Angular momentum operators and non-zero values

[physica]

Hey I have been absent because I am coming to the end of my final year in physics. I recently got an 89% in my last quantum mechanics. I’m hoping to get a 1^st so I can do a masters in physics at UCL. However, I do not understand a certain concept. Why do angular momentum operators (including the Hamiltonian) always leave non-zero answers? Looking at hydrogen (simplest for this sort of thing) is there a certain characteristic that these operators have that result in a definite non-zero state? Does this change with state? So do hydrogen atoms have definite non-zero values no matter what state they are? I’ve tried to read round the subject but I never get a straight answer. I’ve read a few pieces stating that it’s due to the fact that the operators do not commute. Is this right or am I being led down the garden path? If it is right then why do non-commuting operators give a non-zero value? Sorry for all the questions but I am really in a muddle. Many thanks for your time.

[studiot]

Well the angular momentum number, L is related to the orbital momentum number l by the relation

 

edit formula corrected

 

[math]L = \hbar \sqrt {l\left( {l + 1} \right)} [/math]

Since l takes the vlues 0, 1,2, 3..... (n-1) where n is the principal quantum no

 

the smallest value that L can have is with l = 0 and is non zero.

Edited April 2, 2015 by studiot

[swansont]

The particles in a hydrogen atom are spin 1/2. Individually, they are never going to leave you with zero angular momentum.

 

I'm wondering if some of this is a specific case that's not generally true, or meant to be generally true. Your wording is a bit vague.

Well the angular momentum number, L is related to the orbital momentum number l by the relation

 

[math]L = \sqrt {\hbar l\left( {l + 1} \right)} [/math]

Since l takes the vlues 0, 1,2, 3..... (n-1) where n is the principal quantum no

 

the smallest value that L can have is with l = 0 and is non zero.

 

l(l+1) is zero for l=0 (and hbar goes outside of the root)

[studiot]

 

swansont

(A) l(l+1) is zero for l=0 (B) (and hbar goes outside of the root)

 

 

(A) Yes I came back because I realised that after I left, but you beat me to it. Thanks.

(B) and I would like to say I hate LaTex.

[swansont]

 

(B) and I would like to say I hate LaTex.

 

I figured it was a LaTex error and not a conceptual one.

[studiot]

There that should be better, but I hate splling as well

 

[ajb]

Why do angular momentum operators (including the Hamiltonian) always leave non-zero answers?

You mean eigenvalues.

 

You can for sure have zero as eigenvalues of L^2 as pointed out above. Zero eigenvalues of a Hamiltonian are particularly interesting, especially in 1 dimension where it is related to supersymmetry and shape invariant potentials etc. Of course zero energy is not an eigenvalue for the Hamiltonian of the hydrogen atom.

 

 

An important thing that I expect you know, is that you can only 'simultaneously diagonalise' operators if they commute. For example, you can look for eigenvectors of L^2 and Lz because they commute. The same for the energy of hydrogen-like atoms, H commutes with L^2 and Lz. This is the role of commutativity in building the Hilbert space.

 

Operators that do not commute will have a generalised uncertainty principle, this could be what is confusing you. If the operators do not commute then we cannot look for simultaneous eigenvector of both operators. In the case of the hydrogen-like atoms I cannot look for eigenvectors of Lx, Lx and Ly. I need to pick one and then look for eigenvectors of that and L^2.