Since the Hamiltonian for atoms shows rotational invariance, the angular momentum is a conserved
property. In addition the spin is directly related to angular momentum; it is a form of angular
momentum without a classical equivalent. Thus the quantum mechanical description of the angular
momentum will be discussed here as one of the first examples, although it is mathematically quite
involved.
Classically the angular momentum is defined as
$$\overrightarrow{L}=\overrightarrow{r}\times \overrightarrow{p}$$  (3.27) 
Following the correspondence principle we therefore get
$$\widehat{L}=\widehat{r}\times \widehat{p}$$  (3.28) 
for the quantum mechanical operator $\widehat{L}$
where $\widehat{r}$
and $\widehat{p}$ are
the operators for position and momentum.
One can easily show (see exercises) that the following relation holds
$$\left[{\widehat{L}}_{x},{\widehat{L}}_{y}\right]=\frac{\hslash}{i}{\widehat{L}}_{z}$$  (3.29) 
the same commutator relations hold for the cyclic permutation of
$x,y$, and
$z$.
We introduce now a general operator $\widehat{A}$
with this commutator relations
This property of a vectorial observable $\widehat{A}$ may be summarized as
$$\left[\widehat{A}\times \widehat{A}\right]=i\hslash \widehat{A}$$  (3.31) 
The 3 components ${\widehat{A}}_{x},{\widehat{A}}_{y}$, and ${\widehat{A}}_{z}$ are scalar observables (i.e., square matrices with Hermitian symmetry). Let’s introduce another scalar observable:
Unlike $\widehat{A}$,
${\widehat{A}}^{2}$ is just a square
matrix. It would be classically associated with the square of the length of a classical 3vector associated with
$\widehat{A}$ (if there’s one). We
will show now, that ${\widehat{A}}^{2}$
commutates with ${\widehat{A}}_{z}$:
Proof: Since the commutator $\left[{\widehat{A}}_{z}{\widehat{A}}_{z},{\widehat{A}}_{z}\right]$
is clearly zero, we have:
Each of those two terms can be evaluated using the above commutation relations:
Therefore, those two terms add up to zero and we obtain:
$\left[{\widehat{A}}^{2},{\widehat{A}}_{z}\right]=0$
The above definition of ${\widehat{A}}^{2}$
ensures that $<\psi \left{\widehat{A}}^{2}\right\psi >$ is
nonnegative for any ket $\psi >$
(HINT: this is the sum of 3 real squares).
Therefore, this operator can only have nonnegative Eigenvalues, which (for the sake of future
simplicity) we may as well put in the following form, for some non negative number
$l$.
$$l\left(l+1\right){\hslash}^{2}$$  (3.36) 
The punch line will be that $l$
is restricted to integer or halfinteger values. For now however, we may just accept
this expression because it spans all non negative values once and only once when
$l$ goes
from zero to infinity.
So, we may use $l$ as an index
to denote each Eigenvalue of ${\widehat{A}}^{2}$ .
Similarly, we may use another index $m$
to identify the Eigenvalue $m\hslash $
of ${\widehat{A}}_{z}$. For now, nothing
special is assumed about $m$
(we’ll show later that $2m$
is an integer).
Since those two observables commute, there is an orthonormal Hilbertian basis consisting entirely
of Eigenvectors common to both of them. We may specify it by introducing a third index
$n$
(needed to distinguish between kets having identical Eigenvalues for each of our two observables). Those
conventions are summarized by the following relations, which clarify the notation used for base kets:
To determine the restrictions that $l$ and $m$ must obey, we introduce the following two nonHermitian operators, which are conjugate of each other. They are collectively known as ladder operators; and are respectively called lowering operator (or anihilation operator) and raising operator (or creation operator) because it turns out that each transforms an Eigenvector into another Eigenvector corresponding to a lesser or greater Eigenvalue, respectively.
$${\widehat{A}}_{}={\widehat{A}}_{x}i{\widehat{A}}_{y}\phantom{\rule{2em}{0ex}}\text{and}\phantom{\rule{2em}{0ex}}{\widehat{A}}_{+}={\widehat{A}}_{x}+i{\widehat{A}}_{y}$$  (3.38) 
Both commute with ${\widehat{A}}^{2}$ (because ${\widehat{A}}_{x}$ and ${\widehat{A}}_{y}$ do). The following holds:
$$\left\right{\widehat{A}}_{+}n,l,m>{}^{2}=<n,l,m\left{\widehat{A}}_{}{\widehat{A}}_{+}\rightn,l,m>$$  (3.39) 
Where
So
$$\left\right{\widehat{A}}_{+}n,l,m>{}^{2}=\left[l\left(l+1\right){m}^{2}m\right]{\hslash}^{2}\phantom{\rule{2em}{0ex}}.$$  (3.41) 
As the non negative square bracket is equal to
$l\left(l+1\right)m\left(m+1\right)$ we see that
$m$ cannot exceed
$l$. We would find
that $\left(m\right)$ cannot exceed
$l$ by performing the
same computation for $\left\right{\widehat{A}}_{}n,l,m>{}^{2}$.
Therefore, all told:
$$l\le m\le l\phantom{\rule{2em}{0ex}}.$$  (3.42) 
Note that the above also proves that the ket
${\widehat{A}}_{+}n,l,m>$ vanishes
only when $m=l$.
Likewise, ${\widehat{A}}_{}n,l,m>$ is
nonzero unless $m=l$.
Except in the aforementioned cases where they vanish, such kets are Eigenvectors of
${\widehat{A}}_{z}$ associated with the
Eigenvalue of index $m\pm 1$.
Let’s prove that:
Therefore,
$${\widehat{A}}_{z}{\widehat{A}}_{+}={\widehat{A}}_{+}{\widehat{A}}_{z}+\hslash {\widehat{A}}_{+}\phantom{\rule{2em}{0ex}}.$$  (3.44) 
So, if $\psi >$ is an Eigenvector of ${\widehat{A}}_{z}$ associated with the value $m\hslash $, then:
$${\widehat{A}}_{z}{\widehat{A}}_{+}\psi >=\left(m+1\right)\hslash {\widehat{A}}_{+}\psi >\phantom{\rule{2em}{0ex}}.$$  (3.45) 
Thus, the ket ${\widehat{A}}_{+}\psi >$ is either
zero or an Eigenvector of ${\widehat{A}}_{z}$
associated with the value $\left(m+1\right)\hslash $.
The same is true of ${\widehat{A}}_{}\psi >$
with $\left(m1\right)\hslash $.
Since we know that $m$
is between $l$
and $+l$ , we see
that both $lm$ and
$l+m$ must be
integers (or else iterating one of the two constructions above would yield a nonzero Eigenvector with a value of
$m$ outside of the allowed
range). Thus, $2l$
and $2m$
must be integers (they are the sum and the difference of the integers
$l+m$ and
$lm$). If
$l$ is an integer,
so is $m$. If
$l$ is an halfinteger,
so is $m$
(by definition, an ”halfinteger” is half the value of an odd integer).
The above demonstration is quite remarkable: It shows how a 3component observable is quantized
whenever it obeys the same commutation relation as an orbital angular momentum. Although
halfinteger values of the numbers l and m are allowed, those do not correspond to an orbital momentum
but to a quantum mechanical spin. Only orbital momenta can lead to whole numbers of
$l$ and
$m$ (which
we will not proof here).

$${Y}_{lm}\left(\Theta ,\Phi \right):=\frac{1}{\sqrt{2\pi}}{N}_{lm}{P}_{lm}\left(\mathrm{cos}\Theta \right){e}^{im\Phi}\phantom{\rule{2em}{0ex}}.$$  (3.46) 
Here the adjoined Legendre polynomials are defined as
$${P}_{lm}\left(x\right):=\frac{{\left(1\right)}^{m}}{{2}^{l}l!}{\left(1{x}^{2}\right)}^{\frac{m}{2}}\frac{{d}^{l+m}}{d{x}^{l+m}}{\left({x}^{2}1\right)}^{l}\phantom{\rule{2em}{0ex}},$$  (3.47) 
and the scaling factor
$${N}_{lm}\left(x\right):=\sqrt{\frac{2l+1}{2}\frac{\left(lm\right)!}{\left(l+m\right)!}}$$  (3.48) 
The first spherical harmonic functions are
${Y}_{lm}$  $l=0$  $l=1$  $l=2$  $l=3$ 
$m=3$  $+\sqrt{\frac{35}{64\pi}}{\mathrm{sin}}^{3}\Theta {e}^{3i\Phi}$  
$m=2$  $+\sqrt{\frac{15}{32\pi}}{\mathrm{sin}}^{2}\Theta {e}^{2i\Phi}$  $+\sqrt{\frac{105}{32\pi}}{\mathrm{sin}}^{2}\Theta \mathrm{cos}\Theta {e}^{2i\Phi}$  
$m=1$  $+\sqrt{\frac{3}{8\pi}}\mathrm{sin}\Theta {e}^{i\Phi}$  $+\sqrt{\frac{15}{8\pi}}\mathrm{sin}\Theta \mathrm{cos}\Theta {e}^{i\Phi}$  $+\sqrt{\frac{21}{64\pi}}\mathrm{sin}\Theta \left(5\mathrm{cos}2\Theta 1\right){e}^{i\Phi}$  
$m=0$  $+\sqrt{\frac{1}{4\pi}}$  $+\sqrt{\frac{3}{4\pi}}\mathrm{cos}\Theta $  $+\sqrt{\frac{5}{16\pi}}\left(3{\mathrm{cos}}^{2}\Theta 1\right)$  $+\sqrt{\frac{7}{16\pi}}\left(5{\mathrm{cos}}^{3}\Theta 3\mathrm{cos}\Theta \right)$ 
$m=1$  $\sqrt{\frac{3}{8\pi}}\mathrm{sin}\Theta {e}^{i\Phi}$  $\sqrt{\frac{15}{8\pi}}\mathrm{sin}\Theta \mathrm{cos}\Theta {e}^{i\Phi}$  $\sqrt{\frac{21}{64\pi}}\mathrm{sin}\Theta \left(5\mathrm{cos}2\Theta 1\right){e}^{i\Phi}$  
$m=2$  $+\sqrt{\frac{15}{32\pi}}{\mathrm{sin}}^{2}\Theta {e}^{2i\Phi}$  $+\sqrt{\frac{105}{32\pi}}{\mathrm{sin}}^{2}\Theta \mathrm{cos}\Theta {e}^{2i\Phi}$  
$m=3$  $\sqrt{\frac{35}{64\pi}}{\mathrm{sin}}^{3}\Theta {e}^{3i\Phi}$  
© J. Carstensen (Quantum Mech.)