Question

Definition of Quantum Number?? Thank you.

Answer 1

It is an integer that quantifies in what way the energies of a system are organized in a discrete manner. It is considered a "good" quantum number if the observable it describes is conserved.

Besides the atomic quantum numbers, other examples are shown here:
https://socratic.org/questions/schr-dinger-s-wave-equation-gives-which-type-of-concept-of-quantam-number?source=search

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ATOMIC QUANTUM NUMBERS

Recall the four quantum numbers for atomic orbitals:

• `n`, the principal quantum number, defines the orbital energy level, and goes as `1, 2, 3, . . .`

• `l`, the angular momentum quantum number, separates the orbitals into their respective shapes; `l = 0, 1, 2, 3, . . .` correspond to `s, p, d, f, . . .` orbital shapes.

• `m_l`, the magnetic quantum number, defines how many distinct orbitals there are of a given shape that have the same energy, and `m_l = {-l, -l+1, . . . , 0, . . . , l-1, l}`. In fact, there are `2l+1` numbers in the set of `m_l` values.

• `m_s`, the spin quantum number, defines what spins are appropriate for an electron, and is `pm1/2`. This forces all orbitals to hold at most `2` electrons.

These four quantum numbers are "good" quantum numbers as long as you are looking at atoms and they are not subject to perturbations that leave UNCONSERVED what events they correspond to.

That means that upon observation, whatever is observed does not change on us.

WHEN CERTAIN QUANTUM NUMBERS ARE NO LONGER "GOOD"

For example, in a magnetic field, total orbital angular momentum `L = sum_i m_(l,i)` and total spin angular momentum `S = sum_i m_(s,i)` are no longer conserved, and `L` and `S` are no longer good quantum numbers.

Instead, `J = {L - S_max, L - S_max + 1, . . . , L + S_max - 1, L + S_max}`, the total angular momentum, becomes a good quantum number.

Lastly, the quantum commutator relationship with the Hamiltonian operator must hold if a quantum number is "good":

`[hatH, hatA] = hatHhatA - hatAhatH stackrel(?" ")(=) 0`,

where `hatH` is the Hamiltonian operator that corresponds to the energy eigenvalue of the system, and `hatA` is some other quantum mechanical operator that corresponds to some other eigenvalue of the system.

For example, the orbital angular momentum in the `x,y,z` directions are given by

`hatL_x = haty hatp_z - hatp_z haty = [haty,hatp_z]`
`hatL_y = hatz hatp_x - hatp_x hatz = [hatz,hatp_x]`
`hatL_z = hatx hatp_y - hatp_y hatx = [hatx,hatp_y]`

where `hatx` is the position operator `xcdot` and `hatp_q = -iℏ(del)/(delq)` is the linear momentum operator in the `q` direction.

If we define `hatH = hatp^2/(2m)` for a free particle, then it can be shown that `[hatH, hatL_z] = 0`, meaning that angular momentum is conserved in the `z` direction for the free particle moving in any direction.

But I won't show that here. If you really want to see that, you can look here (pp. 9 - 11).