Paschen back effect and commutator [J^2,Lz]

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Homework Statement


I have been given a question on how the commutator relates to the paschen back effect the exact question is as follows

Calculate the commutator ##[J^2,L_z]## where ##\vec{J}=\vec{L}+\vec{S}## and explain the relevance of this with respect to the paschen back effect

Homework Equations

The Attempt at a Solution


I normally it forum rules to preferable use latex, but it would take me sometime to write all my working out so I have photo in my working for the ##[J^2,L_z]## part
Handwritten derivation on lined paper titled Step 1, showing [L^2, Lz] = 0 using commutator expansions of Lx, Ly, Lz with highlighted terms and a final line [L^2, Lz] = 0 shown.
Handwritten derivation on lined paper titled Step 2, showing that the commutator of S-squared with Lz equals zero, using commutator identities and symmetry.
Handwritten derivation on lined paper of the commutator [L.S, Lz] using operator components, concluding that it is nonzero and the quantities cannot be measured simultaneously.


So my final result for the commutator was ##[J^2,L_z]=2i\bar{h}[\hat{L_x}\hat{S_y}-\hat{L_y}\hat{S_x}]##

So how this relates to paschen back I am not sure exactly I have been looking at, how spin orbit coupling works in weak and strong mag fields and my working theory is as follows.

In a weak mag field the spin/orbit couple form the constant total angular momentum vector or total angular quantum number honestly I am always confused why vectors suddenly become operators I know how to work them out but everything iv watch on you-tube or even read never really explains it in full, maybe someone could expand on this?

Anyway, so whilst in a weak mag field we would see that the commutator ##[J^2,L_z]## would not be commutable as J is partly formed by L so there would be no way to know simultaneously the measurement of ##J## or a component of ##L## in this case ##L_x##.

With respect to the panache effect the magnetic field is strong enough to decouple the spin orbit and so the two act independently precessing around the magnetic field in the direction in which the magnetic field point in, then as there has been a decoupling then there is no total angular moment and the square magnitudes of the angular momentum and intrinsic spin angular momentum become commutable with ##L_z##.

In honestly I am still unsure of my explanation and have only come up with by looking at a diagram of spin/orbit coupling in a magnetic field. Would it be possible if someone maybe could explain, if I am at least on the correct lines or well off the mark.

Any advice would be much appreciated.
 
Attachments
  • Handwritten derivation on lined paper titled Step 1, showing [L^2, Lz] = 0 using commutator expansions of Lx, Ly, Lz with highlighted terms and a final line [L^2, Lz] = 0 shown.
    P1.jpg
    39.1 KB · Views: 2,567
  • Handwritten derivation on lined paper titled Step 2, showing that the commutator of S-squared with Lz equals zero, using commutator identities and symmetry.
    P2.jpg
    37.5 KB · Views: 1,376
  • Handwritten derivation on lined paper of the commutator [L.S, Lz] using operator components, concluding that it is nonzero and the quantities cannot be measured simultaneously.
    P3.jpg
    35.7 KB · Views: 1,845
I don't know what the precise answer to your question is, but try reading this: http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/paschen.html It seems to be a pretty good write-up. ## \\ ## Additional item: The energy eigenstates in the weak magnetic field are eigenstates of the ## J_z ## operator, at least to first order, if I understand it correctly. Meanwhile, with a strong magnetic field, the eigenstates switch to being eigenstates of ## L_z ## and ## S_z ##.
 
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