This document is an exercise that shows can it can be checked that the squared angular momentum operator L^2, which is equal to l(l+1) can be obtained by summing the average value of Lx^2+Ly^2+Lz^2 which initially seems counter-intuitive. It is evaluated on the Hydrogen wavefunctions for the n=2 and l=1 states, the exercise is extended to the case where spin is included.

1. Exercise: Verify that the 𝑳𝟐
= 𝒍(𝒍 + 𝟏) operator leads to a result
consistent with the probabili es for the 𝑳𝒙, 𝑳𝒚 and 𝑳𝒛 operators
in the Hydrogen atom (2,1,m) wave func ons
Sergio Prats López, December 2023
SergioPL81@gmail.com
Introduc on
It is a well-known fact that the squared angular momentum, 𝐿 = 𝐿 + 𝐿 + 𝐿 is equal to
𝑙(𝑙 + 1), where 𝑙 is the eigenvalue for the angular momentum operator 𝑳 = [𝐿 , 𝐿 , 𝐿 ].
The demonstra on for this can be found in public sites such as Wikipedia, the demonstra on is
based in the use of the ladder operators, 𝐿 and 𝐿 , and the assump on that the values for a
single dimension such as 𝑙 must never exceed 𝑙 in absolute value: |𝑙 | ≤ 𝑙. This means that
𝐿 |𝜓 > and 𝐿 |𝜓 > lead to the no-func on and therefore are equal to zero.
Although the former demonstra on is clear, it is counter-intui ve because one may state that
for a given wavefunc on whose angular momentum is in one direc on such as Z, the other
direc ons should have no angular momentum at all and therefore 𝐿 should be 𝐿 = 𝑙 = 𝑙
and not 𝑙(𝑙 + 1).
The reason of this discrepancy lies in the fact that operators 𝐿 , 𝐿 and 𝐿 do not commute
between them, therefore they do not share the same eigenvectors. Because of this, we can
have a wavefunc on equal to an eigenstate in the Z direc on with 𝑙 = 𝑙 but that does not
mean that the 𝐿 and 𝐿 eigenstates will be necessarily 𝑙 = 𝑙 = 0 since we may have some
probability of having momentum in other direc ons than the one which is well defined, this is
roughly the reason why the 𝐿 operator has a value greater than 𝑙 .
Evaluate it on the n=2, l=1 Hydrogen wavefunc ons
I want to verify on the Hydrogen atom wavefunc ons for 𝑛 = 2 and 𝑙 = 1 that 𝐿 = 2, as
expected for the 𝐿 = 𝑙(𝑙 + 1) formula, let it be the wavefunc ons:
𝜓 =
1
√2
∗
√6
2
cos(𝜃) ∗
1
2√6𝑎
𝑟
𝑎
𝑒 /
𝜓 =
1
√2
𝑒 ∗
√6
2
cos(𝜃) ∗
1
2√6𝑎
𝑟
𝑎
𝑒 /

2. 𝜓 =
1
√2
𝑒 ∗
√6
2
cos(𝜃) ∗
1
2√6𝑎
𝑟
𝑎
𝑒 /
The previous wavefunc ons have been wri en in the form 𝜓 = 𝐹(𝜙) ∗ 𝑃(𝜃) ∗ 𝑅(𝑟), we do not
need the 𝑅(𝑟) for the calcula ons done here since it is enough with integra ng over the polar
coordinates. We do not need to consider any other wave func on, since there is no projec on
to X or Y states with 𝑛 ≠ 2 or 𝑙 ≠ 1.
I expect the value of 𝐿 to be equal to the sum of the squared angular momentums for each
state and each direc on, mul plied by the probability of having this state in this direc on, this
way 𝐿 acts like an expected value for the squared angular momentum on each direc on. This
could be formulated as:
𝐿 = 𝑃 𝐿
{ , , }
= 𝑙(𝑙 + 1)
Where 𝑃 is the probability of having the state 𝑛 in the direc on 𝑚. I am assigning 𝑚 = −1
for the (2,1, −1) state, 𝑚 = 0 for (2,1,0) and 𝑚 = 1 for (2,1,1). Of course, 𝐿 = 1 for the
𝑚 = ±1 states and 𝐿 = 0 for the 𝑚 = 0 state.
For example, if I have the 𝜓 wavefunc on this is the eigenstate for 𝑃 , therefore 𝑃 = 1
and 𝑃 = 𝑃 = 0. For the states in the X and Y direc ons the probabili es are obtained
through the projec on of the current wavefunc on over the different states:
𝑃 = | < 𝜓 | 𝜓 > |
Because of the space isotropy, the results obtained for the X component should be equal to the
ones for the Y component, this means 𝑃 = 𝑃 .
The way I have obtained these probabili es is by doing numerical integra on over a spherical
surface with a simple Matlab script. I have calculated the probabili es only for the X states
since 𝑃 = 𝑃 .
The way I implemented the script is by assuming that for the “X states”, the 𝜃 coordinate was
comparing X against the YZ plane and Z was taking the role of X for the 𝜙 coordinate.
Therefore, for a given SO(3) point (𝑋, 𝑌, 𝑍) = (cos(𝜙) sin(𝜃) , sin(𝜙) sin(𝜃) , 𝑐𝑜𝑠(𝜃) ) we can
obtain the angles for the “X states” this way:
cos(𝜃 ) = 𝑋
sin(𝜃 ) = 𝑌 + 𝑍
sin(𝜙 ) =
𝑌
𝑌 + 𝑍
cos(𝜙 ) =
𝑍
𝑌 + 𝑍
Where I use 𝜃 and 𝜙 to refer the polar coordinates that would use the X coordinate as the
main axis instead of Z. For example, to project 𝜓 against 𝜓 we have to integrate:
cos(𝜃) ∗ cos(𝜃 ) = 𝑍 ∗ 𝑋 over the spheric surface.

3. I have obtained the following probabili es on the 𝜓 state for each of the states
𝜓 states:
Wavefunc on 𝑃 (𝜓 ) 𝑃 (𝜓 ) 𝑃 (𝜓 )
𝜓 0 1/2 1/2
𝜓 1/2 1/4 1/4
𝜓 1/2 1/4 1/4
The probabili es for the 𝑃 states would be the same. With these results is straigh orward to
measure 𝐿 , by adding the contribu ons for each state in each direc on:
Wavefunc on < 𝐿 > < 𝐿 > < 𝐿 >
𝜓 0 1 1
𝜓 1 1/2 1/2
𝜓 1 1/2 1/2
It can be seen that for each of the three eigenfunc ons:
𝐿 =< 𝐿 > +< 𝐿 > +< 𝐿 >= 2 = 𝑙(𝑙 + 1)
This is in accordance with the expected result.
Expected overall angular momentum J (orbital momentum plus spin)
Now I will discuss the results when we include the electron spin in the angular momentum. The
spin can only take the ±1/2 values. I will add to the previous wave func ons an extra index to
iden fy the spin, for example 𝜓 is a Z-defined (2,1,1) wavefunc on with posi ve spin,
with 𝑗 = 𝑙 + 𝑠 = 1 + = , 𝜓 is the same func on but with nega ve spin, therefore it
has 𝑗 = 1 − = .
The overall momentum 𝑗 is defined this way:
It can take the following range of values, jumping only in integer steps:
|𝑙 − 𝑠| ≤ 𝑗 ≤ 𝑙 + 𝑠
This means that for the 𝜓 states 𝑗 could be equal either to 1/2 or to 3/2. Given a value for
𝑗, all the states with |𝑗 | > 𝑗 must be forbidden.
The previous statement is enough to assure that, for the 𝜓 and 𝜓 states, which
have 𝑗 = , 𝑗 = , otherwise the exis ng state would be forbidden, which would be a
contradic on.
On the other hand, it does not seem a contradic on for any of the other states: 𝜓 ,
𝜓 ± and 𝜓 to have 𝑗 = , however, calcula ons show that in order to achieve for
these states 𝐽 = 𝑗(𝑗 + 1) we need 𝑗 to be ½ in these cases.

4. The non-spined wavefunc ons in X and Y direc ons such as 𝜓 will have to split their
probability between the two spin states by the following rule:
If none of the spin states has to |𝑗 | > 𝑗 each spin state will have half of the
probability, however, if there is one state with |𝑗 | > 𝑗, that state will be forbidden and
the other one will get all the probability.
With these assump ons, I have evaluated first the 𝜓 wavefunc on, which requires 𝑗 = ,
and leads to the following probabili es for other states:
Wavefunc on 𝑃 𝑃 𝑃 𝑃 𝑃 𝑃
𝜓 1/4 1/4 1/8 1/8 1/8 1/8
The square of the angular momentum for each of these states is
State 𝜓 𝜓 𝜓 𝜓 𝜓 𝜓
𝑗 1/4 1/4 9/4 1/4 1/4 9/4
If we sum the 𝜓 squared angular momentum (9/4) with all the other states mul plied by
its probability and then mul plied by two (to include the X and Y direc ons) we obtain:
𝐽 =
9
4
+ 2 ∗
9
4
∗ 2 ∗
1
8
+
1
4
∗ 2 ∗
1
4
+
1
4
∗ 2 ∗
1
8
=
15
4
=
3
2
(
3
2
+ 1)
Which confirms that the 𝑗(𝑗 + 1) rule also works with spin states.
The next I want to evaluate is the 𝜓 state, I do not have a way to determine if 𝑗 = or
𝑗 = , however, if I choose the first op on it turns out that the probabili es for all the X an Y
states will remain the same than in the previous case with the 𝜓 state, however, the
overall value for 𝐽 will not be the same since 𝜓 contributes with 9/4 to 𝐽 while 𝜓
contributes only with 1/4. Therefore, 𝑗 must be equal to 1/2.
With 𝑗 = the probabili es for each of the states will be:
Wavefunc on 𝑃 𝑃 𝑃 𝑃 𝑃 𝑃
𝜓 1/4 1/4 0 1/4 1/4 0
We can see that all the states with non-zero probability contribute to 𝐽 with 1/4, therefore
𝐽 = = ( + 1).
Last, I will check the 𝜓 state, which is equivalent to 𝜓 , I want to evaluate 𝐽 for both
𝑗 = and 𝑗 = , if any of these does not lead to a correct 𝐽 , I will assume that value for 𝑗 is not
possible having this state.
If 𝑗 = , we would have these probabili es:
Wavefunc on 𝑃 𝑃 𝑃 𝑃 𝑃 𝑃
𝜓 0 0 1/4 1/4 1/4 1/4

5. This leads to 𝐽 = + ∗ ∗ 2 ∗ 2 + ∗ ∗ 2 ∗ 2 = ≠ ∗
Therefore, it seems it is not possible to have the 𝜓 state with 𝑗 = .
On the other hand, if 𝑗 = , we have these probabili es:
Wavefunc on 𝑃 𝑃 𝑃 𝑃 𝑃 𝑃
𝜓 0 0 0 1/2 0 1/2
Since all the valid states have 1/2 momentum it is obvious that 𝐽 = 3 ∗ = ∗
Therefore, when the wave is in the 𝜓 state, the overall momentum must be 𝑗 = .
References
h ps://en.wikipedia.org/wiki/Angular_momentum_operator
Condon & Shortley 1935, pp. 46–47
h p://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hydwf.html
h ps://en.wikipedia.org/wiki/Total_angular_momentum_quantum_number