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Understanding Angular Momentum in 3D Motion

Learn about the concept of angular momentum in 3D motion and how it can be measured. Explore the algebra and eigenstates of angular momentum, as well as the significance of ladder operators.

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Understanding Angular Momentum in 3D Motion

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  1. q f R ^ ^ ^ q f R To spell out 3D motion, need to understand its angular momentum 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

  2. 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

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

  4. 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 !!!

  5. 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

  6. 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)

  7. ^ f Let’s consider Lz first eimj f

  8. 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

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

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

  11. 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.

  12. 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-

  13. 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>)

  14. 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

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

  16. 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>

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

  18. 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

  19. 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>

  20. 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

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