In: Physics
(a) Under what condition, does the angular momentum become a good quantum number? (Explain your answer with the mathematical description.)
(b) In that case, what do Hamiltonian and angular momentum operators share? (Briefly explain why you came up with the answer)
(c) In general, boundary conditions give discrete eigenvalues. What is the 'boundary condition' for a rotation problem? (What is the physical reason for discrete angular momentum values?)
(d) Can you explain why the orbital angular momentum operator cannot have half-integer eigenvalues?
(e) Based on your answer for (d), can you find the meaning of 'half-integer' spin quantum number?
a) If the expectation value of the angular momentum over a stationary state remains constant, then it becomes a good quantum number. This happens if the Hamiltonian and the z component of the angular momentum commutes
From Ehrenfest theorem
Since the angular momentum is not an explicit function of time,
and since it commutes with the Hamiltonian,
This means the angular momentum is conserved and hence it is a good quantum number.
b) Since the angular momentum and the Hamiltonian commute with each other, this means they share a simultaneous eigenstate.
and
here is the eigenstate of both the Hamiltonian and the angular momentum operator.
c) For a rotational problem, the boundary condition is that the eigenfunction remains invariant if the angle of rotation is rotated by 2 Pi radians or 360 degree
d) The eigenfunction for rotational problem is
where c is a constant and m is the eigenvalue of the angular momentum
The angular momentum operator is
Thus
Therefore under the rotation of the angle
the state becomes
For the eigenfunction to remain invariant,
This is possible if m is an integer
Hence the orbital angular momentum have only integer eigenvalues and NOT half integer eigenvalues
e) For spin angular momentum, the eigenvalues are half-integer. This means the spin eigenfunctions are invariant if the angle of rotation is rotated by 4Pi
where the suffix s is for spin.