Lectures 7-8. x y z p x p y p z L =

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1 S.K. Saikin October 1, Angular Momentum I Content: Definition of Orbital Angular Momentum. Commutation Rules. Lectures 7-8 Matrix Representation of Angular Momentum. 7.1 Definition Let us use the same definition as in classical mechanics L = r p, (1) where r and p are the position and the momentum operators respectively. The vector product (1) can be written as ê x ê y ê z L = x y z p x p y p z. (2) In the spatial basis we need to replace the momentum operator as p = i. The Cartesian components of the orbital angular momentum are ( ) L x = i y z z, y ( ) L y = i z x x, (3) z ( ) L z = i x y y. x Exercise: How it can be written in the momentum basis? We should introduce one more operator L 2 = L 2 x + L 2 y + L 2 z. (4) The angular momentum defined by Eq. (1) has units of. Sometimes it is easier to operate with a unitless angular momentum assuming = 1. This is only a mathematical trick. A physical quantity should have units. 1

2 7.2 Commutation Rules Commutators of an orbital angular momentum operator L with a position operator r, a momentum operator p and itself may be written in a compact form [L j, A k ] = i ɛ jkl A l, (5) where j, k, l are for the Cartesian components of the vectors, and the antisymmetric tensor ɛ jkl is defined as 1 j, k, l cyclic ɛ jkl = 1 j, k, l anticyclic 0 otherwise Example 1: [L z, x] = [xp y yp x, x] = xp y x yp x x xxp y + xyp x = x[p y, x] + y[x, p x ] = i y. (6) Example 2: [L z, p y ] = [xp y yp x, p y ] = xp y p y yp x p y p y xp y + p y yp x = [p y, y]p x + [x, p y ]p y = i p x. (7) One more useful commutator: [L 2, L i ] = 0. (8) The vector of the angular momentum operator doesn t commute with itself, but it commutes with its absolute value. We can define a basis ψ, where two of the operators L 2 and L i, i = x, y, z are simultaneously diagonal. Usually, the basis states are chosen to diagonalize L z. In this case L x and L y are non-diagonal. 7.3 Matrix representation of the angular momentum operator 1. Let us introduce operator a mode general operator of angular momentum J. It satisfies all the commutation rules given above, but not Eq. 1. For the sake of simplicity we assume that = We also define the rising J + and the lowering J operators as These operators are non-hermitian J ± = J x ± J y. (9) J = J +. (10) 2

3 Components of the angular momentum operator in terms of J ± are And the square of angular momentum is J x = 1 2 (J + + J ), (11) J y = i 2 (J + J ). J 2 = J 2 z (J +J + J J + ). (12) 3. Commutation rules and useful relations: [J 2, J ± ] = 0, (13) [J z, J ± ] = ±J ±, (14) Combining Eq. (14) with Eq. (12) we get [J +, J ] = 2J z. (15) J + J = J 2 J 2 z + J z, (16) J J + = J 2 J 2 z J z. (17) Operators J ± are non-diagonal in the chosen basis. However, J ± J are diagonal. 4. Let us define the basis states µ, ν that satisfy to two eigenvalue problems J 2 µ, ν = ν µ, ν, (18) J z µ, ν = µ µ, ν. (19) We will construct the matrices of the operators J 2, J z, and J ± in this basis. Because J ± commutes with J 2 we get J 2 J ± µ, ν = J ± J 2 µ, ν = νj ± µ, ν (20) The state J ± µ, ν is an eigenstate of the operator J 2 with the same eigenvalue. Similarly, using the commutation relations for J z and J ± we get that the state J ± µ, ν is an eigenstate of the operator J z with the eigenvalue µ ± 1 respectively. Thus, we can write that J ± µ, ν = γ µ,ν µ ± 1, ν, (21) 3

4 where γ µ,ν is a complex coefficient, which may depend on the eigenvalues of J 2 and J z. 5. Let us impose the condition that the norm of the eigenvector is not negative µ, ν µ, ν 0. (22) It has been shown above that if we apply the operator J + to the state µ, ν the resulting state should be proportional to µ + 1, ν. Its norm also should be not negative. This condition can be written as µ, ν J J + µ, ν 0. (23) We can substitute the J J + operator with Eq. 17 and obtain the relation for eigenvalues of J 2 and J z which set up the upper boundary on the value of µ ν µ(µ 1) 0, (24) µ max ν. (25) 4 Let assume that µ max = j, then ν = j(j + 1). Actually, from the the norm inequality we can get only a weaker condition ν j(j + 1). To get the equality we need to use the definition of J 2, see Eq. (4) and matrices of all the components of J. Another way to get ν is using the relation µ max, ν J J + µ max, ν = 0. (26) This is true because J + µ max, ν = 0. Then, we can substitute J J + with Eq. (17) and get ν = j(j + 1). Applying J + to the state with a maximal value of µ we should get 0. But we can apply J. If we apply it k-times the final state will be proportional to j k, ν. This state also should satisfy the condition of Eq. 22. In the way similar to the discussed above we can get that k max = 2j, (27) where k max is a maximal number of J operators that we can apply to the state j k, ν. Excercise: Derive the lower boundary for µ. In Eq. (27) k m ax is an integer number. Therefore, j can be and for each j the value of m can be j = 0, 1 2, 1, 3,... (28) 2 m = j, j + 1,..., j 1, j (29) 4

5 Now, for each j we can construct matrices of J 2 and J z. and For example for j = 3/2 m, j J z m, j = mδ jj δ mm (30) m, j J 2 m, j = j(j + 1)δ jj δ mm (31) J z = 3/ / / /2 (32) and J 2 = (33) For the orbital angular momentum operator the the values of m can be integer only. This can be shown using a spatial representation of L. Half-integer values may appear when we study spin. For the operators J ± we know that or J + m, j = λ m,j m + 1, j (34) m + 1, j J + m, j = λ m,j (35) In a most general case the coefficient λ m,j can be a complex number. If we take a Hermitian conjugate of the last equation we get Now, we can calculate matrix elements in the following way or J m + 1, j = λ m,j m, j (36) J J + m, j = λ m,j 2 m, j = (J 2 J 2 z J z ) m, j, (37) λ m,j = j(j + 1) m(m + 1) = (j m)(j + m + 1) (38) 5

6 The matrix elements of the operators J ± are defined by the following relations and For example, for j = 3/2 m, j J + m, j = (j m)(j + m + 1)δ j jδ m m+1 (39) m, j J m, j = (j m + 1)(j + m)δ j jδ m m 1. (40) J + = J = (41) At this point we can come back and prove that the eigenvalue of the operator J 2 is j(j + 1). We just need to write it in terms of the components of the angular momentum. Exercise 1: Show that ν = j(j + 1) (j = 3/2) using the matrix form for J +, J, J z, and J 2. Exercise 2: Write matrices J x,y,z,, J ±, J 2 for j = Spin Now, it is easy to introduce a spin S as an intrinsic angular momentum of a particle. It possesses all the properties of J. Its basis states may be written as m s, s. Because the spin of a particle does not depend on its orbital motion for a given type of particles s = const. For electrons, protons and neutrons s = 1/2. References [1] W. H. Louisell, Quantum statistical properties of radiation. (Wiley, New York, 1990), Part II, Chapters 2.6. [2] J. J. Sakurai, Modern Quantum Mechanics (Addison-Wesley, New York, 1994), Chapters