Lz: A Convenient but Arbitrary Choice in Quantum Mechanics

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SUMMARY

The discussion centers on the role of the angular momentum operator Lz in quantum mechanics, particularly in the context of the hydrogen atom. It establishes that while Lz is a convenient choice for measuring angular momentum about the z-axis, it is fundamentally arbitrary, as one could just as easily choose Lx or Ly. The operators Lx, Ly, and Lz do not commute with each other but do commute with the total angular momentum operator L^2, allowing for a complete basis of eigenvectors. The measurement of Lz results in the electron being in an eigenstate of Lz, which limits knowledge about Lx and Ly.

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mattlorig
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In the hydrogen atom, I believe most people are familiar with the following three equations:
L^2 (psi) = l*(l+1)*hbar^2*(psi)
Lz (psi) = ml*hbar*(psi)
H (psi) = -1/n^2*junk*(psi)
where L^2, Lz, and H are the linear operators for total angular momentum squared, angular momentum about the z-axis, and energy. I'm comfortable with eigenvalues, eigenvectors, etc. The thing I don't understand, however, is what Lz really is. Since our choice of axes is completely arbitrary, I could have just as easily chosen Lx to be Lz. But of course, if I know Lz, I can not possibly know anything about Lx (other than, perhaps, its maximum value).

I gues what I'm asking is the following: is Lz just a way to say, that if we were to measure the angular momentum of an electron about a certain axis (which we'll call z) it can only have values of ml*hbar, and once we know what Lz is, we can't possibly know anything about Lx and Ly?

I'd really appreciate it if somebody could straighten this out for me.
 
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Yes, our choice of axes is of course completely arbitrary.

L_x,L_y and L_z do not commute, but they DO commute with L^2. So we can find a complete basis of eigenvectors common to L^2 and an axis. So we have to make a choice, L_z is used for convenience.

If you measured L_z, then the electron will be in an eigenstate of L_z. Now try to see what the probabilities are of getting m_l\hbar when measuring L_x.
(Use L_x=\frac{1}{2}(L_++L_-))
 
Galileo said:
Yes, our choice of axes is of course completely arbitrary.
L_x,L_y and L_z do not commute, but they DO commute with L^2. So we can find a complete basis of eigenvectors common to L^2 and an axis. So we have to make a choice, L_z is used for convenience.
If you measured L_z, then the electron will be in an eigenstate of L_z. Now try to see what the probabilities are of getting m_l\hbar when measuring L_x.
(Use L_x=\frac{1}{2}(L_++L_-))


Sorry,Galileo,but i just couldn't help myself. :-p
So,this convention is one of the many more encountered in physics.Think about the old famous conventions regarding the magnetic field (induction) \vec{B}.Both in electrodynamics (charged particle in magnetic/electromagnetic field) and in QM (Zeeman effect (normal/anomal)) it's always chosen along "Oz(=Ox_{3})" axis.I don't know why,i never met the guys who did that. :wink: You'll have to accept it,the same way you accepted those wicked conventions in geometrical optics,that convention for the sign of work in thermodynamics and many more.

A physicist's mind is twisted in uncountable ways... :cool:

Daniel.
 
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