Common sets of eigenfunctions in angular momentum

[Turtle492]

Hi,

I'm a second year physics undergrad currently revising quantum mechanics, and I came across a phrase about angular momentum which has confused me, so I was wondering if anyone could help.

We looked at different components of angular momentum (in Cartesian) and decided that they did not commute with one another, but that the square of the magnitude (L^2) commutes with all of them, meaning 'that L^2 and one of the L components have a common set of eigenfunctions'. What I don't understand is how L^2 can have eigenfunctions in common with, Lx, Ly and Lz, but then Lz doesn't have any in common with say Ly.

I think maybe the problem is that I don't really understand what's meant by a 'common set of eigenfunctions'. What makes up a set? So far we've mainly looked at spin where the eigenfunctions form a complete, orthogonal set. Does the same happen with angular momentum?

Thanks for your help

 

[SpectraCat]

Hi,

I'm a second year physics undergrad currently revising quantum mechanics, and I came across a phrase about angular momentum which has confused me, so I was wondering if anyone could help.

We looked at different components of angular momentum (in Cartesian) and decided that they did not commute with one another, but that the square of the magnitude (L^2) commutes with all of them, meaning 'that L^2 and one of the L components have a common set of eigenfunctions'. What I don't understand is how L^2 can have eigenfunctions in common with, Lx, Ly and Lz, but then Lz doesn't have any in common with say Ly.

I think maybe the problem is that I don't really understand what's meant by a 'common set of eigenfunctions'. What makes up a set? So far we've mainly looked at spin where the eigenfunctions form a complete, orthogonal set. Does the same happen with angular momentum?

Thanks for your help

First of all, when observables commute, it means that they can simultaneously be measured to arbitrary precision. Another was to say this is that the eigenstates are simultaneous eigenstates of both operators. When operators don't commute, then the HUP says that you cannot simultaneously know both of them to arbitrary precision. Furthermore, say you have measured a state and know it to be in an eigenstate of some operator A. If you then make a second measurement of another property corresponding to an operator B that does NOT commute with A, the system will no longer be in an eigenstate of A after the second measurement, and you will have lost knowledge about the property corresponding to A.

An important point missing from your analysis is that, for a quantum system, only one of Lx, Ly and Lz can be known at a time (because they don't commute with one another). Another way of saying this is that you can only ever know the projection of L on one cartesian axis (the axis of quantization) at a time. Whichever one of these is known (we usually take Lz by default), commutes with L². However, if you suddenly change your axis of quantization to the x-axis, then you will know Lx and it will commute with L², but you will have lost all knowledge of Lz. A good way to learn more about this is to study the triple Stern Gerlach experiment. (http://en.wikipedia.org/wiki/Stern–Gerlach_experiment. ) There are also Java applets out there to help you understand it. http://www.if.ufrgs.br/~betz/quantum/SGtext.htm

 

[Turtle492]

So if you measure Lx and L², say, then it will be in an eigenstate that is common to both of them? What if I measure Lz and L² and I get the same value for L² as I did when I was using Lx, would it not then be in the same eigenstate for L² as it was before, and is now sharing that one with Lz? Does that not mean that Lz and Lx should also have that eigenstate in common?

Or is there some kind of degeneracy whereby L² has 3 eigenstates that give the same eigenvalue, so that it can share one with each of the other components of L? I'm a little confused.

 

[SpectraCat]

So if you measure Lx and L², say, then it will be in an eigenstate that is common to both of them? What if I measure Lz and L² and I get the same value for L² as I did when I was using Lx, would it not then be in the same eigenstate for L² as it was before, and is now sharing that one with Lz? Does that not mean that Lz and Lx should also have that eigenstate in common?

No ...

Or is there some kind of degeneracy whereby L² has 3 eigenstates that give the same eigenvalue, so that it can share one with each of the other components of L? I'm a little confused.

Yes .. that is more like it, although not precisely correct. For all states with non-zero angular momentum, L² has 2l+1 degenerate eigenstates (where l is the angular momentum quantum number), each of which have different eigenvalues of Lz (the default choice for the axis of quantization). If you change the axis of quantization to say, x, you transform to a different set of 2l+1 eigenstates. These all still correspond to the same eigenvalue of L², but now they are eigenstates of Lx, instead of Lz.

 

[Turtle492]

OK, I think I understand. Thanks a lot, that's really helpful.