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Chapter 4 SU(2)

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Chapter 4 SU(2) Chapter 4 SU(2) 4.1 Representations of SU(2) We will now work out in detail the properties of SU(2) and its representations. We have already seen that the generators may be chosen to be 1 L = σ , with σ = the Pauli matrices. i 2 i i k Then cij = ǫijk are the structure constants, and β = ǫ ǫ =2δ . ij − aib bja ij Thus our generators are not quite canonically normalized, but are all nor- malized equally, and β is positive definite. This is related to the fact, which we have already seen, that the group is compact. In writing down our generators, we have chosen, arbitrarily, one direction to make diagonal. Any rotation can be, by similarity transformation with at rotation, rotated into the z direction, so as we have a choice as to which of the infinite number of equivalent representations to choose, we may choose L3 to be diagonal. If we had several generators which commuted with each other, we could have chosen all of them to be diagonal. In general, we will take a maximal set of commuting generators, called the Cartan subalgebra, and represent them as diagonal. For SU(2), no rotations about any other axis commute with L3, so the Cartan subalgebra is one dimensional. 67 68. Last Latexed: April 25, 2017 at 9:45 Joel A. Shapiro We want to look for finite dimensional irreducible representations, and we have chosen to make Γ(L3) diagonal. Consider a basis vector em with L3em = mem, or Γm′m(L3) = mδm′ m, where so far we are not restricting m other than to say it is real. However, we know that e4πiL3 = 1I, so e4πim = 1, and m = n/2 for some integer n, for us to have a true representation of the SU(2) group. But even without assuming this, we will find it anyway, from requiring the representation to be finite dimensional. From L1 and L2 we can form the two operators L1 iL2 L± = ± . √2 These are not actually part of the Lie algebra, because that is defined over the reals, but any representation of the algebra is automatically a representation of the complexification, i.e. v L , v C. { a a} a ∈ L± are called raising and lowering operators respectively because P 1 i i 1 [L3, L±]= [L3, L1] [L3, L2]= L2 L1 = L±. √2 ± √2 √2 ± √2 ± Thus if L3em = mem, L (L e )=(L L L ) e =(m 1)L e , 3 ± m ± 3 ± ± m ± ± m so L±em is a new basis vector of the representation, unless it vanishes. Applying L+ an arbitrary number of times generates an arbitrary number of vectors unless at some point it gives 0. All of these vectors are linearly independent, so we would generate an infinite-dimensional space unless there p exists some state ej L+em with L3ej = jej on which L+ej = 0. ej is called the highest weight∝ state. 1 Let us form a normalized vector proportional to ej called j, j , and write the inner product in quantum mechanical form | i j, j j, j =1. h | i Now we generate a sequence of orthonormal states j, m = N Lj−m j, j , | i j,m − | i 1If there are several eigenvectors with eigenvalue m, add a label α to the states generated by L+, so the highest weight ones will be j, j, α , but the states with different α will turn out to be in different irreducible representations,| i as shown below. 618: Last Latexed: April 25, 2017 at 9:45 69 where Nj,m is a real normalization constant not quite the same as Georgi’s. The state j, m is an eigenstate of L with eigenvalue m, | i 3 L j, m = m j, m 3 | i | i 2 because each application of L− lowers the eigenvalue of L3 by one . The normalization factors can be found by observing that Nj,m j, m = L− j, m +1 | i | i Nj,m+1 −1 Nm+1 in Georgi or L− j, m +1 = Nm+1 j, m . | {z } On| the otheri hand, | i L j, m = N L Lk j, j with k = j m + | i j,m + − | i − k−1 = N Lk−1−r [L , L ] Lr j, j + Lk L j, j . j,m − + − − | i − + | i r=0 ! X 1 r But [L+, L−] = 2 [L1 + iL2, L1 iL2] = i [L2, L1] = L3 and L3L− j, j = (j r)Lr j, j as L is the lowering− operator. So as L j, j = 0, | i − − | i − + | i k−1 L j, m = N (j r)Lk−1 j, j + | i j,m − − | i r=0 X k(k 1) = kj − N Lk−1 j, j − 2 j,m − | i k 1 = k j − N N −1 j, m +1 − 2 j,m j,m+1 | i 1 = (j m)(j + m + 1)N −1 j, m +1 . 2 − m+1 | i † Now L+ = L−, so j, m +1 L j, m = j, m L j, m +1 ∗ h | + | i h | − | i = = 1 (j m)(j + m + 1)N −1 = N 2 − m+1 m+1 2 The state j, m, α so generated will be proportional to the original em because | i [L−,L+] = L3 and em is an eigenstate of L3. Thus all the states generated have the same α. − 70. Last Latexed: April 25, 2017 at 9:45 Joel A. Shapiro or 1 Nm+1 = (j m)(j + m + 1) √2 − 1 p Nm = (j m + 1)(j + m). √2 − p Now the set of states j, m m = j, j 1, j 2,... must terminate somewhere | i − − if the representation is to be finite dimensional, so for some m < j, Nm =0, so j + m = 0, or m = j. But j m is an integer, so 2j is an integer, and hence 2m is an integer− as well. − We have formed a representation with orthonormal basis j, m , m = j, j 1, j 2, , j , which is 2j + 1 dimensional, with 2j an{| integer.i We − − ··· − } 3 found the representation of L3 and L±, from which L1 and L2 follow . There is exactly one irreducible representation of SU(2) for each dimen- 1 3 sion and for each j = 0, 2 , 1, 2 ,.... Each representation has just one eigen- state of L3 with each eigenvalue m [ j, j +1,...,j]. The representation is given by ∈ − − L j, m = m j, m 3 | i | i 1 L+ j, m = (j m)(j + m + 1) j, m+1 | i √2 − | i 1 p L− j, m = (j + m)(j m + 1) j, m 1 | i √2 − | − i p the representation of the group elements is, of course, given by the exponen- tial of the representation of the generators. The 2j + 1 dimensional repre- sentation is usually denoted j (g) Dm1,m2 1 The j = 2 representation is just the group elements themselves. Therefore it is called the defining representation. 3 Notice that if we started with 2 orthogonal states em,α and em,β, all the states we generated from em,α would be orthogonal to all those from em,β, and the first set by itself would be an irreducible representation. 618: Last Latexed: April 25, 2017 at 9:45 71 4.2 Reduction of Direct Products If Γ1 and Γ2 are two representations of a Lie group, and Γ is their direct product, then 1 2 Γ(ix),(jy)(g)=Γij(g)Γxy(g). Recalling that4 Γ(L )= i d Γ(g(v )) , a dva a v=0 − =1 g 1 2 1 2 1 2 Γ(ix),(jy)(L)=Γij(L)Γxy(1I) + Γij(1I)Γxy(L)= δxyΓij(L)+ δijΓxy(L). Now for SU(2), consider the direct product states with bases j , j ; m , m = j , m j , m . | 1 2 1 2i | 1 1i⊗| 2 2i This representation can in principle be decomposed into a set of irreducible representations j, m . To see which, we count the dimensionality of the | i eigenspaces of L3, L j , j ; m , m = (L j , m ) j , m + j , m (L j , m ) 3 | 1 2 1 2i 3 | 1 1i ⊗| 2 2i | 1 1i ⊗ 3 | 2 2i = (m + m ) j , m ; j , m . 1 2 | 1 1 2 2i Fortunately our states are already eigenstates of L3. The largest eigenvalue is j1 + j2, which can occur in only one way. So the representation j = j1 + j2 occurs exactly once in the direct product. This representation accounts for one of the basis vectors for each m (= eigenvalue of L ) with m j + j . 3 | | ≤ 1 2 Now consider m = m + m = j + j 1. There are two ways to build 1 2 1 2 − it up, with m1 = j1 1, m2 = j2, or with m1 = j1, m2 = j2 1, as long as each j is at least 1 .− The next higher m, m = j + j 2, can− be built three 2 1 2 − different ways. The growth stops with m = j1 +j2 k, which can be written in k + 1 ways, with m = j r, m = j k + r, r−= 0,...,k. What stops 1 1 − 2 2 − it is that we must require m j , and m j , so k 2 min(j , j ). 1 ≥ − 1 2 ≥ − 2 ≤ 1 2 After that, the m corresponding to the smaller j (say jmin = k/2)) takes on all 2j + 1 values, until we get to m < j j , where the possible min −| max − min| values of m for the smaller j are limited.
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