Angular Momentum. Classical. J r p. radius vector from origin. linear momentum. determinant form of cross product iˆ xˆ J J J J J J

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1 Angular Momentum Classical r p p radius vector from origin linear momentum r iˆ ˆj kˆ x y p p p x y determinant form of cross product iˆ xˆ ˆj yˆ kˆ ˆ y p p x y p x p y x x p y p y x x y Copyright Michael D. Fayer, 07

2 Q.M. Angular Momentum In the Schrödinger Representation, use Q.M. operators for x and p, etc. P x = i x = x x Substituting i ˆ ˆ ˆ i j k x y x y x i y y ix y y x y i x x x y Copyright Michael D. Fayer, 07

3 Commutators Consider, x y x y y x substituting operators in units of x y y x y x Keep track of what commutes. y y x x x y x y Similarly y x y x y x x x y y Subtracting x, y y x x y Copyright Michael D. Fayer, 07

4 x, y y x x y y x, x y x y, y x i, But, because P i Therefore,, i x y i i P,, P i i Using P, i, i x y in conventional units Copyright Michael D. Fayer, 07

5 The commutators in units of are, i x y, i y x, i. x y Using these it is found that,, x, y 0 Components of angular momentum do not commute. commutes with all components. Copyright Michael D. Fayer, 07

6 Therefore, and one component of angular momentum can be measured simultaneously. Call this component. Therefore, and can be simultaneously diagonalied by the same unitary transformation. Furthermore, Therefore, H, 0 H, 0 ( looks like rotation) H,, are all simultaneous observables. Copyright Michael D. Fayer, 07

7 Diagonaliation of and and commute. Therefore, set of vectors m are eigenvectors of both operators. Labeling kets with eigenvalues. and are simultaneously diagonal in the basis m m m m m m (in units of ) Copyright Michael D. Fayer, 07

8 Form operators i x i y x y From the definitions of and and the angular momentum commutators, the following commutators and identities can be derived. Commutators Identities,,, Copyright Michael D. Fayer, 07

9 Expectation value m m m m Because x y m m m m m m m m Positive numbers because s are Hermitian give real numbers. Square of real numbers positive. Therefore, the sum of three positive numbers is greater than or equal to one of them. Now m m m m m Therefore, m Eigenvalues of are greater than or equal to square of eigenvalues of. Copyright Michael D. Fayer, 07

10 Using, Consider m m m m m m m m eigenvalue eigenvector Furthermore,, 0 commutes with + because it commutes with x and y. Then m m m eigenvalue eigenvector Copyright Michael D. Fayer, 07

11 Thus, m is eigenvector of with eigenvalue m + and of with eigenvalue. + is a raising operator. It increases m by and leaves unchanged. Copyright Michael D. Fayer, 07

12 Repeated applications of to m gives new eigenvectors of values of m. (and ) with larger and larger But, because this must stop at a largest value of m, m max m. (m increases, doesn t change) Call largest value of m (m max ) j. m max =j For this value of m, that is, m = j j 0 with j 0 Can t raise past max value. Copyright Michael D. Fayer, 07

13 In similar manner can prove m is an eigenvector of with eigenvalues m and of with eigenvalues. Therefore, is a lowering operator. It reduces the value of m by and leaves unchanged. Operating repeatedly on j j m j, j, j, largest value of m gives eigenvectors with sequence of m eigenvalues Copyright Michael D. Fayer, 07

14 But, m Therefore, can t lower indefinitely. Must be some j such that j 0 with j 0 Smallest value of m. Can t lower below smallest value. Thus, largest value of m j = j' + an integer. smallest value of m Went from largest value to smallest value in unit steps. Copyright Michael D. Fayer, 07

15 We have largest value of m j 0 j 0 smallest value of m Left multiplying top equation by and bottom equation by j j 0 0 identities Then and operating 0 j j 0 j j j 0 j j j 0 j j j j Copyright Michael D. Fayer, 07

16 0 j j 0 j j j j Because j 0 and j 0 the coefficients of the kets must equal 0. Therefore, Because j > j' and j j j( j ) and ( j)( j ) j j j = an integer j = integer/; j can have integer or half integer values. because we go from j to j' =-jin unit steps with lowering operator. Thus, the eigenvalues of are 3 j( j) and j 0,,,, (largest m for a ) The eigenvalues of are m j, j,, j, j largest m change by unit steps smallest value of m Copyright Michael D. Fayer, 07

17 Final results jm j( j) jm jm m jm There are (j + ) m-states for a given j, going from j to j in integer steps. Can derive jm jm jm jm jm jm jm jm Copyright Michael D. Fayer, 07

18 Angular momentum states can be grouped by the value of j. Eigenvalues of, = j(j + ). j 0, /,, 3/,, j 0 m 0 00 j j j / m /, / m, 0, / m 3/, /, /, 3/ j m,, 0,, 0 etc. Copyright Michael D. Fayer, 07

19 Eigenvalues of are the square of the total angular momentum. The length of the angular momentum vector is j( j ) or in conventional units j( j ) Example m = j = Eigenvalues of are the projections of the angular momentum on the axis. m = 0 m = - Copyright Michael D. Fayer, 07

20 The matrix elements of are ( ) jj m, m jm jm j j jm jm m jj m, m jj m, m jm jm jm jm jj m, m jm jm jm jm The matrices for the first few values of j are (in units of ) j = 0 j = / (0) (0) (0) (0) / / Copyright Michael D. Fayer, 07

21 j = The jm are eigenkets of the and operators diagonal matrices. The raising and lowering operators and have matrix elements one step above and one step below the principal diagonal, respectively. Copyright Michael D. Fayer, 07

22 Particles such as atoms m RrY ( ) (, ) spherical harmonics from solution of H atom m The Y (, ) are the eigenvectors of the operators L and L. The m Y (, ) jm m LY LY m m m (, ) ( ) Y (, ) m (, ) my (, ) Copyright Michael D. Fayer, 07

23 Addition of Angular Momentum Examples Orbital and spin angular momentum - and s. These are really coupled spin-orbit coupling. ESR electron spins coupled to nuclear spins Inorganic spectroscopy unpaired d electrons Molecular excited triplet states two unpaired electrons Could consider separate angular momentum vectors j and j. These are distinct. But will see, that when they are coupled, want to combine the angular momentum vectors into one resultant vector. Copyright Michael D. Fayer, 07

24 Specific Case j m j m Four product states j m jm mm j and j omitted because they are always the same. Called the m m representation The two angular momenta are considered separately. Copyright Michael D. Fayer, 07

25 mm jjmm m m representation Want different representation Unitary Transformation to coupled rep. Angular momentum vectors added. New States labeled jm jjjm jm jm representation Copyright Michael D. Fayer, 07

26 jm where Eigenkets of operators in jm representation. and jm j j jm jm m jm vector sum of j and j Want unitary transformation from the m m representation to the jm representation. Copyright Michael D. Fayer, 07

27 Want jm Cmm m m mm C mm mm jm Cmm mm are the Clebsch-Gordan coefficients; Wigner coefficients; vector coupling coefficients are the basis vectors N states in the m m representation N states in the jm representation. N ( j )( j ) and obey the normal commutator relations. Prove by using and cranking through commutator relations using the fact that and and their components commute. Operators operating on different state spaces commute. Copyright Michael D. Fayer, 07

28 Finding the transformation m m m or coupling coefficient vanishes. To see this consider jm Cmm m m mm Operate with equal jm m jm Cmm m m mm These must be equal. Other terms 0 C mm if m m m mm mm m m C m m Copyright Michael D. Fayer, 07

29 Largest value of m since largest max m j + j m m max m j and m j Then the largest value of j is j j j because the largest value of j equals the largest value of m. There is only one state with the largest j and m. There are a total of (j + ) m states associated with the largest j j j. Copyright Michael D. Fayer, 07

30 Next largest m (m ) But m j j m m m Two ways to get m - m j and m j m j and m j Can form two orthogonal and normalied combinations. One of the combinations belongs to j j j Because this value of j has m values m ( j j ),( j j ),,( j j ) Other combination with m j j with j j j m ( j j ), ( j j ),,( j j ) largest smallest Copyright Michael D. Fayer, 07

31 Doing this repeatedly j values from j j j to j j in unit steps Each j has associated with it, its j + m values. Copyright Michael D. Fayer, 07

32 Example j values j, j j j j to j j in unit steps. j j 0 j j m, 0, 0 m 0 jm rep. kets m m rep. kets Know jm kets,0,00,,,, still need correct combo s of m m rep. kets Copyright Michael D. Fayer, 07

33 Generating procedure Start with the jm ket with the largest value of j and the largest value of m. m = But m m m Therefore, m m m because this is the only way to get m Then jm m m Clebsch-Gordan coefficient = Copyright Michael D. Fayer, 07

34 Use lowering operators jm m m jm m m from lowering op. expression Then 0 0 Clebsch-Gordan Coefficients from lowering op. expression (Use correct j i and m i values.) Copyright Michael D. Fayer, 07

35 Plug into raising and lowering op. formulas correctly. jm jm jm jm jm jm jm jm For jm rep. plug in j and m. jm For m m rep. mm mm means j jmm For must put in an d j when operating with and m and j and mwhen operating with Copyright Michael D. Fayer, 07

36 Lowering again 0 m m m m 00 Therefore, jm m m Have found the three m states for j = in terms of the m m states. Still need 00 m 0 m m Copyright Michael D. Fayer, 07

37 Need jm 00 m 0 m m 0 Two m m kets with mm 0, The 00 is a superposition of these. Have already used one superposition of these to form orthogonal to 0 and normalied. Find combination of normalied and orthogonal to 0. 00, Clebsch-Gordan Coefficients Copyright Michael D. Fayer, 07

38 Table of Clebsch-Gordan Coefficients j m m m j =/ j =/ Copyright Michael D. Fayer, 07

39 Next largest system j m j,0, m, m m kets 0 0 jm states j j j m,,, j j j m, jm kets Copyright Michael D. Fayer, 07

40 0 3 0 m m j m j = j = / Table of Clebsch-Gordan Coefficients jm m m m m Example Copyright Michael D. Fayer, 07