Question

In representing the eigenstates of angular momentum, eigenfunctions can be found using the ladder operators listed below:
L+ = Lx + iLy
L- = Lx - iLy
In all the sources that I read, these ladder operators seem to be defined. Is there a reason to define the ladder operators this way? In other words, where do these two ladder operators come from?

My guess is that since L^2 and Lz are hermitian, the difference L^2 - Lz^2 must be real. Since L^2 is defined as:
L^2 = Lx^2 + Ly^2 + Lz^2
we can subtract Lz^2 from both sides to get
L^2 - Lz^2 = Lx^2 + Ly^2
which has to equal a real value.

Answer 1

And L+ and L- are the attempted factorizations of Lx^2 and Ly^2 in the above equation.

Answer 2

L+ and L- are not terms of factoring Lx^2 + Ly^2. Lx and Ly are operators, and do not commute to allow this to be so.
\[(L_x + i L_y)(L_x - iL_y) = L_x^2 + L_y^2 +iL_yL_x - iL_xL_y\]
but
\[[L_x,L_y] = i \hbar L_z\]
\[(L_x + i L_y)(L_x - iL_y) = L_x^2 + L_y^2 - i[L_x,L_y] = L_x^2+L_y^2+\hbar L_z\]

As for why they are of this form, it's so they show the required behavior of ladder operators.

If [O,L] = cL, where c is a scalar, allow |n> to be an eigenvector of O with eigenvalue n:
\[O|n> = n|n>\]Then
\[OL|n> = (OL+LO-LO)|n> = (LO + [O,L])|n> = (n+c)L|n>\]
If c is real and positive, L is a raising operator. If c is real and negative, we call it a lowering operator.

Often, it's convenient to also require that
\[L_+^\dagger = L_-\]

Answer 3

Small correction:
\[L_+^\dagger = L_-\]
Is a consequence of the operator being hermitian, along with flipping the sign on the scalar, c.

Answer 4

Such operators are introduced, since they are very useful. I can show the physical meaning of those operators:
let's consider an eigenstate of L^2 and L_z:
\[\begin{gathered}
{L^2}\left| {a,b} \right\rangle = a\left| {a,b} \right\rangle \hfill \\
{L_z}\left| {a,b} \right\rangle = b\left| {a,b} \right\rangle \hfill \\
\end{gathered} \]

Answer 5

then we can write:
\[\Large \begin{gathered}
{L_z}{L_ + }\left| {a,b} \right\rangle = \left( {\left[ {{L_z},{L_ + }} \right] + {L_ + }{L_z}} \right)\left| {a,b} \right\rangle = \hfill \\
\hfill \\
= \left( {b + \hbar } \right)\left( {{L_z}\left| {a,b} \right\rangle } \right) \hfill \\
\end{gathered} \]
since:
\[\Large \left[ {{L_z},{L_ + }} \right] = \hbar {L_z}\].
Similarly for operator:
\[\Large {L_ - }\]

Answer 6

oops.. I have made an error, here are the right formulas:
\[\Large \begin{gathered}
{L_z}{L_ + }\left| {a,b} \right\rangle = \left( {\left[ {{L_z},{L_ + }} \right] + {L_ + }{L_z}} \right)\left| {a,b} \right\rangle = \hfill \\
= \left( {b + h} \right){L_ + }\left| {a,b} \right\rangle \hfill \\
\end{gathered} \]
where:
\[\Large \begin{gathered}
\left[ {{L_z},{L_ + }} \right] = {L_z}{L_ + } - {L_ + }{L_z}, \hfill \\
\hfill \\
\left[ {{L_z},{L_ + }} \right] = \hbar {L_ + } \hfill \\
\end{gathered} \]
similarly goes, for the operator \[\Large {L_ - }\]