Problem 2:

Suppose the potential energy between an electron and a proton had a term U₀(a₀ /r)² in addition to usual electrostatic potential energy -e²/r, where e² = q_e²/(4πε₀).
To the first order in U₀, where U₀ = 0.01 eV, by how much would the ground state energy of the hydrogen atom be changed?

Solution:

• Concepts:
  First order perturbation theory for non-degenerate states
• Reasoning:
  The ground state of the hydrogen atom is non-degenerate (neglecting spin).  We add the perturbation H' = U₀(a₀ /r)².
• Details of the calculation:
  The ground state wave function of the hydrogen atom is ψ(r) = (πa₀³)^-½exp(-r/a₀).
  The first order correction to the ground state energy is
  E₀¹ = [4πU₀a₀²/(πa₀³)] ∫₀^∞dr r² exp(-2r/a₀) (1/r²)  = 2U₀.
  The ground state energy is shifted up by 0.02 eV.

Problem 3:

Consider a system with a 4-dimensionl state space.  The Hamiltonian of the system is H₀.  The normalized eigenbasis of H₀ is {|1>, |2>, |3>, |4>}.  In that basis the matrix of H₀ is

The system is perturbed and the Hamiltonian becomes

with ε << 1.
Use perturbation theory to find the first order corrections to the eigenvalues and the zeroth order eigenstates of H.

Solution:

• Concepts:
  Stationary perturbation theory for degenerate states
• Reasoning:
  The eigenvalue E₀ is 3-fold degenerate.
• Details of the calculation:
  H = H₀ + H'.
  The eigenvalues of H₀ are E₀ with eigenvectors |1>, |2>, |3> and -3E₀ with eigenvector |4>.
  The eigenvectors |2> and |4> are also eigenvectors of H with eigenvalue E₀ ad -3E₀ respectively.
  We now diagonalize the perturbation H' in the subspace spanned by |1> and |3>.
  
  γ² - ε² = 0, γ = ±ε.  The first order corrections to the eigenvalues are εE₀ and -εE₀.
  The corresponding eigenstates are 2^-½(|1> + |3>) and 2^-½(|1> - |3>) respectively.

Problem 4:

Do a variational calculation.  A particle of mass M is subjected to a potential energy function of the form U = infinite for x < 0 and U = bx for x > 0
Let the trial wave function be ψ(x) = N x exp(-cx), where "c" is a variational parameter.  Using this function, construct a best estimate of the ground state energy.

Solution:

• Concepts:
  The variational method
• Reasoning:
  We are asked to use the variational method to estimate the ground state energy.
• Details of the calculation:
  H = p²/(2m) + bx,  x > 0.
  The eigenfunctions of H must be zero at x = 0 and x = ∞.  The derivative dψ/dx must be continuous at all x except x = 0.  Any function that satisfies these boundary conditions can be written as a linear combination of eigenfunctions of H and is an acceptable trial function giving <H> > E₀.
  We choose
  ψ(x) = N x exp(-cx),  N²∫₀^∞ x² exp(-2cx) dx = N²/(4c³) = 1,  N² = 4c³.
  Here c is the adjustable parameter.
  <H> =  N²∫₀^∞ x exp(-cx)[(-ħ²/2m)(∂²/∂x²) + bx]x exp(-cx) dx
  = N²(-ħ²/2m)∫₀^∞ (-2cx + c²x²) exp(-2cx) dx + N²b∫₀^∞ x³ exp(-2cx) dx
  = N²(ħ²/4mc  - ħ²/8mc + 6b/(2c)⁴).
  d<H>/dc = 0 --> c³ = 3bm/(2ħ²).
  Insert c back into the equation for <H>.
  E = 1.96 b^(2/3)ħ^(2/3)/m^(1/3).
  
  Exact solutions:
  E = 1.85 F^(2/3)ħ^(2/3)/m^(1/3).