Problem 1:

In one dimension, the potential energy of a particle of mass m as a function of x is given by
U(x) = (b²|x|)^½.  Here b is a positive constant with units energy/length^½.
(a)  Use the WKB method to estimate the energy of the particle in the ground state.
(b)  Use the variational method to estimate the energy of the particle in the ground state.
(c)  Which estimate is closer to the true ground-state energy?

∫₀^∞exp(-x²) √x dx = Γ(3/4)/2 = 0.612708

Solution:

• Concepts:
  The WKB and variational methods
• Reasoning:
  The variational method often yields a very good estimate for the ground state energy of a system.
• Details of the calculation:
  (a)  For this problem the WKB approximation requires that  ∫_xmin^xmax pdx = ½(h/2),
  or ∫_xmin^xmax pdx = πħ/2, or ∫₀^xmax pdx = πħ/4.
  p = (2m)^½(E - b√x)^½.  ∫₀^xmax pdx = (2m)^½∫₀^{(E/b)^2} (E - b√x)^½dx,
  since E = b√(x_max).
  Let u = b√x.  du = (b/2)dx/√x = (b²/2)dx/u.  dx = (2/b²)u du.
  ∫₀^xmax pdx = (2/b²)(2m)^½∫₀^E (E - u)^½u du = (2/b²)(2m)^½ 4E^5/2/15.
  (2/b²)(2m)^½ 4E^5/2/15 = πħ/4.  E^5/2 = 15πħb²/(32*(2m)^½) = 1.041 (ħb²/m^½).
  E =  1.01632 (ħ²b⁴/m)^1/5.
  
  (b)  H = (-ħ²/2μ)(∂²/∂x²) + (b²|x|)^½.
  Choose Φ(x) = Aexp(-a²x²).  Here a is an adjustable parameter.
  Φ(x) = 0 as x --> ±∞, and dΦ/dx is continuous at all x.
  If we let a² = ½mω/ħ, then Φ(x)  is the ground state wave function for a harmonic oscillator potential U(x) = ½mω²x².
  The normalization constant is A = (mω/(πħ))^¼ = (2a²/π)^¼,
  and the expectation value of the kinetic energy is
  <T> = <p²>/(2m) = ħω/4 = ħ²a²/(2m).
  <H> = <T> + <U>,  <U> = (2a²/π)^½ 2∫₀^∞ exp(-2a²x²) b √x dx.
  Let x' = √2 a x,  then dx = dx'/(a√2) and √x = √x'/(2^¼√a).  Therefore
  <U> = (2^3/4b/π^½) a^-½ ∫₀^∞ exp(-x'²) √x' dx' = (2^3/4b/π^½) a^-½ * X,  where X = 0.612708.
  <H> =  ħ²a²/(2m) + (2^3/4b/π^½) a^-½ * X.
  d<H>/da² = ħ²/(2m) - (1/4) (2^3/4b/π^½) (a²)^-5/4 * X = 0.
  a² = (mbX/(2^¼ħ²π^½))^4/5 = (mb/ħ²)^4/5 * 0.372116.
  E = 0.186083 ħ^2/5 m^-1/5 b^4/5 + 0.744332*ħ^2/5 m^-1/5 b^4/5.
  E =  0.930415 (ħ²b⁴/m)^1/5.
  
  (c)  The variational method yields an upper limit for the ground state energy.  Since this limit for E is lower than the value for E obtained using the WKB approximation, the variational estimate is closer to the true ground-state energy than the WKB estimate.

Problem 2:

The hydrogen atom states ²S_½ and ²P_½ are still degenerate, even after including the spin-orbit interaction.  However, they are split by a small "Lamb shift" energy ħω₀ = 2πħ*1.06 GHz.
The ²P_½ state has the lower energy.
Consider only the subspace spanned by these two states.  With the appropriate choice of the zero of the energy scale the Hamiltonian in the absence of an external field is

(a)  Find the eigenvalues of H in an applied electric field E = E_field k.
(Use the matrix element <²P_½|z|²S_½> = 3a₀, where the Bohr radius a₀ ≈ 0.053 nm.)
Write down the low-field and the high-field approximations for the eigenvalues.
Sketch these eigenenergies vs. the field strength E_field.
What are the approximate low-field and high-field eigenstates of H?

(b)  How large should the electric field be for the energy shifts (i.e. Stark shifts) to become linear?
(c)  What is the value of the linear Stark shift (in MHz/(V/cm)) at large electric fields?

Solution:

• Concepts:
  A two-state system
• Reasoning:
  The dipole moment of the hydrogen atom is p = -q_er, where r is the vector pointing from the proton to the electron.
  The energy of a dipole in an external field is U = -p∙E.
  In the presence of an electric field H = H₀ + H₁ = H₀ + q_eE_fieldz.

H₁ = ħ 

with U = 3a₀q_eE_field/ħ.
(The dipole moment operator q_ez only couples states of opposite parity, i.e. S and P.)

• Details of the calculation:
  (a)  det(H - EI) = 0 yields
  E_± =  ħω₀/2 ± ħ((ω₀/2)² + U²)^½ = ½ħω₀(1 ± (1 + 4U²/ω₀²)^½).
  Limiting cases:
  |U| << ω₀,  E_± ≈ ½ħω₀(1 ± (1 + 2U²/ω₀²)).
  E₊ = ħω₀ + ħU²/ω₀,  E_- = -ħU²/ω₀.
  |ψ₊> = a|S> + b|P>,  b ≈ aU/ω₀ ≈ 0,  |ψ₊> ≈ |S>.
  |ψ_-> = a|S> + b|P>,  b ≈ -a( ω₀/U + U/ω₀), |b| >> |a|,  | ψ_-> ≈ |P>.

  |U| >> ω₀,  E_± ≈ ½ħω₀(1 ±  2U/ω₀).
  E₊ = ½ħω₀ + ħ|U|,  E_- = ½ħω₀ - ħ|U|.
  |ψ₊> = a|S> + b|P>,  b ≈ a,  |ψ₊> ≈ (|S> + |P>)/√2.
  |Ψ_-> = a|S> + b|P>,  b ≈ -a,  |ψ_-> ≈ (|S> - |P>)/√2.
  The Stark shift is quadratic in the field strength at low fields and linear at high fields.
  

  (b) E_± changes behavior from quadratic to linear in U when 4U²/ω₀² ≈ 1.
  Then U = 3a₀q_eE_field/ħ  = ω₀/2,  E_field = ħω₀/(6a₀q_e) ≈ 1.4*10⁴ V/m = 140 V/cm.
  
  (c)  At large fields the shift is proportional to U =  3a₀q_eE_field/ħ.
  ħU = hf.  f = ħU/h = U/(2π).
  (2π)^-1dU/dE_field = 3a₀q_e/h  ≈ 3.8 MHz/(V/cm).