In the case of Hydrogen atom for example we use {n,l,m} (ignoring spin) ,because $H$,$L^2$,$L_Z$ operators commute with each other .

Does this mean if we find another operator which commutes with above operators ,we will have to introduce another quantum number to describe the system (like we do for spin for example)?

If I am correct how do I find out minimum number of operators(which commutes with each other) needed to describe the quantum system,what if there are more such operators? I mean given a set of operators how do I tell that these are enough to describe the state? Also can I use a different set of such operators to describe the state,if yes how do I decide which is better?

  • $\begingroup$ back ground - I am studying spin-orbit coupling ,L-S coupling etc and I am confused how they find the quantum numbers and decide which is better? $\endgroup$
    – Paul
    Commented Aug 19, 2018 at 11:20
  • $\begingroup$ But that's another topic. The "best" basis is the one where the spin-orbit matrix is diagonal. You'll find that it is $|n,l,s,j,m_j\rangle$; and not $|n,l,m,s,\mu\rangle$. Nevertheless, they both have the same number of quantum numbers, as it must be, since all basis have the same dimension. $\endgroup$
    – FGSUZ
    Commented Aug 19, 2018 at 11:26

2 Answers 2


Okay, this is all about what is known as CSCO, complete set of commuting observables.

You are right: a set of observables that commute with each other is $H, L^2, L_z$.

The key is wether it is complete or not. And what does that mean? A set is complete if their eigenvectors cover all space.

  1. If so, specifiying their quantum numbers determines one and only one wavefunction, except for a global phase factor.
  2. It is also useful to decide the basis order.

So that's the way you know it. The common eigenvectors of $L^2$ and $L_z$ are the spherical harmonics $Y_l^m$. However, there are many wavefunctions with the same harmonic, but they are different because they have different energies.

Conclusion: you need one more quantum number. Then, specifying $|n, l, m\rangle$ does provide a full basis.

This, of course, ignoring spin. You actually need $|n,l,m,s,\mu\rangle $

  • 1
    $\begingroup$ Saying a set of observables is complete if it specifies exactly one wavefunction is really just renaming the problem. How can you tell if a set of quantum numbers specifies exactly one wavefunction? $\endgroup$ Commented Aug 19, 2018 at 12:37
  • $\begingroup$ Typo: $L_z$ and $L^2$ $\endgroup$
    – kηives
    Commented Aug 19, 2018 at 14:07
  • $\begingroup$ You decide them according to what you are willing (and able) to measure. If you wanted to use $L_x$ instead, then your SCOC would be $H,L^2, L_x,(S^2,S_z)$, so that the basis would be $|n, l, m_x, (s),(\mu)\rangle$, because you want Energy and x angular momentum (and spin). $\endgroup$
    – FGSUZ
    Commented Aug 19, 2018 at 15:31
  • 1
    $\begingroup$ @BySymmetry A system of commuting Hermitian operators forms a complete set if their simultaneous eigenvalues uniquely define (up to an overall normalization) a common eigenvector, i.e. their joint spectrum is non-degenerate. $\endgroup$ Commented Mar 17, 2019 at 19:02

The central problem of quantum mechanics is one of labelling; if the labels are sufficiently descriptive, we have the required information about the states we're interested in, and we can calculate what we want with these states.

One can phrase this in the context of the following analogy: "Is it enough to refer to user "Zero" to uniquely identify this user?". If the answer is yes, then no other label is required. If the answer is no, then we need more labels; "Is it enough to refer to use "Zero TheHero" to uniquely identify this user?" and so forth.

Mathematically, the labels are eigenvalues of operators . Of course ideally the labels do not change with time - like the username would ideally not change with time - so we choose these labels to be operators to commute with the energy operator $H$ because the energy of a state a very useful quantity and usually it is conserved for time-independent problems. We also want these operators to commute amongst themselves so that the labels are simultaneously good labels, i.e. we don't give eigenvalues of $S_x$ and $S_z$ because eigenstates of $S_z$ do not have well defined eigenvalues of $S_x$.

Thus, were we to discover tomorrow that we need to refine our labelling of hydrogen states, we'd need another label and thus another operator so its eigenvalues could provide this additional label. In the other direction, we may find operators that commute with the three labelling operators, but there might not be a need to use the labels associated with such operators. It kinda depends on the information you're looking for.

An example of this would be the 3D harmonic oscillator. One can label the states by $n,\ell,m$ (eigenvalues of $H$, $L^2$ and $L_z$) or by $n_x,n_y,n_z$ (related to the number of quanta in each direction). Depending on the problem at hand, one or the other set of commuting operators

  • $\begingroup$ This is a great answer. In other words, the question is "how much information do you need to completely describe a quantum state?". This requires some a-priori knowledge. If you decide that you can describe your system completely with the energy, angular momentum, projection of that angular momentum in some direction, and the spin, then that's all the quantum numbers (and associated operators) that you need! But maybe next year a huge discovery is made and it turns out that all atoms also have some other measurable quantity that we didn't know about. So now we want to add that quantum number! $\endgroup$
    – TanyaR
    Commented Aug 19, 2018 at 23:09

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