Angular Momentum PDF

Chapter 5 Angular Momentum 5.1 Introduction After treating one-dimensional problems in Chapter 4, we now should deal with three-dimensional problems. However, the study of three-dimensional systems such as atoms cannot be under- taken unless we first cover the formalism of angular momentum. The cu...

Chapter 5 Angular Momentum 5.1 Introduction After treating one-dimensional problems in Chapter 4, we now should deal with three-dimensional problems. However, the study of three-dimensional systems such as atoms cannot be under- taken unless we first cover the formalism of angular momentum. The current chapter, therefore, serves as an essential prelude to Chapter 6. Angular momentum is as important in classical mechanics as in quantum mechanics. It is particularly useful for studying the dynamics of systems that move under the influence of spherically symmetric, or central, potentials, V ;r   V r , for the orbital angular momenta of these systems are conserved. For instance, as mentioned in Chapter 1, one of the cornerstones of Bohr’s model of the hydrogen atom (where the electron moves in the proton’s Coulomb potential, a central potential) is based on the quantization of angular momentum. Additionally, angular momentum plays a critical role in the description of molecular rotations, the motion of electrons in atoms, and the motion of nucleons in nuclei. The quantum theory of angular momentum is thus a prerequisite for studying molecular, atomic, and nuclear systems. In this chapter we are going to consider the general formalism of angular momentum. We will examine the various properties of the angular momentum operator, and then focus on de- termining its eigenvalues and eigenstates. Finally, we will apply this formalism to the determi- nation of the eigenvalues and eigenvectors of the spin and orbital angular momenta. 5.2 Orbital Angular Momentum In classical physics the angular momentum of a particle with momentum p; and position r; is defined by L;  r;  p;  ypz  zp y ;i  zpx  x pz  ;j  x p y  ypx k ; (5.1) The orbital angular momentum operator L; can be obtained at once by replacing r; and p; by the corresponding operators in the position representation, R; and P;  i h V: ; L;  R;  P;  i h R;  V ;  (5.2) 283 284 CHAPTER 5. ANGULAR MOMENTUM The Cartesian components of L; are t u " " L x  Y Pz  Z Py  i h Y Z  (5.3) "z "y tu " " L y  Z Px  X Pz  i h Z X  (5.4) "x "z t u " " L z  X Py  Y Px  i h X Y  (5.5) "y "x Clearly, angular momentum does not exist in a one-dimensional space. We should mention that ; the components L x , L y , L z , and the square of L, 2 L;  L 2x  L 2y  L 2z  (5.6) are all Hermitian. Commutation relations Since X, Y , and Z mutually commute and so do Px , Py , and Pz , and since [ X Px ]  i h , [Y  Py ]  i h , [ Z Pz ]  i h , we have [ L x  L y ]  [Y Pz  Z Py  Z Px  X Pz ]  [Y Pz  Z Px ]  [Y Pz  X Pz ]  [ Z Py  Z Px ]  [ Z Py  X Pz ]  Y [ Pz  Z ] Px  X [ Z  Pz ] Py  i h  X Py  Y Px   i h L z  (5.7) A similar calculation yields the other two commutation relations; but it is much simpler to infer them from (5.7) by means of a cyclic permutation of the x yz components, x  y  z  x: [ L x  L y ]  i h L z  [ L y  L z ]  i h L x  [ L z  L x ]  i h L y  (5.8) As mentioned in Chapter 3, since L x , L y , and L z do not commute, we cannot measure them simultaneously to arbitrary accuracy. Note that the commutation relations (5.8) were derived by expressing the orbital angular momentum in the position representation, but since these are operator relations, they must be valid in any representation. In the following section we are going to consider the general formalism of angular momentum, a formalism that is restricted to no particular representation. Example 5.1 (a) Calculate the commutators [ X  L x ], [ X L y ], and [ X L z ]. (b) Calculate the commutators: [ Px  L x ], [ Px  L y ], and [ Px  L z ]. (c) Use the results of (a) and (b) to calculate [ X  L; 2 ] and [ Px  L; 2 ]. 5.3. GENERAL FORMALISM OF ANGULAR MOMENTUM 285 Solution (a) The only nonzero commutator which involves X and the various components of L x , L y , L z is [ X Px ]  i h. Having stated this result, we can easily evaluate the needed commutators. First, since L x  Y Pz  Z Py involves no Px , the operator X commutes separately with Y , Pz , Z, and Py ; hence [ X L x ]  [ X Y Pz  Z Py ]  0 (5.9) The evaluation of the other two commutators is straightforward: [ X  L y ]  [ X  Z Px  X Pz ]  [ X  Z Px ]  Z [ X Px ]  i h Z (5.10) [ X  L z ]  [ X  X Py  Y Px ]  [ X Y Px ]  Y [ X  Px ]  i h Y  (5.11) (b) The only commutator between Px and the components of L x , L y , L z that survives is again [ Px  X]  i h. We may thus infer [ Px  L x ]  [ Px  Y Pz  Z Py ]  0 (5.12) [ Px  L y ]  [ Px  Z Px  X Pz ]  [ Px  X Pz ]  [ Px  X ] Pz  i h Pz  (5.13) [ Px  L z ]  [ Px  X Py  Y Px ]  [ Px  X Py ]  [ Px  X] Py  i h Py  (5.14) (c) Using the commutators derived in (a) and (b), we infer [ X L; 2 ]  [ X L 2x ]  [ X  L 2y ]  [ X L 2z ]  0  L y [ X  L y ]  [ X  L y ] L y  L z [ X L z ]  [ X  L z ] L z  i h  L y Z  Z L y  L z Y  Y L y  (5.15) ;2 [ Px  L ]  [ Px  L 2x ]  [ Px  L 2y ]  [ Px  L 2z ]  0  L y [ Px  L y ]  [ Px  L y ] L y  L z [ Px  L z ]  [ Px  L z ] L z  i h  L y Pz  Pz L y  L z Py  Py L y  (5.16) 5.3 General Formalism of Angular Momentum Let us now introduce a more general angular momentum operator J; that is defined by its three components Jx  Jy  and Jz , which satisfy the following commutation relations: [ Jx  Jy ]  i h Jz  [ Jy  Jz ]  i h Jx  [ Jz  Jx ]  i h Jy  (5.17) or equivalently by J;  J;  i h J; (5.18) Since Jx , Jy , and Jz do not mutually commute, they cannot be simultaneously diagonalized; that is, they do not possess common eigenstates. The square of the angular momentum, J;2  Jx2  Jy2  Jz2  (5.19) 286 CHAPTER 5. ANGULAR MOMENTUM is a scalar operator; hence it commutes with Jx  Jy  and Jz : [ J;2  Jk ]  0 (5.20) where k stands for x y, and z. For instance, in the the case k  x we have [ J;2  Jx ]  [ Jx2  Jx ]  Jy [ Jy  Jx ]  [ Jy  Jx ] Jy  Jz [ Jz  Jx ]  [ Jz  Jx ] Jz  Jy i h Jz   i h Jz  Jy  Jz i h Jy   i h Jy  Jz  0 (5.21) because [ Jx2  Jx ]  0, [ Jy  Jx ]  i h Jz , and [ Jz  Jx ]  i h Jy. We should note that the operators Jx , Jy , Jz , and J;2 are all Hermitian; their eigenvalues are real. Eigenstates and eigenvalues of the angular momentum operator Since J;2 commutes with Jx , Jy and Jz , each component of J; can be separately diagonalized (hence it has simultaneous eigenfunctions) with J;2. But since the components Jx , Jy  and Jz do not mutually commute, we can choose only one of them to be simultaneously diagonalized with J;2. By convention we choose Jz. There is nothing special about the z-direction; we can just as well take J;2 and Jx or J;2 and Jy. Let us now look for the joint eigenstates of J;2 and Jz and their corresponding eigenvalues. Denoting the joint eigenstates by : ;O and the eigenvalues of J;2 and Jz by h 2 : and h ;, respectively, we have J;2 : ;O  h 2 : : ;O (5.22) Jz : ;O  h ; : ;O (5.23) The factor h is introduced so that : and ; are dimensionless; recall that the angular momentum has the dimensions of h and that the physical dimensions of h are: [h ]  energy  time. For simplicity, we will assume that these eigenstates are orthonormal: N: )  ; ) : ;O  =:)  : =; )  ;  (5.24) Now we need to introduce raising and lowering operators J and J , just as we did when we studied the harmonic oscillator in Chapter 4: J  Jx  i Jy  (5.25) This leads to 1 1 Jx   J  J  Jy   J  J  (5.26) 2 2i hence 1 2 1 Jx2   J  J J  J J  J2  Jy2    J2  J J  J J  J2  (5.27) 4  4 Using (5.17) we can easily obtain the following commutation relations: [ J;2  J ]  0 [ J  J ]  2h Jz  [ Jz  J ]  h J  (5.28) 5.3. GENERAL FORMALISM OF ANGULAR MOMENTUM 287 In addition, J and J satisfy J J  Jx2  Jy2  h Jz  J;2  Jz2  h Jz  (5.29) J J  Jx2  Jy2  h Jz  J;2  Jz2  h Jz  (5.30) These relations lead to J;2  J Jb  Jz2 b h Jz  (5.31) which in turn yield 1 J;2   J J  J J   Jz2  (5.32) 2 Let us see how J operate on : ;O. First, since J do not commute with Jz , the kets : ;O are not eigenstates of J. Using the relations (5.28) we have Jz  J : ;O   J Jz  h J  : ;O  h ;  1 J : ;O (5.33) hence the ket ( J : ;O) is an eigenstate of Jz with eigenvalues h ;  1. Now since Jz and J 2 commute, ( J : ;O) must also be an eigenstate of J;2. The eigenvalue of J;2 when acting ; on J : ;O can be determined by making use of the commutator [ J;2  J ]  0. The state ( J : ;O is also an eigenstate of J;2 with eigenvalue h 2 :: J;2  J : ;O  J J;2 : ;O  h 2 : J : ;O (5.34) From (5.33) and (5.34) we infer that when J acts on : ;O, it does not affect the first quantum number :, but it raises or lowers the second quantum number ; by one unit. That is, J : ;O is proportional to : ;  1O:  J : ;O  C:; : ;  1O (5.35)  We will determine the constant C:; later on. Note that, for a given eigenvalue : of J;2 , there exists an upper limit for the quantum number ;. This is due to the fact that the operator J;2  Jz2 is positive, since the matrix elements of J;2  Jz2  Jx2  Jy2 are o 0; we can therefore write N: ; J;2  Jz2 : ;O  h 2 :  ; 2  o 0 >" : o ; 2 (5.36) Since ; has an upper limit ;max , there must exist a state : ;max O which cannot be raised further: J : ;max O  0 (5.37) Using this relation along with J J  J;2  Jz2  h Jz , we see that J J : ;max O  0 or  J;2  Jz2  h Jz  : ;max O  h 2 :  ;max 2  ;max  : ;max O (5.38) 288 CHAPTER 5. ANGULAR MOMENTUM hence :  ;max ;max  1 (5.39) After n successive applications of J on : ;max O, we must be able to reach a state : ;min O which cannot be lowered further: J : ;min O  0 (5.40) Using J J  J;2  Jz2  h Jz , and by analogy with (5.38) and (5.39), we infer that :  ;min ;min  1 (5.41) Comparing (5.39) and (5.41) we obtain ;max  ;min  (5.42) Since ;min was reached by n applications of J on : ;max O, it follows that ;max  ;mi n  n (5.43) and since ;mi n  ;max we conclude that n ;max   (5.44) 2 Hence ;max can be integer or half-odd-integer, depending on n being even or odd. It is now appropriate to introduce the notation j and m to denote ;max and ;, respectively: n j  ;max   m  ; (5.45) 2 hence the eigenvalue of J;2 is given by :  j  j  1 (5.46) Now since ;mi n  ;max , and with n positive, we infer that the allowed values of m lie between  j and  j:  j n m n j (5.47) The results obtained thus far can be summarized as follows: the eigenvalues of J;2 and Jz corresponding to the joint eigenvectors j mO are given, respectively, by h 2 j  j  1 and h m: J;2 j mO  h 2 j  j  1 j mO and Jz j mO  h m j mO (5.48) where j  0, 12, 1, 32    and m   j,  j 1,   , j 1, j. So for each j there are 2 j 1 values of m. For example, if j  1 then m takes the three values 1, 0, 1; and if j  52 then m takes the six values 52, 32, 12, 12, 32, 52. The values of j are either integer or half-integer. We see that the spectra of the angular momentum operators J;2 and Jz are discrete. Since the eigenstates corresponding to different angular momenta are orthogonal, and since the angular momentum spectra are discrete, the orthonormality condition is N j ) m) j mO  = j )  j =m )  m  (5.49) 5.3. GENERAL FORMALISM OF ANGULAR MOMENTUM 289 Let us now determine the eigenvalues of J within the j mO basis; j mO is not an eigenstate of J. We can rewrite equation (5.35) as J j mO  C  jm j m  1O (5.50) We are going to derive C   j m and then infer C j m. Since j mO is normalized, we can use (5.50) to obtain the following two expressions:  J j mO†  J j mO  C  2  2 j m N j m  1 j m  1O  C j m  (5.51) n n n  n2 nC j m n  N j m J J j mO (5.52) But since J J is equal to  J;2  Jz2  h Jz , and assuming the arbitrary phase of C  j m to be zero, we conclude that T  S C jm  N j m J;2  Jz2  h Jz j mO  h j  j  1  mm  1 (5.53) By analogy with C   j m we can easily infer the expression for C j m : S Cjm  h  j  j  1  mm  1 (5.54) Thus, the eigenvalue equations for J and J are given by S J j mO  h j  j  1  mm  1 j m  1O (5.55) or S J j mO  h  j b m j  m  1 j m  1O (5.56) which in turn leads to the two relations: 1 Jx j mO   J  J  j mO 2 h KS S L   j  m j  m  1 j m  1O   j  m j  m  1 j m  1O  2 (5.57) 1 Jy j mO   J  J  j mO 2i h KS S L   j  m j  m  1 j m  1O   j  m j  m  1 j m  1O  2i (5.58) The expectation values of Jx and Jy are therefore zero: N j m Jx j mO  N j m Jy j mO  0 (5.59) We will show later in (5.208) that the expectation values N j m Jx2 j mO and N j m Jy2 j mO are equal and given by 1K L h2 K  L N Jx2 O  N Jy2 O  N j m J;2 j mO  N j m Jz2 j mO  j  j  1  m 2  2 2 (5.60) 290 CHAPTER 5. ANGULAR MOMENTUM Example 5.2 Calculate [ Jx2  Jy ], [ Jz2  Jy ], and [ J;2  Jy ]; then show N j m Jx2 j mO  N j m Jy2 j mO. Solution Since [ Jx  Jy ]  i h Jz and [ Jz  Jx ]  i h Jy , we have [ Jx2  Jy ]  Jx [ Jx  Jy ]  [ Jx  Jy ] Jx  i h  Jx Jz  Jz Jx   i h 2 Jx Jz  i h Jy  (5.61) Similarly, since [ Jz  Jy ]  i h Jx and [ Jz  Jx ]  i h Jy , we have [ Jz2  Jy ]  Jz [ Jz  Jy ]  [ Jz  Jy ] Jz  i h  Jz Jx  Jx Jz   i h 2 Jx Jz  i h Jy  (5.62) The previous two expressions yield [ J;2  Jy ]  [ Jx2  Jy2  Jz2  Jy ]  [ Jx2  Jy ]  [ Jz2  Jy ]  i h 2 Jx Jz  i h Jy   i h 2 Jx Jz  i h Jy   0 (5.63) Since we have 1 1 Jx2   J2  J J  J J  J2  Jy2    J2  J J  J J  J2  (5.64) 4 4 and since N j m J2 j mO  N j m J2 j mO  0, we can write 1 N j m Jx2 j mO  N j m J J  J J j mO  N j m Jy2 j mO (5.65) 4 5.4 Matrix Representation of Angular Momentum The formalism of the previous section is general and independent of any particular representa- tion. There are many ways to represent the angular momentum operators and their eigenstates. In this section we are going to discuss the matrix representation of angular momentum where eigenkets and operators will be represented by column vectors and square matrices, respec- tively. This is achieved by expanding states and operators in a discrete basis. We will see later how to represent the orbital angular momentum in the position representation. Since J;2 and Jz commute, the set of their common eigenstates j mO can be chosen as a basis; this basis is discrete, orthonormal, and complete. For a given value of j, the orthonormal- ization condition for this base is given by (5.49), and the completeness condition is expressed by j ; j mON j m  I  (5.66) m j where I is the unit matrix. The operators J;2 and Jz are diagonal in the basis given by their joint eigenstates N j )  m ) J;2 j mO  h 2 j  j  1= j )  j =m )  m  (5.67) N j ) m) Jz j mO  h m= j )  j =m )  m  (5.68) 5.4. MATRIX REPRESENTATION OF ANGULAR MOMENTUM 291 Thus, the matrices representing J;2 and Jz in the j mO eigenbasis are diagonal, their diago- nal elements being equal to h 2 j  j  1 and h m, respectively. Now since the operators J do not commute with Jz , they are represented in the j mO basis by matrices that are not diagonal: S N j )  m ) J j mO  h j  j  1  mm  1 = j )  j =m )  m1  (5.69) We can infer the matrices of Jx and Jy from (5.57) and (5.58): h KS N j )  m ) Jx j mO  j  j  1  mm  1=m )  m1 2 L S  j  j  1  mm  1=m )  m1 = j )  j  (5.70) h KS N j )  m ) Jy j mO  j  j  1  mm  1=m )  m1 2i L S  j  j  1  mm  1=m )  m1 = j )  j  (5.71) Example 5.3 (Angular momentum j  1) Consider the case where j  1. (a) Find the matrices representing the operators J;2 , Jz , J , Jx , and Jy. (b) Find the joint eigenstates of J;2 and Jz and verify that they form an orthonormal and complete basis. (c) Use the matrices of Jx , Jy and Jz to calculate [ Jx  Jy ], [ Jy  Jz ], and [ Jz  Jx ]. (d) Verify that Jz3  h 2 Jz and J3  0. Solution (a) For j  1 the allowed values of m are 1, 0, 1. The joint eigenstates of J;2 and Jz are 1 1O, 1 0O, and 1 1O. The matrix representations of the operators J;2 and Jz can be inferred from (5.67) and (5.68):   N1 1 J;2 1 1O N1 1 J;2 1 0O N1 1 J;2 1 1O % & J;2  % # N1 0 J; 1 1O 2 N1 0 J;2 1 0O N1 0 J;2 1 1O & $ N1 1 J;2 1 1O N1 1 J;2 1 0O N1 1 J;2 1 1O   1 0 0  2h 2 # 0 1 0 $ (5.72) 0 0 1   1 0 0 Jz  h # 0 0 0 $ (5.73) 0 0 1 Similarly, using (5.69), we can ascertain that the matrices of J and J are given by     T 0 0 0 T 0 1 0 J  h 2 # 1 0 0 $  J  h 2 # 0 0 1 $  (5.74) 0 1 0 0 0 0 292 CHAPTER 5. ANGULAR MOMENTUM The matrices for Jx and Jy in the j mO basis result immediately from the relations Jx   J  J 2 and Jy  i J  J 2:     0 1 0 0 i 0 h # h Jx  T 1 0 1 $ Jy  T # i 0 i $  (5.75) 2 0 1 0 2 0 i 0 (b) The joint eigenvectors of J;2 and Jz can be obtained as follows. The matrix equation of Jz j mO  m h j mO is      1 0 0 a a h a  m h a h # 0 0 0 $ # b $  m h # b $ >" 0  m h b  (5.76) 0 0 1 c c h c  m h c The normalized solutions to these equations for m  1, 0, 1 are respectively given by a  1, b  c  0; a  0, b  1, c  0; and a  b  0, c  1; that is,       1 0 0 1 1O  # 0 $  1 0O  # 1 $  1 1O  # 0 $  (5.77) 0 0 1 We can verify that these vectors are orthonormal: N1 m ) 1 mO  =m )  m m )  m  1 0 1 (5.78) We can also verify that they are complete:       ;1 0 0 1 1 mON1 m  # 0 $ 0 0 1  # 1 $ 0 1 0  # 0 $ 1 0 0 m1 1 0 0   1 0 0  # 0 1 0 $ (5.79) 0 0 1 (c) Using the matrices (5.75) we have      2 0 1 0 0 i 0 i 0 i h # h2  # Jx Jy  1 0 1 $# i 0 i $  0 0 0 $ (5.80) 2 0 1 0 0 i 0 2 i 0 i      0 i 0 0 1 0 i 0 i h 2 # 2 h # Jy Jx  i 0 i $ # 1 0 1 $ 0 0 0 $ (5.81) 2 0 i 0 0 1 0 2 i 0 i hence     2i 0 0 1 0 0 h 2 # Jx Jy  Jy Jx  0 0 0 $  i h 2 # 0 0 0 $  i h Jz  (5.82) 2 0 0 2i 0 0 1 where the matrix of Jz is given by (5.73). A similar calculation leads to [ Jy  Jz ]  i h Jx and [ Jz  Jx ]  i h Jy. 5.5. GEOMETRICAL REPRESENTATION OF ANGULAR MOMENTUM 293 Jz Jz 6 6 h T j  j  1  m h @ ¡ @ I µ ¡ @ ¢̧ ¡ @ ¡ @ J;¢¢¡¡ @ J;¡¡ @ @ @ ¢¡ @ ¡ ©¡ @¢ - Jy @¡ - Jx y © ¡@ © ©© ¡ @ ¼ © ¡ J; @ Jx ¡ @ ¡ @ ª ¡ m h R @ T h j  j  1 Figure 5.1 Geometrical representation of the angular momentum J;: the vector J; rotates along the surface of a cone about its axis; the cone’s height is equal to m h , the projection of J; on the T cone’s axis. The tip of J; lies, within the Jz Jx y plane, on a circle of radius h j  j  1. (d) The calculation of Jz3 and J3 is straightforward:  3   1 0 0 1 0 0 Jz3  h 3 # 0 0 0 $  h 3 # 0 0 0 $  h 2 Jz  (5.83) 0 0 1 0 0 1  3   T 0 1 0 T 0 0 0 J3  2h 3 2 # 0 0 1 $  2h 3 2 # 0 0 0 $  0 (5.84) 0 0 0 0 0 0 and  3   T 0 0 0 T 0 0 0 J3  2h 3 2 # 1 0 0 $  2h 3 2 # 0 0 0 $  0 (5.85) 0 1 0 0 0 0 5.5 Geometrical Representation of Angular Momentum At issue here is the relationship between the angular momentum and its z-component; this relation can be represented geometrically as follows. For a fixed value of j, the total angular momentum T J; may be represented by a vector whose length, as displayed in Figure 5.1, is given 2 T by N J; O  h j  j  1 and whose z-component is N Jz O  h m. Since Jx and Jy are separately undefined, only their sum Jx2  Jy2  J;2  Jz2 , which lies within the x y plane, is well defined. 294 CHAPTER 5. ANGULAR MOMENTUM Jz Jz 6T 6T 6h 6h 2h KA ¢¢̧ ¢¢̧ A ¢ ¢ A ¢ ¢ Y H HH A ¢ ©* © h ¢ * © © HH A ¢ ©©© ¢ ©©© HHA ¢©© ¢©© H ©A¢© H - Jx y ¢© - Jx y AHHH 0 © © ¢¢AAHHH © A HH © HH ©© ¼ © ¢ ¢ A A Hj H h A A HH j H ¢ A A ¢ A A ®¢ AUA 2h AUA Figure 5.2 Graphical representation of the angular j  2 for the state 2 mO with T momentum T m  2 1 0 1 2. The radius of the circle is h 22  1  6h. In classical terms, we canT think of J; as representable graphically by a vector, whose endpoint lies on a circle of radius h j  j  1, rotating along the surface of a cone of half-angle t u 1 m A  cos T  (5.86) j  j  1 such that its projection along the z-axis is always m h. Notice that, as the values of the quantum number m are limited to m   j,  j  1,   , j  1, j, the angle A is quantized; the only possible values of A consist of a discrete set of 2 j  1 values: t u t u t u 1 j 1 j  1 1 j 1 A  cos T  cos T      cos T  j  j  1 j  j  1 j  j  1 t u j cos1 T  (5.87) j  j  1 Since all orientations of J; on the surface of the cone are equally likely, the projection of J; on both the x and y axes average out to zero: N Jx O  N Jy O  0 (5.88) where N Jx O stands for N j m Jx j mO. As an example, Figure 5.2 shows the graphical representation for the j  2 case. As specified in (5.87), A takes only a discrete set of values. In this case where j  2, the angle A takes only five values corresponding respectively to m  2 1 0 1 2; they are given by A  3526i  6591i  90i  6591i  3526i  (5.89) 5.6. SPIN ANGULAR MOMENTUM 295 Sz 6             h 2 S;             Spin up ¡ µ             ©* © ¡ T  A ¡ 3 h 2 - 0 ¡ - Sx y Beam of @ T             H j H A @ 3 h 2 silver atoms                         Spin down @R ; @  h 2 S Magnet Screen (a) (b) Figure 5.3 (a) Stern–Gerlach experiment: when a beam of silver atoms passes through an inhomogeneous magnetic field, it splits into two distinct components corresponding to spin-up and spin-down. (b) Graphical representation of spin 12 : the tip of S; lies on a circle of radius T S;  3h 2 so that its projection on the z-axis takes only two values, h 2, with A  5473i. 5.6 Spin Angular Momentum 5.6.1 Experimental Evidence of the Spin The existence of spin was confirmed experimentally by Stern and Gerlach in 1922 using silver (Ag) atoms. Silver has 47 electrons; 46 of them form a spherically symmetric charge distrib- ution and the 47th electron occupies a 5s orbital. If the silver atom were in its ground state, its total orbital angular momentum would be zero: l  0 (since the fifth shell electron would be in a 5s state). In the Stern–Gerlach experiment, a beam of silver atoms passes through an inhomogeneous (nonuniform) magnetic field. If, for argument’s sake, the field were along the z-direction, we would expect classically to see on the screen a continuous band that is symmet- ric about the undeflected direction, z  0. According to Schrödinger’s wave theory, however, if the atoms had an orbital angular momentum l, we would expect the beam to split into an odd (discrete) number of 2l  1 components. Suppose the beam’s atoms were in their ground state l  0, there would be only one spot on the screen, and if the fifth shell electron were in a 5p state (l  1), we would expect to see three spots. Experimentally, however, the beam behaves according to the predictions of neither classical physics nor Schrödinger’s wave theory. Instead, it splits into two distinct components as shown in Figure 5.3a. This result was also observed for hydrogen atoms in their ground state (l  0), where no splitting is expected. To solve this puzzle, Goudsmit and Uhlenbeck postulated in 1925 that, in addition to its orbital angular momentum, the electron possesses an intrinsic angular momentum which, un- like the orbital angular momentum, has nothing to do with the spatial degrees of freedom. By analogy with the motion of the Earth, which consists of an orbital motion around the Sun and an internal rotational or spinning motion about its axis, the electron or, for that matter, any other microscopic particle may also be considered to have some sort of internal or intrinsic spinning motion. This intrinsic degree of freedom was given the suggestive name of spin angular mo- mentum. One has to keep in mind, however, that the electron remains thus far a structureless or pointlike particle; hence caution has to be exercised when trying to link the electron’s spin to an internal spinning motion. The spin angular momentum of a particle does not depend on 296 CHAPTER 5. ANGULAR MOMENTUM z B; 6 L; 6 E;L  q ; 2mc L ¡µ ¡ Lz ¡; ¡ L q w³³³ PP m 1 P ); ¡ q © ¡ -y © ©©© (a) ¼ © (b) x Figure 5.4 (a) Orbital magnetic dipole moment of a positive charge q. (b) When an external magnetic field is applied, the orbital magnetic moment precesses about it. its spatial degrees of freedom. The spin, an intrinsic degree of freedom, is a purely quantum mechanical concept with no classical analog. Unlike the orbital angular momentum, the spin cannot be described by a differential operator. From the classical theory of electromagnetism, an orbital magnetic dipole moment is gen- erated with the orbital motion of a particle of charge q: q ; ;L  E L (5.90) 2mc where L; is the orbital angular momentum of the particle, m is its mass, and c is the speed of light. As shown in Figure 5.4a, if the charge q is positive, E ; L and L; will be in the same direction; for a negative charge such as an electron (q  e), the magnetic dipole moment E ; ; L  e L2m e c and the orbital angular momentum will be in opposite directions. Similarly, if we follow a classical analysis and picture the electron as a spinning spherical charge, then we obtain an intrinsic or spin magnetic dipole moment E ; ; S  e S2m e c. This classical derivation of E ; S is quite erroneous, since the electron cannot be viewed as a spinning sphere; in fact, it turns out that the electron’s spin magnetic moment is twice its classical expression. Although the spin magnetic moment cannot be derived classically, as we did for the orbital magnetic moment, it can still be postulated by analogy with (5.90): e ; ; S  gs E S (5.91) 2m e c where gs is called the Landé factor or the gyromagnetic ratio of the electron; its experimental value is gs 2 (this factor can be calculated using Dirac’s relativistic theory of the electron). When the electron is placed in a magnetic field B; and if the field is inhomogeneous, a force will be exerted on the electron’s intrinsic dipole moment; the direction and magnitude of the force depend on the relative orientation of the field and the dipole. This force tends to align E;S along B,; producing a precessional motion of E ; S around B; (Figure 5.4b). For instance, if E; S is ; the electron will move in the direction in which the field increases; conversely, if parallel to B, ; the electron will move in the direction in which the field decreases. For ; S is antiparallel to B, E hydrogen-like atoms (such as silver) that are in the ground state, the orbital angular momentum will be zero; hence the dipole moment of the atom will be entirely due to the spin of the electron. 5.6. SPIN ANGULAR MOMENTUM 297 The atomic beam will therefore deflect according to the orientation of the electron’s spin. Since, experimentally, the beam splits into two components, the electron’s spin must have only two possible orientations relative to the magnetic field, either parallel or antiparallel. By analogy with the orbital angular momentum of a particle, which is characterized by two quantum numbers—the orbital number l and the azimuthal number m l (with m l  l, l  1,   , l  1, l)—the spin angular momentum is also characterized by two quantum numbers, the spin s and its projection m s on the z-axis (the direction of the magnetic field), where m s  s, s1,   , s1, s. Since only two components were observed in the Stern–Gerlach experiment, we must have 2s  1  2. The quantum numbers for the electron must then be given by s  21 and m s   21. In nature it turns out that every fundamental particle has a specific spin. Some particles have integer spins s  0, 1, 2    (the pi mesons have spin s  0, the photons have spin s  1, and so on) and others have half-odd-integer spins s  21 , 23 , 25     (the electrons, protons, and neutrons have spin s  12 , the deltas have spin s  32 , and so on). We will see in Chapter 8 that particles with half-odd-integer spins are called fermions (quarks, electrons, protons, neutrons, etc.) and those with integer spins are called bosons (pions, photons, gravitons, etc.). Besides confirming the existence of spin and measuring it, the Stern–Gerlach experiment offers a number of other important uses to quantum mechanics. First, by showing that a beam splits into a discrete set of components rather than a continuous band, it provides additional confirmation for the quantum hypothesis on the discrete character of the microphysical world. The Stern–Gerlach experiment also turns out to be an invaluable technique for preparing a quantum state. Suppose we want to prepare a beam of spin-up atoms; we simply pass an unpolarized beam through an inhomogeneous magnet, then collect the desired component and discard (or block) the other. The Stern–Gerlach experiment can also be used to determine the total angular momentum of an atom which, in the case where l / 0, is given by the sum of the orbital and spin angular momenta: J;  L;  S. ; The addition of angular momenta is covered in Chapter 7. 5.6.2 General Theory of Spin The theory of spin is identical to the general theory of angular momentum (Section 5.3). By analogy with the vector angular momentum J;, the spin is also represented by a vector operator S; whose components Sx , Sy , Sz obey the same commutation relations as Jx , Jy , Jz : [ Sx  S y ]  i h Sz  [ S y  Sz ]  i h Sx  [ Sz  Sx ]  i h S y  (5.92) In addition, S;2 and Sz commute; hence they have common eigenvectors: S; 2 s m s O  h 2 ss  1 s m s O Sz s m s O  h m s s m s O (5.93) where m s  s, s  1,   , s  1, s. Similarly, we have S S s m s O  h ss  1  m s m s  1 s m s  1O (5.94) where S  Sx  i S y , and 1 ;2 h 2 K L N Sx2 O  N Sy2 O  N S O  N Sz2 O  ss  1  m 2s  (5.95) 2 2 298 CHAPTER 5. ANGULAR MOMENTUM where N AO denotes N AO  Ns m s A s m s O. The spin states form an orthonormal and complete basis ; s Ns )  m )s s m s O  =s )  s =m )s  m s  s m s ONs m s  I (5.96) m s s where I is the unit matrix. 5.6.3 Spin 12 and the Pauli Matrices 1 1 1 For a particle with spin the quantum number m s takes only two values: mn s   2 ( 2 and n 2. The ( n n particle can thus be found in either of the following two states: s m s O  n 21  12 and n 21   21. The eigenvalues of S; 2 and Sz are given by n  n  n  n  n1 1 3 n1 1 n1 1 h n1 S;2 nn    h 2 nn    Sz nn     n  1  (5.97) 2 2 4 2 2 2 2 2 n2 2 HenceTthe spin may be represented graphically, as shown Tin Figure 5.3b, by a vector of length S;  3h 2, whose endpoint lies on a circle of radius 3h 2, rotating along the surface of a cone with half-angle t u t u t u ms h 2 1 A  cos1 T  cos1 T  cos1 T  5473i  (5.98) ss  1 3h 2 3 The projection of S; on the z-axis is restricted to two values only: h 2 corresponding to spin- up and spin-down. Let us now study the matrix representation of the spin s  21. Using (5.67) and (5.68) we can represent the operators S;2 and Sz within the s m s O basis by the following matrices: ‚  t u 1 1 ;2 1 1 1 1 ; 2 1 1O 3h 2 1 0 ; 2 N  S  O N  S S  2 2 2 2 2 2 2 2   (5.99) N 21   12 S;2 12  12 O N 21   12 S;2 12   12 O 4 0 1 t u h 1 0 Sz   (5.100) 2 0 1 The matrices of S and S can be inferred from (5.69): t u t u 0 1 0 0 S  h  S  h  (5.101) 0 0 1 0 and since Sx  12  S  S  and S y  2i  S  S , we have t u t u h 0 1 h 0 i Sx   Sy   (5.102) 2 1 0 2 i 0 The joint eigenvectors of S;2 and Sz are expressed in terms of two-element column matrices, known as spinors: n  t u n  t u n1 1 1 n1 1 0 n  n n2 2  0  n2 2  1  (5.103) 5.6. SPIN ANGULAR MOMENTUM 299 It is easy to verify that these eigenvectors form a basis that is complete, 1 n ~ n t u t u t u ; 2 n1 n n  m s 1  m s n  0 0 1  1 1 0  1 0  (5.104) n2 2 n 1 0 0 1 1 m s  2 and orthonormal, ~ n  t u 1 1 nn 1 1 1    1 0  1 (5.105) 2 2 n2 2 0 ~ n  t u 1 1 n1 1 0   nn    0 1  1 (5.106) 2 2 2 2 1 ~ n  ~ n  1 1 nn 1 1 1 1 nn 1 1         0 (5.107) 2 2 n2 2 2 2 n2 2 Let us now find the eigenvectors of Sx and S y. First, note that the basis vectors s m s O are eigenvectors of neither Sx nor S y ; their eigenvectors can, however, be expressed in terms of s m s O as follows: vn  n w 1 n1 1 n1 1 Ox O  T nn   nn    (5.108) 2 2 2 2 2 vn  n w 1 n1 1 n1 1 O y O  T nn   i nn    (5.109) 2 2 2 2 2 The eigenvalue equations for Sx and S y are thus given by h h Sx Ox O   Ox O  S y O y O   O y O  (5.110) 2 2 Pauli matrices When s  12 it is convenient to introduce the Pauli matrices Jx , J y , Jz , which are related to the spin vector as follows: h S;  J;  (5.111) 2 Using this relation along with (5.100) and (5.102), we have t u t u t u 0 1 0 i 1 0 Jx   Jy   Jz   (5.112) 1 0 i 0 0 1 These matrices satisfy the following two properties: J 2j  I  j  x y z (5.113) J j Jk  Jk J j  0  j / k (5.114) where the subscripts j and k refer to x y, z, and I is the 2  2 unit matrix. These two equations are equivalent to the anticommutation relation j k J j  Jk  2 I = j k  (5.115) 300 CHAPTER 5. ANGULAR MOMENTUM We can verify that the Pauli matrices satisfy the commutation relations [J j  Jk ]  2i  jkl Jl  (5.116) where  jkl is the antisymmetric tensor (also known as the Levi–Civita tensor)   1 if jkl is an even permutation of x y z  jkl  1 if jkl is an odd permutation of x y z (5.117)  0 if any two indices among j k l are equal. We can condense the relations (5.113), (5.114), and (5.116) into ; J j Jk  = j k  i  jkl Jl  (5.118) l Using this relation we can verify that, for any two vectors A; and B; which commute with J; , we have ; J  B J  A; ; ;   A;  B ; I  i J;   A;  B ; (5.119) where I is the unit matrix. The Pauli matrices are Hermitian, traceless, and have determinants equal to 1: † Jj  Jj TrJ j   0 detJ j   1  j  x y z (5.120) Using the relation Jx J y  iJz along with Jz2  I , we obtain J x J y Jz  i I  (5.121) From the commutation relations (5.116) we can show that ei:J j  I cos :  iJ j sin :  j  x y z (5.122) where I is the unit matrix and : is an arbitrary real constant. Remarks  Since the spin does not depend on the spatial degrees of freedom, the components Sx , S y , Sz of the spin operator commute with all the spatial operators, notably the orbital angular momentum L, ; the position and the momentum operators R; and P: ; [ S j  L k ]  0 [ S j  Rk ]  0 [ S j  Pk ]  0  j k  x y z (5.123)  The total wave function O of a system with spin consists of a product of two parts: a spatial part O;r  and a spin part s m s O: O  OO s m s O (5.124) This product of the space and spin degrees of freedom is not a product in the usual sense, but a direct or tensor product as discussed in Chapter 7. We will show in Chapter 6 that the four quantum numbers n, l, m l , and m s are required to completely describe the state of an electron moving in a central field; its wave function is nlml m s ;r   Onlml ; r  s m s O (5.125) 5.7. EIGENFUNCTIONS OF ORBITAL ANGULAR MOMENTUM 301 Since the spin operator does not depend on the spatial degrees of freedom, it acts only on the spin part s m s O and leaves the spatial wave function, Onlm l ; r , unchanged; conversely, the spatial operators L, ; R, ; and P; act on the spatial part and not on the spin part. For spin 12 particles, the total wave function corresponding to spin-up and spin-down cases are respectively expressed in terms of the spinors: t u t u 1 Onlm l ;r  nlm l 1 ;r   Onlml ; r   (5.126) 2 0 0 t u t u 0 0 nlml  1 ;r   Onlml ; r   (5.127) 2 1 Onlm l ;r  Example 5.4 3 Find the energy levels of a spin s  2 particle whose Hamiltonian is given by : 2 ; H 2  Sx  Sy2  2 Sz2   Sz  h h : and ; are constants. Are these levels degenerate? Solution Rewriting H in the form, : r ;2 s ; H 2 S  3 Sz2  Sz  (5.128) h h we see that H is diagonal in the s mO basis: : K 2 2 2 L ; 15 E m  Ns m H s mO   h ss  1  3  h m  h m  :m3:m; (5.129) h 2 h 4 where the quantum number m takes any of the four values m   32 ,  21 , 12 , 23. Since E m depends on m, the energy levels of this particle are nondegenerate. 5.7 Eigenfunctions of Orbital Angular Momentum We now turn to the coordinate representation of the angular momentum. In this section, we are going to work within the spherical coordinate system. Let us denote the joint eigenstates of L; 2 and L z by l mO: L; 2 l mO  h 2ll  1 l mO (5.130) L z l mO  h m l mO (5.131) 302 CHAPTER 5. ANGULAR MOMENTUM The operators L z , L; 2 , L  , whose Cartesian components are listed in Eqs (5.3) to (5.5), can be expressed in terms of spherical coordinates (Appendix B) as follows: " L z  i h  (5.132) " v t u w 1 " " 1 "2 L; 2  h 2 sin A  2  (5.133) sin A "A "A sin A " 2 v w " cos A " L  L x  i L y  h ei i  (5.134) "A sin A " Since the operators L z and L; depend only on the angles A and , their eigenstates depend only on A and. Denoting their joint eigenstates by NA l mO  Ylm A  (5.135) where1 Ylm A  are continuous functions of A and , we can rewrite the eigenvalue equations (5.130) and (5.131) as follows: L; 2 Ylm A   h 2ll  1Ylm A  (5.136) L z Ylm A   m h Ylm A  (5.137) Since L z depends only on , as shown in (5.132), the previous two equations suggest that the eigenfunctions Ylm A  are separable: Ylm A   lm Am   (5.138) We ascertain that S L  Ylm A   h ll  1  mm  1 Yl m1 A  (5.139) 5.7.1 Eigenfunctions and Eigenvalues of L z Inserting (5.138) into (5.137) we obtain L z lm Am    m h lm Am  . Now since L z  i h "" , we have "m   i h lm A  m h lm Am   (5.140) " which reduces to "m   i  mm   (5.141) " The normalized solutions of this equation are given by 1 m    T ei m  (5.142) 2H 1 For notational consistency throughout this text, we will insert a comma between l and m in Y A  whenever m lm is negative. 5.7. EIGENFUNCTIONS OF ORBITAL ANGULAR MOMENTUM 303 T where 1 2H is the normalization constant, = 2H d `m )  m    =m )  m  (5.143) 0 For m   to be single-valued, it must be periodic in with period 2H , m   2H  m  ; hence eim 2H  ei m  (5.144) This relation shows that the expectation value of L z , l z  Nl m L z l mO, is restricted to a discrete set of values l z  m h  m  0 1 2 3     (5.145) Thus, the values of m vary from l to l: m  l l  1 l  2     0 1 2     l  2 l  1 l (5.146) Hence the quantum number l must also be an integer. This is expected since the orbital angular momentum must have integer values. 5.7.2 Eigenfunctions of L; 2 Let us now focus on determining the eigenfunctions lm A of L; 2. We are going to follow two methods. The first method involves differential equations and gives lm A in terms of the well-known associated Legendre functions. The second method is algebraic; it deals with the operators L  and enables an explicit construction of Ylm A , the spherical harmonics. 5.7.2.1 First Method for Determining the Eigenfunctions of L; 2 We begin by applying L; 2 of (5.133) to the eigenfunctions 1 Ylm A   T lm Aeim  (5.147) 2H This gives v t u w ; 2 h 2 1 " " 1 "2 L Ylm A   T sin A  lm Aeim 2H sin A "A "A sin2 A " 2 h 2ll  1  T lm Aeim  (5.148) 2H which, after eliminating the -dependence, reduces to t u v w 1 d dlm A m2 sin A  ll  1  2 lm A  0 (5.149) sin A dA dA sin A This equation is known as the Legendre differential equation. Its solutions can be expressed in terms of the associated Legendre functions Plm cos A: lm A  Clm Plm cos A (5.150) 304 CHAPTER 5. ANGULAR MOMENTUM which are defined by dm Plm x  1  x 2  m 2 Pl x (5.151) dx m This shows that Plm x  Plm x (5.152) where Pl x is the lth Legendre polynomial which is defined by the Rodrigues formula 1 dl 2 Pl x  x  1l  (5.153) 2l l! dx l We can obtain at once the first few Legendre polynomials: 1 dx 2  1 P0 x  1 P1 x   x (5.154) 2 dx 1 d 2 x 2  12 1 1 d 3 x 2  13 1 P2 x  2  3x 2  1 P3 x  3  5x 3  3x 8 dx 2 48 dx 2 (5.155) 1 1 P4 x  35x 4  30x 2  3 P5 x  63x 5  70x 3  15x (5.156) 8 8 The Legendre polynomials satisfy the following closure or completeness relation: * 1; 2l  1Pl x ) Pl x  =x  x )  (5.157) 2 l0 From (5.153) we can infer at once Pl x  1l Pl x (5.158) A similar calculation leads to the first few associated Legendre functions: S P11 x  1  x 2  (5.159) S P21 x  3x 1  x 2  P22 x  31  x 2  (5.160) 3 S P31 x  5x 2  1 1  x 2  P32 x  15x1  x 2  P33 x  151  x 2 32  (5.161) 2 where Pl0 x  Pl x, with l  0 1 2 3   . The first few expressions for the associated Legendre functions and the Legendre polynomials are listed in Table 5.1. Note that Plm x  1lm Plm x (5.162) The constant Clm of (5.150) can be determined from the orthonormalization condition = 2H = H Nl )  m ) l mO  d dA sin ANl )  m ) A ONA l mO  =l )  l =m )  m  (5.163) 0 0 which can be written as = 2H = H d dA sin A Yl`) m ) A Ylm A   =l )  l =m )  m  (5.164) 0 0 5.7. EIGENFUNCTIONS OF ORBITAL ANGULAR MOMENTUM 305 Table 5.1 First few Legendre polynomials and associated Legendre functions. Legendre polynomials Associated Legendre functions P0 cos A  1 P11 cos A  sin A P1 cos A  cos A P21 cos A  3 cos A sin A P2 cos A  21 3 cos2 A  1 P22 cos A  3 sin2 A P3 cos A  21 5 cos3 A  3 cos A P31 cos A  32 sin A5 cos2 A  1 P4 cos A  81 35 cos4 A  30 cos2 A  3 P32 cos A  15 sin2 A cos A P5 cos A  18 63 cos5 A  70 cos3 A  15 cos A P33 cos A  15 sin3 A This relation is known as the normalization condition of spherical harmonics. Using the form (5.147) for Ylm A , we obtain = 2H = H = = H Clm 2 2H d dA sin A Ylm A  2  d dA sin A Plm cos A 2  1 (5.165) 0 0 2H 0 0 From the theory of associated Legendre functions, we have = H 2 l  m! dA sin A Plm cos APlm) cos A  =l l )  (5.166) 0 2l  1 l  m! which is known as the normalization condition of associated Legendre functions. A combina- tion of the previous two relations leads to an expression for the coefficient Clm : Vt u m 2l  1 l  m! Clm  1 m o 0 (5.167) 2 l  m! Inserting this equation into (5.150), we obtain the eigenfunctions of L; 2 : Vt u m 2l  1 l  m! m lm A  1 P cos A (5.168) 2 l  m! l Finally, the joint eigenfunctions, Ylm A , of L; 2 and Jz can be obtained by substituting (5.142) and (5.168) into (5.138): Vt u m 2l  1 l  m! m Ylm A   1 P cos Aeim m o 0 (5.169) 4H l  m! l These are called the normalized spherical harmonics. 5.7.2.2 Second Method for Determining the Eigenfunctions of L; 2 The second method deals with a direct construction of Ylm A ; it starts with the case m  l (this is the maximum value of m). By analogy with the general angular momentum algebra developed in the previous section, the action of L  on Yll gives zero, NA L  l lO  L  Yll A   0 (5.170) 306 CHAPTER 5. ANGULAR MOMENTUM since Yll cannot be raised further as Yll  Ylm max. Using the expression (5.134) for L  in the spherical coordinates, we can rewrite (5.170) as follows: v w h ei " " T  i cot A ll Aei l  0 (5.171) 2H "A " which leads to 1 "ll A  l cot A (5.172) ll "A The solution to this differential equation is of the form ll A  Cl sinl A (5.173) where Cl is a constant to be determined from the normalization condition (5.164) of Yll A : Cl Yll A   T eil sinl A (5.174) 2H We can ascertain that Cl is given by U 1l 2l  1! Cl  l  (5.175) 2 l! 2 The action of L  on Yll A  is given, on the one hand, by T L  Yll A   h 2lYll1 A  (5.176) and, on the other hand, by U 1l 2l  1! il1 d L  Yll A   h l e sin A1l [sin A2l ] (5.177) 2 l! 4H dcos A where we have used the spherical coordinate form (5.134). Similarly, we can show that the action of L lm  on Yll A  is given, on the one hand, by V 2l!l  m! L lm  Yll A   h lm Ylm A  (5.178) l  m! and, on the other hand, by U 1l 2l!2l  1! im 1 d lm L lm  Yll A  h lm l e m sin A2l  (5.179) 2 l! 4H sin A dcos Alm where m o 0. Equating the previous two relations, we obtain the expression of the spherical harmonic Ylm A  for m o 0: Vt u 1l 2l  1 l  m! im 1 d lm Ylm A   l e sin A2l  (5.180) 2 l! 4H l  m! sinm A dcos Alm 5.7. EIGENFUNCTIONS OF ORBITAL ANGULAR MOMENTUM 307 5.7.3 Properties of the Spherical Harmonics Since the spherical harmonics Ylm A  are joint eigenfunctions of L; 2 and L z and are ortho- normal (5.164), they constitute an orthonormal basis in the Hilbert space of square-integrable functions of A and. The completeness relation is given by ; l l mONl m  1 (5.181) ml or ; ; NA l mONl m A ) ) O  Yl`m A )  ) Ylm A   =cos A  cos A ) =  )  m m =A  A )  )  =   (5.182) sin A Let us mention some essential properties of the spherical harmonics. First, the spherical har- monics are complex functions; their complex conjugate is given by [Ylm A ]`  1m Ylm A  (5.183) We can verify that Ylm A  is an eigenstate of the parity operator P with an eigenvalue 1l : PYlm A   Ylm H  A  H  1l Ylm A  (5.184) since a spatial reflection about the origin, r; )  ;r , corresponds to r )  r , A )  H  A, and )  H  , which leads to P m cos A )   P m  cos A  1lm P m cos A and ei m )  l l l eimH ei m  1m eim. We can establish a connection between the spherical harmonics and the Legendre polyno- mials by simply taking m  0. Then equation (5.180) yields U U 1l 2l  1 dl 2l 2l  1 Yl0 A   l sin A  Pl cos A (5.185) 2 l! 4H dcos Al 4H with 1 dl Pl cos A  cos2 A  1l  (5.186) 2l l! dcos Al From the expression of Ylm , we can verify that U 2l  1 Ylm 0   =m 0  (5.187) 4H The expressions for the spherical harmonics corresponding to l  0 l  1, and l  2 are listed in Table 5.2. Spherical harmonics in Cartesian coordinates Note that Ylm A  can also be expressed in terms of the Cartesian coordinates. For this, we need only to substitute x y z sin A cos   sin A sin   cos A  (5.188) r r r 308 CHAPTER 5. ANGULAR MOMENTUM Table 5.2 Spherical harmonics and their expressions in Cartesian coordinates. Ylm A  Ylm x y z Y00 A   T1 Y00 x y z  T1 4H 4H T T 3 3 z Y10 A   cos A 4H Y10 x y z  4H r T T 3 i 3 xi y Y11 A   b 8H e sin A Y11 x y z  b 8H r T T 5 5 3z r 2 2 Y20 A   16H 3 cos2 A  1 Y20 x y z  16H r2 T T 15 i 15 xi yz Y21 A   b 8H e sin A cos A Y21 x y z  b 8H r2 T T 15 2i 15 x 2 y 2 2i x y Y22 A   32H e sin2 A Y22 x y z  32H r2 in the expression for Ylm A . As an illustration, let us show how to T derive the Cartesian expressions for Y10 and Y11. Substituting cos A  zr into Y10 A   34H cos A Y10 , we have U U 3 z 3 z Y10 x y z   S  (5.189) 4H r 4H x 2  y 2  z 2 Using sin A cos  xr and sin A sin  yr , we obtain x  iy  sin A cos  i sin A sin  sin A ei  (5.190) r T which, when substituted into Y11 A   b 38H sin A ei , leads to U 3 x  iy Y11 x y z  b  (5.191) 8H r Following the same procedure, we can derive the Cartesian expressions of the remaining har- monics; for a listing, see Table 5.2. Example 5.5 (Application of ladder T operators to spherical harmonics) (a) Use the relation Yl0 A   2l  14H Pl cos A to find the expression of Y30 A . (b) Find the expression of Y30 in Cartesian coordinates. (c) Use the expression of Y30 A  to infer those of Y31 A . Solution (a) From Table 5.1 we have P3 cos A  12 5 cos3 A  3 cos A; hence U U 7 7 Y30 A   P3 cos A  5 cos3 A  3 cos A (5.192) 4H 16H 5.7. EIGENFUNCTIONS OF ORBITAL ANGULAR MOMENTUM 309 (b) Since cos A  zr, we have 5 cos3 A 3 cos A  5 cos A5 cos2 A 3  z5z 2 3r 2 r 3 ; hence U 7 z Y30 x y z  5z 2  3r 2  (5.193) 16H r 3 (c) To find Y31 from Y30 , we need to apply the ladder operator L  on Y30 in two ways: first, algebraically S T L  Y30  h 33  1  0 Y31  2h 3 Y31 (5.194) and hence 1 Y31  T L  Y30  (5.195) 2h 3 then we use the differential form (5.134) of L  : v w i " cos A " L  Y30 A   h e i Y30 A  "A sin A " U v w 7 i " cos A "  h e i 5 cos3 A  3 cos A 16H "A sin A " U 7  3h sin A5 cos2 A  1ei  (5.196) 16H Inserting (5.196) into (5.195) we end up with U 1 21 Y31  T L  Y30   sin A5 cos2 A  1ei  (5.197) 2h 3 64H Now, to find Y31 from Y30 , we also need to apply L  on Y30 in two ways: S T L  Y30  h 33  1  0Y31  2h 3Y31 (5.198) and hence 1 Y31  T L  Y30  (5.199) 2h 3 then we use the differential form (5.134) of L  : v w " cos A " L  Y30 A   h ei i Y30 A  "A sin A " U v w 7 i " cos A "  h e i 5 cos3 A  3 cos A 16H "A sin A " U 7  3h sin A5 cos2 A  1ei  (5.200) 16H Inserting (5.200) into (5.199), we obtain U 1 21 Y31  T L  Y30  sin A5 cos2 A  1ei  (5.201) 2h 3 64H 310 CHAPTER 5. ANGULAR MOMENTUM 5.8 Solved Problems Problem 5.1 T 2 (a) Show that Jx Jy  h [ j  j  1  m ]2, where Jx  N Jx2 O  N Jx O2 and the same 2 for Jy. (b) Show that this relation is consistent with Jx Jy o h 2 N Jz O  h 2 m2. Solution (a) First, note that N Jx O and N Jy O are zero, since 1 1 N Jx O  N j m J j mO  N j m J j mO  0 (5.202) 2 2 As for N Jx2 O and N Jy2 O, they are given by 1 1 N Jx2 O  N J  J 2 O  N J2  J J  J J  J2 O (5.203) 4 4 1 1 N Jy2 O  N J  J 2 O   N J2  J J  J J  J2 O (5.204) 4 4 Since N J2 O  N J2 O  0, we see that 1 N Jx2 O  N J J  J J O  N Jy2 O (5.205) 4 Using the fact that N Jx2 O  N Jy2 O  N J;2 O  N Jz2 O (5.206) along with N Jx2 O  N Jy2 O, we see that 1 ;2 N Jx2 O  N Jy2 O  [N J O  N Jz2 O] (5.207) 2 Now, since j mO is a joint eigenstate of J; 2 and Jz with eigenvalues j  j  1h 2 and m h , we can easily see that the expressions of N Jx2 O and N Jy2 O are given by 1 ;2 h 2 K L N Jx2 O  N Jy2 O  [N J O  N Jz2 O]  j  j  1  m 2  (5.208) 2 2 Hence Jx Jy is given by T h 2 N Jx2 ON Jy2 O  [ j  j  1  m 2 ] Jx Jy  (5.209) 2 (b) Since j o m (because m   j  j  1     j  1 j), we have j  j  1  m 2 o mm  1  m 2  m (5.210) from which we infer that Jx Jy o  h 2 m2, or h Jx Jy o N Jz O  (5.211) 2 5.8. SOLVED PROBLEMS 311 Problem 5.2 Find the energy levels of a particle which is free except that it is constrained to move on the surface of a sphere of radius r. Solution This system consists of a particle that is constrained to move on the surface of a sphere but free from the influence of any other potential; it is called a rigid rotator. Since V  0 the energy of this system is purely kinetic; the Hamiltonian of the rotator is L; 2 H  (5.212) 2I where I  mr 2 is the moment of inertia of the particle with respect to the origin. In deriving this relation, we have used the fact that H  p2 2m  r p2 2mr 2  L 2 2I , since L  r;  p;  r p. The wave function of the system is clearly independent of the radial degree of freedom, for it is constant. The Schrödinger equation is thus given by L; 2 H OA   OA   EOA  (5.213) 2I Since the eigenstates of L; 2 are the spherical harmonics Ylm A , the corresponding energy eigenvalues are given by h 2 El  ll  1 l  0 1 2 3     (5.214) 2I and the Schrödinger equation by L; 2 h 2 Ylm A   ll  1Ylm A  (5.215) 2I 2I Note that the energy levels do not depend on the azimuthal quantum number m. This means that there are 2l  1 eigenfunctions Yl l , Yl l1 ,    , Yl l1 , Yll corresponding to the same energy. Thus, every energy level El is 2l  1-fold degenerate. This is due to the fact that the rotator’s Hamiltonian, L; 2 2I , commutes with L;. That is, the Hamiltonian is independent of the orientation of L; in space; hence the energy spectrum does not depend on the component of L; in any particular direction. Problem 5.3 Find the rotational energy levels of a diatomic molecule. Solution Consider two molecules of masses m 1 and m 2 separated by a constant distance r;. Let r1 and r2 be their distances from the center of mass, i.e., m 1r1  m 2r2. The moment of inertia of the diatomic molecule is I  m 1r12  m 2r22 k Er 2  (5.216) 312 CHAPTER 5. ANGULAR MOMENTUM where r  r;1  r;2 and where E is their reduced mass, E  m 1 m 2 m 1  m 2 . The total angular momentum is given by L;  m 1r1r1   m 2r2r2   I   Er 2  (5.217) and the Hamiltonian by L; 2 L; 2 H   (5.218) 2I 2Er 2 The corresponding eigenvalue equation L; 2 ll  1h 2 H l mO  l mO  l mO (5.219) 2Er 2 2Er 2 shows that the eigenenergies are 2l  1-fold degenerate and given by ll  1h 2 El   (5.220) 2Er 2 Problem 5.4 (a) Find the eigenvalues and eigenstates of the spin operator S; of an electron in the direction of a unit vector n;; assume that n; lies in the x z plane. (b) Find the probability of measuring Sz  h 2. Solution (a) In this question we want to solve h n;  S; DO  D DO (5.221) 2 where n; is given by n;  sin A ;i  cos A k,