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Example 04 Group SU(2)

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Example 04 Group SU(2) QCD lecture 02 - Example 04 Group SU(2) - generators and construction of irreducible representations. Solution: SU(2) generators Group SU(2) - 2 × 2 unitary (U) matrices with unit determinant (S - special). This is also a representation of the group on two dimensional Hilbert space. Generators should be Hermitian matrices 2 × 2 with zero trace. Such general Hermitian matrix can be parametrized with three real numbers a, b, and c c a − ib : a + ib −c The SU(2) has therefor three generators which can be chosen as 0 1 0 −i 1 0 σ = ; σ = ; σ = : 1 1 0 2 i 0 3 0 −1 These are the well known Pauli matrices. It is strightforward to show that these matrices satisfy relation σiσj = δij1 + i"ijkσk ; with "ijk being the Levi-Civita symbol: 0 1 0 1 1 0 0 1 0 −i i 0 σ σ = · = ; σ σ = · = = iσ ; 1 1 1 0 1 0 0 1 1 2 1 0 i 0 0 −i 3 0 1 1 0 0 −1 σ σ = · = = −iσ ; 1 3 1 0 0 −1 1 0 2 0 −i 0 1 −i 0 0 −i 0 −i 1 0 σ σ = · = = −iσ ; σ σ = · = ; 2 1 i 0 1 0 0 i 3 2 2 i 0 i 0 0 1 0 −i 1 0 0 i σ σ = · = = iσ ; 2 3 i 0 0 −1 i 0 1 1 0 0 1 0 1 σ σ = · = = iσ ; 3 1 0 −1 1 0 −1 0 2 1 0 0 −i 0 −i 1 0 1 0 1 0 σ σ = · = = −iσ ; σ σ = · = : 3 2 0 −1 i 0 −i 0 1 3 3 0 −1 0 −1 0 1 In general, any complex 2×2 matrix can be expressed as a (complex) linear combintaion of the Pauli matrices and unit matrix 1. This is very easy to see. In particlar, for the product of two 2 × 2 Hermitian matrices with zero trace one can explicitly find that the combination is a3 a1 − ia2 b3 b1 − ib2 a1 + ia2 −a3 b1 + ib2 −b3 a b + a b + a b + ia b − ia b a b − a b + ia b − ia b = 1 1 2 2 3 3 1 2 2 1 3 1 1 3 2 3 3 2 a1b3 − a3b1 + ia2b3 − ia3b2 a1b1 + a2b2 + a3b3 − ia1b2 + ia2b1 = (a1b1 + a2b2 + a3b3)1 + i(a2b3 − a3b2)σ1 + i(a3b1 − a1b3)σ2 + i(a1b2 − a2b1)σ3 = ~a ·~b 1 + i ~a ×~b · ~σ; which leads immediately to the above formula for the product of two Pauli matrices. From this follows directly: [σi; σj] = 2i"ijkσk : It is also easy to see from the formula that the base is orthogonal: + Tr σiσj = Tr (σiσj) = 2δij : Canonical normalization (1=2δij) is then satisfied by matrices 1 J = σ : i 2 i These generators then satisfy commutation relations [Ji;Jj] = i"ijkJk : In the canonical normalization, the structure constants of SU(2) group are Levi-Civita symbols fabc = "abc. SU(2) roots and shifting operators Group roots ~α and shifting operators E~α are defined by relation [Hi;E~α] = αiE~α ; 8 i 2 1; : : : ; m: Number m represents the rank of the group and it corresponds to maximal number of commuting generators of the group, here denoted as Hi 8 i; j 2 1; : : : ; m :[Hi;Hj] = 0 ; so called Cartan subalgebra. Such operators E~α do indeed represent the shifting operators. Let ~µ represent some proper state of operators Hi. Since all Hi commute one can find the basis in the Hilbert space where the basis vectors are proper vectors of all Hi: Hi j~µi = µi j~µi ; 8 i 2 1; : : : ; m: Then HiE~α j~µi = [Hi;E~α] j~µi + E~αHi j~µi = αiE~α j~µi + µiE~α j~µi = (µi + αi)E~α j~µi ; (1) i.e. E~α j~µi / j~µ + ~αi. Group SU(2) is of rank one, as none of the generators commute with other SU(2) 0 generators. Therefore the dimension of root vector~α is one. Let's take H1 = J3 and build from J1 and J2 shifting operators E~α corresponding to the roots α. The general approach is to rewrite the above definition in the adjoint representation, where Hi jE~αi = j[Hi;E~α]i = αi jEαi : In other words, we can find the roots as the proper numbers of commuting generators Hi in adjoint representation and the proper vectors are than shifting operators. Generator H1 = J3 in adjoint representation reads ((Ta)bc = −ifabc): 0 0 1 01 00 −i 01 T3 = −i @−1 0 0A = @i 0 0A : 0 0 0 0 0 0 In general, the columns and raws that correspond to the possition of commuting generator Hi are filled with zeros, as follows from the construction of the adjoint representation. These empty lines are not relevant for finding proper numbers and vectors. In case of SU(2) we are left with two linear equations: 0 −i x x = α ) −iy = αx; and ix = αy : i 0 y y Getting rid of x from the first equation (using the second equation) we see that 2 y = α y ) α± = ±1: We have found two roots of SU(2) group which are α+ = 1 and α− = −1. Corresponding proper vectors are then 1 1 E / ; E / : α+ i α− −i We get canonical normalization hEα jEβi = δαβ by chosing normalization 1 1 1 1 Eα+ = p ; Eα− = p : 2 i 2 −i Going back to general representation of group SU(2), we found that J1 + iJ2 J1 − iJ2 J+ ≡ Eα=1 = p ;J− ≡ Eα=−1 = p : 2 2 It is straightforward to check that indeed those operators satisfy the original definition [J3;J+] = J+; [J3;J−] = −J−: + + We also observe that (J+) = J− and (J−) = J+. In general, one can prove that if ~α is + root then −~α is root as well and that (E~α) = E−~α. It follows from the antisymmetry of structure constants. Another useful relation, simple to verify, is [J+;J−] = J3: Again, it follows from the general relation [E~α;E−~α] = αiHi: Irreducible representaions of group SU(2) Let's form the basis of the Hilbert space from the proper vectors of operator J3: J3 jmi = m jmi : From the general proposition, the irreducible representation of SU(2) group is of finite dimension. It means that there must be a state with maximal weight j such that J+ jji = 0: We will start from this state and we will reconstruct the all other states using the lowering operator J−. In such a way, we will get entire irreducible representation. In principle, we have 3 generator, one (J3) is doing nothing to the basic vector. And since [J+;J−] = J3, by combing J+J− we will get again to original state. Don't know how to prove it elegantly. The problem is with degenerate states. If for example there are two states jj1i and jj2i corresponding to maximal weight j, it is easy to show that operator J+J− does not mix them, simply because J+J− jj1i = [J+;J−] jj1i + J−J+ jj1i = J3 jj1i. By induction we can show that the lower states don't mix as well. In other words if we start from two states with maximal weight j, we construct two subsets of base vectors that are not mixed with the J+ and J− operators (i.e. by all members of the group) and the representation is reducible then. As this state with maximal weight defines the representation, we will write in general for the basic vectors of the Hilbert space J3 jj; mi = m jj; mi : Number k = j − m will characterize how many times we have lowered the state with maximal weight. Let's denote the normlization of the state lowered by J− as Nk: J− jj; m = j − ki = Nk jj; m = j − k − 1i ) Nk = hj; j − k − 1j J− jj; j − ki : Complex conjugation gives ? ? Nk = hj; j − kj J+ jj; j − k − 1i ) J+ jj; m = j − k − 1i = Nk jj; m = j − ki : The square of modulus of the normalization is then ? Nk Nk = hj − kj J+J− jj − ki = hj − kj [J+;J−] jj − ki + hj − kj J−J+ jj − ki ? = hj − kj J3 jj − ki + hj − kj J−J+ jj − ki = (j − k) + Nk−1Nk−1 ; or ? ? j − k = Nk Nk − Nk−1Nk−1 : Starting from the state with maximal weight we get series ? k = 0 : j = N0 N0 ? ? k = 1 : j − 1 = N1 N1 − N0 N0 . ? ? k : j − k = Nk Nk − Nk−1Nk−1 : The sum of all terms then leads to formula for the normalization k + 1 N ?N = (j + j − k) : k k 2 The finite dimension of irreducible representation implies that for some value of k the state is lowered to zero vector, i.e. Nk = 0. This can be satisfied only if k = 2j which means that j is half integer. Irreducible representation of SU(2) is than characterized by this half-integer maximal weight j and such representation has 2j + 1 states with m = −j; −j + 1; : : : ; j − 1; j:.
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