Quantum Field Theory III

Transcription

1 Quantum Field Theory III Prof. Erick Weinberg January 19, Lecture Structure We will start with a bit of group theory, and we will talk about spontaneous symmetry broken. Then we will talk about anomalies and grand unification. Then we will cover solitions & duality instantons. Finally we wil talk about supersymmetry. 1.2 Group theory Suppose we have a group G with elements g 1, g 2,... with operation g 1 g 2 = g 3. We require the operation to be associative, there is an identity and every element has an inverse gi = Ig = g, gg 1 = g 1 g = I (1) A group can have finite number of elements or infinite. For the infinite case we can have discrete elements (group of integers) or continuous (real numbers). The most important kind of groups in high energy physics is Lie groups, which is a continuous group. A group element will be labeled by a number of labels g(x 1, x 2,... ) = g(x). We have the group multiplication law g(x)g(y) = g(z) (2) where z is a continuous differentiable function of x and y. We can also think of the Lie group as a manifold. The dimension of the manifold, which is also the dimension of the Lie group, is equal to the number of parameters necessary to specify each group element. The manifold can be compact or non-compact. For example, a rotation group is compact, and if you keep rotation you will come back to the original position. However the Lorentz group is not compact, and you can boost forever. The Lorentz group will likely be the only non-compact group that we will make reference to. References can be found in the book by Georgi. Another one written by Gilmore. Let s look at some examples. Consider the group SO(N). This can be defined as the group of N N orthogonal matrices with determinant 1, or can be defined as the rotation group in N dimensions. From N N we have N 2 parameters, but from orthogonality we have the constraint R ij R jk = δ ik (3) This gives N conditions when i = k and N(N 1)/2 conditions when i k. So in the end we have N(N 1)/2 free parameters. So the group has dimension N(N 1)/2. Now if we erase S and consider 1

2 O(N). It has the same dimension as SO(N) but has two disconnected parts. The one containing identity is the same as SO(N). Now we consider U(N) which is the group of complex N N unitary matrices. Because it is unitary we have det U = 1. By the same argument as above we can find the dimension of the group to be of dimension N 2. Now if we consider SU(N), then we should have det U = 1. This subtracts one parameter as the determinant is a continuous variable. So the dimension of SU(N) is N 2 1. The Lorentz group, which is non-compact, has 3 boosts and 3 rotations. So this has 6 dimensions. These are the examples we want to consider. But these groups are not all distinct. We can write an element in U(1) as e iθ, and an element in SO(2) as a matrix with parameter θ. These two groups are essentially identical U(1) = SO(2) (4) We also have relations between SU(2) and SO(3). Each element in SO(3) corresponds to 2 elements in SU(2). In general we have the group Spin(N) which has two to one correspondence with SO(N). We also have correspondences between Lie groups and Lie algebras. A Lie algebra is related to the neighborhood of identity in the Lie group. If we think of the Lie group as a manifold, then the structure around the identity is almost enough to determine the whole group (apart from some double-cover things). To do this we think of the group as group of matrices and choose the coordinates so that the identity is I = (0, 0,... ). Near identity we have g = I + i j α j T j + O(α 2 ) (5) The α s are the coordinates and T j are generator matrices. The factor of i is just convention. In SO(3) the conventional generators will be the angular momenta J 1, J 2, J 3. An important result in the theory of Lie groups is that any element continuously connected to I can be written as g = exp (i ) α j T j (6) Now let s get some conditions on the T j generators. Let s look at the element g = e iλta e iλt b e iλta e iλt b (7) We assume λ 1. By group multiplication law this is an element of the group, so we can write ( g = exp i ) α k T k k (8) Now we need to do some manipulations ) g = e (e iλta iλta e iλt b + [e iλt b, e iλta ] e iλt b = I + e iλta [e iλt b, e iλta ]e iλt b (9) Now we can expand everything for small λ and keep up to terms of order λ 2, we get g = I + [iλt b, iλt a ] + O(λ 3 ) = I + λ 2 [T b, T a ] + O(λ 3 ) (10) Now this should be a group element, so the commutator of two generators should be a linear combination of all the generators [T a, T b ] = if c ab T c (11) 2

3 It is apparent that f c ab = f c ba. If we choose T s correctly then f abc will be completely antisymmetric. In SO(3) for conventional generators we have f abc = ε ijk. The f s are called structure constants. Now we come back to Lie algebra. A Lie algebra is defined to be a vector space with a bracket operation satisfying that i[a, B] is inside the algebra for any A, B in the algebra, and we have to satisfy the Jacobian identity [A, [B, C]] + [B, [C, A]] + [C, [A, B]] = 0 (12) The Poisson bracket is a form of Lie algebra operation {f(q, p), g(q, p)} = ( f g f ) g q j j p j p j q j (13) It satisfies the Jacobian identity but it is nontrivial to show. If we consider hermition matrices, in general (AB) AB, but the commutator (i[a, B]) = i[b, A ] = i[a, B] (14) So the commutator is inside the Lie algebra. The structure constants determines the whole algebra, so the algebras for SO(3) and SU(2) are the same. Let s consider SO(N). The elements satisfy R T R = I. So we have ( ) I + i j α j Tj T +... I + i k α k T k +... = I + i α j (T j + T T j ) + = I (15) So the generators must be N N antisymmetric matrices. This also applies to O(N). Same procedure shows that generators of U(N) are N N hermitian matrices. Now for SU(N) we need a condition on the determinant det [ e iα jt j ] = 1 = det [I + iαj T j +... ] = 1 + iα j Tr T j (16) So the generators are traceless. We can define direct products of different groups G = H K, with elements g = (h, k) and product g 1 g 2 = (h 1 h 2, k 1 k 2 ). The dimension is obviously the sum of two dimensions. We can choose a basis where the generators of G is the union of generators of H and generators of K with the two subsets commuting with each other. Conversely if the set of generators {T j } can be written as { } { } {T j } = t a (1) t b (2) (17) with the two subsets mutually commuting then any g can be written as g = e iα jt j (1)e iβ kt k (2) (18) Then locally G = H K, G is a direct product. For example we consider SO(4), the generators are J ij = J ji. Rotations are associated with a plane. We can define two sets of generators H i = 2 (J 23 + J 14 ), 1 2 (J 31 + J 24 ), 1 } 2 (J 12 + J 34 ), K i = 2 (J 23 J 14 ), 1 2 (J 31 J 24 ), 1 } 2 (J 12 J 34 ) 3 (19)

4 We can check that the H i commute with K j and they individually form SO(2) algebra. So the SO(4) algebra is the same as SU(2) SU(2) algebra. The correspondence looks like ( e iα H, e iβ K) e iα H e iβ K (20) Suppose we rotate by 2π we can have ( I, I) or (I, I) on the left side, but only I on the right side. So we have the relation SO(4) SU(2) SU(2) = (21) Z 2 Similar procedure can be carried out for the Lorentz group SO(3, 1). Because the commutation relation for the SO(3, 1) algebra is obtained by just replacing the δ ij in the commutation relation of SO(4) by g µν, the corresponding modification in H and K is just adding i in front of all the J with index 1: H i = 2 (J 23 + ij 14 ), 1 2 (ij 31 + J 24 ), 1 2 (ij 12 + J 34 ) }, K i = 2 (J 23 ij 14 ), 1 2 (ij 31 J 24 ), 1 2 (ij 12 J 34 ) (22) Now because there is a factor of i in front of some of the original generators, we would expect the exponential e iα jh j to have some non-periodic behavior in one direction. This is what makes the Lorentz group noncompact. A group is simple if there is not subgroup N such that gng 1 = N. There is no normal subgroups. A semisimple group is a group without an abelian normal subgroup. A Lie group is simple if it can t (even locally) be written as a product H K. A Lie group is semisimple if it has no U(1) factors. SO(3) is simple and semisimple. SU(4) SU(2) SU(2) is semisimple but not simple. U(N) U(1) SU(N) is neither simple nor semisimple. Any compact Lie group can be build up from simple compact Lie groups either by direct products or by direct products with some quotients. So if we know the simple compact groups we know all the compact groups. So we will only care about simple Lie groups. These are classified as SU(N), SO(N), Sp(N), E 6, E 7,... Now we consider group representations. We associate with every group element a matrix D(g) which satisfies D(g 1 )D(g 2 ) = D(g 1 g 2 ) (23) Two g s can have the same matrix, so there is aways the trivial representation D(g) = 1 which exists for any group. The dimension of the representation is the dimension of D which has nothing to do with the dimension of the group. Matrices correspond to linear transformations, so an r-dimensional representation corresponds to a set of r objects that mix under these linear transformations. Two representations are equivalent D (1) (g) = D (2) (g) if for every g we have D (1) (g) = SD (2) (g)s 1 (24) and S does not depend on g. This is just like changing basis. A unitary representation is where all D(g) are unitary matrices. A finite or compact group implies that all representations are equivalent to a unitary representation. A noncompact group is not so, i.e. we can find representations not equivalent to unitary ones. The Lorentz group has unitary representations, but they are infinite dimensional. It also has finite dimensional representations, but they are not unitary. A reducible representation means that by a change of basis we can write [ ] D(g) = S 1 D (1) (g) D (2) (g) S (25) } 4

5 so the representation is equivalent to a block-diagonal one. We can build up all representations this way, so we will be only interested in irreducible representations. For example, for SO(3) we have vectors which transform like V i V i = R ij V j (26) But we also have rank-2 tensors which transform like T ij T ij = R ik R jl T kl (27) But we can write any tensor into symmetric part and antisymmetric part. A symmetric tensor remains symmetric under rotation, so does the antisymmetric tensor. It can be further decomposed because trace is invariant under rotation. So T ij = T δ ij + A ij + S ij (28) 5