13. Applications of Approximation Methods What Approximations Have We Made? When describing hydrogen, we started with this Hamiltonian: Some things that this doesn’t include: Recoil of nucleus Handled in chapter 7 by replacing electron mass me with reduced mass  Finite nuclear size Relativistic corrections Nuclear spin and magnetic field Other, external effects Background electric or magnetic fields Van-der-Waals interactions

Electric Field from a Finite Nucleus 13A. Finite Nuclear Size Electric Field from a Finite Nucleus We need the electric potential from a finite nucleus Imagine the nucleus is a sphere of uniform charge density Radius a Charge density is We will use Gauss’s Law to find electric field everywhere By spherical symmetry, the electric field points radially outwards The electric field depends only on the amount of charge closer than the radius where you measure it For r > a, it looks like a point charge at the origin of magnitude e But for r < a, we only see the field from the charge closer than r The charge contained within a sphere of radius r < a is So the electric field inside this is

Electric Potential from Electric Field We now have the electric field We need the electric potential, related by We therefore integrate: Problem: Keep careful track of constant of integration! Solution: Potential is continuous, and vanishes at infinity Integrate in each region Use fact that U() = 0 Use fact that U is continuous at a

The Perturbation We have Hamiltonian: We know how to find eigenstates of Our Hamiltonian differs from this only in the tiny region r < R We therefore anticipate that we can use perturbation theory We write We therefore have Unperturbed states have wave function: Unperturbed energies are

First Order Correction Finite Nuclear Size First order correction to the energy is given by Nucleus is much smaller than the atom Wave function hardly changes on the scale of the nucleus We therefore approximate Comment: Rnl(0) = 0 for l  0 Only non-vanishing for l = 0

Magnitude of Nuclear Size Corrections Wave function at origin is of order Energy is of order If we do hydrogen-like atom with nuclear charge Z, then Size of atom decreases by factor of Z Unperturbed energy increases by factor of Z2 Nuclear size a increases by factor of A1/3 ~ Z1/3 Even with Z ~ 100, it is around a 10-7 size effect In atoms with multiple electrons, this effect is not that important 2s and 2p states are not degenerate anyway

Charge Radius of Proton Puzzle Another way to increase the effect is to replace the electron by a muon The size of muonic hydrogen is decreased by We can measure the “charge radius” of the proton three different ways: Electromagnetic scattering of electrons by a proton The split between the 2s and 2p levels of conventional hydrogen The split between the 2s and 2p levels of muonic hydrogen The first two methods are less exact, but lead to charge radius of This is not exactly the same as a from previous parts Muonic hydrogen gives a more precise value These disagree! At present, an unsolved mystery

Sample Problem (1) Suppose we model a proton as a hollow sphere of outer radius a and inner radius a/2 with total charge +e. Find the resulting shift in energy of all levels of hydrogen. a a/2 The volume of this region is The charge density is: Easiest way to handle this: Full sphere of radius a and charge + Charge of this sphere is Anti-sphere of radius a/2 and charge – Note that total charge is +e

Sample Problem (2) Suppose we model a proton as a hollow sphere of outer radius R and inner radius R/2 with total charge +e. Find the resulting shift in energy of all levels of hydrogen. a a/2 The total potential will be the sum of the potentials from the sphere and the anti-sphere The perturbation is then First term is 8/7 of the result we found before for a sphere of radius a Second term is –1/7 of the result we found before, if we modify for radius a/2 Our final answer, therefore is just

13B. Relativistic Corrections Types of Corrections The electron in hydrogen has speed of order This causes relativistic corrections: This implies corrections to the energy (eg. 2p vs. 2s) of order To fully understand these, need a relativistic theory of the electron The Dirac equation, chapter 16 For hydrogen-like atoms, we will solve this exactly For other atoms, relativistic corrections must be approximated Since states 2s/2p are not degenerate for these atoms, corrections not important But corrections that depend on spin are important

Spin-Orbit Coupling There is a term in the Hamiltonian caused by the magnetic dipole moment of the electron g  2 But is there any magnetic field? In the rest frame of the nucleus, no magnetic field But according to special relativity, particles moving in an electric field experience a magnetic field This suggests a perturbation of the form: Which one is correct? Answer turns out to be the average of these two answers Note that we can write: So we have

Spin Orbit Coupling – Why It’s Hard We have spin in the perturbation, so must include spin in the unperturbed eigenstates These are eigenstates of Energy for arbitrary atom depends on n, l, s The states with different m and ms are all degenerate Must use degenerate perturbation theory We need to calculate the matrix elements Because LS doesn’t commute with Lz or Sz, there will be non-diagonal terms Means we will have to deal with matrices We need to find a better way

A Better Way Instead of working with eigenstates of Lz and Sz, we can work with eigenstates of J2 and Jz Eigenstates will now look like We therefore need matrix elements As proven in a homework problem: We therefore have There is no angular dependence in the matrix element, so it is diagonal

Spin-Orbit Coupling Effects Expectation value of the Coulomb potential is of order the unperturbed energy We therefore have, approximately So we have Same size as other relativistic effects But other relativistic effects don’t split on the basis of spin Consider, for example, Sodium Ground state, the outermost electron in 3s1/2 First excited state has electron in 3p3/2 or 3p1/2 state Transition between 3p  3s causes wavelengths at 389.0 nm and 389.6 nm 3p1/2 3p3/2 3s1/2

Sample Problem An electron (with spin ½) is trapped in a Coulomb potential VC(r) = m2r2/2. Find the energy shift of the electron due to spin orbit coupling. The unperturbed Hamiltonian is a 3D harmonic oscillator Solved in homework problem, ignoring spin Spin orbit coupling adds a shift The matrix element we need is Spin is ½ for a single electron, so Total angular momentum ranges from j = |l – s| to l + s If l > 0, this means j = l – ½ or j = l + ½. Simply substitute these two expressions for j into this expression

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Beyond Spin-Orbit Coupling? Atomic states look like Energy is non-degenerate based on: n: governs overall electronic configuration l: for multiple electrons, energy is different due to screening s: spin state of multiple electrons affects symmetry and hence energy j: due to spin-orbit coupling Can anything break the remaining degeneracy? Hamiltonian rotationally invariant As we perform a rotation of the atom, mj changes Therefore energy independent of mj Unless: The nucleus is not rotationally invariant (spin) There are external forces breaking the degeneracy Such as a magnetic field

13C. The Hyperfine Splitting Nuclear Magnetic Dipole Moment The proton is itself a rotating electric charge We would expect it would have a magnetic dipole moment in the direction it is spinning: Where I is the spin of the proton If the proton were elementary, we would predict gp 2, but actually We will approximate the proton as a uniformly magnetized sphere of radius a: Magnetization: Vector potential, Jackson eq. (5.111): 0 is magnetic permeability of free space Magnetic field: a

Shrinking the Nucleus Again, the nucleus is small, so we’d like to take the limit a  0 Can we just always use the lower formula? How large is the magnetic field integrated just over the volume of the nucleus? In the limit a  0, must add a contribution of this magnitude right at the origin No comparable contribution from A, since it is smaller by factor of r near origin

Shrinking the Nucleus Again, the nucleus is small, so we’d like to take the limit a  0 Can we just always use the lower formula? How large is the magnetic field integrated just over the volume of the nucleus? In the limit a  0, must add a contribution of this magnitude right at the origin No comparable contribution from A, since it is smaller by factor of r near origin

The Hyperfine Perturbation for Hydrogen Recall the Hamiltonian for electromagnetic interactions: Drop A2, and approximate g = 2: Compare to unperturbed Hamiltonian The perturbation is therefore:

Hyperfine Perturbation for s-Waves Replace R  P = L: We first need unperturbed states Have to include nuclear spin now! mI is eigenvalue associated with Iz Because of spin-orbit coupling, it is better to add L and S to get total electron angular momentum J = L + S So better choice of basis states would be Degenerate perturbation theory again So we will need matrix elements For l = 0 (s-waves), the angular momentum operator always vanishes Less obvious: for l = 0, the next two terms also vanish Next slide from now, plus proof by homework problem

Sample Problem Prove that for s-wave states, only the final term contributes to hyperfine splitting All s-waves (l = 0) have no angular dependence, so So contributions to perturbation theory from this term will be For the other two terms, write it out as Because the wave function has no angular dependence, the angular integral is This can be shown to always vanish (nine expressions, six independent) For example, let j = k = z, then

Total Atomic Angular Momentum We are still working in basis states of J2 and Jz, where J = L + S But as before, we could then combine the electron angular momentum with the proton’s spin to get the total internal angular momentum of the atom: Eigenstates of H0 will now look like: These are eigenstates of Since we have l = 0, ignore L and so F = S + I Since electron has s = ½ and proton has i = ½, total spin is f = 0 or f = 1 Use addition of angular momentum trick to write Therefore,

Hyperfine Splitting for Hydrogen s-waves We need matrix elements Energy is now diagonal Most important effect is difference between f = 0 and f = 1 energy

Comments on Hyperfine Splitting Permittivity of free space is related to Coulomb’s constant by Therefore By comparison, spin-orbit coupling is of order 2E1s So hyperfine is suppressed by m/mp But this is the only contribution that distinguishes these spin states Transition from f = 1 to f = 0 generates electromagnetic radiation with a wavelength of 21 cm. The 21 cm line is used to track atomic hydrogen throughout the galaxy The comparable transition in Cesium is used for atomic clocks

13D. The Zeeman Effect Weak Magnetic Fields For an isolated atom in vacuum, including the hyperfine interaction, general state now looks something like Since rotating the atom changes mf different mf values must be truly degenerate Consider a uniform magnetic field acting on an atom Ignore hyperfine splitting, since it’s so small Unperturbed states are therefore Assume a weak field, so that spin-orbit coupling dominates magnetic effects So energy in absence of magnetic field depends on n, l, s, and j Presence of magnetic field adds to the Hamiltonian a perturbation: See chapter 9 Lz and Sz are sums over all electrons

Zeeman Effect Computation We need Fortunately, Lz and Sz both commute with Jz = Lz + Sz Therefore, Jz eigenvalues can’t change Insert complete basis |n, l, s, ml, ms These matrix elements are just CG coefficients Though it looks like a double sum, it really isn’t Recall, CG coefficients vanish unless mj = ml + ms Can show in general that shift is proportional to mj For example, if s = 0, then ms = 0 and CG coefficients are 1 And mj = ml

13E. Van Der Waals Interaction The Hamiltonian Consider two atoms that are close, but not too close to each other For definiteness, two hydrogen atoms Treat the nuclei as fixed point positive charges separated by a Assume they are far enough apart that the electron wave functions are not overlapping Then we don’t have to worry about anti-symmetrizing the wave function Treat the electrons quantum mechanically, with positions R1 and – R2 relative to their nuclei Then the Hamiltonian is: R2 R1 a

The Perturbation The unperturbed Hamiltonian is: This has energy eigenstates: The perturbation is given by: For definiteness, choose: We want to calculate this in the limit that a >> R

The Perturbation Simplified Substitute and simplify: Let’s do first order perturbation theory on the ground state |100;100:

Second Order Perturbation Theory Let’s try second order perturbation theory on ground state: This sextuple sum (!) should exclude the state |100,100 Technically, it also contains unbound states As previously shown, matrix elements like nlm|R|n'l'm' only non-vanishing if l and l' differ by exactly 1 Therefore, only l, l' = 1 contributes (which means n, n'  2) Note that energy denominator is always negative, times a positive number

A Lower Limit on the Energy Shift The largest matrix elements come from n = 2, or n' = 2 All terms negative, so dropping any terms leads to an overestimate of the energy Let’s include only n = n' = 2 (and recall that only l = l' =1 contributes): For example, if m = 0, then only 210|Z|100  0, and then forced to pick m' = 0 Previous homework problem: Substitute this in Find energy denominator: Include m, m' = 1 Put it all together

A Lower Limit on the Energy Shift The largest matrix elements come from n = 2, or n' = 2 All terms negative, so dropping any terms leads to an overestimate of the energy Let’s include only n = n' = 2 (and recall that only l = l' =1 contributes): Non-vanishing matrix elements: Energy denominator: Put it all together

An Upper Limit on the Energy Shift (1) The smallest magnitude energy denominator is n = n' = 2 Small energy denominator causes large (negative) contribution to the energy shift If we replace energy denominator by 11 – 22 on every term, we are overestimating the (negative) energy shift, so Technically, sum excludes ground state, but this matrix element vanishes Use completeness

An Upper Limit on the Energy Shift (2) Substitute in: Because of spherical symmetry, cross terms do not contribute

Combining the Limits Combining the limits, we have Sophisticated analysis yields  = 6.50 Attractive potential that goes like a–6 This is for hydrogen-hydrogen Will it apply in general? For neutral atoms, since they are in states of definite l, we will have So always have to do second order perturbation theory So we generally get attractive a–6 potential Only the factor of  changes Larger if electron easily excited to state with different l value