Commutator of ##L^2## with ##L_x,L_y,L_z##..

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Summary:

##[L^2,L_i]=0##.....where ##i=x,y,z##

Main Question or Discussion Point

For a given state say ##{l,m_l}## where ##l## is the orbital angular momentum quantum no. and ##m_l## be it's ##z## component.....a given state ##|l,m_l> ## is an eigenstate of ##L^2## but not an eigenstate of ##L_x##....therefore all eigenstates of ##L_x## are eigenstates of ##L^2## but the reverse is not true... isn't it??
 

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  • #2
DrClaude
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therefore all eigenstates of ##L_x## are eigenstates of ##L^2## but the reverse is not true... isn't it??
Correct. When you have a degeneracy, an eigenstate corresponding to the degenerate eigenvalue is not necessarily an eigenstate of another operator that commutes. However, it is always possible to find a linear combination of these degenerate eigenstates that will be eigenstates of both operators.
 
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  • #3
strangerep
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@DrClaude : Sorry -- I must be missing something...

If we are given only that ##L_x \Psi = m \Psi##, how does that imply ##\Psi## is necessarily an eigenstate if ##L^2## ?

Maybe I'm overlooking a tacit assumption?
 
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  • #4
anuttarasammyak
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OP uses ##m_l## to mean as common eigenstate of Lz, usually z is taken, and L^2.

Otherwise, eigenstates of m in general includes states of l=0,1,2,3,... and eigenstates of l in general includes the states of m=−l,−l+1,0,1,2,3,...,l
One can play a role of sub-index of the other.

For an example, s electron and p_z electron are both eigenstates of m=0. We can distinguish them by value of l as sub-index because of the commutation of L^2 and Lz.
 
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  • #5
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@DrClaude : Sorry -- I must be missing something...

If we are given only that ##L_x \Psi = m \Psi##, how does that imply ##\Psi## is necessarily an eigenstate if ##L^2## ?

Maybe I'm overlooking a tacit assumption?
For a given ##l## the eigenstates of ##L_x## are some linear combinations of ##|l,m_l>## which are also the eigenstates of ##L^2##...
 
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  • #6
DrClaude
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@DrClaude : Sorry -- I must be missing something...

If we are given only that ##L_x \Psi = m \Psi##, how does that imply ##\Psi## is necessarily an eigenstate if ##L^2## ?

Maybe I'm overlooking a tacit assumption?
OP was looking at states that are eigenstates of ##L^2##.

Of course, this works both ways. eigenstates of ##L_x## are not necessarily eigenstates of ##L^2##, but there exists a linear combinations of the degenerate ##L_x## states that form a common basis for ##L_x## and ##L^2##.
 
  • #7
PeroK
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@DrClaude : Sorry -- I must be missing something...

If we are given only that ##L_x \Psi = m \Psi##, how does that imply ##\Psi## is necessarily an eigenstate if ##L^2## ?

Maybe I'm overlooking a tacit assumption?
Starting from ##L_+ = L_x +iL_y## and ##L_- = L_x - iL_y##, we get:
$$L^2 = L_x^2 + L_y^2 + L_z^2 = L_+L_- -\hbar L_z + L_z^2$$
And any eigenstate of ##L_z## is an eigenstate of ##L^2##. And, by symmetry, the same is true for ##L_x## and ##L_y##.
 
  • #8
vanhees71
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No! There are eigenstates of ##L_z## that are not eigenstates of ##L^2##, e.g., ##|l_1,m \rangle+|l_2,m \rangle## with ##l_1 \neq l_2##.
 
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  • #9
PeroK
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No! There are eigenstates of ##L_z## that are not eigenstates of ##L^2##, e.g., ##|l_1,m \rangle+|l_2,m \rangle## with ##l_1 \neq l_2##.
Yes, of course. I didn't think of that.
 
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  • #10
anuttarasammyak
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And any eigenstate of LzLzL_z is an eigenstate of L2L2L^2. And, by symmetry, the same is true for LxLxL_x and LyLyL_y.
More precise saying is
we can find common set of eigenvectors for the two operators, i.e.,
[tex]L^2|l,m>=l^2|l,m>[/tex]
[tex]L_z|l,m>=m|l,m>[/tex]

May I remind that
[tex]\sum_m c_m|l,m>[/tex] is an eigenvector of L^2 with eigenvalue l^2, but not an eigenvector of L_z anymore ?
 
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  • #11
A. Neumaier
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how does that imply \Psi is necessarily an eigenstate of L^2 ?
L^2 is a central element (Casimir) of the universal enveloping algebra, hence in an irreducible unitary representation, it has every vector as eigenvector.
 
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  • #12
strangerep
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L^2 is a central element (Casimir) of the universal enveloping algebra, hence in an irreducible unitary representation, it has every vector as eigenvector.
In a specific unirep, yes, but not in general, as Hendrik pointed out in his post #8.
 
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  • #13
A. Neumaier
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In a specific unirep, yes, but not in general, as Hendrik pointed out in his post #8.
In every irreducible unitary representation. The rep specified in post #1 is of this form, while that in #8 is not.
 
  • #14
vanhees71
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We were talking about all eigenvectors of ##L^2## and ##L_z##, and there an eigenvector of ##L_z## is not necessarily also an eigenvector of ##L^2##.

Of course if you restrict yourself to a subspace, which builds an irreducible representation of the rotation group, than it's an eigenspace of ##L^2## to only one eigenvalue ##\hbar^2 l(l+1)## and thus in this subspace all vectors are eigenvectors of ##L^2## with this eigenvalue.
 
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  • #15
strangerep
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In every irreducible unitary representation. The rep specified in post #1 is of this form, while that in #8 is not.
Yes, of course. But I was trying to clarify Dr Claude's post#2, which omits some of the OP's context. (Anyway, I think this thread has run its course, and then some. <Exit>)
 
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