtain This shows that the operators form a Lie algebra , called the extended Heisenberg algebra. . So we are interested in the representation of the Lie algebra in Suppose we have an eigenvector of with eigenvalue and norm 1. Since is adjoint to , we have The equality holds if and only if . Clearly any vector annihilated by is an eigenvector of with minimal possible absolute value of its eigenvalue. A vector of norm one with such a property is called a vacuum vector. Denote a vacuum vector by . Because of the relation , we have This shows that is a new eigenvector with eigenvalue . Since eigenvalues are bounded from below, we get that for some . Thus and is a vacuum vector. Thus we see that the existence of one eigenvalue of is equivalent to the existence of a vacuum vector. Now if we start applying to the vacuum vector , we get, as above, eigenvectors with eigenvalue . So we are getting a countable set of eigenvectors
F G eigenvectors ) a with eigenvalues G It is easy to see, using induction on that that After renormalization we obtain a countable set of orthonormal y v d y y d v v y y d v F{ p F t F p H v d v p F v Hp d D v B v wF t tu1ttu ~1o y d 5 y v o d g t F g g F r X v G F i y d v t F { G G G G G F This implies that all eigenvalues are real and satisfy the inequality (3.11) (3.12) v t g T F r g T v d F H v v d r I g G p o s d v 5 r o s d v v v ~o t v F d D v q v Hp d D v B B F F d D v v q y d D v v B d D v v B B G ) G ) G F F G G F d ) v d v y G d (3.9) Y G G F F F t 9F d g T T d H d g T v G G v F qFp d D B G G v d (3.10a) (3.10b) 36 LECTURE 3. QUANTUM MECHANICS a One can show that the closure of the subspace of spanned by the vectors is an irreducible representation of the Lie algebra . The existence of a vacuum vector is proved by a direct computation. We solve the differential equation and get (3.13) In fact, we can find all the eigenvectors a where
F is a Hermite polynomial of degree . It is knlet action in the case when the metrics and are flat (i.e. and are constant functions). We get the equations , and this is just the Laplace equation with In the case when respect to the metric . When , its solutions are harmonic functions. In general, the EulerLagrange equation for the Lagrangian (4.3) can be written in an invariant form: where is the covariant derivative of a section of with respect to the natural Riemannian connection defined on the bundle . Consider the special case when , with coordinates , and diag . Take . Then the Lagrangian density becomes The EulerLagrange equation is D uout1tut~1o~o p B We can view as the displacement of the particle located at position at time . The EulerLagrange equation for the scalar field can be thought as the motion equation for infinitely many harmonic oscillators arranged at each point of the straight line. If is the flat Lorentzian metric in defined by the diagonal matrix diag the EulerLagrange equation for a scalar field with Dirichlet action is Notice the analogy with the Lagrangian for a harmonic oscillator (with o d d t h h D g 8 p g 8 B d g g I d I d d i p 8 ! d 8 h y d 8 h @ P g g g t y d 8 g p 8 g p d 8 d 8 I 8 8 g t h g p g h h d g g t y d 8 g p 8 g d I G I d d t @R1tutut'o d } y d 8 g 8 v v t h h R o p d I d I d td @
Tr
s Tr { . and the metrics are the standard Euclidean metrics. { { s { { Uo p ~o (4.8) d d { @ ) 45 The operator is called the D'Alembertian operator or relativistic Laplacian. A little more general, if we take the Lagrangian the EulerLagrange equation is the KleinGordon equation In many quantum field theories is a fibre bundle over and is a section. When is a bundle with some structure group a map is called a classical field, otherwise is called a nonlinear field. For example, when is the trivial vector bundle of rank , a classical field is called a scalar field. Of course any map can be considered a section of a fibre bundle, the trivial bundle . Example 4.2. An example of a classical field is a gauge field or a connection on a principal bundle over . It is defined by a 1form on with values in the adjoint affine bundle Ad . In other words, it is a section of the bundle Ad . For example when GL , a gauge field is a map of vector bundles End where is a smooth vector bundle of rank over . It satisfies where is a local smooth function and is a local section of . It is clear that the difference of two connections is a morphism of vector bundles and thus a connection is a section of an affine bundle. Each connection defines the Lie valued 2form, the curvature form, Here is the staroperator on the space of differential forms with values in a vector bundle equipped with a metric . It is determined by the property where is a natural bilinear form on the space of such forms (determined by the Riemannian metric on and the metric on ) and is the volume form defined by the metric on . The EulerLagrange equation for the gauge fields is the YangMills equation: 'h d } {s v v v v t h R1tut1t~o d y d D r B g h } 'h s s s n h n Here is a local function on density on by setting
P s with values in Lie . We define the Lagrangian ۡ (4.9) g q 8 r 8 X n g q p w r wE d Rw v v d h t d v v D t h h d ) B o r h d g s n ty d 8g r 8 p p g 8 g Χ8 8 o d
p q u w s n n n p n o p Y n s n 8 g g h g 8 g p W v s 46 LECTURE 4. THE DIRICHLET ACTION There are two approaches to quantization in higherdimensional QFT. First uses functional integrals which generalize the path integrals. Consider a space of fields Map on a dimensional manifold . We assume that , where ( a "time factor"). For each we denote by the restriction of a field to . Let be the space of fields on obtained by restrictions of fields from . Fix two fields Consider an action and set where we integrate over the space of fields on such that . We use some measure on . Observe the obvious analogy with our previous definition where we take point and is the set of maps . Now if we consider some Hilbert space of functions on , the integral operator with kernel (4.10) defines a linear map h } i h h 3 ! The kernel has special meaning for Tr
d . The integral Map is the trace of the operator . It is called the partition function of the theory. More generally, let be a local quantum field at a point on (a local field is a functional on which depends only on and derivatives of at ). An example of a local field is the functional , where is a function on . We set l h i This leads to the correlation function R h d which can be used to define a Hermitian map d F It also defines a selfadjoint Hamiltonian operator h v v h t k k b b d 1tut t v v v v v h DB x8 [email protected] w8 U"TS $vQ h d k b Iv ڧ` A 98 8 v v 8 AR1tut1t~o d & 5h $v " " & h v & $ D #! B [email protected] b d " h g h d & h v & $ g g t h " d & h v & $ h " d h h h " w D [email protected] # w #! d a dw v x8 [email protected] DB 8 d 8 g $vQ DE8 [email protected] U"ST h d w v I t ~o d w d o Y8 w d G d d T
Map such that , h h @ 3 ! b h d h j h } h } 8 8 (4.10) (4.11) (4.12) 47 We can use (4.11) to define a Hermitian form on the space Map
G G u7 of functions on We can get rid of the parameter
G by letting it go to infinity, i.e. define G d d G Another approach to quantization generalizes the one we used for the harmonic oscillator. Again we assume that . For any field we denote by the partial derivative in the time variable. By analogy with classical mechanics we introduce the conjugate momentum field Then the EulerLagrange equation is equivalent to the Hamiltonian equations for fields where the dot means the derivative with respect to the time variable. Here we consider and as independent variables in the functional and use the partial derivatives of To quantize the fields and we have to reinterpret them as Hermitian operators in some Hilbert space which satisfy the commutator relations (remembering that is an analog of and is an analog of ). (4.13)
F (4.14) 8 8 p d u u d ة 8 t y d D E 5 B d D E 5 B 5E pi F d D E p B u F 6 For example, when introduce the Hamiltonian functional : F t h p 8 u 98e o d g g S zp g S A d 98 S du pt w8 d u In this way we get a local operator operator associated to a functional in the Hilbert space . . It is called the vertex we obtain . We can also This is still linear in such that h } and halflinear in
G G w d a` 8 v & $ g g Uv$ pt T d g & h $v & h v & $ g p d t " " d g h h g g w g t D [email protected] D [email protected] b w B B . Thus it defines a linear operator G d d i G F G d h u H g H G 8 T g g g d w h H G G } G 8 tu h } F 48 LECTURE 4. THE DIRICHLET ACTION Here we have to consider as operator valued distributions, i.e. a continuous linear functionals on the space of test functions on equipped with some measure with values in the space of operators in a Hilbert space . Any function on with values in the space of operators in which is integrable with respect to some operatorvalued measure defines a distribution The commutator of two operator valued distributions is a bilinear form on the space of test functions: Thus the meaning of (4.13) is
H Example 4.3. Assume that and is the KleinGordon Lagrangian . A solution of the KleinGordon equation can be written as a Fourier integral r 7 r h 7 r i H where Similarly we have a Fourier integral for To quantize we replace with an operator and with the adjoint operator and consider the above expansions as operator integrals. This implies that the operators and are Hermitian. The commutator relations (4.13) will be satisfied if we require the commutator relations
r h r r i r i This is in complete analogy with the case of the harmonic oscillator, where we had only one pair of operators satisfying (or operators satisfying ). There is a big difference however. In our case the Heisenberg Lie algebra generated by is infinitedimensional. s s s s v v 'o v v v v v v y D B D 1B D B v v y d D v v B dd D v v v 1B'v o d d D vv v Bv d v v v v t y d D v vB d D v vB p p w s d D v v B # v v v v v &'$ v '$ $v & $v u t h v r v Ap s d t g r g d g & $v & $v u h " v r " v s o d h h g8g p S h p S h I dw 5 w8 B : } r r i t }'h W i8 d D g t D A5 98 B t 'h i8 i 8 } Q G G r i r G G 6 98 H Q } 'h 49 Exercises 4.1 Let maps be a Lagrangian on with a metric . Define the energymomentum tensor by (i) Show that this definition agrees with the one given for the Dirichlet action; 4.2 Consider a system of arranged on a segment . Write the Lagrangian goes to zero is equal to
e where are some positive constants and is the function which measures the displacement of the particle located at position at time . 4.3 Let be a Riemannian manifold of dimension and be a vector bundle over equipped with a Riemannian metric. Show that there exists a unique linear isomorphism such that, for any one has is defined vol . Here locally by extending via linearity the product . Also is the inverse metric on extended to by the formula 4.4 Using the staroperator defined in the previous problem show that the Dirichlet action can be rewriten in the form
s w C q 8 q s C @ s q Y where 8h 8 u 8 h 8 h d 98 R& $ $ $ $ R& 5tu v w d w & v $ $ $ $ v w d w v & r } & d p E g 8 x g t h D g 8 p g 8 w w D v B 8 y d U"S' is considered as a section of the bundlle u (ii) Prove that if satisfies the EulerLagrange equations. harmonic oscillators viewed as a finite set of masses , each connected to the next one via springs of length describing this system. Show that the limit of when w8
. p w8 d w8 I 8 h 98 d { q' defined on the space of 50 LECTURE 4. THE DIRICHLET ACTION Lecture 5 Bosonic strings
From now on we stick with dimension of our QFT. This is where strings appear. Our manifold will be a smooth 2manifold with a pseudoRiemannian metric . It could be the plane or a cylinder , or a torus , or a sphere , or a compact Riemann surface of genus . Of course each time we should specify a metric on . We shall begin with the case when is a cylinder (closed strings) or (an open string). We use the coordinate in the circle direction and the coordinate (time) in the direction. A map can be considered as a map where is the loop space of , i.e. the space of smooth maps from a circle to . In the case of open strings must be replaced with the space of paths in . We shall consider only closed strings, however occasionally we state the corresponding results for open strings. with the Lorentzian flat metric We shall also assume in the beginning that , where diag . We will write vectors in as and denote by the vector equal to . Later on we will of course consider more general target spaces . We consider the Dirichlet action Here is a certain parameter of a string (the string tension). It is equal to for open strings, where is a certain other constant called the Regge slope. For closed strings . We use the subscript to emphasize the dependence of the action on . ` h p w8 d A 8 w w u Uo d w u Uo t 'h 8 8 d ~h g 8 h d 98 } } RRut 1tutut1tut W W Rut1ttu g Ap 1tut1t i fuo1tutut'uo'o qp d d d I d (5.1) (5.2) 51 g 5 8 % I ܨ p ~ 8 I 8 I { o j I dd @ gI { q D yB I d { 52 LECTURE 5. BOSONIC STRINGS where is a diffeomorphism of . This means that the action is invariant with respect to smooth reparametrizations of the maps. Also, for any , we have This means that the action is conformally invariant. It is known (see, for example, [Modern Geometry] by Dubrovin, Fomenko and Novikov) that there exists a unique diffeomeorphism such that , where is a smooth function and is a flat metric given locally by the diagonal matrix diag . By (5.2) and (5.3) We shall fix the metric on by equipping with the metric and taking with the metric induced by the standard metric on . Then we have two constraints on . One comes from the EulerLagrange equation for the action and another comes from the vanishing of the energymomentum tensor. Since the Lagrangian function for the action is equal to
e So our field satisfies the KleinGordon massless equation. The value of the energymomentum tensor at is equal to To solve the wave equation (5.6) we introduce the lightcone coordinates Let denote the partial derivatives with respect to these coordinates. We have Thus we can rewrite (5.6) in the form This easily implies that a general solution of (??)EQ) can be written as as sum q ~ q ty d 8 8 r & tp 8 8 d 8 r & ty d 8 & t5 p o d p r o d & r d t p d e the EulerLagrange equation for the action is d $ d k Y Ig p h 8 8 o d d xy d 8 t @R1tutut'o d } y d 8 g p g 8 8 p 8 8 d g h I 5 8 d w8 $ 5t 98 d w8 ` q 8 8 d d 98 f1o~o Ap & m u I (5.3) (5.4) w e (5.5) (5.6) (5.7) 53 Using the boundary conditions, we see that, in the case of a closed string, the functions , and are periodic with period , so that we can use the Fourier expansion to write and . We shall see in a moment a reason for the choice of where the constant . Also, since we want to be real, ' z { The field (resp. describes the "leftmoving" modes (resp. "rightmoving" modes) of a closed string. Note that ~ It is clear that and can be interpreted as the centerofmass coordinates. By analogy with QFT the momentum field is defined to be The expression is the total momentum coordinate of the string at equations (5.11) in the form
w v&$ & V r g r d 8 v & $ o o o V o r g o r o d 8 y d t d td h wy h 8 h g d u t 8 d 8 d o d @ @R1tut1t~o d } h Wy 8 g d v&$ & tV v & $ d 8 V d 8 & 8 8 & t d d 8 g d d r & r o d 8 r r R o d 8 z v&$ V v&$ o V o z z z . Now we can rewrite the z z w u w w w w {$ {$ z { ~ z { S ~ ~ S 8 r 8 ~ (5.8a) (5.8b) (5.9a) (5.9b) (5.10) (5.11a) (5.11b) 54 LECTURE 5. BOSONIC STRINGS Remark 5.1. If we choose the Riemannian metric on instead of pseudoRiemannian, we will be able to identify the cylinder with the punctured complex plane by means of the transformation The EulerLagrange equation (3.7) gives The equation of a string becomes This is a familiar expression from the conformal field theory. The Hamiltonian of our theory is equal to Observe that it vanishes on a string which satisfies the constraint that the energymomentum tensor vanishes. Plugging in the expressions for in terms of , we obtain (5.14) h F Now it is clear the introduction of the constant . It made our formulas not depend on . Observe that we could simplify the sum by getting rid of but we don't do it, since in a moment the coefficients will become operators. Using the new coordinates we can also rewrite the constraints (5.7) in the form This immediately gives t y d w8 98 d w8 w8 & & t y d p o d 8 8 y d er o d 8 8 g & & r o d 8 t h 8 8 r 8 8 g d h p 8 4 {$ z { {$ 4 ~ {$ ~ { The stresstensor can be rewritten in the new coordinates too. We have & & tg d 8 8 d g ʯ o o 4 4 r r t
w z w g o r o r g Ro r Ro r y d ~ a f v i& $ VWf" d I d 4 4 m t m ( 4 4 4 d 8 8 d ٩ d ` 8 d ` 8 4 4 ~ z % g d 3 y ! F d (5.12a) (5.12b) (5.13) 55 This can be restated in terms of the Fourier coefficients as follows: Observe that We need Plugging in the mode expansions, we see that this is equivalent to the following commutator relations all other commutators between, are equal to zero. We can also quantize the expressions for . In order they make sense as operators in some Hilbert spaces we will require (by analogy with the harmonic oscillator) that in our representation kills any state provided that is large enough. Thus the sum makes sense. Let us set, for any operators with indices in an ordered set , if ; otherwise.
s n n F s n n s n n  u T  u The situation with we know that is more complicated since do not commute. Of course so we can write It is called the normal order for the composition of operators. Since we can rewrite in the form 9 & p @ o r & r o d& & d r & o d & y d y dD B v v v { d v & D D d d D B B d B p w i p d D w 5 B q { t d { R & n s {$ d n q Hilbert space. Since we want to be Hermitian we require 8 & Now we quantize as in the previous lecture by taking as operators in some ty d y d & & {$ {$ r d v g & $ d h $ o v g & o d h F ~ w8 w8 w8 98 ~ 8 d (5.15a) (5.15b) d (5.16) (5.17a) (5.17b) (5.18) , 56 LECTURE 5. BOSONIC STRINGS Here we get because when we sum with respect to , the contributions corresponding to cancel each other. We shall deal with the last sum later. Now we define and by dropping out the infinite sum. Similarly we define the operators . The operators are called the Virasoro operators. Notice that and (resp. and ) are adjoint of each other. The expression of the Hamiltonian operator is now straightforward: Since the last sum obviously does not make sense, we regularize it by setting
v So, finally we get Let us find the commutator relations between the operators following wellknown identity:
o n o n o n o n This gives Here we skip the upper index . This easily implies h r & & p pt W r o d D B & & & & } & p W r D i B & p & & t & & & & D & & r r D & & & d & B & r & B r D B r D ~ d D B B D B r D t D @ B r @ r D @ B [email protected] B d D @ B & t r o d & r o d p o @ p r d pA d r r d From now on o n )r t o o p d fo t p @ & r r T T )r } & F F r 1o y d } p @ (5.19) (5.20) . First we use the 57
r r o Changing id This shows that for some constants . We will fix the constants when we consider the representation of the Lie algebra generated by in some Hilbert space. Exercises
5.1 Let Vect be the Lie algebra of complex vector fields on the circle. Each field is given by a convergent series , where . Let . (i) Show that . d d 5.2 Let
u R& 5t R& 5 8 D R& B v d A D D dB R& B v d p w p p w p y d C y d v Gar d v d r r D t y d DU B r D B % B w pA D% B d D u 5 v B w d p $v i5 d D B v w where is a bilinear form on Vect . Show that this defines a structure of a Lie algebra on Vir if and only if satisfies p v o p v o o v p o o (iv) Prove that two bilinear forms and define isomorphic Lie algebras if and only if is a linear function in . be the algebra of Laurent polynomials in one variable. For any let Res . Define a bilinear form on Res
o o o o o o (iii) Let
n . Show that for some . p q o p Y p q (ii) Let Vir Vect . Set unless $v d n v n {$ u n n r Setting and gives r n r i n r r i n n for some scalar . Using the Jacobi identity, we find that, for y d & n {$ u For we have a problem since , we see that the difference
r is not defined. Since dD , and by v v ar v d is G v t fo U o p p d U o r r o r o pp d o d p r t y d W # i p p r rr W & i r D B & td p y d r B t y d r i d D B p y d er p s
to in the first sum we obtain for if
9 9 58 (i) Show that LECTURE 5. BOSONIC STRINGS (iv) Show that any central extension of the Lie algebra Vect with onedimensional center is isomorphic to the Lie algebra defined by the commutator relations as in (iii). (iii) Let . Show that d d d h (ii) Let w & t p i o o r i d D B p Uuo wy d p y qp d 5t 8 D h h D h h h h B d p h 5 v h B & D & B d D & B Dd B t y d 8 8 r 8 r
Der . Show that
d } } d } d 8 is skewsymmetric and satisfies be the Lie algebra of derivations of is a Lie algebra with respect to the Lie bracket Lecture 6 Fock space
. Recall the construction Let be a Lie algebra over a field with the Lie bracket of the envelopping algebra . It is an associative algebra over which is universal with respect to homomorphisms of associative algebras such that . It is constructed as the quotient of the tensor algebra where is the ideal generated by elements , where . For example, if is a commutative Lie algebra (i.e. for all ) is isomorphic to the symmetric algebra Sym ), i.e. a free commutative algebra generated by the vector space ( isomorphic to the polynomial algebra in variables indexed by a basis of ). In general, has a basis consisting of ordered products where is an ordered basis of the vector space . Recall that a linear representation of in a vector space is a homomorphism of the Lie algebras End , where the latter is equipped with a structure of a Lie algebra by setting . We say that is a module. By definition of the envelopping algebra, this is equivalent t that has a basis consisting of elements and , where . This allows us to identify with the linear space . We have an isomorphism of linear spaces Sym (6.3) @ 0 The vector is called the vacuum vector. The structure of a module on s is given by a Note that carries a natural grading defined by j where and . s s n.o.d s n.o.d We call the above the normal ordering decomposition of the monomial write it as n.o.d . For example, the normal ordering decomposition of is equal to v& v C D ' E v v yDC d y q Ur1tutdr d u1 t v v & v 7 v & v 5t& y u1 uu v & v & o o d u1 u1 d U 1u u1 v 7 8' o d o d y & 5t 8' & R& d v 7 D g v v y q ut1tut o v B v v v o o v v v 7 v 7 ut1tt v o v C uu1 u v ov v v `do u1 v u vu o qu 1 g & 1u& v g & 7 g & & g & t& g 1& u R t ` d r d v v r &g & v u1 v v v v u1 v t r % uu d u1 g i " ut1tut " d q v 1tut1t v i ut1tt v v 1u d 1u R& ut1tt g D u1 1u 1tutut g x p v 7 ut1tut g ut 1t3tB utD1tut D g B v v v v % v 1tutut B & & y j r d R& v $ $ tD utut1t ut1t3tB 5' D g 1tutut utut1t 1tutut B w 4 w 4 v r p `vv v d r v v v Ud and . Using the relations (6.2), we can write for some permutation r normally ordered monomials of degree less than a U j (6.2) and 62 LECTURE 6. FOCK SPACE Remark 6.1. More explicitly the representation of can be described as follows. We identify with the polynomial algebra and assign to the operator , to the operator , and to the id and hence we get a representation obvioulsy scalar operator id. Then isomorphic to . There is an inner product on the space defined as follows. Let be any linear space over a field of characteristic 0 equipped with a symmetric bilinear form . First we define the bilinear form in by W W W W 4 and then extend it to the whole by requiring that and are mutually orthogonal. Using the polarization process, we identify with equal to the subspace of symmetric tensors in and then restrict to to get a symmetric bilinear form sym . One can show that this inner product is nondegenerate if is nondegenerate. Recalling the polarization isomorphism we see that s s s s sym where the sum is taken with respect to all permutations of letters. Here we identify Sym with the space of polynomials in a basis of . This defines a symmetric bilinear form sym on Sym . Following the physics agreement we shall drop in this formula. A similar construction can be given for any hermitian bilinear form. In fact, if we choose a positive definite hermitian form on we can complete the tensor algebra with respect to the corresponding norm and obtain a Hilbert space . This space is called the Fock space associated to the unitary space . The completion of the subspace Sym is called the bosonic Fock space. Similarly we can restrict ourselves with the exterior algebra identified with the subspace of alternating tensors in . Its completion is called the fermionic Fock space. We will deal with it later. We apply the construction of the Fock space to the Heisenberg algebra over by taking , where is equipped with a structure of a unitary space. Let us consider the Lie algebra with a linear basis with Lie bracket defined by commutator relations (5.17b): { Let be the graded Heisenberg algebra over with be a basis in such that, for any
4 { Consider the direct sum of Lie algebras , where is viewed as an abelian Lie algebra. Let be a basis of . I claim that is isomorphic to . To see
& Wۮ w v HI W $ $ v $ $ v $ $ 5 " ut1t " o d 1tutt t u w F w F u W w W W d w d W u1 d 1tutd ut1t G t t F W v 7 v 7 v v d D yn one checks that q s { i s Here corresponds to the matrices Define the operators
s , where if and if & where is a fixed linear form. It is easy to see that, in the case get a representation isomorphic to . i Y a r Y p d D `Y B & & ~ p o p ip d v $ v v$ $ o d o p o d vo d v v p v $ v$ $ $ $ $ r $ v p p v p v D % $ d D $ $ $ B v d $ B @R1tutut'o d y d D B v v @ 1o $ $ y I @ Rut1tutv ' o d d Id d 7 v r v f1o 7 ~o p @ R& I 7 d 7 7 I d & & & 5 d 1tutt o o & & & & & & & & & t 1tutt 1tutt & & & & d y j R& WR& W' & wB & & d W & ut1tt & y i j d wrt1utRrt & A R& 1ttut ' d W' I B d W d & t x d D B m Y C D B A d W' E i act by
W 9 s 4 W s s W W s W s W W 4 s W s W 4 6 s W { W W 7 i ' (6.8) , we (6.9a) (6.9b) (6.9c) . (6.10) 65 This shows that the correspondence is a representation of the Poincar Lie algebra in the Fock spaces e (one has only to replace the commutators with ). Here the operators correspond to the matrices in the natural linear representation of in the space of vector fields in as the vector fields . We can view a state as a polynomial function on Sym with values in Sym so that the vector field acts naturally on such states. The translation part of the Lie algebra of the Poincar group acts via the e operators . Remark 6.3. Recall that an irreducible linear representation of the Poincar group is e described by the following data. First one restricts the representation to the translation subgroup . Since the latter is an abelian group the linear space decomposes into the , where direct sum of eigensubspaces The Lorentz group SO acts on . It is easy to see that the set is an orbit of . Let be the isotropy subgroup of some . Then the restriction of the representation to defines an irreducible representation of in . Now the natural action of on lifts to an action on the vector bundle , where acts on the product by . There is a natural action of on the space of sections of this bundle and the representation is realized as an irreducible subrepresentation of . For example, consider the irreducible representation which contains a vacuum vector . The translation group acts via the operators . This shows that is an eigenvector corresponding to the character . This shows that the fibres of the vector bundle are onedimensional and the group acts identically on the fibre over . Thus the data describing the representation consists of the orbit of determined by (if the norm is positive SO is compact) and the trivial representation then the group SO of SO . Physicists say that transforms like a scalar. We can define the similar space corresponding to right movers. Its vacuum state is denoted by . Then we consider the tensor product . Its vacuum state is . Its vectors look like this i ~ i h ~ ~ h i r is symmetric in and (resp. in One defines the norm on degenerate inner product on i h and ) separately. similar to the norm on and then gets a non. Finally we complete this space to get the s r r r where 7 v 7 7 7 W } R& W d & & & 1tutt & & ut 1tt ututt 1tutt 7 7 7 f1o~o p fo p d f1o~o p Tg T Uuo'o p d f1o~o p d a I 7 w & 7 w d ` h7 d h7 d d h 7 y h p d 7 Y p t f1o~o p p d 7 d 7 5p i7 d h h 7 and and the polarization tensor q r W R gV U I Iv $ 7 f1op ~o $ 7 $ Y Y t `Y p a r a r DY p W F { s p i g F F F 7 { i W ! F & g g F { e i g h g e d 3 %3 g e g d D Y B ! W 9 F { & fe 8' p @ %db p d c D B W ! { 3 D B p g d g T R& {$ 5t q {$ n Finally, let us see the representation of the Virasoro algebra generated by the operators (resp. ) in the space (resp. ). Recall that and a similar formula holds for . Recall that id. We have to find the constant . One applies to some ground states to compute these constants. Notice that r r h 9 T T T T Q Q R& & d D B d g T T d & & & g& & T d g T &T o d p d R& 5&R& & R& d r o d o d g T T o d o d G& y j y d o d & D B W i r i D ar v d p & d & B & & ut1tt frut1tr d ututt t & & y d d d & & t d d r o d r o d 7 7 i h $ pt 7 7 td s i i i i i h The operators where Fock space of the closed bosonic string theory 66 and and we obtain Here we used that the operators Also
n are called level operators. It is easy to check that closed and are adjoint to each other. Thus LECTURE 6. FOCK SPACE (6.12) (6.11) u 6.1 Let be a graded Heisenberg Lie algebra. Let Sp be the symplectic group of linear automorphisms of which preserve the alternating from . Construct a linear projective representaion of Sp in the space which is compatible with the representation of in . 6.2 Compute the norm of the state . This gives , hence . Finally we obtain that for all , we have the following commutator relation for the Virasoro operators acting in the space . 6.3 Compute the norm of any state . 6.4 Let denote the dimension of the space of eigenvectors of the level operator with eigenvalue . Compute the generating function Tr . Explain the notation. ~ d & & utt1t v 7 A v 7 rw 5' E i 8' ' ' ' d ' h n [email protected] d R& d h d
67 & p t p W @ o o r i d D B 7 m g o @ d v d r v vd ` @ d v u t [email protected] o r g T T d r g T T g & g g& g d r d d D B R& R& R& R& td & d & R& r & R& d g T r d & & R& T o r g g T [email protected] o r g T d g R& & g& g & g & g Tg r T o d g T T d y d er v d fo s pt g fo ro g T T d Uo s r 3 r n id n n This gives n Thus Here we used (6.7) and . Also id n Exercises (6.13) 68 LECTURE 6. FOCK SPACE Lecture 7 Physical states for bosonic string
The expression for the Hamiltonian provides the masssquared formula . Recall that in the special relativity theory the mass is defined as the negative of the norm of the moment vector in the Minkowski spacetime. Let us explain it. We use the metric in the spacetime with coordinates defined by . Here , where is time and is a constant equal to the speed of light. To describe the motion we use the Lagrangian density where and is a constant called the mass. The energy and the moment for this Lagrangian are equal to
W We have so if we set we obtain that 69 & where we use the Minkowski norm defined by the matrix diag is called the total momentum vector. In our situation p g h p g h D uo'Ruo ut'utuot1~ o p B d y g T p d g y 5u g g w d y g d g kx g p g g vv t g p o y g d p d v g v E 'o d g p o y g d d g W d g h g p o y g d h g w p g p d g d I i d i W W { dw g h p gg h
y . The vector so we can 70 LECTURE 7. PHYSICAL STATES FOR BOSONIC STRING define the quantum masssquare operator by We shall scale the masses to assume that Recall that in the prequantized theory we had Virasoro constraints . It follows from (5.15) that the analogs of these constraints in the quantum string theory are the conditions that for any element of the Fock space. However, because id, this would imply that . Thus we have to require that only for positive and for some and similarly for the operators . We set i h
phys Thus the intersection of phys with the sum spur belongs to the nullspace of phys . The elements of this space are called spurious states. The Hilbert space which we want will be the quotient
spur Note that the operators defined in (6.9) commute with Virasoro operators, so that the Poincar Lie algebra acts in the spaces of physical states. e Remark 7.1. An abstract Lie algebra is called a Virasoro algebra if it can be defined by generators with commutator relations It can be shown that any any Lie algebra obtained as a central extension with onedimensional center of the algebra of vector fields on a circle is isomorphic to a Virasoro algebra. A representation of the Virasoro algebra in a vector space is called a representation with highest weight and charge if acts as a scalar operator id and there exists a vector (called a highest weight vector) such that 7 & t y d D x Bqx p o r pW m 6 4 t v d y j y d 4 W 4 closed closed phys closed phys p A state satisfying these conditions is called physical. Also for any state physical state , we have t d & & d y y d p 98 d p 8 j h i  h W h phys a h h h h $ closed phys h h v W h { W h h dD B A x 6 6 4 h phys phys 3 3 y y dd { G G R pt v p ֩ y v p y H y d G & G " j y d j y d v d G G G Thus the masssquare of the ground state y d g t g p ut1tr g p g t 5t r Ap drg o ro u g 0d 5t rd p d o ro
so that is equal to U h ~ U Q G G d k d g p r qp d g h h h U o pd o dd y & dD y d r d B b G ! G " ! G p dg 8 h h h G (7.1a) (7.1b) (7.1c) and a 71 A universal representation with this property is called a Verma module and is denoted . It can be constructed by using a similar construction as the representations by we constructed for a Heisenberg algebra. One considers the subalgebra Vir generated by the operators , then defines a onedimensional representation by and finally takes the induced representation Vir Vir . Its elements are linear combinations of monomials with positive 's. Any irreducible representation with highest weight and charge is isomorphic to a quotient of . So, we see that each nonzero generates a phys representation space for the Virasoro algebra with highest weight and charge . Its highest weight vector is . As we have seen before any physical state belongs to . Since all physical states are eigenvalues of massformula for physical states: with eigenvalue , we obtain the
p Let us see which ground states in closed are physical. Since For this vacuum state We shall see from the next discussion that must be equal to . Thus the vacuum vectors have negative mass. Such states are called tachyons (they travel faster than light!). The existence of such states will force us to abandon bosonic strings and consider superstrings. Let us look for of level 1. Each such has the form . We phys have Thus is physical if and only if If , we may choose , and then satisfies (7.4) but . This means that we have ghosts, i.e. states of negative norm. This should be avoided since the the quantum mechanics deals only with Hilbert spaces with unitary inner product. This forces us to take . G o v y R1tut1t y 'o d y ut1tut y ~1o Wy t p v td g y d Ur g d 'j y d o o H R& o G o G and the same is true for the right mode operators is physical if and only if ~ v & v& ututt 6 i W t v td g g o d y j y d y d e ֩ t v p td p p d g , we see that the ground state v h G v v p d g G u 7 v 7 y y y { j d o
h G G i d d o p d g d T o j v h G 6 G & 7 @ d v v d o u v 7v 7 d G & 6 d G (7.2) (7.3) (7.4) 72 If LECTURE 7. PHYSICAL STATES FOR BOSONIC STRING , we may take and so we have dimensional space of physical states of positive norm and no states of nonpositive norm. If , we may take and hence . This shows that we have a dimensional space of states of positive norm and a onedimensional space of states of norm 0. The state is spurious and is physical if . Thus, if , phys contains a onedimensional space of spurious states of norm 0. Factoring this space out we get a dimensional space , each element of which can be represented by a state of phys spur phys positive norm. So far, we find that and no restriction on appears. Let where phys phys phys phys Applying we see that and applying we see that . This implies that . Hence i G h G & h h C G h h h h G i G h i G h G G i u h 0 i h We have
G (7.5) Similarly, (7.6) In view of (7.5) and (7.6), we get The norm of the state is equal to If , we have no restriction on and hence we have physical states of negative norm. So . If , we may assume that so that guarantees that is physical. But if we take for , we get a state of negative norm, a ghost. So { T u If , we may assume that , and then the condition is so that the norm is equal to . We see that y d v d g g y 1tut1t y 'o p 'o t 'o v dv y y y ututd t1 y ~o d d v v t d 9 2i { { s s { s { { { d d v v v o p @Rut1tut y d dy d o v y d y d 0 Also so that (7.7) h G G { { G GR { G G G { G r G where of level 1, i.e.
r . Let us look at the physical states in . They are of the form & & 5 ut1tt v v vo pt v d 5 @ p @ & d 1tutut wy d h
closed phys ty d d t p v d g g o g & H Ur d d d t ~o j y d 5 & R& R& d W d t'tj y d o 5 & R& d tp d d r t & & ut1t tr d frut1tdr 1tutt v v p v d vd v o v 5 v d dp o v y d R& o d v d g d p @o v y t1ut1t y 'o p 'o d d o p @ o v G G 9 G 9 y ut1tut y ~o d 73 if all physical states are of nonnegative norm. The states of zero norm satisfy . Note that the states and are spurious. Since , these states are also physical because . It is easy to see now that any physical state of norm 0 is spurious and we can factor it out. Thus we obtain that for any nonzero lightlike closed the space phys is of dimension and all its elements can be represented by physical states of positive norm. For any of norm 0, the space of solutions of (7.5) and (7.6) is the direct sum of onedimensional space of matrices with nonzero trace, the dimensional space of traceless symmetric matrices and dimensional space of antisymmetric matrices. The corresponding physical states are called dilatons, ). gravitons and antisymmetric tensors. These are massless particles (i.e. Let us go to the second level, i.e. consider the physical states in phys of the form { i G h { { G u This implies The norm of this state is equal to { , where s Choose a system of coordinates in such that (7.8) we can eliminate 's and so that
& & &  { G R  s G R  G G h G { h d { G q s s where . We have (7.8) . Using (7.9) v v Tvv v t g R& r g & o p d v I zd g 5 y R1tutut y d t g r g p g r g p g d r t p d p v td g y d r y d r t d r r g ro g o d R& g y d r W d o r d & g & y d r W dd r & R& g& d p r y d g g p g p fo p @ @ ' @ @ W d p @ @ d @ p @ p g @ & R& R& & & R& y p R& y d g & d d v d { d y d g d v v d g d g h h u { W } o 74 LECTURE 7. PHYSICAL STATES FOR BOSONIC STRING Applying the CauchySchwarz inequality, we obtain , all states are of nonnegative norm. If , the state with and has the norm equal to . Now if we take a physical state from phys of level 2 and of positive norm, then closed is an element of phys of negative norm. So we have ghosts. If ,a state of zero norm must satisfy . The states
0 Thus, if are spurious. It is easy to see that any norm 0 physical state is equal to a physical spurious state. So we can factor them out and obtain only the space with only postive norms. Thus we have shown that . The proof that consists of analyzing states of the next level. We skip it. closed The result that phys does not contain ghosts if and only if is called the No Ghost Theorem. It was proven by R. Brower, P. Goddard and C. Thorn. Observe that and agrees with the definition of the Hamiltonian operator using the regularization of the sum . So physical states satisfy .
closed Remark 7.2. One can show that one can choose a subspace in phys invariant with respect to SO such that its states represent all states of positive norm modulo spurious states. This is achieved by a "lightcone gauge" which consists of fixing the first and the last coordinate of the string. The group SO acts in the space in closed closed via its induced representation. Thus defines a linear representation of the group . It also leaves the homogeneous parts invariant and hence defines a finite dimensional representation in each space of given level. Elements of this space which belong to an irreducible component are interpreted as elementary particles. For exam. ple, the antisymmetric tensors of level 1 define the adjoint representation of SO The dilatons define the trivial representation and gravitons define the standard representation of SO on the space . Exercises 7.1 Find physical states of level 2 in the Fock space of a closed bosonic string. 7.2 By analyzing physical states of level 3 in finish the proof of the No Ghost Theorem. d @ y Tvv @ qp Uo p @ j r d Tvv v @ t g R& p @ p h g & G 8 'o g & g p y d d @ o d vr d d @~o d v @ d @ & R& & R& g& r T r W d v A T W r d ~o { y d v o d y d y d t{ ~o v v v Tvv v v v @ g g d g & o p @ p g g { & u h s { 9 i s Q h h s h g I g h s G 9 h & (7.10) 9 #i & h s G h s s F F G G Lecture 8 BRSTcohomology
We shall discuss another approach to defining physical states which is called the BRSTquantization. In this approach one introduces an operator in a Fock space of a given string theory such that and
phys Ker Im We shall start with reminding the definition of the cohomology group of a Lie be algebra with coefficients in its linear representation . Let the tensor product of the exterior algebra of the space and the space . This will be an analog of our Fock space. If Lie for some Lie group , then can be identified with the space inv of leftinvariant smooth differential forms on with values in the trivial vector bundle defined by the space . Let be a basis of and be the dual basis of . Let so that . Elements of are linear combinations of the decomposable tensors Let us assign to the operator and to the contraction operator such that if for some and otherwise. If Lie the operator is the exterior product with the differential and . Here are local coordinates on corresponding to the basis . We have
g W 3 s i f ! 3 s i ! g h s W 5 3 s ! r It is called the anticommutator or the Poisson bracket. The structure of is determined by the constants such that
s s 75 n 3 ! n Here, for any associative algebra $ v $ v$ v t d 1D B d r v v v v t y d v v d v v d v v v v v v v v& utv 1tut d v v h v y v d uttut 1tv tut v $ Uo v qp v d v w utv 1tt dd vv d v 7 1tutt 1tutt 7 ! 7 d v $ 7 f d 7 ! 7 d and , 7 7 1 ! d
h g C g h h h W d 7 d 0 g q 0 d W d y dg h d (8.1) v 9 a a s a Q s a s s s s s s a s s s s s a s s s s s s h s v v v y d r r v v v to ~d v v v v v v r v v v v v v Urd v v v v v v r v v v v v v r o v v v Urd v v v v v v r v v v v v v o v v v g Urd v v v v v v r v v v v v v d g o v d v g d v v v t ~od v v v v v v r v v v v v v d d v v v g o v v v v y dd g y gg v g r rg p p d v v v v v v d vv v d g g 3 t v v p v v d v v v v r v v v v o v p d v r v v v d v v r v v v v v d v v v v r v v v v d y vd v v v v v r v v v v v g g t v v v v r v v v v o p d r v v v v v v v 3 3 t p w˩ v v v v d v v v v d vv r vv dg v v g v v ot v v v d v d y dg v v v v v v tp 7 o v v v o p v d v $ v % d s s s s s s s s s s n 9 9 r s s d s s h Using the anticommutator relations we see that In the latter case s s s s s Here we used that and . It remains to show that s s s s n d Proof. Let We have Lemma 8.1. Let 76 . Define the BRSToperator in , n n Also . This shows that . We have n n n it is easy to see that each of the four sums is equal to zero.
s s n s s s s n Using the Jacoby identity
s s n n s n n n n s n n n If
n , the expression in the bracket is equal to zero. So by End LECTURE 8. BRSTCOHOMOLOGY , where we skip the summation sign. unless so that . 77 Ker Im Example 8.1. Let be an abelian Lie algebra of dimension . Its linear representation is a module over . Let with the action of defined by . Then can be identified with the space of smooth differential forms of degree The BRSToperator coincides with the exterior derivative . We know that
F F
DR and In our situation we want to take for the Virasoro algebra Vir and its representation in the Fock space of bosonic string. The space is called the space of ghost fields. We will be dealing with a version of the BRST complex which uses the semiinfinite cohomology. Instead of differential forms we consider semiinfinite forms. Let be any strictly decreasing sequence of integers such that the sets and are finite. A semiinfinite form is a formal expression of the form where Its degree is equal to . We extend the operators to semidefinite forms in the obvious manner. Note that any form of degree be obtained from by applying operators . Also observe that
G j r G G v $ G G  8 The number Let is called the degree of G 7 1 ! v ututt C D H j y d v vv 5tutut1t~o p W d v $ d n m v Wҩ v n m p m d tut & U g V g V o r g g V V d d y gg d @ d r t i s p d v v v v r v v v v # p p g g r v v v v r v v v v i p p d r & t v v p @ r v v ip o v v d g B d D v v o p & t v v er Ap o p d A p x1D G B v v d v v r v v er & t v v p p D G B v v d v v r v v & t v p p v d v v v ip d v 3 g 5t v v r v v g o d d % y dg dd @ d d h F h r 7 ` r r r d Proof. Let Let be the linear space of semiinfinite forms of degree . It is clear that maps to . Let Vir Ker Im , where . We set rel Vir be the subspace of Vir generated by the cosets of forms which do not contain and which are annihilated by . 7 r r n n n n n n Changing the index Using this we get We use that Theorem 8.1. If 80 to r r So Finally, For any graded vector space U d a h d h n On the other hand, if . We have , then char . in the second sum, and applying (8.3), we get, . LECTURE 8. BRSTCOHOMOLOGY with we set d n h 81 If So it is easy to see that where where . Let us assume that . Let denote the Verma module for the representation of Vir with central charge and character (see the previous Lecture). The Verma module has the universal property with respect to all representations of Vir with central charge and character . Any such representations is a quotient of the Verma module. One can show that is spanned by the elements , . The grading of is defined by taking to be the subspace spanned by the monomials as above with . We have char 6 i It is known that the Verma module is irreducible if and irreducible and . Considering as a representation of Vir with character unitary for and central charge . Comparing the characters, we find that
0 h h i h 6 r r h T h char r The character of the irreducible module This shows that
h is equal to h char h The charge of the representation . We have
h is equal to and the character is equal to 9 h We have already noticed that the representation of Vir in to decompose it into irreducible modules. First we write i h a h h 5 h h & 9 h  l h i char Tr i We shall apply this to the case when g 5 I r g w " w o g d A R& h dh 8 & h r g dw h t g 8 d A w g w g w v 5tUuo g w w o Ap d ww g w w g p d o ww Uuo d o C 'l j o j y R& t 8 d u t v & & d fr1tutdr y j y 1tutt u u 1 u g d ww p ut1tut d 1 y d w w y d w 5 ww w d pt p o d 8 & h Ih t 8 d d k & & t5 efr1tuttdr r g o d v ut1tt v d d 7 t5 d 7 d 7 d 7 7 7 such that , then and
' char Tr 7 d 7 j W 3 u e W h 2 W " ! Tr and . Recall that h is reducible. Let us try . s si i s s h G W G W h h 9 h 9 V q r V W V ! r h W r r h T a i h where y g g i d d g p o " d a T 5tA d % R& 5tA R& R& v d ` ty v v v o p d p v d & & R& R & o t v p d R& Utr v v Ap Rv & d y v y v v y y d i d y { y { d & v v y v d CR & v y vd v v v v t 1tutt y { g & r R & y r d & v y A& y d y j A y d ` v d d % H g g p o g g o p o " d A y d y d A y d g t y Y y d p r g w ro C " d j g f1o g w w qp g 5t I r g w o " f1o g ww o qp d & t 8 d " i h h h h W d W h W W G W G { G G R G G 0 Let We con...
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