Consider an electron in a hydrogen atom in the quantum state: [imath]\psi = \sqrt{\frac{1}{6}} \psi_{321} + \sqrt{\frac{1}{2}} \psi_{320} + \sqrt{\frac{1}{3}} \psi_{3 2 -1}[/imath]

(Where the subscripts are the quantum numbers n, l, m respectively for the basic states of a hydrogen atom, such that [imath]\psi_{nlm} = R_{nl} Y^{m}_{l}[/imath].)

The energy of the electron, L

^{2}and L

_{Z}are measured. What are the value(s) of these quantities that can be measured, and what is the expectation value of L

_{Z}?

Now;

Energy: The energy of the electron is determined by the quantum number n, and there is no ambiguity in the measured value of the energy, since n =3 in all of the terms in the wavefunction. E = - 13.6 eV * (1/3

^{2}), which is equal to about -1.51 eV.

L

^{2}: The eigenvalue of L

^{2}is l(l+1)[imath]\hbar^{2}[/imath], and l = 2 for all of the terms, therefore, the measured value of L

^{2}is simply equal to 6[imath]\hbar^{2}[/imath].

The eigenvalue of L

_{Z}is [imath]\hbar[/imath]m, so we have three different possible values corresponding to each of the three terms, L

_{Z}= [imath]\hbar[/imath] with probability 1/6, L

_{Z}= 0 with probability 1/2 and L

_{Z}= -[imath]\hbar[/imath] with probability 1/3.

Add up those three terms to get <L

_{Z}>, and we have (1/6 - 1/3)[imath]\hbar[/imath] = -(1/6)[imath]\hbar[/imath].

Does that all sound like it's on the right track?

Thanks.