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Azimuthal quantum number $\ell$, and magnetic quantum number $m$ are defined when we do derivation for $L^2f=\ell(\ell+1){\hbar}^2f$ and $L_zf=\ell{\hbar}f$.

This is my own conclusion after studying introduction to quantum mechanics by Griffiths. Correct?

But I don't know why does $L_zf=\ell{\hbar}f$ become $L_zf=m{\hbar}f$? I wonder why when we operator $L_+$ on an eigenstate, the only change is that $m$ increases by one not for $\ell$, e.g.$$L_+ Y_{2}^{1}= AY_{2}^{2},$$ why not $\ell$ change?

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This is not generally(as in, it can be but is not always) true: $$L_zf=l{\hbar}f.$$

What is true is: $$L_z f = m\hbar f $$

But $m$ can take values between $l$ and $-l$ in integer steps. Look through the derivation again now that you know this and see if that helps.

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Your second equation is not always true, as Bobby said. What is true is this:

$L^2$ is an operator which when acting on an eigenstate $|\ell,m\rangle$ gives $\ell(\ell+1)\hbar^2|\ell,m\rangle$, while $L_z$ is an operator which when acting on an eigenstate $|\ell,m\rangle$ gives $m\hbar|\ell,m\rangle$. Here the familiar quantum numbers are really just eigenvalues of the operators (important!), and what you're saying is the eigenvalue corresponds to the physics you're interested in.

The answer to your second question is provided by the definition of the angular momentum ladder operators: $L_{\pm} := L_x\pm iL_y$. The only way to evaluate these is to make it into a combination of $L_z$ operators which imply the final state is raised/lowered in the spin number.

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  • $\begingroup$ I'm a little rusty on the exact calculation, but it should involve using commutators and seeing if you can subtract out the $L_x$ and $L_y$ parts. A bit of a long but important exercise. If you want you can also think that the angular momentum state fixes the angular momentum number--you'd need some sort of interaction to take place to change that (conservation of angular momentum). $m$ on the other hand is not well defined from angular momentum alone, so this can change without an outside interaction. $\endgroup$ – ALB Nov 28 '13 at 21:27
  • $\begingroup$ Thanks.I thing i have one more problem. [$L_z$,$L_+$] =${\hbar} L_+$, let $L_zf$=m ${\hbar}f$ Where f is a eigenstate. $$L_z(L_+f)= (L_zL_+-L_+L_z+L_+L_z)f$$ $$L_z(L_+f)={\hbar} L_+f+ (L_+L_z)f$$ $$L_z(L_+f)={\hbar} (m+1)(L_+f)$$ Then my question is : how should I write next? $$L_z(L_+f)=L_+{\hbar} ((m+1)f)$$ Because ((m+1)f) tell us that the m of eigenstate increase by one like $$L_+ Y_{2}^{1}= AY_{2}^{2},$$how to write in this form? $\endgroup$ – Outrageous Nov 29 '13 at 3:13
  • $\begingroup$ These are typically easier to do the other way: evaluate $[L_z,L_{+}]f = L_zL_{+}f-L_{+}L_zf$. Note that the first term has an $m+1$ coefficient due to the $L_z$ action on a raised state and the second only an $m$, so these cancel to give the desired result, $\hbar L_{+}$. Also, not to be petty, but if you like my answer you are happily encouraged to 'accept' it ;) $\endgroup$ – ALB Nov 29 '13 at 10:58

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