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To show that the eigenvalue to $L^2$ is proportional to $\hbar^2$ is shown from

$L_z=xP_y-yP_x$

$p_y=-i\hbar\frac{\partial}{\partial y}$

$p_x=-i\hbar\frac{\partial}{\partial x}$

$L_z=-i\hbar(x\frac{\partial}{\partial y}-y\frac{\partial}{\partial x})$

$L_z=\hbar^2(x\frac{\partial}{\partial y}-y\frac{\partial}{\partial x})^2$

In the same way can $L_x^2$ and $L_y^2$ be obtained and since

$L^2=L_x^2+L_y^2+L_z^2$ (I) we see that $L^2$ is proportional to $\hbar^2$

So we can establish this eigenvalue equation

$L^2|\alpha,\beta>=\hbar^2\alpha|\alpha, \beta>$

From solving this it is obtained that:

$\alpha=l(l+1)$

My problem is that in the derivations that I have found they use the defintion of Raisng and lowering operators:

$L_+=Lx+iLy$ Raising

$L_-=Lx-iLy$ Lowering

They increase and decrease respectively $\alpha$ with 1 integer value. But why can we know that the square of the angular momentum eigenvalue can be increased or decreased in inger values? The rest of the derivation makes sense to me. But I need one of two explanations:

1: Prior to the ladder operators are there a way to show that $\alpha$ can only have integer values?

or 2: A way to derive the ladder operators and why they are how they are and deriving why they work on $\alpha$ the undefined part of the the square of the angualar momentum eigenvalue to give solutions to $\alpha$

Here is an example of a derivation of the type I am refering to

http://en.wikipedia.org/wiki/Angular_momentum_operator#Derivation_using_ladder_operators

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    $\begingroup$ Start from the commutation relations $[L_i,L_j]=i\epsilon_{ijk}L_k$ and calculate $L_zL_{\pm}\vert \alpha,\beta\rangle$ $\endgroup$ – Nephente Feb 27 '15 at 13:48
  • $\begingroup$ I am just wondering why you can use the lowering or raising ladder this way? Thanks! $\endgroup$ – fisher garry Feb 27 '15 at 13:52
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$\newcommand{\ket}[1]{\lvert #1 \rangle}$There is no "reason" you can use the ladder operators. Rather, they are the reason angular momentum is quantized in integer steps.

You can define them, there's no inconsistency, so you can use them, and using them leads you to conclude that the angular momentum is raised/lowered in integer steps by them, in the way nephente indicates in a comment: Simply start from the $\mathfrak{so}(3)$ commutation relations

$$ [L_i,L_j] = \mathrm{i}e_{ijk}L_k$$

and an eigenstate $L_z \ket{j} = j\ket{j}$ of $L_z$ and calculate what $L_z L_\pm \ket{j}$ gives you.

There is no reason to be suspicious of the operators. You have to just do the math.

The ladder operators are the most elegant way to do representation theory of $\mathrm{SO}(3)$, or rather, $\mathrm{SU}(2)$, which is essentially what we are doing here.

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  • $\begingroup$ L can also be written: $L=\frac{nh}{2\pi}$. Where n are the energy levels. Can one not use this defintion of L to solve the equation for L above? If not why not? $\endgroup$ – fisher garry Feb 28 '15 at 22:07

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