Title: 7. Angular Momentum
17. Angular Momentum
7A. Angular Momentum Commutation
Why it doesnt commute
- The order in which you rotate things makes a
difference, ?1?2 ? ?2?1 - We can use this to work out commutation relations
for the Ls - It can be done more easily directly
- Recall
- Also recall
- We will calculate the following to second order
in ? - If rotations commuted, both sides would be the
identity relation
2Calculating the Left Side
- To second order in ?
- Second half is same thing with ? ? ?
3Calculating the Right Side
- To second order in ?
- Second half is same thing with ? ? ?
4Matching the Two Sides
- To second order in ?
- Now match the two sides
5Levi-Civita Symbol
- Generalizing, we have
- Define the Levi-Civita symbol
- Then we write
- We will call any three Hermitian operators Jthat
work this way generalized angular momentum
6 7B. Generalized Angular Momentum
J2 and the raising/lowering operators
- What can we conclude just from the commutation
relations? - If J commutes with the Hamiltonian, than we
cansimultaneously diagonalize H and one
component of J - Normally pick Jz
- Define some new operators
- Reverse these if we want
- These satisfy the following properties
- Proof by homework problem
7Eigenstates
- Since J2 commutes with Jz, we can diagonalize
them simultaneously - We will (arbitrarily for now) choose an odd way
to write the eigenvalues - Note that j and m are dimensionless
- Note that J2 has positive eigenvalues
- We can choose j to be non-negative
- We can let J? act on any state j,m? to produce a
new state J? j,m? - This new state must be proportional to
8Eigenstates (2)
- To find proportionality, consider
- This expression must not be negative
- When it is positive, then we have
- Choose the phase positive
- Conclusion given a state j,m?, we can produce a
series of other states - Problem if you raise or lower enough times, you
eventually get m gt j - Resolution You must have
9Summary
- Eigenstates look like
- The values of m are
- There are 2j 1 of them
- Since 2j1 is an integer
- We can use these expressions to write out Js as
matrices of size (2j 1) ? (2j 1) - First, pick an order for your eigenstates,
traditionally - The matrix Jz is trivial to write down, and is
diagonal - The matrix J is a little harder, and is just
above the diagonal - You then get J- J and can find Jx and Jy
10Sample Problem
- Basis states
- Jz is diagonal
- J just above the diagonal
Write out the matrix form for J for j 1
11Sample Problem (2)
Write out the matrix form for J for j 1
- Now work out Jx and Jy
- As a check, find J2
12Special Cases and Pauli Matrices
- The matrices for j 0 are really simple
- We sometimes call this the scalar representation
- The j ½ is called the spinor representation,
and is important - There are only two states
- Often these states abbreviated
- The corresponding 2?2 matrices are written in
terms of the Pauli matrices - The Pauli matrices are given by
- Useful formulas
13 7C. Spherically Symmetric Problems
Spherical Coordinates
- Consider this Hamiltonian
- All components of L commute with H, because they
commute with R2 - It makes sense to choose eigenstates of H, L2 and
Lz - It seems sensible to switch to spherical
coordinates - We write Schrödingers equation in spherical
coordinates
14L in Spherical coordinates (1)
- We need to write L in spherical coordinates
- Start by writing angular derivatives out
- Its not hard to get Lz from these equations
15L in Spherical coordinates (2)
- Now its time to get clever consider
- And we get clever once more
16Other Operators in Spherical coordinates
- It will help to get the raising and lowering
operators - And we need L2
- Compare to Schrödinger
17Solving Spherically Symmetric Problems
- Rewrite Schrödingers equation
- Our eigenstates will be
- The angular properties are governed by l and m
- This suggests factoring ? into angular and radial
parts - Substitute into Schrodinger
- Cancel Y
18The Problem Divided
- It remains to find and normalize R and Y
- Note that Y problem is independent of the
potential V - Note that the radial problem is a 1D problem
- Easily solved numerically
- Normalization
- Split this up how you want, but normally
19 7D. The Spherical Harmonics
Dependence on ?, and m restrictions
- We will call our angular functions spherical
harmonics and label them - We previously found
- For general angular momentum we know
- We can easily determine the ? dependence of the
spherical harmonics - Also, recall that ? 0 is the same as ? 2?
- It follows that m (and therefore l) is an integer
20Finding one Spherical Harmonic
- We previously found
- For general angular momentum
- If we lower m l, we must get zero
- Normalize it
21Finding All Spherical Harmonics
- To get the others, just raise this repeatedly
- Sane people, or those who wish to remain so, do
not use this formula - Many sources list them
- P. 124 for l 0 to 4
- Computer programs can calculate them for you
- Hydrogen on my website
22Properties of Spherical Harmonics
- They are eigenstates of L2 and Lz
- They are orthonormal
- They are complete any angular function can be
written in terms of them - This helps us write the completion relation
23More Properties of Spherical Harmonics
- Recall parity commutes with L
- It follows that
- Hence when you let parity act, youmust be
getting essentially the same state - Recall L2 is real but Lz is pure imaginary
- Take the complex conjugate of our relations
above - This implies
- It works out to
24The Spherical Harmonics
25 7E. The Hydrogen Atom
Changing Operators
- Hamiltonian for hydrogen (SI units)
- These operators have commutators
- Classically, what we do
- Total momentum is conserved
- Center of mass moves uniformly
- Work in terms of relative position
- Quantum mechanically Lets try
- Find commutation operators for these
- Proof by homework problem
- Find the new Hamiltonian
- Proof by homework problem
26Reducing the problem 6D to 3D
- Note that for actual hydrogen, ? is essentially
the electron mass - Split the Hamiltonian into two pieces
- These two pieces have nothing to do with each
other - It is essentially two problems
- Hcm is basically a free particle of mass M,
- It is trivial to solve
- The remaining problem is effectively a single
particle of mass ? in a spherically symmetric
potential
27Reducing the problem 3D to 1D
- Because the problem is spherically symmetric, we
will have states - These will have wave functions
- The radial wave function will satisfy
- Note that m does not appear in this equation, so
R wont depend on it - We will focus on bound states E lt 0
28The Radial Equation For r Large and Small
- Lets try to approximate behavior at r 0 and r
? - Large r keep dominant terms, ignore those with
negative powers of r - Define a such that
- Then we have
- Want convergent
- Now, guess that for small r we have
- Substitute in, keeping smallest powers of r
- Want it convergent
29The Radial Equation Removing Asymptotic
- Factor out the expected asymptotic behavior
- This is just a definition of f(r)
- Substitute in, multiply by 2?/?2
- Define the Bohr radius
30The Radial Equation Taylor Expansion
- Write f as a power series around the origin
- Recall that at small r it goes like rl
- Substitute in
- Gather like powers of r
- On right side, replace i ? i 1
- On left side, first term vanishes
- These must be identical expressions, so
31Are We Done?
- It looks like we have a solution for any E
- Pick fl to be anything
- Deduce the rest by recursion
- Now just normalize everything
- Problem No guarantee it is normalizable
- Study the behavior at large i
- Only way to avoid this catastrophe is to make
sure some f vanishes, say fn
32Summarizing Everything
- Because the exponential beats the polynomial,
these functions are now all normalizable - Arbitrarily pick fl gt 0
- Online Hydrogen or p. 124
- Note that n gt l, n is positive integer
- Include the angular wave functions
- Restrictions on the quantum numbers
- Another way of writing the energy
- For an electron orbiting a nucleus, ? is almost
exactly the electron mass, ?c2 511 keV
33Radial Wave Functions
34Sample Problem
What is the expectation value for ?R? for a
hydrogen atom in the state n,l,m? 4,2,-1??
- The spherical harmonics are orthonormal over
angles
gt integrate(radial(4,2)2r3,r0..infinity)
35Degeneracy and Other Issues
- Note energy depends only on n, not l or m
- Not on m because states related by rotation
- Not on l is an accident accidental degeneracy
- How many states with the same energy En?
- 2l 1 values of m
- l runs from 0 to n 1
- Later we will learn about spin, and realize there
are actually twice as many states - Are our results truly exact?
- We did include nuclear recoil, the fact that the
nucleus has finite mass - Relativistic effects
- Small for hydrogen, can show v/c ? 1/137
- Finite nuclear size
- Nucleus is 104 to 105 times smaller
- Very small effect
- Nuclear magnetic field interacting with the
electron
36 7E. Hydrogen-Like Atoms
Other Nuclei
- Can we apply our formulas to any other systems?
- Other atoms if they have only one electron in
them - The charge on the nucleus multiplies potential by
Z - Reduced mass essentially still the electron mass
- Just replace e2 by e2Z
- Atom gets smaller
- But still much larger than the nucleus
- Relativistic effects get bigger
- Now v/c Z?
37Bizarre atoms
- We can replace the nucleus or the electron with
other things - Anti-muon plus electron
- Anti-muon has same charge as proton, and much
more mass than electron - Essentially identical with hydrogen
- Positronium anti-electron (positron) plus
electron - Same charge as proton
- Positrons mass electrons mass
- Reduced mass and energy states reduced by half
- Nucleus plus muon
- Muon 207 times heavier than electron
- Atom is 207 times smaller
- Even inside a complex atom, muon sees essentially
bare nucleus - Atom small enough that for large Z, muon is
partly inside nucleus - Anti-hydrogen anti-proton plus anti-electron
- Identical to hydrogen