1 Theta functions and their modular properties

Transcription

1 Week 3 Reading material from the books Polchinski, chapter 7 Ginspargs lectures, chapter 7 Theta functions and their modular properties The theta function is one of the basic functions that appears again and again when describing modular invariants. There are multiple notations. I will use mostly the same notation as Polchinski s book, but not quite everywhere. The basic definition is in terms of a sum θ(z; τ) := θ 3 (z; τ) := θ 00 (z; τ) := exp(πiτn + πinz) () This sum is absolutely convergent for Im(τ) > 0 and is well defined for any complex variable z. It can be interpreted as a Fourier series in the variable z (if it were real) for periodic functions in the interval 0, ]. Because of this, it is obvious that the θ function is periodic It is also obviously an even function of z, given by θ(z + ; τ) = θ(z : τ) () θ( z; τ) = θ(z; τ) (3) The next property is to understand what happens when we send z z + τ.

2 That is, let us compute θ(z + τ; τ) = exp(πiτn + πinz + πinτ) = = exp(πiτn + πinτ + iπτ iπτ + πi(n + )z πiz) exp(πiτ(n + ) iπτ + πi(n + )z πiz) = exp( iπτ πiz)θ(z; τ) (4) This means that it is almost periodic with period τ as well (it transforms simply). It follows that under arbitrary shifts with a, b integers we get that θ(z + a + bτ) = exp( πibz πib τ)θ(z; τ) (5) Now, let us study what happens when we shift the variable τ τ +, one of the main modular transformations. We get that θ(z; τ + ) = exp(πiτn + πin + πinz) (6) But we notice that the term exp(πin ) = exp(iπn) is a phase that only depends on the parity of n. Thus we make that substitution to find that θ(z; τ + ) = exp(πiτn + πin + πinz) = θ(z + /; τ) (7) We will call this shifted θ function θ 0. That is θ 0 (z; τ) := θ 00 (z + ; τ) (8) Similarly, we define shifted θ functions by the half period τ, in such a way that we compensate for part of the phase in (4). These are ( ) θ 0 (z; τ) = exp πiτ + πiz θ(z + τ; τ) (9) 4 ( ) θ (z; τ) = exp πiτ + πi(z + /) θ(z + τ + /; τ) (0) 4

3 Basically, the, 0 indicate the shifts by half the lattice determined by τ. Now let us show the modular transform of θ, when we send τ /τ. The transformation is θ(z/τ; /τ) = ( iτ) / exp(πiz /τ)θ(z; τ) () Let us prove this. We start with an identity that converts a sum into an integral exp(πiτn +πinz) = exp(πiτx + πixz) ] δ(x n)dx () Now the function δ(x n) is a delta function that is periodic with period one. We can represent it in terms of the complete set of orthogonal functions on the circle with period. The complete such set is exp(πikx), where k =.... These are orthonormal, so we have that f(x) = k exp(πikx) 0 exp( πikx)f(x) (3) so it follows that δ(x n) = k= exp(πikx) (4) Here, the label k should be thought of as being in the dual lattice to the lattice of shifts of x by integers. This is because the dual lattice is defined by the set of numbers such that k.n is an integer for each integer n. This set is again the set of integers. Using this identity we find that θ(z; τ) = exp(πiτx + πixz) ] k= exp(πikx)dx (5) Now we exchange the sum and the integral, and we complete the square. This is done by identifying A(x + B) AB = Ax + ABx + AB AB = πiτx +πixz +πik, so that A = πiτ, AB = πi(z +k). Thus B = (z +k)/τ. With this shift we find that θ(z; τ) = exp(πiτ(x + (z + k)/τ) πi(z + k) /τ) ] dx (6) k= 3

4 So manipulating a bit we have that θ(z; τ) = k= exp( πi(k + kz + z )/τ) exp(πiτ( x) ) ] d x (7) where we have shifted x. It is clear that the integral is now independent of k, and we recognize a sum that is a θ function again. This one has τ = /τ. That is, we get that θ(z; τ) = θ( z/τ; /τ) exp( πiz /τ)) exp(πiτ( x) ) ] d x (8) The integral is a Gaussian integral equal to π/ iπτ, so we get that θ(z; τ) = θ( z/τ; /τ) exp( πiz /τ))/( iτ) / (9) A quick manipulation proves the modular transformation property (). The triple product formula and bosonization The theta function can also be written as an infinite product θ 3 (w; τ) = ( q n )( + q n / w)( + q n / w ) (0) n= where w = exp(πiz), and q = exp(πiτ). While with these variables we also have that θ 3 (w; τ) = q n / w n () The product is an absolutely convergent product, so the terms can be reordered at will. We want to show that the two are the same. We will give a physics proof. Let us massage the formulas a little bit. What we want to prove is ( ) ( + q n / w)( + q n / w ) = q k / w k ( q n ) () n= 4 k= n=

5 The left hand side is a product formula, and can be interpreted as a partition function for two types of fermions of opposite charge. The factor of q is like exp( βh), and the factor of w is like exp( µ): a chemical potential for a charge. It adds to positive charge objects, and it substracts on negative charged objects. The total charge number is N + N. The sum on the right hand side is summing over the different charge sectors. There is a correction to the energy given by q k /, and then there is a product that is familiar: it is related to the Dirichlet eta function. It can also be interpreted as a partition function over bosons in statistical mechanics. The identity basically states that two fermions is equal to one boson. What we are seeing is that this product formula expresses a very deep result in conformal field theory: it is the bosonization of a pair of charged fermions. The energy shift in charge sector n is also the energy of the corresponding tachyon operator in the boson theory. So how do we show the equality? We want to consider the partition function on the fermion side at fixed charge k. A system of particles with charge, and particles with opposite charge can be interpreted as system of particles and holes in a Dirac sea. This is not different, at least conceptually, from having an electronic material where the physics is determined by the Fermi sea. The issue is to count states at fixed charge. The first thing that happens is that at fixed charge the Fermi sea moves. At charge k, we have to sum the lowest energies that we are able to excite: this is given by k = k (3) For negative k it works the same. This explains the sum on the right hand side. Now, to excite the system without changing the total number of particles, we want to write the excitations by moving fermions from a lower lying state to a higher state. This also includes the process of creating holes: removing an electron. We do this in the following way: we take the topmost fermion in the Fermi sea and we excite it by n units for it to become the highest excited state in the new configuration of fermions. We do the same thing with the 5

6 next one, and we excite it n units, but we have to make sure that it is below the first state. From here we get that n n. We do this various times (let us say s times). In the end, we get a collection of s integers n n n s 0 (4) and we get an energy equal to E = n + + n s. For fixed E, the different number of ways in which we can do this is the set of collections of the ordered n i that sum to E. This is, each fermion configuration corresponds to a partition of E into integers. Thus, we have a sum of the form P (E)q E = E ( q n ) (5) as we have proved already (see previous notes). Corollary From the triple product we find that the theta function vanishes (has a simple zero) for w = q ±(n /) (6) This is the same as n= πiz = ±πi(n /)τ + πi mod (πi) (7) Or equivalently, z /+τ/ in the fundamental domain of the torus. That is, there is only a single simple zero in the torus (at a particular value of z). Together with the quasi-double periodicity this determines the θ function uniquely. 6