The Stark Effect. a. Evaluate a matrix element. = ee. = ee 200 r cos θ 210. r 2 dr. Angular part use Griffiths page 139 for Yl m (θ, φ)

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1 The Stark Effect a. Evaluate a matrix element. H ee z ee r cos θ ee r dr Angular part use Griffiths page 9 for Yl m θ, φ) Use the famous substitution to find that the angular part is dφ Y θ, φ) cos θ Y θ, φ) µ cos θ dφ R r)y θ, φ) r cos θ R r)y θ, φ) 4π π) dµ dφ Y θ, φ) cos θ Y θ, φ) dφ cos θ + µ dµ µ dµ ] ] cos θ 4π 4π cos θ µ ] Radial part use Griffiths page 4 for R r) and R r): r dr Rr) r R r) r dr a / ) ] ) ] r e r/a)/ r a / r e r/a)/ 4 Use the substitution r r/a : r dr R r) r R r) a 4 a 4 a r 4 d r r ) e r d r r 4 e r d r r e r ] According to Dwight 86.7 these two integrals evaluate to 4! 4 and to! respectively. Thus r dr Rr) r R r) a 4) 4 ] ) 9 a a All together now: H ee 9 ) ) a eea.

2 The Stark Effect b. Average positions. Find r via x, y, z. State,, + has a wavefunction proportional to Y + θ, φ). State,, has a wavefunction proportional to Y θ, φ). So both states have angular distributions as suggested by y x From which it s clear that x y z. The two remaining states of interest are,, ±,, ] R r)y θ, φ) ± R r)y θ, φ)] 4π R r) ± R r) cosθ)]. This is independent of φ, so x y. We evaluate z with the help of the matrix element calculated in part a): z,, ±,, ) z,, ±,, )],, z,, ±,, z,, +,, z,, ] }{{}}{{}}{{} a a Thus, for the latter two states, r a ẑ. c. Escape from contradiction. If two different probability densities both have z, then any combination of those densities has z. But if two wavefunctions or probability amplitude densities ) both have z, then a combination of those wavefunctions might have z. What is the physical principle behind this mathematical fact? My favorite answer is Interference, because in interference experiments finite probability to go thorough one

3 The Stark Effect slit) plus finite probability to go through another slit) can sum to zero probability to go through both slits) thereby showing that the physically controlling entity is not probability density, but something deeper that we eventually found to be wavefunction. But answers of Superposition or The wavefunction has phase as well as amplitude or The wavefunction contains more information than the probability density or Probability density alone doesn t specify a state or Square of sum, not sum of squares are also fine. e. Stark effect for n : find the Hamiltonian matrix. This is a 9 9 matrix... 8 integrals... looks pretty formidable! BUT... general rules nlm z nlm nlm z n l m unless m m For nlm z n l m, the part involving angle φ is are pure real! So we arrange the states in the order dφ e imφ e +imφ π, so the matrix elements,,,,,,,,. The general rules find our non-zero matrix elements: A B A C B C D D E E The kets and bras at the borders of this matrix are lm or lm... the value n is not shown because it is the same in all cases. The value is shown as because otherwise it messes up the spacing.) Some of these matrix elements are easy: B z + r dr R r) r R r) dφ Y θ, φ) }{{} cos θ Y θ, φ) }{{} cos θ ) cos θ ) cos θ Then use µ cos θ... ] µ )µ dµ But integrand is odd so... ]

4 4 The Stark Effect Moral of the story: Evaluate the angular integral first... it s more likely to vanish.] Other relationships are also easy to find: D, z, E, z, r dr R r) r R r) r dr R r) r R r) But Y Y ) and Y Y ), so D E. dφ Y θ, φ) cos θ Y θ, φ) dφ Y θ, φ) cos θ Y θ, φ) Material below this line uses atomic units, plus the tables on Griffiths pages 9 and 4.] But some matrix elements are hard: A z A angular A radial r dr R r) r R r) dφ Y θ, φ) cos θ Y θ, φ) }{{}}{{} A radial A angular ] ] dφ 4π cos θ cos θ 4π π 4π π) cos θ Use µ cos θ... ] + µ dµ µ dµ µ] 4 8 r dr 7 6 ] ] 6 r) re r/ r 7 r + 7 r) e r/ dr r 4 6 r) r + 7 r) e r/ Use u r... ] 4 ) du u 4 4 u) u + 6 u) e u du u 4 4 u + u6 4 u7] e u 4! 4! + 6! 4 7!] 8 A 8 ) ) 6

5 The Stark Effect C z r dr R r) r R r) dφ Y θ, φ) cos θ Y θ, φ) }{{}}{{} C radial C angular ] ] C angular dφ 6π cos θ ) cos θ 4π cos θ + 8π π) µ )µ dµ Using µ cos θ. ] µ 4 µ ) dµ µ µ] C radial C D z D angular 4 r dr 4 8 r e r/ ] 8 r r) re r/ ] 8 dr r 6 6 r) e r/ Use u r... ] 4 ) 7 8 du u 6 4 u) e u 6! 4 7!] 9 8π π) 4 9 ) ) r dr R r) r R r) dφ Y θ, φ) cos θ Y θ, φ) }{{}}{{} C radial D angular π ] ] dφ sin θ cos θe iφ cos θ sin θeiφ 8π 8π + sin θ cos θ Use µ cos θ, sin θ µ... ] µ )µ dµ µ µ 4 ) dµ µ µ4]

6 6 The Stark Effect Whence, remembering the value of C radial, End use of atomic units.] D 9 ) ) 9 Thus the matrix of the perturbation Ĥ is 6 6 eea 9/ 9/ 9/ 9/ f. Diagonalize this matrix to find the eigenvalues and degeneracies. Each of the submatrix blocks can be diagonalized independently. ] D Diagonalize a submatrix of the form : D ] λ D det λ D λ ±D. D λ Or, in our case, λ ±eea 9/). Diagonalize a submatrix of the form λ det Resulting in λ, ± A + C. Or, in our case, λ, ±eea 9). λ C A A C C det : ] C A det λ λ A A λ C C λ A C λ ] λλ C ] A Aλ] λ λ + C + A )

7 The Stark Effect 7 Thus the first-order energy shifts are +9eEa degeneracy ) + 9 eea degeneracy ) degeneracy ) 9 eea degeneracy ) 9eEa degeneracy )