7. Angular Momentum 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 L’s –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 Why it doesn’t commute 7A. Angular Momentum Commutation

To second order in  : Second half is same thing with   –  Calculating the Left Side

To second order in  : Second half is same thing with   –  Calculating the Right Side

To second order in  : Now match the two sides: Matching the Two Sides

Generalizing, we have Define the Levi-Civita symbol Then we write: We will call any three Hermitian operators J that work this way generalized angular momentum Levi-Civita Symbol

What can we conclude just from the commutation relations? If J commutes with the Hamiltonian, than we can simultaneously diagonalize H and one component of J –Normally pick J z Define some new operators: Reverse these if we want: These satisfy the following properties –Proof by homework problem J 2 and the raising/lowering operators 7B. Generalized Angular Momentum

Eigenstates Since J 2 commutes with J z, 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 J 2 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:

Eigenstates (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| > j Resolution: You must have:

Summary Eigenstates look like The values of m are –There are 2j + 1 of them Since 2j+1 is an integer: We can use these expressions to write out J’s as matrices of size (2j + 1)  (2j + 1): First, pick an order for your eigenstates, traditionally The matrix J z 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 J x and J y

Sample Problem Write out the matrix form for J for j = 1 Basis states J z is diagonal: J + just above the diagonal:

Sample Problem (2) Now work out J x and J y : As a check, find J 2 Write out the matrix form for J for j = 1

Special 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:

Consider this Hamiltonian: All components of L commute with H, because they commute with R 2 It makes sense to choose eigenstates of H, L 2 and L z It seems sensible to switch to spherical coordinates: We write Schrödinger’s equation in spherical coordinates Spherical Coordinates 7C. Spherically Symmetric Problems

L in Spherical coordinates (1) We need to write L in spherical coordinates Start by writing angular derivatives out: It’s not hard to get L z from these equations:

L in Spherical coordinates (2) Now it’s time to get clever: consider And we get clever once more:

Other Operators in Spherical coordinates It will help to get the raising and lowering operators: And we need L 2 : Compare to Schrödinger:

Solving Spherically Symmetric Problems: Rewrite Schrödinger’s 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:

The 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:

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 Dependence on , and m restrictions 7D. The Spherical Harmonics

Finding one Spherical Harmonic: We previously found: For general angular momentum: If we lower m = – l, we must get zero: Normalize it:

Finding 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

Properties of Spherical Harmonics: They are eigenstates of L 2 and L z They are orthonormal: They are complete; any angular function can be written in terms of them: This helps us write the completion relation

More Properties of Spherical Harmonics: Recall: parity commutes with L It follows that Hence when you let parity act, you must be getting essentially the same state Recall: L 2 is real but L z is pure imaginary Take the complex conjugate of our relations above: This implies It works out to

The Spherical Harmonics

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: Let’s try Find commutation operators for these –Proof by homework problem Find the new Hamiltonian –Proof by homework problem Changing Operators 7E. The Hydrogen Atom

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 H cm 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 Reducing the problem: 6D to 3D

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 won’t depend on it We will focus on bound states E < 0 Reducing the problem: 3D to 1D

Let’s 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 The Radial Equation: For r Large and Small

Factor out the expected asymptotic behavior: –This is just a definition of f(r) Substitute in, multiply by 2  /  2 Define the Bohr radius: The Radial Equation: Removing Asymptotic

Write f as a power series around the origin –Recall that at small r it goes like r l 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: The Radial Equation: Taylor Expansion

It looks like we have a solution for any E: –Pick f l 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 f n Are We Done?

Because the exponential beats the polynomial, these functions are now all normalizable –Arbitrarily pick f l > 0 Online “Hydrogen” or p. 124 Note that n > 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,  c 2 = 511 keV Summarizing Everything

Radial Wave Functions

Sample 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 1 > integrate(radial(4,2)^2*r^3,r=0..infinity);

Degeneracy 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 E n ? 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 10 4 to 10 5 times smaller –Very small effect Nuclear magnetic field interacting with the electron

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 e 2 by e 2 Z Atom gets smaller –But still much larger than the nucleus Relativistic effects get bigger –Now v/c ~ Z  Other Nuclei 7E. Hydrogen-Like Atoms

Bizarre “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 Positron’s mass = electron’s 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