More Problems

Problem 1:

At t < 0 a particle with mass m is located in a one-dimensional potential described by
U(x) = ∞,  x < 0,  x > L,
U(x) = 0,  0 ≤ x ≤ L.
A potential of ΔV(x) = αx is added to 0 ≤ x ≤ L from t = 0 with α << 1.

(a)  What are the energy eigenvalues and eigenstates for the system at t < 0?
(b)  Calculate the first order correction to the lowest two energy eigenvalues at t --> ∞ by treating  ΔV(x) as a perturbation.

Solution:

• Concepts:
  First order perturbation theory for non-degenerate states
• Reasoning:
  We are asked to solve the unperturbed problem and find the energy correction due to the perturbation.
  H = H₀ + H'.  H₀ is the Hamiltonian of the infinite 1D well, H' = αx, 0 ≤ x ≤ L.
• Details of the calculation:
  (a)  The eigenfunctions of H₀ are Φ_n = (2/L)^½sin(nπx/L), the associated eigenvalues are
  E₀ = n²π²ħ²/(2mL²).
  
  (b)  The first order correction to the ground state energy is
  ΔE₁ = < Φ₁ |H'| Φ₁ > = (2/L)∫₀^L sin²(πx/L) αx dx = (2αL/π²)∫₀^π sin²(y) y dy = αL/2.
  Ground state energy to first order:  E₁ = π²ħ²/(2mL²) + αL/2.
  
  The first order correction to the energy of the first excited state is
  ΔE₂ = < Φ₂ |H'| Φ₂ > = (2/L)∫₀^L sin²(2πx/L) αx dx = (αL/(2π²))∫₀^2π sin²(y) y dy = αL/2.
  First excited state energy to first order:  E₂ = 4π²ħ²/(2mL²) + αL/2.

Problem 2:

In the WKB approximation, find the allowed energies that a ball of mass m, bouncing due to gravity on a perfectly reflecting surface, can have.  You can use the fact that for this problem the WKB approximation gives
∮p dq  = (n - ¼)h, ( n = 1, 2, ...), where p(q) is the momentum of the ball at the height q and the integral is over a full periodic path.

Solution:

• Concepts:
  The WKB approximation
• Reasoning:
  For this problem the WKB approximation requires that  ∫_q1^q2 pdq = ∫_q1^q2 ħkdq= (n - ¼)h/2.
  Here k² = (2m/ħ²)(E - U(q)).
• Details of the calculation:
  For stationary bound states we want
  ∫_qmin^qmax ħkdq = (n - ¼)πħ, n = 1, 2, ... .
  In our problem U(q) = mgq,  k² = (2m/ħ²)(E - mgq). 
  (Here q is the vertical coordinate.  The q-axis points upward and the origin lies at the surface.)
  We need (2m)^½∫₀^qmax (E - mgq)^½dq = (n - ¼)πħ,
  with mgq_max = E,  q_max = E/(mg).
  Therefore (2m)^½∫₀^E/(mg) (E - mgq)^½dq = (n - ¼)πħ.
  
  This integral evaluate to E^3/2 = (3/2)(n - ¼)πħmg/(2m)^½.

Problem 3:

Estimate the eigenvalues of this Hamiltonian using perturbation theory.  Do you have to use degenerate or non-degenerate perturbation theory?

Solution: 

• Concepts:
  Stationary perturbation theory for degenerate states
• Reasoning:
  Let H = H₀  + W.
  

  H₀ = ε  

  	2	0	0
  	0	3	0
  	0	0	2

  .  

   

  
  

  
  

  Let   

  	1
  	0
  	0

  ,   

  	0
  	1
  	0

  ,   

  	0
  	0
  	1

     denote the eigenbasis of H₀.

   

  
  

   

  	0
  	1
  	0

    

   is an eigenvector of H with eigenvalue 3ε.

   

   

  
  To find the other eigenvalues of H using first order perturbation theory we diagonalize the matrix

  	0	0.1
  	0.1	0

   .

   

  
  

 
• Details of the calculation:
  (-λ)² - 0.01 = 0,  λ = ± 0.1.
  

  The energy eigenvalues are 2.1 ε with eigenvector 2^-½

  	1
  	0
  	1

  ,   

  
  

  
  

  and 1.9 ε with eigenvector 2^-½

  	1
  	0
  	-1

  .