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

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

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

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 !!!

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

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)

Let’s consider Lz first eimj f ^

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

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

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

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.

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-

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>)

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

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

= 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>

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)

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

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>

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