# How is angular momentum quantized?

Mar 10, 2018

L^2= (l(l+1) h^2 and $L z = m h$
where $m = - l , - l + 1 , . - 1. .0 \ldots .1 , 2. . . l$ and h is h/2pi

#### Explanation:

L^2 and Lz are here representing the operators representing square of the Angular momentum and Z- component of L.

These operators are defined after separation of radial and angular part of the Schrodinger equation and its eigenfunctions are the spherical harmonics.

For a quantum system the Angular momentum space is quantized means the ang. momentum can not take continuous values .

If l is the angular momentum of a state of particle, in that case the L^2 and Lz operators simultaneously commute with

Hamiltonian (the total energy operator) of the system which have energy as their eigenvalues .

**(L^2 )Y(l,m) = (l(l+1) (h/(2pi)^2**

and
Lz . Y(l,m) = m,(h/2pi)

and only eigen values of L^2 and Lz are well defined or can be measured simultaneously.

They commute with each other and H-the Hamiltonian.

Suppose a particle is in p-state ; so its l=1

then the possible eigen value of L^2 will be

$1. \left(1 + 1\right) . {h}^{2}$ = $2. {h}^{2}$

(here we are writing h for h/2.Pi)

And the z-component can have only three possible Values
 m= -1 , 0, +1

So , $L z = - \frac{h}{2 \pi}$ , 0 , $+ \frac{h}{2 \pi}$

If plotted on a Vector model The Lz will appear as intercept of

projection of L on z-Axis with length of L vector as sqrt(2.(h/(2pi))#.
For l=0 i.e. a S-state L=0(0+1)hbar=0

For higher angular momentum states the Lz values will be more spread

for l=2 . Lz will have five allowed states..and so on..