
#1
Jun1811, 03:40 PM

P: 261

Hi,
I am having a little trouble with the concept of finding out the maximal set of commuting observables. Suppose I have n commuting operators. Then the wavefunction I use must have n parameters also. For instance, [itex] L_{3}, L^{2}[/itex] and [itex] H[/itex] where H is the Hamiltonian and L is the angular momentum operator all commute. So the common eigenstates have three variables inside i.e [itex]n l m>[/itex]. What I can't figure out is how to ensure that these operators are "distinct" from each other. For example, if I take a one of the above operators and multiply it by a scalar I will get a different operator, say something like [itex]2H[/itex]. Now I have four commuting operators instead of three. Obviously, this is not going to give me another independent variable inside the wavefunction. This was a trivial example but in general, how should one ensure that the commuting operators that one has are all distinct? My best guess so far is to check that the matrix representations form a linearly independent set but is there a) a weaker condition and b) a way to do this without resorting to matrix representation? Thank you. 



#2
Jun1811, 05:27 PM

Sci Advisor
P: 1,395

The answer is that, in order completely specify an eigenstate, you must specify its eigenvalues (quantum numbers will typically suffice) with respect to all commuting observables of which it is an eigenfunction. Thus, if you find degenerate eigenstates, it suggests (perhaps guarantees) that you do not have a complete specification of your quantum state yet.
In the example of the oneelectron atomic states you mentioned, if you specify only the energy (eigenvalue of H, quantum number [itex]n[/itex]), then you have an n^2fold degeneracy to deal with. Specifying the angular momentum (eigenvalue of L^{2}, quantum number [itex]l[/itex]), improves that, but now each state for which you know [itex]n[/itex] and [itex]l[/itex] still has a degeneracy of [itex]2l+1[/itex], which you rectify by specifying the zprojection of angular momentum (eigenvalue of L_{z}, quantum number [itex]m_l[/itex]). Now you have a unique designation completely specifying a particular atomic orbital. All of these wavefunctions are orthogonal and uniquely specified by their set of quantum numbers. As you point out, you can create arbitrary operators such as 2H for which they are also eigenfunctions, but doing so gives you no further information about the quantum states. 



#3
Jun1911, 01:03 AM

P: 261

SpectraCat, my problem is like this: Let's say I have three commuting operators. Now, when I specify the common eigenstate, I should put in three quantum numbers. But how can I be sure that my three operators are all distinct and not trivial variations of one another like H and 2H are?




#4
Jun1911, 03:12 AM

P: 679

Maximal set of commuting observables
It's probably easier to check the eigenvalues over the simultaneous eigenvectors, let's say if [itex]L_3[/itex] is a linear combination of [itex]L^2[/itex] and [itex]H[/itex], then their eigenvalues must satisfy the same linear combination, that means if you can find a definite linear relation among those eigenvalues, then you only have 2 effective commuting observables;if you can't, then you have 3.




#5
Jun1911, 08:09 AM

Sci Advisor
P: 1,395

I guess my point is that I don't think this issue comes up in practice ... can you give an example of what you are talking about? I mean one that doesn't rely on contrived operators like 2H, but rather on operators corresponding to physical observables that have a linear dependency that is not immediately obvious. 



#6
Jun1911, 09:32 AM

P: 261

I agree, it hasn't come up in practice yet but I'm only a couple of semesters into QM so I thought perhaps there might be situations where it would be necessary to verify that the operators were all independent.
Just one more thing: Suppose we had a set of commuting operators where one of the elements was not a linear combination but had some other relationship with the other elements in the set. For instance let's pretend [itex]L_{1}[/itex], [itex]L_{2}[/itex], [itex]L_{3}[/itex] and [itex]L^{2}[/itex] did commute. Now [itex]L^{2}=L_{1}^{2}+L_{2}^{2}+L_{3}^{2}[/itex]. How many quantum numbers should the common eigenstate have? So my question is, in general, when do we know that every one of the commuting operators gives us a new quantum number? Not just for this example but more generally. Sorry if this sounds very contrived. I admit I've never (yet) faced a problem but I just want to know exactly how to deal with it if I ever need to. Thank you 



#7
Jun1911, 10:16 AM

Sci Advisor
P: 1,395





#8
Jun1911, 10:25 AM

P: 261

SpectraCat, in your example, isn't it just a linear relation? The sum of the 1D Hamiltonians is the 3D Hamiltonian. If one operator is a linear combination of the others, then we already know by kof95's example that the operator in question will not result in a new quantum number.
I know my example is nonphysical but I can't think of a physically correct one where there is a non linear relationship between the commuting operators. Could you perhaps think of one? Even otherwise, what would you say about the angular momentum example? Thank you for taking the time to reply to my questions. I appreciate it! 



#9
Jun1911, 10:41 AM

P: 684

You can consider the case of a free particle. The Hamiltonian reads [itex]H \propto P_1^2 + P_2^2 + P_3^2[/itex], so you have 4 commuting observables. Of the four quantum numbers [itex]E, p_1, p_2, p_3[/itex] you can chose three to characterize your state.
There is no difference between linear and nonlinear combinations of observables, since the relation [itex] f(A_1, A_2, ...) eigenstate \rangle = f(a_1, a_2,...) eigenstate \rangle [/itex] (with eigenvalues [itex]a_i[/itex]) always holds. 



#10
Jun1911, 10:45 AM

P: 261

So as long as I can express one operator as some function of the others, it will not result in an additional quantum number. That makes sense.
Thank you very much for all the replies. 


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