1. MODULE 6 ANGULAR MOTION AND ANGULAR MOMENTUM In Module 3 we solved the Schrödinger equation for motion in a circular trajectory about a central, fixed point. The wavefunction and energy solutions were shown to be the integer m l is a quantum number that specifies individual wavefunctions and energy levels but now we show that it has a wider significance.

2. MODULE 6 Angular Momentum Considerations The particle of mass m rotates in the xy plane that is perpendicular to the defined z-axis. The system rotates clockwise when viewed from below with an angular velocity  (radians per second). The radial vector is defined by r the linear momentum vector (p) is directed tangentially to the radial vector.

3. MODULE 6 Classical physics tells us that such motion produces an angular momentum vector (L) in the direction orthogonal to the plane of the motion. The cross (vector) product of the vectors yields a third vector that is orthogonal to the components. when the motion is in three dimensions we use the determinant i, j, k, are the unit vectors in x, y, z directions

4. MODULE 6 The magnitude of the z-component of angular momentum can be picked out of the determinant whence the operator for the z-component of angular momentum becomes: in polar coordinates this becomes

5. MODULE 6 Operating on the wavefunction for rotary motion with this operator consider motion in one direction by putting B = 0 This is an eigenvalue equation the (REAL) eigenvalue for the z-component of AM is

6. MODULE 6. Thus we note that the z-component of the angular momentum is quantized and we see that m l is the quantum number for the z-component of the angular momentum due to motion about a central point. Normalization of the wavefunction yields

7. MODULE 6 Angular momentum operators If i, j, and k are the orthogonal unit vectors, then the position and linear momentum vectors of a particle can be expressed as And similarly

8. MODULE 6 Note that each of these identities can be generated by cyclic permutation of its predecessor. Also recognize that vectors are in bold and scalars (e.g. the magnitudes of the vectors) are in normal type.

9. MODULE 6 The magnitude (L, a scalar quantity) of the angular momentum is related to its components: In classical mechanics, no constraints are placed on the energy of a rotating system Nor on L, or any of its components. What does the quantum treatment give? We need first to construct the operators.

10. MODULE 6 Recall that we construct a position operator from q by replacing it with the multiplier q, and a momentum operator by replacing p q by

11. MODULE 6 Commutation relationships of the AM operators In the first line of the above sequence we replaced the operators by the magnitudes of the observables they represent To generate line 2 we multiplied out term by term. The zeros in line 3 appear because p z commutes with both y and x, and z commutes with p x and p y

12. MODULE 6 The other two commutators can be found in the same way, or can be simply written down using cyclical permutation. See Module 6 to see how to find the commutation relationships for the operator for the square of the magnitude of the AM

13. MODULE 6 The Particle on a Sphere a particle moving in a circle (2-dimensions) a particle moving on the surface of a sphere (3-dimensions). This is equivalent to the motion of a particle constrained to move at a constant radius from a central point, or the motion of a point in a solid body that represents the motion of the whole body. the results of solving the equations of motion for this seemingly irrelevant system are very useful when we come to look at the motion of electrons in spherically symmetrical Coulomb fields, and at the energy states of rotating diatomic molecules.

14. MODULE 6 We define the potential energy of the particle on the sphere to be constant and we equate it to zero. Then the hamiltonian is To convert the Laplacian to spherical polar coordinates we use  2 is the Legendrian operator

15. MODULE 6 The legendrian contains no terms in r It is the angular part of the Laplacian operator. Since the particle is confined to a spherical surface of fixed radius we can ignore the radial derivatives in the Laplacian. Thus the legendrian becomes the operator we need

16. MODULE 6 the wavefunction is a function of  and  only  r   x y z

17. MODULE 6 a sphere can be regarded as simply a stack of rings. Thus the solution should resemble that for the case of the ring except that now the particle can travel from ring to ring always maintaining its same distance from the center of the sphere. In this analogy we would expect the  equation to be separable and of the form This is so and the  component is the same as that for the particle on the ring.

18. MODULE 6 The cyclical boundary conditions are the same, the solutions are the same, and are specified by the quantum number m l. The quantum number has the same values as before. It specifies the magnitude of the AM component on the z-axis. We still have to worry about the  function, which is not so simple. The particle on the ring had to obey cyclical boundary conditions which led to the quantum number m l, now there will be a circum-polar cyclical boundary condition also to be satisfied This will no doubt lead to another quantum number.

19. MODULE 6 the solutions to the  equation depend on a set of functions called the spherical harmonics,

20. MODULE 6 The spherical harmonics are a set of functions that satisfy the EV equation The labels l and m l have integral values: l = 0, 1, 2, …. And m l = l, l-1, l-2, … -l Thus the solutions behave in a quantized manner and for a given value of l there are 2l+1 values of m l.

21. MODULE 6

22. The spherical harmonics are complex functions except for the cases where m l = 0. To envisage their shapes we plot the boundary surfaces of The boundary surface for l = m l = 0, has spherical symmetry The surfaces for l, m l > 0 have nodes.

23. MODULE 6 Note that the number of nodes increases as l increases

24. MODULE 6 Comparing the following We arrive at the set of energy states for the particle The quantum number l specifies the energy of the particle (independent of m l ). Each energy level is (2l+1)-fold degenerate.

25. MODULE 6 Angular Momentum Considerations The quantum number m l specified the magnitude of the z- component of AM of the particle moving in a circle, we might anticipate that the quantum number l likewise has implications to the angular momentum of the particle on the sphere. Classical physics tells us that the rotational energy of a spherical body of moment of inertia I and angular velocity  is given by E = I  2 /2. This is analogous to the equation for the kinetic energy of a moving particle of mass m and velocity v, i.e. E kin = mv 2 /2.

26. MODULE 6 the angular momentum of the spherical rotator is given as L = I  cf p = mv) E = L 2 /2I. Comparing this with Thus the magnitude of the AM is restricted by l – it is quantized l is the angular momentum quantum number.

27. MODULE 6 The spherical harmonics are eigenfunctions of the operator for the z-component of the angular momentum, The eigenvalues are Thus m l specifies the component of the angular momentum that can be ascribed to rotation around the z-axis. But, since m l is restricted to certain values, governed by l, the z- component of the angular momentum is restricted to 2l+1 discreet values for any value of l.

28. MODULE 6 The AM can be represented as a vector of length that is proportional to its magnitude ( {l(l+1)} 1/2 units), The vector must be oriented with respect to the (arbitrary) z-axis such that its projection thereon is equal to a length of m l units. This result means that the plane of rotation of the particle on the sphere is restricted in its spatial orientation Thus, contrary to what classical physics tells us, quantum mechanics requires that the orientation of a rotating body is subject to quantum restrictions. This is called space quantization.

29. MODULE 6 The L z component is defined absolutely, but L x and L y do not commute with L z. Therefore, if L z is precisely defined, the other components are highly uncertain. In the Figure the vector arrows are pointing to a specific direction in space, which implies that all 3 components of AM are precisely defined. Thus the Figure, while useful, is not a true representation of the QM situation.

30. MODULE 6 This often seen Figure is in error for the same reason, viz., it implies that all 3 components of AM are precisely defined.

31. MODULE 6 This Figure is an improvement. The cones represent the uncertainty in the directions of the x- and y- components of the angular momentum. The vector can be considered as being of length {l(l+1)} 1/2 units and laying along the cones with its tip at the cone mouth. Each cone has a z-projection of m l units on the z-axis. Each vector however lies at an unspecified azimuthal angle on its cone.

32. MODULE 6 Thus we see that the quantization of AM forbids the vector to be exactly parallel to the specified z-axis. The maximum value of the magnitude of the z-component occurs when m l = l, so the maximum co-latitude angle (  max ) will be given by for low values of l, cos  max 0. Only when l is very large (the classical limit) is {l(l+1)} 1/2 ~ l Then the co-latitude will become zero and the axis of rotation will lie along the z-axis. This corresponds to classical behavior.