General Formalism Of Angular Momentum - Binghamton University

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Angular momentum: general formalism Masatsugu Sei Suzuki Department of Physics, SUNY at Binghamton (Date, October 13, 2014) Here we consider the general formalism of angular momentum. We will discuss the various properties of the angular momentum operator including the commutation relations. The eigenvalues and eigenkets of the angular momentum are determined. The matrix elements of the angular momentum operators are evaluated for angular momentum with integers as well as half-integers. 1. Commutation relations The commutation relations for the orbital angular momentum [ Lˆ x , Lˆ y ] i Lˆ z , [ Lˆ y , Lˆ z ] i Lˆ x , [ Lˆ z , Lˆ x ] i Lˆ y Generalization: definition of an angular momentum. The origin of the above relations lies in the geometric properties of rotations in three-dimensional space. Now we define an angular momentum Jˆi (i x, y, z) as any set of three observables satisfying [ Jˆ x , Jˆ y ] i Jˆ z , [ Jˆ y , Jˆ z ] i Jˆ x , [ Jˆ z , Jˆ x ] i Jˆ y 2 2 2 Jˆ 2 Jˆ x Jˆ y Jˆ z [ Jˆ 2 , Jˆ x ] 0̂ , [ Jˆ 2 , Jˆ y ] 0̂ , [ Jˆ 2 , Jˆ z ] 0̂ ((Proof)) 2 2 2 [ Jˆ 2 , Jˆ x ] [ Jˆ x , Jˆ x ] [ Jˆ y , Jˆ x ] [ Jˆ z , Jˆ x ] 2 2 [ Jˆ y , Jˆ x ] [ Jˆ z , Jˆ x ] 2i ( Jˆ y Jˆ z Jˆ z Jˆ y ) 2i ( Jˆ y Jˆ z Jˆ z Jˆ y ) 0̂ since 1

2 [ Jˆ y , Jˆ x ] Jˆ y Jˆ y Jˆ x Jˆ x Jˆ y Jˆ y Jˆ y ( Jˆ y Jˆ x Jˆ x Jˆ y ) ( Jˆ y Jˆ x Jˆ x Jˆ y ) Jˆ y Jˆ y [ Jˆ x , Jˆ y ] [ Jˆ x , Jˆ y ]Jˆ y 2i ( Jˆ y Jˆ z Jˆ z Jˆ y ) 2 [ Jˆ z , Jˆ x ] Jˆ z Jˆ z Jˆ x Jˆ x Jˆ z Jˆ z Jˆ z ( Jˆ z Jˆ x Jˆ x Jˆ z ) ( Jˆ z Jˆ x Jˆ x Jˆ z ) Jˆ z Jˆ [ Jˆ , Jˆ ] [ Jˆ , Jˆ ]Jˆ z z x z x z 2i ( Jˆ y Jˆ z Jˆ z Jˆ y ) 2. General theory of angular momentum (a) Jˆ and Jˆ Jˆ Jˆx iJˆy where Jˆ Jˆ , Jˆ Jˆ [ Jˆ z , Jˆ ] Jˆ , [ Jˆ z , Jˆ ] Jˆ , [ Jˆ 2 , Jˆ ] [ Jˆ 2 , Jˆ ] [ Jˆ 2 , Jˆ z ] 0̂ 2 Jˆ Jˆ Jˆ 2 Jˆ z Jˆ z 2 Jˆ Jˆ Jˆ 2 Jˆ z Jˆ z Thus we have 1 2 Jˆ 2 ( Jˆ Jˆ Jˆ Jˆ ) Jˆ z 2 Formula: [a Jˆ , b Jˆ ] i (a b) Jˆ ((Proof)) 2 [ Jˆ , Jˆ ] 2 Jˆ z

I [a Jˆ , b Jˆ ] aib j [ Jˆi , Jˆ j ] i, j Since [ Jˆi , Jˆ j ] i ijk Jˆk then I aib j i ijk Jˆk i (a b) Jˆ i, j (b) Notation for the eigenvalues of Ĵ 2 and Jˆz For any ket Jˆ 2 Jˆ x 2 Jˆ y 2 Jˆ z 2 Jˆ x Jˆ x Jˆ y Jˆ y Jˆ z Jˆ z 0 For an eigenket Ĵ 2 Jˆ 2 0 We shall write Jˆ 2 2 j ( j 1) 2 where j(j 1) 0 ((Note)) Levi-Civita symbol In three dimensions, the Levi-Civita symbol ijk is defined as follows: ijk is 1 if (i, j, k) is an even permutation of (1,2,3). ijk is -1 if it is an odd permutation, ijk is 0 if any index is repeated. 123 1 , 132 1 , 213 1 , 3

231 1 , 3. 312 1 , 321 1 . Eigenvalue equations for Jˆ2 and Jˆz j, m j.m is the simultaneous eigenket of Ĵ 2 and Jˆz , since [ Jˆ 2 , Jˆ z ] 0̂ Jˆ 2 j , m 2 j ( j 1) j , m Jˆ z j , m m j , m We discuss the eigenvalues of Ĵ 2 and Jˆz Lemma 1 (Properties of eigenvalues of Ĵ 2 and Jˆz ) j and m satisfy the inequality -j m j ((Proof)) j , m Jˆ Jˆ j , m 0 , 4

j , m Jˆ Jˆ j , m 0 . We find 2 j , m Jˆ Jˆ j , m j , m Jˆ 2 Jˆ z Jˆ z j , m 2 [ j ( j 1) m(m 1)] 0 , and 2 j , m Jˆ Jˆ j , m j , m Jˆ 2 Jˆ z Jˆ z j , m 2 [ j ( j 1) m(m 1)] 0 . Then we have [ j ( j 1) m (m 1)] ( j m )( j m 1) 0 , [ j ( j 1) m (m 1)] ( j m 1)( j m ) 0 . Then ( j 1) m j , and j m j 1. If j m j , these two conditions are satisfied simultaneously. Lemma II (Properties of the ket vector of Jˆ j , m ) (i) If m -j, Jˆ j , m 0̂ (ii) If m -j, Jˆ j , m is a non-null eigenket of Ĵ 2 and Jˆz with the eigenvalues j(j 1) 2 and (m -1) . [Proof of (i))] Since j , m Jˆ Jˆ j , m 2 [( j m)( j m 1)] 0 for m -j, we get Jˆ j , m 0̂ 5

for m -j. Conversely, if Jˆ j , m 0̂ then we have j , m Jˆ Jˆ j , m 2 [( j m)( j m 1)] 0 Then we have j -m. [Proof of (ii)] Since [ Jˆ 2 , Jˆ ] 0̂ [ Jˆ 2 , Jˆ ] j , m 0̂ or Jˆ 2 Jˆ j , m Jˆ Jˆ 2 j , m 2 j ( j 1) Jˆ j , m So Jˆ j , m is an eigenket of Ĵ 2 with the eigenvalue 2 j( j 1) . Moreover, [ Jˆ z , Jˆ ] j , m Jˆ j , m or Jˆ z Jˆ j , m Jˆ Jˆ z j , m Jˆ j , m (m 1) Jˆ j , m So Jˆ j, m is an eigenket of Jˆz with the eigenvalue ( m 1) . Lemma III (i) (Properties of the ket vector of Jˆ j , m ) If m j, Jˆ j , m 0̂ . 6

(ii) If m j, Jˆ j , m is a non-null eigenket of Ĵ 2 and Jˆz with the eigenvalues j(j 1) 2 and (m 1) . [Proof of (i)] j, m Jˆ Jˆ j, m 2 [( j m)( j m 1)] 0 If m j, then Jˆ j , m 0̂ . [Proof of (ii)] [ Jˆ 2 , Jˆ ] 0̂ , or [ Jˆ 2 , Jˆ ] j , m 0̂ , or j , m Jˆ 2 Jˆ j , m 2 j ( j 1) Jˆ j , m . 2 Jˆ j, m is an eigenket of Jˆ with an eigenvalues 2j(j 1). [ Jˆ z , Jˆ ] j , m Jˆ j , m or Jˆ z Jˆ j , m Jˆ Jˆ z j , m Jˆ j , m (m 1) j , m Jˆ j , m is an eigenket of Jˆz with an eigenvalues (m 1). 4. Determination of the spectrum of Jˆ2 and Jˆz . There exists a positive or zero integer p such that m - p -j (1) j , m : [eigenvalues m, 2j(j 1)] 7

Jˆ j , m : [ (m-1), 2j(j 1)] ( Jˆ ) 2 j , m : [ (m-2), 2j(j 1)] . ( Jˆ ) p j , m : [ (m-p), 2j(j 1)] ( Jˆ ) p 1 j , m 0̂ There exists a positive or zero integer such that m q j, (2) j , m : [eigenvalues m, 2j(j 1)] Jˆ j , m : [ (m 1), 2j(j 1)] ( Jˆ ) 2 j , m : [ (m 2), 2j(j 1)] . ( Jˆ ) q j , m : [ (m q), 2j(j 1)] ( Jˆ ) q 1 j , m 0̂ Combining Eqs.(1) and (2), p q 2j Since p and q are integers, j is therefore an integer or a half-integer. j 0, 1/2, 1, 3/2, 2, . m -j, -j 1,., j-1, j. (i) (ii) 5. If j is an integer, then m is an integer. If j is a half-integer, then m is a half-integer. j, m representation 8

2 j , m Jˆ Jˆ j , m j , m Jˆ 2 Jˆ z Jˆ z j , m 2 [( j m)( j m 1)] Since Jˆ j , m j , m 1 we have 2 j , m Jˆ Jˆ j , m 2 [( j m)( j m 1)] . Thus Jˆ j , m ( j m)( j m 1) j , m 1 . Similarly 2 j , m Jˆ Jˆ j , m j , m Jˆ 2 Jˆ z Jˆ z j , m 2 [( j m)( j m 1)] . Since Jˆ j , m j , m 1 , we have j , m Jˆ Jˆ j , m 2 [( j m)( j m 1)] , 2 or Jˆ j, m ( j m )( j m 1) j, m 1 . In summary: Jˆ 2 j , m 2 j ( j 1) j , m Jˆ z j, m m j, m Jˆ j, m ( j m )( j m 1) j, m 1 Jˆ j, m ( j m)( j m 1) j, m 1 9

6. Matrix element of the angular momentum Using Mathematica, you determine the matrix elements of Jˆx , Jˆy , and Jˆz Jˆ 2 j.m 2 j ( j 1) j.m Jˆ z j , m m j , m Jˆ j , m ( j m)( j m 1) j , m 1 Jˆ j , m ( j m)( j m 1) j , m 1 Since Jˆ Jˆ x iJˆ y Jˆ Jˆ x iJˆ y we get 1 Jˆ x ( Jˆ Jˆ ) 2 1 i Jˆ y ( Jˆ Jˆ ) ( Jˆ Jˆ ) 2i 2 1 Jˆ x j , m ( Jˆ Jˆ ) j , m 2 ( ( j m)( j m 1) j , m 1 ( j m)( j m 1) j , m 1 ) 2 i Jˆ y j , m ( Jˆ Jˆ ) j , m 2 i ( ( j m)( j m 1) j , m 1 ( j m)( j m 1) j , m 1 ) 2 Jˆ z j , m m j , m Thus the matrix elements are expressed by 10

j , m' Jˆ x j , m ( ( j m)( j m 1) j , m' j , m 1 ( j m)( j m 1) j , m' j , m 1 ) 2 ( ( j m)( j m 1) m ', m 1 ( j m)( j m 1) m ', m 1 ) 2 i j , m' Jˆ y j , m ( ( j m)( j m 1) j , m' j , m 1 ( j m)( j m 1) j , m' j , m 1 ) 2 i ( ( j m)( j m 1) m ',m 1 ( j m)( j m 1) m ',m 1 ) 2 j.m' Jˆ z j , m m j , m j , m' m m ,m ' Using this formula, we can get the matrix of Ĵ x , Ĵ y , and Ĵ z for each j ( 1/2, 1, 3/2, 2, 5/2, .) and m ( j, j-1, j-2,., -j 1, -j). 7. Effect of fluctuation in the direction of J We now consider the fluctuation of the angular momentum, J x 2 J 2 y 2 j , m ( Jˆ x Jˆ x 1̂) 2 j , m Jˆ x Jˆ x 2 j, m ( Jˆ y Jˆ y 1̂) 2 j, m Jˆ y Jˆ y 2 2 Jˆ x 2 2 Jˆ y Then we have Jˆ 2 2 j ( j 1) 2 2 2 Jˆ x Jˆ y Jˆ z 2 2 2 J x J x Jˆ z J x J x 2 m 2 2 2 or J x 2 J x 2 2 [ j( j 1) m 2 ] The fluctuation in the component of the angular momentum which is normal to the z axis will be a minimum when m j. We therefore obtain [ J x J x ]min j 2 2 2 11

This means in a rough manner of speaking that the minimum angle between the direction of the J vector and the z axis is given by sin min j j ( j 1) 1 j 1 or cos min j j ( j 1) ((D. Bohm, quantum theory, p.319)) It is not correct to imagine that the angular momentum points in some definite direction which we do not happen to be able to measure with complete precision. Instead, whenever Ĵ 2 and Ĵ z have definite values, one should imagine that the entire cone of directions corresponding to those values of Jx and Jy consistent with the given j2 and m are covered simultaneously because important physical consequence may follow from the effects of interference of wave functions corresponding to different components of angular momentum. 8. Vector representation of allowed angular momentum We consider a case which j is some fixed number (j 1/2, 1, 3/2, 1,.). Then the total angular momentum may be represented by a vector of length j ( j 1) The component m in the z direction is m j, j - 1, j - 2, ., -j 1, -j The vector J should be thought of as covering a cone, with vector angle given by cos m m j ( j 1) (a) j 1/2 m 1/2, -1/2 12

1 2 1 2 1 m 1 2 m -1 2 (b) j 1 m 1, 0, -1 13

1 1 1 m 1 m 0 m -1 (c) j 3/2 m 3/2, 1/2, -1/2, -3/2 14

3 2 3 2 1 m 3 2 m 1 2 m -1 2 m -3 2 (d) j 3 m 3, 2, 1, 0, -1, -2, -3 15

3 3 1 m 3 m 2 m 1 m 0 m -1 m -2 m -3 (e) j 6 m 6, 5, 4, 3, 2, 1, 0, -1, -2, -3, -4, -5, -6 16

6 6 1 m 6 m 5 m 4 m 3 m 2 m 1 m 0 m -1 m -2 m -3 m -4 m -5 m -6 (a) j integers 17

(b) j half integer 18

REFERENCES J.J. Sakurai, Modern Quantum Mechanics, Revised Edition (Addison-Wesley, Reading Massachusetts, 1994). C. Cohen-Tannoudji and B. Diu, and F. Laloe, Quantum Mechanics, vol.1 and vol. 2 (John Wiley & Sons, New York, 1977). A.R. Edmonds, Angular Momentum in Quantum Mechanics (Princeton University Press, 1957). 19

M.E. Rose, Elementary Theory of Angular Momentum (John Wiley & Sons, 1957, New York, Dover Publications, New York, 1957). L.C. Biedenharn and J.D. Louck, Angular Momentum in Quantum Physics (AddisonWesley, Reading, 1981). D.M. Brink and G.R. Satchler, Angular Momentum, 2nd edition (Clarendon Press, Oxford, 1968). D.H. McIntyre, Quantum Mechanics: A paradigms Approach (Pearson, 2012). V. Devanthan, Angular Momentum Techniques in Quantum Mechanics (Kluwer Academic). 20

Here we consider the general formalism of angular momentum. We will discuss the various properties of the angular momentum operator including the commutation relations. The eigenvalues and eigenkets of the angular momentum are determined. The matrix elements of the angular momentum operators are evaluated for angular

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