1. Particle on a Ring An introduction to Angular Momentum Quantum Physics II Recommended Reading: Harris, Chapter 6

2. Particle confined to a circular ring In this problem we consider a particle of mass m confined to move in a horizontal circle of radius. This problem has important applications in the spectroscopy of molecules and is a good way to introduce the concept of ANGULAR MOMENTUM in quantum mechanics. Can specify the position of the particle at any time by giving its x and y coordinates. r m  BUT this is a one dimensional problem and so we only need one coordinate to specify the position of the particle at any time t. The position is specified by the angle  (t), the angle made by the vector r with the horizontal. If orbit is horizontal then there potential energy of the particle is constant and can be taken as zero: U = 0 What is the potential if the orbit is vertical? Then we must include gravity, much more difficult problem.

3. Classical Treatment r m  v s m = mass of particle v = instantaneous velocity = ds/dt r = radius vector s = arc length  = angle = s/r Can then define angular velocity  The kinetic energy is constant and equal to ½mv 2 In terms of the angular velocity this can be expressed as The quantity I = mr 2 is the moment of inertia of the particle, then The angular momentum L of the particle is defined as Can write the energy of the particle in terms of the angular momentum

4. Note that the angular momentum L is perpendicular to both r and v (since L = mr  v ). can have two directions for the velocity, clockwise or anti-clockwise, r L v r v L Magnitude of L is the same in both situations, but direction is different. Must remember that L is a vector

5. TISE for particle confined to a circular ring The time independent Schrodinger equation is (in (x,y) coordinates) (1) If we express this in terms of the angle  then x = r.cos(  ), y = r.sin(  ) or tan(  ) = y/x and Schrodinger’s equation becomes (2) (3) but mr 2 = I, the moment of inertia of the particle, So TISE is (4) want to solve this equation for the wavefunctions and the allowed energy levels

6. (5) Solutions to TISE for particle on a circular ring rearrange (4) A general solution of this differential equation is (6) where Check this (7) The wavefunction must be single valued for all values of . In particular if we rotate through an angle 2  the wavefunction must return back to the same value it started with: from (7) we then get Energies are quantised. Allowed values are given by eqn (8) (8)

7. Energy Spectrum where m L =0,  1,  2,  3, …. m L =0 m L =1 m L =2 m L =3 m L =4 m L =5 E 0 =0 E 1 =1 E 2 =4 E 2 =9 E 2 =16 E 2 =25 Energies in units of

8. Energy Spectrum where m L =0,  1,  2,  3, …. Energies in units of m L =0 m L =1 m L =2 m L =3 m L =4 m L =5 E 0 =0 E 1 =1 E 2 =4 E 2 =9 E 2 =16 E 2 =25 Energy all states are doubly degenerate except the ground state (m = 0) which is singly degenerate. - m Lc mLmL + m L

9. Degeneracy of Solutions Equation (6) shows that there are two solutions for each value of m L, (except m L = 0) [ exp(im L  ) and exp(-im L  )]  doubly degenerate system. The two solutions correspond to particles with the same energy but rotating in opposite directions r m  r m  Compare this with linear motion of a free particle, where the solutions are also doubly degenerate; Aexp(ikx) (+x direction) and Aexp(-ikx) (-x direction)

10. Normalisation From normalisation condition Hence And the normalised wavefunctions are (9) (10)

11. Probability Distribution What is it probability P, that the particle will be found at a particular angle  ? Note that this is independent of the angle  and the quantum number m L  equal probability of finding the particle anywhere on the ring no matter what state it is in! 0 22 Angle  Probability 1/2  However, note that the wavefunctions are complex. They have a real and an imaginary part

12. The Wavefunctions Recall that so the wavefunctions can be written as Real part of the wavefunction = cos(m L  ) Imaginary part of the wavefunction = sin(m L  ) and again we see that We can visualise the real and imaginary parts of the wavefunction as follows.

13. sin(1.  ) cos(1.  ) 11 1111 0 22 Angle 

14. sin(2.  ) cos(2.  ) 2222 0 22 Angle 

15. 3333 sin(3.  )cos(3.  ) 0 22 Angle 

16. sin(1.  ) cos(1.  ) 11

17. 2222