Let ${\displaystyle O}$  be an operator which commutes with the Hamiltonian ${\displaystyle H}$ . This implies that we can have common eigenstates of ${\displaystyle O}$  and ${\displaystyle H}$ .^[2] Assume that our system is in one of these common eigenstates. If we measure of ${\displaystyle O}$ , it will definitely yield an eigenvalue ${\displaystyle q}$  (the good quantum number). Also, it is a well-known result that an eigenstate of the Hamiltonian is a stationary state,^[3] which means that even if the system is left to evolve for some time before the measurement is made, it will still yield the same eigenvalue.^[4] Therefore, If our system is in a common eigenstate, its eigenvalues of ${\displaystyle O}$  (good quantum numbers) won't change with time.

States which can be labelled by good quantum numbers are eigenstates of the Hamiltonian. They are also called stationary states.^[5] They are so called because the system remains in the same state as time elapses, in every observable way.

Such a state satisfies:

${\displaystyle {\hat {H}}|\Psi \rangle =E_{\Psi }|\Psi \rangle }$ ,

where ${\displaystyle |\Psi \rangle }$  is a quantum state, ${\displaystyle {\hat {H}}}$  is the Hamiltonian operator, and ${\displaystyle E_{\Psi }}$  is the energy eigenvalue of the state ${\displaystyle |\Psi \rangle }$ .

The evolution of the state ket is governed by the Schrödinger equation:

${\displaystyle i\hbar {\frac {\partial }{\partial t}}|\Psi \rangle =E_{\Psi }|\Psi \rangle }$ 

It gives the time evolution of the state of the system as:

${\displaystyle |\Psi (t)\rangle =e^{-iE_{\Psi }t/\hbar }|\Psi (0)\rangle }$ 

The time evolution only involves a steady change of a complex phase factor, which can't be observed. The state itself remains the same.

The hydrogen atom: no spin-orbit coupling edit

In the case of the hydrogen atom (with the assumption that there is no spin-orbit coupling), the observables that commute with Hamiltonian are the orbital angular momentum, spin angular momentum, the sum of the spin angular momentum and orbital angular momentum, and the ${\displaystyle z}$  components of the above angular momenta. Thus, the good quantum numbers in this case, (which are the eigenvalues of these observables) are ${\displaystyle l,j,m_{\text{l}},m_{s},m_{j}}$ .^[6] We have omitted ${\displaystyle s}$ , since it always is constant for an electron and carries no significance as far the labeling of states is concerned.

However, all the good quantum numbers in the above case of the hydrogen atom (with negligible spin-orbit coupling), namely ${\displaystyle l,j,m_{\text{l}},m_{s},m_{j}}$  can't be used simultaneously to specify a state. Here is when CSCO (Complete set of commuting observables) comes into play. Here are some general results which are of general validity :

1. A certain number of good quantum numbers can be used to specify uniquely a certain quantum state only when the observables corresponding to the good quantum numbers form a CSCO.
2. If the observables commute, but don't form a CSCO, then their good quantum numbers refer to a set of states. In this case they don't refer to a state uniquely.
3. If the observables don't commute they can't even be used to refer to any set of states, let alone refer to any unique state.

In the case of hydrogen atom, the ${\displaystyle L^{2},J^{2},L_{z},J_{z}}$  don't form a commuting set. But ${\displaystyle n,l,m_{\text{l}},m_{s}}$  are the quantum numbers of a CSCO. So, are in this case, they form a set of good quantum numbers. Similarly, ${\displaystyle n,l,j,m_{\text{j}}}$  too form a set of good quantum numbers.

The hydrogen atom: spin-orbit interaction included edit

To take the spin-orbit interaction is taken into account, we have to add an extra term in Hamiltonian^[7]

${\displaystyle \Delta H_{\text{SO}}={\frac {\alpha }{r^{3}}}\,{\boldsymbol {S}}\cdot {\boldsymbol {L}}}$ ,

where the prefactor ${\displaystyle \alpha }$  determines the strength of the spin-orbit coupling. Now, the new Hamiltonian with this new ${\displaystyle \Delta H_{\text{SO}}}$  term does not commute with ${\displaystyle {\boldsymbol {L}}}$  and ${\displaystyle {\boldsymbol {S}}}$ . It only commutes with ${\displaystyle L^{2}}$ , ${\displaystyle S^{2}}$ , ${\displaystyle J^{2}}$  and ${\displaystyle {\boldsymbol {J}}={\boldsymbol {L}}+{\boldsymbol {S}}}$  , which is the total angular momentum operator. In other words, ${\displaystyle m_{\text{l}},m_{s}}$  are no longer good quantum numbers, but ${\displaystyle l,s,j,m_{\text{j}}}$  are (in addition to the principal quantum number ${\displaystyle n}$ ).

And since, good quantum numbers are used to label the eigenstates, the relevant formulae of interest are expressed in terms of them.^{[dubious – discuss]} For example, the expectation value of the spin-orbit interaction energy is given by^[8]

${\displaystyle \langle n,l,s,j,m_{j}|\Delta H_{\text{SO}}|n,l,s,j,m_{j}\rangle ={\beta \over 2}(j(j+1)-l(l+1)-s(s+1))}$ 

where

${\displaystyle \beta =\beta (n,l)=Z^{4}{\mu _{0} \over 4{\pi }^{4}}g_{\text{s}}\mu _{\text{B}}^{2}{1 \over n^{3}a_{0}^{3}l(l+1/2)(l+1)}}$ 

The above expressions contain the good quantum numbers characterizing the eigenstate.

• Complete set of commuting observables
• Hamiltonian (quantum mechanics)
• Stationary state
• Constant of motion
• Quantum number
• Measurement in quantum mechanics
• Ehrenfest theorem
• Operator (physics)