2. L = r x p To spell out 3D motion, need to understand its angular
momentum R q f ^
A general 3D rigid body motion
is always a translation + rotation.
In spherical coordinates this is
easy to describe. The translation is
along the radius, while the rotation
is decomposable around two axes,
one along the latitude and one along
the longitude.
L = r x p

3. But can’t measure x and p together !
[x, px]y = x(-iħy/x) + iħ(xy)/x = iħy
x,p ⌃ = [ ] iħ
So can’t specify L fully

4. Cannot measure all angular momentum components simultaneously !
[x, px] = iħ
[y, py] = iħ
[z, pz] = iħ
[x, py] = 0
[y, pz] = 0
[z, px] = 0
[x, pz] = 0
[y, px] = 0
[z, py] = 0

5. Angular Momentum Algebra
[Lx, Ly] = iħLz and
Circular permutations
e.g. [Lx, Ly]y = (-iħ)[y/z- z/y] (-iħ)[z/x- x/z]y
- (-iħ)[z/x- x/z] (-iħ)[y/z- z/y]y
= (-iħ)[y/z] (-iħ)[z/x]y
- (-iħ)[- x/z] (-iħ)[- z/y]y (The other terms cancel !)
= (-iħ)(-iħ)[yy/x - xy/y]
= iħLz
Cannot measure all components of
L simultaneously !!!

6. Just how much of L is measurable?
we can easily show that
[L2, Lx] = [L2, Ly] = [L2, Lz] = 0
So we can measure L2 and any one component
We choose that component to be Lz

7. So an angular momentum state is specified with two indices..
|l,m >
l: Azimuthal Quantum Number
(eigenstate of L2)
m: Magnetic Quantum Number
(eigenstate of Lz)

8. Let’s consider Lz first
eimj f ^

9. Let’s consider Lz first
First try a translation !!
Y(x+a) = Y(x) + aY/x + (a2/2)2Y/x2 + …
= [exp(a/x)]Y(x)
= [exp(iapx/ħ)]Y(x), where px = -iħ/x
displacement
operator
By analogy:
Y(q,j+j0) = [exp(ij0Lz/ħ)]Y(q,j), where Lz = -iħ/j

10. Eigenstates of Lz Q = eimj Lz = -iħ/j LzQ = (-iħ/j)Q = mħQ
m = 0, ±1, ±2, ±3, … due to periodic boundary conditions
So Lz values separated by ħ

11. Lz  Lz ˆ
z-component
Quantized mħ
m = -l, -(l-1),...
(l-1), l

12. What about Lx and Ly ? Since Lx upsets Lz measurements, it can take
a z-up state into mixtures of up and down
But let’s focus on a one-way jump, ie,
operators that ONLY take say, down to up.
Clearly this must be a mixture of Lx and Ly.

13. More Angular Momentum Algebra
Define “Ladder” Operators: (They will move z-spins
ONLY up or ONLY down, as we will see)
L+ = Lx + iLy
L- = Lx - iLy
Using [Lx,Ly] = iħLz, can readily show:
[L+, L-] = 2ħLz
[L+, Lz] = -ħL+
[L-, Lz] = ħL-

14. Why “Ladder” Operators?
Act L+ on state |l,m>. What happens?
Act Lz on state L+|l,m>. What happens? L+ L-
Lz (L+|l,m>) = -[L+,Lz]|l,m> + L+Lz|l,m>
= ħL+|l,m> + mħL+|l,m>
= (m+1)ħ (L+|l,m>)

15. Why “Ladder” Operators?
So state L+|l,m> is an eigenstate of
Lz with eigenvalue (m+1)ħ L+ L-
Thus, L+|l,m> = C|l,m+1>
Step up ladder !! Takes it one notch up
Must have L+|l,l> = 0

16. Using the ladder expressions
L2 = Lx2 + Ly2 + Lz2
= L+L- - ħLz + Lz2
= L-L+ + ħLz + Lz2

17. = L-L+|l,l> + ħLz|l,l> + Lz2|l,l> = (0 + ħ2l + ħ2l2)|l,l>
Act on largest m = l
L2 |l,l>
= L-L+|l,l> + ħLz|l,l> + Lz2|l,l>
= (0 + ħ2l + ħ2l2)|l,l>
= l(l+1)ħ2|l,l>

18. Angular Momentum eigenstates
Lz  Lz ˆ
L2  L2 ˆ
z-component
Quantized mħ
m = -l, -(l-1),...
(l-1), l
Angular Mom
Quantized
l(l+1)ħ2
l = 0, 1, 2,.. (n-1)

19. Note !! Maximum Lz has length l Radius of sphere has length √l(l+1)
So maximum z is not a spin
completely along z axis

20. What is C? |L+|l,m>|2 = <l,m|L-L+|l,m>
= <l,m|L2 - Lz2 - ħLz|l,m>
= [l(l+1)-m2-m]ħ2 <l,m|l,m>
= (l-m)(l+m+1)ħ2
So, L+ |l,m> = [√(l-m)(l+m+1)ħ] |l,m+1>

21. L+|l,m> = [√(l-m)(l+m+1)ħ] |l,m+1>
Thus
L+|l,m> = [√(l-m)(l+m+1)ħ] |l,m+1>
L-|l,m> = [√(l+m)(l-m+1)ħ] |l,m-1>
Since we reach 0 beyond m = l or -l values,
where L+ takes you up by 1 and L- takes you down by 1,
the m’s must be integers (ie, -l, -l+1, -l+2,….,l-1, l).
ie, 2l+1 values