Angular Momentum. A vector operator J, whose components satisfy the commutation ~ relations. ffl This can also be written

Transcription

1 Angular Momentum A vector operator J, whose components satisfy the commutation ~ relations [ ^J x ; ^J y ] = i ^J z [ ^J z ; ^J x ] = i ^J y [ ^J y ; ^J z ] = i ^J x : This can also be written [ ^J p ; ^J q ] = i" pqr ^J r ; where " is + for pqr an even permutation of three objects (cyclic permutation here), for an odd permutation, and if any indices coincide (Levi-Civita tensor or antisymmetric tensor of rank three). Since components of ~J do not commute, we cannot find a simultaneous eigenfunction of these three components.

2 ^J 2 jjmi = J (J + ) jjmi ^J z jjmi = M jjmi Angular Momentum However, the square ~ of J, that is ~ ~ J: J, denoted,commutes with all components: 2 ^J [ ^J 2 ; ^J x ] = [ ^J 2 ; ^J y ] = [ ^J 2 ; ^J z ] = : Hence we can find a simultaneous eigenfunction of and any 2 one of ^J the p : conventional to choose z. ^J ^J Eigenfunctions of these operators are jjmi, for which The effect of ^J x and ^J y is more complicated.

3 ^J + = ^J x + i ^J y ^J = ^J x i ^J y ; ^J + jjmi = (J (J + ) M (M + )) 2 jjm + i Angular Momentum It is more convenient to use the new operators which also must commute with ^J 2. These shift operators have the effect of changing an eigenfunction jjmi: ^J jjmi = (J (J + ) M (M )) 2 jjm i provided on the RHS we cannot have jm + j > J (the result is then zero).

4 hjmj ^J 2 jj M i = J (J + )ffi JJ ffi MM hjmj ^J z jj M i = Mffi JJ ffi MM hjmj ^J ± jj M i = (J (J + ) M (M ± )) 2 ffi JJ ffi MM ± Angular Momentum We have then the following matrix elements:

5 Coupling of Angular Momentum Suppose we have two angular momentum eigenfunctions jjmi and jj M i. We can expand their product as jj M i = X jjmi J X C J M JMJ M jj M i : M The expansion coefficients C above are the Clebsch-Gordan coefficients (within a phase: Wigner coefficients, 3j-symbols, vector-coupling coefficients) We will revisit this...

6 L p = i" @q Classical Angular Momentum The most obvious example of an angular momentum operator is obtained from the classical angular momentum L = ~r ~p. ~ The quantum mechanical analog is ~r i ~ r with elements L x = y z L y = @x L z = i which could also be x y

7 We therefore will have eigenfunctions jlm L i of ^L 2 and ^L z.

8 Symmetry of Angular Momentum Operators It is straightforward to see that any rotation around any axis must leave unchanged, since is isotropic, or spherically symmetric. 2 ^L 2 ^L The symmetry is the same as rotations of a sphere. Denoted (Herzberg) or R(3) (Chisholm, Lomont). K Rotations in a three-dimensional space are real orthogonal matrices (OO = ) with determinant +. This group is denoted SO(3), thespecial (or unimodular) orthogonal group T in 3-dimensional space.

9 Symmetry of Angular Momentum Operators Inversion through the origin also leaves invariant. This is a 2 transformation with determinant ^L. The full group is thus SO(3) Ω S 2, which is actually just the orthogonal O(3) group K (also h (Herzberg)). We will concentrate here on the subgroup SO(3) of the pure rotations.

10 R jlm L i = LX jlm Li c(l; M L;M L )(R); = L L M Transformation of Eigenfunctions The R SO(3) transformations commute with so in 2 2 general we must ^L have since there is no way for R to mix different L values. If we identify the c with elements of a matrix c(l; M L ;M L)(R) = D L M L M L (R) this looks exactly like a set of basis functions for an irrep...

11 LM L ( ; ffi) = NPjM Lj L (cos ) exp(im L ffi) Y The irreps are of dimension 2L + Transformation of Eigenfunctions In this way we can find basis functions and irreducible representations for SO(3). The eigenfunctions jlm L i are our old friends the spherical harmonics: L i = Y LM L ( ; ffi); jlm except that these are more usually taken in the complex form The phase relations are established by choosing all Y L positive and the effect of the shift operators as giving positive signs (as shown previously)

12 Parametrization of Rotations SO(3) is a three parameter Lie group, since we can rotate around three axes. Alternatively, consider a 3 3 orthogonal matrix: nine elements and six conditions, so three free parameters. Most popular parametrization is via Euler angles, denoted ffffiflg either or fffi χg: counterclockwise rotations by ffi about the original z axis about the new y axis about the new z axis χ

13 R jlm L i = Euler Angles Alternatively, we can carry out the following counterclockwise rotations: about the original z axis χ about the original y axis about the original z axis, ffi and obtain the same result(!) We know already how the Y LM L transform under a rotation R (specified now (ffi; ; by χ)) LX L = L jlm Li D L M L M L (ffi; ; χ) M

14 We state without proof that Irreps of SO(3) D L M L M L (ffi; ; χ) = exp( im L ffi)dl M L M L ( ) exp( im Lχ); where d L M L M L ( ) is a polynomial in cos( =2) and sin( =2). The presence of the half angles here will have profound consequences later.

15 2j + Irreps of SO(3) All rotations through the same angle ' are in the same class: we have the character table E C '... S... P 3 3 ' sin ' sin )' sin(j ' sin

16 2 fl : 2 Non-integer Angular Momentum Experiment has established that there are systems (such as an electron) with properties that can be rationalized only in terms of an intrinsic angular momentum quantized in half-units ( spin ). In the simplest case of J = 2 we have two spin functions ff = fi = fi 2 2 fl fi Hence we have fl 2 fi = 2 2fi ^J z fi 2 2 fl 2 fi = 3 4 2fi ^J 2 fi 2 2

17 2fi ^J + fi 2 2 fl fi = 2 etc.

18 @ A A A Non-integer Angular Momentum One way to write the spin functions is as vectors Then the various J operators are represented by matrices, e.g.,

19 ff = ff ff 2 = A ff i A A If we construct four matrices Pauli Matrices i all our angular momentum operators for the spin functions can be expressed in terms of these matrices. ^J 2 = 3 4 ff, ^J + = 2 (ff + iff 2 ),etc. Spinor basis, Pauli matrices.