angular momentum in QM

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 Joined: Wed Mar 26, 2008 9:14 am
angular momentum in QM
well i am studying from griffith.. and as he has introduced the concept of angular momentum.. the first question that he asked and emphasized on is to find a basis in which both L[sub]x[/sub] and L[sub]y[/sub] or other combinations commute.. so finally coming to L[sup]2[/sup] and L which commute..
my question is why do i need to search for such a basis.. or in other words. why do i need to find a set of common eigenfunctions for both L[sup]2[/sup] and L
my question is why do i need to search for such a basis.. or in other words. why do i need to find a set of common eigenfunctions for both L[sup]2[/sup] and L
Re: angular momentum in QM
States that are eigenstates of both Lx and Ly (or Lx and L^2, etc) won't change if you switch the order in which you apply the operator.
If you choose a random set of eigenstates (say, using the default Lz/L^2 eigenstates), and try to measure Lx then Lz starting from a pure Lz/L^2 eigenstate, you will first get a mixture of eigenstates and you will have no idea which one you will get when you apply Lz (except for the probability). If you measure Lz and then Lx you will then get a mixture of states that make up the original Lz state.
If you choose a random set of eigenstates (say, using the default Lz/L^2 eigenstates), and try to measure Lx then Lz starting from a pure Lz/L^2 eigenstate, you will first get a mixture of eigenstates and you will have no idea which one you will get when you apply Lz (except for the probability). If you measure Lz and then Lx you will then get a mixture of states that make up the original Lz state.

 Posts: 378
 Joined: Wed Mar 26, 2008 9:14 am
Re: angular momentum in QM
the first statement that you made.. how can you have eigenstates for both Lx and Ly simultaneously? since they do not commute?vicente wrote:States that are eigenstates of both Lx and Ly (or Lx and L^2, etc) won't change if you switch the order in which you apply the operator.
If you choose a random set of eigenstates (say, using the default Lz/L^2 eigenstates), and try to measure Lx then Lz starting from a pure Lz/L^2 eigenstate, you will first get a mixture of eigenstates and you will have no idea which one you will get when you apply Lz (except for the probability). If you measure Lz and then Lx you will then get a mixture of states that make up the original Lz state.
wel i do not know this Lz/L^2 eigenstate.. can you please emphasize on it?
Re: angular momentum in QM
You are right, Lx and Ly don't commute, you can't find simultaenous eigenstates for them both.
With Lz and L^2, you can use the  n l m > states
where Lz  n l m > = m hbar  n l m > and L^2  n l m > = l*(l+1)*hbar^2  n l m >.
The spatial representation of the  n l m > states are given by a product of the R_nl states and the Y_lm spherical harmonics
With Lz and L^2, you can use the  n l m > states
where Lz  n l m > = m hbar  n l m > and L^2  n l m > = l*(l+1)*hbar^2  n l m >.
The spatial representation of the  n l m > states are given by a product of the R_nl states and the Y_lm spherical harmonics
Re: angular momentum in QM
May be I missed something, but this statement "'You are right, Lx and Ly don't commute, you can't find simultaenous eigenstates for them both.""" does not seem right to me.
Lx and Ly do not commute does not mean that they do not have any common eigenstate.They DO have common eigenstates, BUT it is just that those common eigenstates DO NOT span the whole Hilbert space. The common eigenstates of two operators span the whole Hilbert space iff two operators(Hermitan) commute.
In the definition of commutation , we should not forget the world "FOR ALL ". [L^2, Lz] commute for all 'shi' but [Lx, Ly] do not commute for all, but there may still be some wave functions for which these operators commute.
I know this is not the answer you are looking for, but I thought I need to point this out so that I could clear up few things for myself.
Lx and Ly do not commute does not mean that they do not have any common eigenstate.They DO have common eigenstates, BUT it is just that those common eigenstates DO NOT span the whole Hilbert space. The common eigenstates of two operators span the whole Hilbert space iff two operators(Hermitan) commute.
In the definition of commutation , we should not forget the world "FOR ALL ". [L^2, Lz] commute for all 'shi' but [Lx, Ly] do not commute for all, but there may still be some wave functions for which these operators commute.
I know this is not the answer you are looking for, but I thought I need to point this out so that I could clear up few things for myself.
Re: angular momentum in QM
VT is right. Exercise: Find a simultaneous eigenstate for Lx and Ly.
Re: angular momentum in QM
I agree that having a nonzero commutation relation does not mean that there does not exist a single common eigenstate, but I don't see how you can find a common eigenstate in this case.
Because finding a common eigenstate betyween Lx and Ly is the same as finding eigenstates between Lz and Lx., except that the direction of m is changed to the xdirection instead of the z. Then  l m > are eigenstates of Lx.
In this case, Ly is then equal to (L+ + L )/2,
so if you apply that to an eigenstate of Lx,  l m>, you get (bla bla)  l (m+1)> + (bla bla)  l (m1) > (I don't remember the coefficients off hand.
So there are no elements that coincide.
I'll have to think about this further, because I'm wondering if you can create some sort of exception for m ~ +/ l values if L+ overflows m beyond l or L throws m under l, or otherwise something trivial like with m_z = 0 so that [Lx , Ly] l 0 > = ihLz l 0> = 0  l 0> gives you zero.
Because finding a common eigenstate betyween Lx and Ly is the same as finding eigenstates between Lz and Lx., except that the direction of m is changed to the xdirection instead of the z. Then  l m > are eigenstates of Lx.
In this case, Ly is then equal to (L+ + L )/2,
so if you apply that to an eigenstate of Lx,  l m>, you get (bla bla)  l (m+1)> + (bla bla)  l (m1) > (I don't remember the coefficients off hand.
So there are no elements that coincide.
I'll have to think about this further, because I'm wondering if you can create some sort of exception for m ~ +/ l values if L+ overflows m beyond l or L throws m under l, or otherwise something trivial like with m_z = 0 so that [Lx , Ly] l 0 > = ihLz l 0> = 0  l 0> gives you zero.
Re: angular momentum in QM
Few comments on blackcat007's post:
First, you cannot call a 'basis' to the common eigenfunctions of Lx and Ly. They do not form a 'basis' of the Hilbert space, cuz they don't commute.
Second, the concept of angular momentum is important in problems ( classical and quantum mechanics) with spherically symmetric potential (central force problem). This is because the angular momentum is the generator of rotation but the Hamiltonian is independent of rotation for central potential. Therefore, it follows from Noether's theorem that such a generator is a conserved quantity.
Now in QM, we can always find a ''complete set of operators'' ( which is NOT unique, btw) for a given Hilbert space. It turns out that H,L^2 and Lz form a complete set of operators of the Hilbert space of central force system. H alone is not complete due to the degeneracy in n. Similarly, L^2 and Lz are not complete either due to their n and l dependence respectively. This is the reason why we want common eigenfunctions of L^2 and Lz ( which will be the eigenfunctions for H too), and they will form a ''complete basis'' of the Hilbert space and you can do all sorts of things after this. ( I hope you are not planning to ask why we want complete set of operators. If in case you are: complete set of operators help us to find nondegerate, complete, commom basis and we can label each basis vector with eigenvalues just like in Hatom, Inlm> )
I do not have Griffiths' book with me and I have never read his book before( neither do I plan to read in the future), so I do not exactly understand your first question.
First, you cannot call a 'basis' to the common eigenfunctions of Lx and Ly. They do not form a 'basis' of the Hilbert space, cuz they don't commute.
Second, the concept of angular momentum is important in problems ( classical and quantum mechanics) with spherically symmetric potential (central force problem). This is because the angular momentum is the generator of rotation but the Hamiltonian is independent of rotation for central potential. Therefore, it follows from Noether's theorem that such a generator is a conserved quantity.
Now in QM, we can always find a ''complete set of operators'' ( which is NOT unique, btw) for a given Hilbert space. It turns out that H,L^2 and Lz form a complete set of operators of the Hilbert space of central force system. H alone is not complete due to the degeneracy in n. Similarly, L^2 and Lz are not complete either due to their n and l dependence respectively. This is the reason why we want common eigenfunctions of L^2 and Lz ( which will be the eigenfunctions for H too), and they will form a ''complete basis'' of the Hilbert space and you can do all sorts of things after this. ( I hope you are not planning to ask why we want complete set of operators. If in case you are: complete set of operators help us to find nondegerate, complete, commom basis and we can label each basis vector with eigenvalues just like in Hatom, Inlm> )
I do not have Griffiths' book with me and I have never read his book before( neither do I plan to read in the future), so I do not exactly understand your first question.

 Posts: 378
 Joined: Wed Mar 26, 2008 9:14 am
Re: angular momentum in QM
VT wrote:Few comments on blackcat007's post:
First, you cannot call a 'basis' to the common eigenfunctions of Lx and Ly. They do not form a 'basis' of the Hilbert space, cuz they don't commute.
Second, the concept of angular momentum is important in problems ( classical and quantum mechanics) with spherically symmetric potential (central force problem). This is because the angular momentum is the generator of rotation but the Hamiltonian is independent of rotation for central potential. Therefore, it follows from Noether's theorem that such a generator is a conserved quantity.
Now in QM, we can always find a ''complete set of operators'' ( which is NOT unique, btw) for a given Hilbert space. It turns out that H,L^2 and Lz form a complete set of operators of the Hilbert space of central force system. H alone is not complete due to the degeneracy in n. Similarly, L^2 and Lz are not complete either due to their n and l dependence respectively. This is the reason why we want common eigenfunctions of L^2 and Lz ( which will be the eigenfunctions for H too), and they will form a ''complete basis'' of the Hilbert space and you can do all sorts of things after this. ( I hope you are not planning to ask why we want complete set of operators. If in case you are: complete set of operators help us to find nondegerate, complete, commom basis and we can label each basis vector with eigenvalues just like in Hatom, Inlm> )
I do not have Griffiths' book with me and I have never read his book before( neither do I plan to read in the future), so I do not exactly understand your first question.
no no..i know that.. they won't form a basis.
i mentioned " and all other combinations" yes i admit it was a bit sloppy.. i meant all combinations including L^2 and since L^2 and L commute thus we can find a basis in which both of them can be diagonalized..
griffith introduces the concept of angular momentum.. and then goes on to find operators which commute, so that he can find a set of common eigenfunctions to construct a common basis.
my question is what is the significance of finding a common set of eigenfunction to form a basis.. why can't we proceed in a different way.. what induced (whoever did it for the first time) to proceed by searching a common set of eigenfunctions.
another question. since the common eigenfunctions of Lx and Ly won't be complete.. then can't they represent any physically realizable state? is it necessary for the eigenstate to span the L2 space?
Re: angular momentum in QM
Ok, a straight answer to your question is: for mathematical simplicity. I mean, if I give you a bunch of matrices, would not you "love" to know the basis on which AS MANY of these matrices are diagonalized? I would def. love to know that.
This way, I can easily know the eigen values of the operators which is what I am looking for.
For your second Q, ofcourse it is necessary for us to take the eigenstates that form a basis of the Hilbert space. This is the reason why we do not consider common eigestates of the operators that do not commute, because we cannot write any physical state as a linear combination of such incomplete eigenstates.
Two things that simplify mathematical calculation in QM are diagonal matrices and (complete) orthogonal eigenfunctions, so we have to hunt them down.
blackcat007, can you tell me how you would proceed the 'other way' without knowing the common (complete)eigenfunctions?
This way, I can easily know the eigen values of the operators which is what I am looking for.
For your second Q, ofcourse it is necessary for us to take the eigenstates that form a basis of the Hilbert space. This is the reason why we do not consider common eigestates of the operators that do not commute, because we cannot write any physical state as a linear combination of such incomplete eigenstates.
Two things that simplify mathematical calculation in QM are diagonal matrices and (complete) orthogonal eigenfunctions, so we have to hunt them down.
blackcat007, can you tell me how you would proceed the 'other way' without knowing the common (complete)eigenfunctions?
Re: angular momentum in QM
The entire hilbert space cannot be represented as a linear superposition of common eigenfunctions to Lx and Ly... But these functions are still in the hilbert space!blackcat007 wrote:another question. since the common eigenfunctions of Lx and Ly won't be complete.. then can't they represent any physically realizable state?

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Re: angular momentum in QM
will wrote: The entire hilbert space cannot be represented as a linear superposition of common eigenfunctions to Lx and Ly... But these functions are still in the hilbert space!
thats fine, but i was asking.. can those eigenfunctions represent physically the state of any particle. is it necessary for physically realizable eigenstates to be complete?

 Posts: 378
 Joined: Wed Mar 26, 2008 9:14 am
Re: angular momentum in QM
nice thanks!!!VT wrote:Ok, a straight answer to your question is: for mathematical simplicity. I mean, if I give you a bunch of matrices, would not you "love" to know the basis on which AS MANY of these matrices are diagonalized? I would def. love to know that.
This way, I can easily know the eigen values of the operators which is what I am looking for.
For your second Q, ofcourse it is necessary for us to take the eigenstates that form a basis of the Hilbert space. This is the reason why we do not consider common eigestates of the operators that do not commute, because we cannot write any physical state as a linear combination of such incomplete eigenstates.
Two things that simplify mathematical calculation in QM are diagonal matrices and (complete) orthogonal eigenfunctions, so we have to hunt them down.
blackcat007, can you tell me how you would proceed the 'other way' without knowing the common (complete)eigenfunctions?
but then the completeness of the eigenfunctions is not merely to simplify stuff, but is a necessary criterion for physical states.. right?Two things that simplify mathematical calculation in QM are diagonal matrices and (complete) orthogonal eigenfunctions, so we have to hunt them down.
Re: angular momentum in QM
No, completeness is NOT a necessary criterion for a physically realizable state. In a given Hilber state, any ray( a normalized vector (with a phase term in it) is called a ray) corresponds to a physical state. To be a physical state, it does not have to be complete( I don't even know what it means to be complete for a single state vector, doesn't make any sense), it does not have to be an eigenstate of ANY physical operators or imaginary operators, it does not have to be anything. As long as it is normalizable and meets the boundary conditions, it is a state vector of a state that can be prepared in a lab. Physics is done now, the rest is Linear algebra( or real analysis). If you want to expand this state vector(ray) on a basis, you do not always have to choose the common basis of complete operators. There are infinite ways of writing a state vector in a basis of a vector space ( but we normally choose the basis of Hamiltonian for variety of reasons. In addition, we would like to pick a common basis of many operators that commute with Hamiltonian. We can always find a bunch of other operators that commute with Hamiltonian if the Energy levels are degenerate.This is how we find complete set of operators whose common bases are no more degenerate). Choosing a common complete basis is, I think, for mathematical simplicity only( cuz on such a basis, many physically relevant operators are simultaneously diagonalized. Now if you want to call this a 'criterion', then you can do that, but I'dn't call it criterion)