Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

How does one determine the members of an simultaneously commuting set (of operators)? For example, I have read that for orbital angular momentum, the set is {$H,L^2,L_z$}. How does one know that these are all of them or that these particular operators should be included? Is it because such a set is sufficient to distinguish all the eigenstates, degeneracies notwithstanding?

Also, why is $$L^2|\ell,m\rangle = \hbar \ell (\ell+1)|\ell,m\rangle$$ and $$L_z|\ell,m\rangle = \hbar m|\ell,m\rangle,$$

where $\ell,m$ are integers?

Is it defined as such or is it the result of the definition of some more fundamental things?

share|improve this question
1  
The set of commuting operators will never be a finite set. Let $|n\rangle$ be a fixed orthonormal basis of your separable Hilbert space, indexed by the natural numbers. Let $f,g:\mathbb{N}\to \mathbb{R}$ be arbitrary functions mapping natural numbers to real numbers, such that they are non-zero for only finitely many inputs. Then $|n\rangle \mapsto f(n)|n\rangle$ and $g$ are commuting compact operators. In general, the set of all compact operators that can be diagonalised by a fixed ON basis is uncountably infinite; because they can all be simultaneously diagonalised, they mutually commute. –  Willie Wong Aug 30 '11 at 12:25
    
@Willie: Thanks! :) I suppose then that somehow {$H,L^2,L_z$} is enough to distinguish the eigenstates? Also, given that {$H,L^2,L_z$} is a commuting set, your nicely argued statement could be realized by forming linear combinations of the components? –  bra-ket Aug 30 '11 at 13:40
1  
Any $f(n)$ that is injective into $\mathbb{R}$ is "enough to distinguish the eigenstates", if by that phrase you mean what I think you mean. (Is the Hamiltonian sufficient to distinguish the eigenstates of the harmonic oscillator?) But no, the set that I wrote down of commuting operators forms in fact a infinite dimensional vector space. So it cannot be given in terms of just the linear combinations of three operators. (For example, the operator $M^2|l,m\rangle = m^2|l,m\rangle$ would commute with $H,L^2,L_z$, but is not a linear combination of the three.) –  Willie Wong Aug 30 '11 at 17:26
2  
bra-ket, you have two separate questions in there. I'd suggest removing the part about why $L^2$ and $L_z$ are defined as they are and posting that as a separate question. –  David Z Aug 30 '11 at 18:29
add comment

1 Answer

up vote 4 down vote accepted

For practical reasons, physicists like to label the states of the system by a set of "quantum numbers". Technically this means that you are looking for a set of mutually commuting Hermitian operators such that: (i) Every vector from the basis of common eigenvectors of these operators is uniquely characterized by the set of eigenvalues (i.e. the above mentioned quantum numbers); (ii) The set of operators is minimal in the sense that by removing any of the operators, you lose the property (i). The first condition implies that your set of operators is complete while the second one that there are no redundancies. Since you typically want to use the quantum numbers to label the stationary states, that is, the eigenstates of the Hamiltonian, you in addition require that your operators commute with the Hamiltonian (the Hamiltonian itself need not be included in the set though).

Now it is obvious that for any given system and its Hilbert space, there are many sets of operators that satisfy the above two conditions. In principle, you can find your own set of operators using the following algorithm. Pick any (Hermitian) operator $A$ and determine its spectrum. If there is no degeneracy, you are done. Otherwise, add another operator $B$ that commutes with $A$ such that there is a pair of common eigenvectors with the same eigenvalues of $A$ but different eigenvalues of $B$. (This guarantees that you improve your "resolution" in the specification of the eigenvectors by the set of eigenvalues.) If there are no two eigenvectors for which both $A$ and $B$ have the same eigenvalue, you are done. Otherwise, repeat the previous step.

It should be emphasized that the notion of a complete set of commuting operators introduced in this way is a bit vague. For example, it does not say how many operators you need to form a complete set in a given Hilbert space. In fact, as pointed out by Willie Wong in his comment, one operator is in principle enough in a separable Hilbert space. Here is maybe a slightly less abstract example. You know that the states of a spin system can be characterized by specifying simultaneously the values of $\vec L^2$ and $L_z$. (I disregard the orbital degrees of freedom so that these two operators are enough.) However, one can as well take the operator $\vec L^2+\alpha L_z$ where $\alpha$ is an arbitrary irrational number. You can easily convince yourself that the eigenvalue of this single operator is sufficient to label uniquely all the states $|l,m\rangle$.

As to your second question, I would simply refer you to any textbook on quantum mechanics. The fact that the eigenvalues of $\vec L^2$ and $L_z$ take the form $\hbar l(l+1)$ (where $l$ is a non-negative integer or half-integer) and $\hbar m$ where $m\in\{-l,-l+1,\dots,l-1,l\}$, is not merely a matter of a definition, but follows from the commutation relation for the angular momentum operator, $$ [L_i,L_j]=i\hbar\varepsilon_{ijk}L_k. $$

share|improve this answer
    
Thank you very much! This is very well explained. :) –  bra-ket Aug 30 '11 at 19:47
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.