Commuting operators and Direct product spaces

[aaa]

Under what conditions is the common eigenspace of two commuting hermitian operators isomorphic to the direct product of their individual eigenspaces?
As I'm not being able to precisely phrase my doubt, consider this example: Hilbert space of a two dimensional particle is the direct product of eigenspaces of cartesian X and Y operators. But as far as I know, the common eigenspace of L_z and L² operators isn't the direct product of their eigenspaces. What is the difference between these two cases?

 

[bhobba]

I am not sure what your doubt is. I will make a guess its the fact derivative operators are only defined on a subset of the two dimensional Hilbert space.

The way its handled is to use what's called Rigged Hilbert Spaces:
http://physics.lamar.edu/rafa/cinvestav/second.pdf

Its basically Hilbert spaces with distribution theory stitched on. Before delving into it, and its also a very good idea for just about any area of applied math, you should become acquainted with distribution theory:
https://www.amazon.com/dp/0521558905/?tag=pfamazon01-20

Thanks
Bill

 

[aaa]

Thanks for the reply. But my doubt is not about rigged Hilbert spaces.
Let me try to rephrase the question:
When can an eigenket |λ1λ2> be considered to be equivalent to the ket |λ1>|λ2>?
For example, the simultaneous eigenkate of cartesian X and Y operators, |xy>, can be equivalently written as |x>|y>. But as far as I know, the eigenket of L_z and L², |lm>, cannot be written as a direct product of two eigenkets.

 

[bhobba]

Thanks for the reply. But my doubt is not about rigged Hilbert spaces.
Let me try to rephrase the question:
When can an eigenket |λ1λ2> be considered to be equivalent to the ket |λ1>|λ2>?
For example, the simultaneous eigenkate of cartesian X and Y operators, |xy>, can be equivalently written as |x>|y>. But as far as I know, the eigenket of L_z and L², |lm>, cannot be written as a direct product of two eigenkets.

I think you need to get across the idea of a complete commuting set of observables:
http://en.wikipedia.org/wiki/Complete_set_of_commuting_observables

Thanks
Bill

 

[The_Duck]

Under what conditions is the common eigenspace of two commuting hermitian operators isomorphic to the direct product of their individual eigenspaces?

You shouldn't expect this to be the case unless the system is explicitly constructed so that it's true. The Hilbert space for a 2D particle is basically defined as the tensor product of two copies of the Hilbert spaces a 1D particle. But in most cases the property you are describing doesn't hold. As another example take the P and H operators for a 1D particle; you shouldn't go taking tensor products here.