Square potentials, bound states

Problem:

Consider the square potential well shown in the figure below.

(a)  Find the most general solution Φ(x) of the eigenvalue equation HΦ(x) = EΦ(x), (E < 0), in regions 1, 2, and 3 and apply boundary conditions.
(b)  Solve the equation that results from part (a) graphically, and find the conditions under which even and odd solutions exist.

Solution:

• Concepts:
  This is a "square potential" problem.  We solve HΦ(x) = EΦ(x) in regions where U(x) is constant and apply boundary conditions.
• Reasoning:
  We are given a piecewise constant potential and are asked to find bound-state solutions.
• Details of the calculation:
  Let -U₀ < E < 0.  The wave function must remain finite at x = ± ∞.  We therefore have:
  Φ₁(x) = B₁e^ρx,   Φ₂(x) = A₂e^ikx + A₂'e^-ikx,  Φ₃(x) = B₃'e^-ρx,
  with k² = (2m/ħ²)(E + U₀) and ρ² = (2m/ħ²)(-E).
  Φ and ∂Φ/∂x  are continuous at x = ± a/2  This implies:
  B₁e^-ρa/2 = A₂e^-ika/2 + A₂'e^ika/2,  B₃'e^-ρa/2 = A₂e^ika/2 + A₂'e^-ika/2.
  ρB₁e^-ρa/2 = ikA₂e^-ika/2 - ikA₂'e^ika/2,  -ρB₃'e^-ρa/2 = ikA₂e^ika/2 - ikA₂'e^-ika/2.
  Solving for A₂ and A₂^' in terms of B₁ we obtain
  A₂ = e^{(-ρ + ik)(a/2)}(ρ + ik)/(2ik)B₁,  -A₂' = e^{-(ρ + ik)(a/2)}(ρ - ik)/(2ik)B₁.
  Solving for A₂ and A₂^' in terms of B₃^' we obtain
  -A₂ = e^{-(ρ + ik)(a/2)}(ρ - ik)/(2ik)B₃',  A₂' = e^{(-ρ + ik)(a/2)}(ρ + ik)/(2ik)B₃'.
  This yields two equations for B₃^' in terms of B₁,
  -B₃' = [(ρ + ik)/(ρ - ik)]e^ikaB₁  and  -B₃' = [(ρ - ik)/(ρ + ik)]e^-ikaB₁,
  which can only simultaneously be satisfied if
  [(ρ - ik)/(ρ + ik)]² = e^2ika.
  There are two possible solutions.
  
  Solution 1:
  [(ρ - ik)/(ρ + ik)] = -e^ika
  But we also have
  [(ρ - ik)/(ρ + ik)] = [(ρ² + k²)^½e^-iθ/(ρ² + k²)^½e^iθ] = e^-i2θ, with cotθ = ρ/k.
  This implies
  e^-i2θ = -e^ika,  -2θ = ka - π,  cotθ = cot(π/2 - ka/2) = tan(ka/2) = ρ/k.
  How do you solve an equation like this for k?
  We can try a graphical solution. 
  Define  k₀² = 2mU₀/ħ² = k² + ρ².
  Then 1/cos²(ka/2) = 1 + tan²(ka/2) = (k² + ρ²)/k² = k₀²/k²,
  or |cos(ka/2)| = k/k₀  for all k for which tan(ka/2) ≥ 0.
  Note: We changed from a tangent to a cosine function for easier graphing.
  
  
  In regions (1), (2), (3,) ...  tan(ka/2) ≥ 0.  Three solutions exist for the given k₀ in the graph.  As we increase k₀ more solutions become possible.  For every k₀ at least one solution is possible.  There exists at least one bound state.  But we have not yet found the complete set of solutions.
  
  Solution 2:
  [(ρ - ik)/(ρ + ik)] = +e^ika
  This implies e^-i2θ = e^ika,  -2θ = ka,  cotθ = cot(-ka/2) = -cot(ka/2) = ρ/k.
  Then 1/sin²(ka/2) = 1 + cot²(ka/2) = (k² + ρ²)/k² = k₀²/k²,
  or |sin(ka/2)| = k/k₀ for all k for which cot(ka/2) ≤ 0. 
  We construct a similar graph to get more solutions.
  
  
   

  After we have found our solutions for k we can substitute  [(ρ - ik)/(ρ + ik)] = ±e^ika  back into the equations giving the relations between the A's and B's.  We find:
  if [(ρ - ik)/(ρ + ik)] = -e^ika  then A₂ = A₂', B₁ = B₃', Φ(-x) = Φ(x), the solutions are even;
  if  [(ρ - ik)/(ρ + ik)] = +e^ika  then A₂ = -A₂', B₁ = -B₃', Φ(-x) = -Φ(x), the solutions are odd.

  For k₀ ≤ π/a only one (even) solution exists.
  If π/a ≤ k₀ ≤ 2π/a  the first odd solution becomes possible. 
  If k₀ is very large, then the slope of the straight line is very small and solutions appear near every k = nπ/a (n = integer).  Consequently 
  E_n = n²π²ħ²/(2ma²) - U₀.
  These are the energy levels of the infinite square well.

Problem:

The one-dimensional square well shown in the figure rises to infinity at x = 0 and has a range "a" and a depth U₀.  Find the bound state solutions.

  Solution:

• Concepts:
  Square potentials
• Reasoning:
  We want to solve for the bound states in a square well.
• Details of the calculation:
  Let region 1 extend from x = 0 to x = a and region 2 from x = a to infinity.
  For bound states we have E < 0. 
  Define k² = (2m/ħ²)(E + U₀), ρ² = (2m/ħ²)(-E), and k₀² = (2m/ħ²)U₀.
  Note: I am assuming that U₀ is a positive number denoting the depth, the potential at the bottom of the well is -U₀.
  In region 1 we have Φ₁(x) = A sin(kx), since  Φ₁(0) = 0.
  In region 2 we have Φ₂(x) = Bexp(-ρx), since Φ₂(∞) = 0.
  At x = a we need hat Φ(a) and (∂/∂x)Φ(x)|_a are continuous.
  A sin(ka) = Bexp(-ρa).
  kA cos(ka) = -ρBexp(-ρa).
  Therefore cot(ka) = -ρ/k.
  1/sin²(ka) = 1 + cot²(ka) = (k² + ρ²)/k²  = k₀²/k².
  We can find a graphical solution by plotting |sin(ka)| and k/k₀ versus k.  The intersections of the two plots in regions where cot(ka) < 0 gives the values of k for which a solution exist.
  
  It is possible to have no solutions, when π/(2k₀a) > 1.

Problem:

Consider a particle with mass m in a potential well in one dimension as shown.  The potential energy U is zero between x = 0 and x = a, U₀ between x = a and x = 2a, and infinite otherwise.
(a)  Assume E = U₀.  Find the smallest value of U₀ for which that state exists.
(b)  Sketch the wave function and the probability density in the whole well for this state.

Solution:

• Concepts:
  This is a "square potential" problem.  We solve Hψ(x) = Eψ(x) in regions where U(x) is constant and apply boundary conditions.
• Reasoning:
  Assume U(x) = U = constant in certain regions of space. 
  In such a region the Schroedinger equation yields (∂²/∂x²)ψ(x) + (2m/ħ²)(E - U)ψ(x) = 0.
  (i)  Let E > U:  (∂²/∂x²)ψ(x) + k²ψ(x) = 0.   E - U = ħ²k²/(2m).
  The most general solution is ψ(x) = Aexp(ikx) + A'exp(-ikx), with A and A' complex constants.
  (ii)  Let E < U:  (∂²/∂x²)ψ(x) - α²ψ(x) = 0.   U - E = ħ²α²/(2m).
  The most general solution is ψ(x) = Bexp(αx) + B'exp(-αx), with B and B' complex constants.
  (iii)  Let E = U: (∂²/∂x²)ψ(x) = 0.  ψ(x) = Cx + C', with C and C' complex constants.
  (Note: A solution exists in the classically forbidden region.)
  At a finite step the boundary conditions are that ψ(x) and (∂/∂x)ψ(x) are continuous.
• Details of the calculation:
  (a)  Let 0 < x < a define region 1, a < x < 2a define region 2, and let E = U₀ = ħ²k²/(2m).  Then, to satisfy the boundary conditions, we need
  ψ₁(x) = Asin(kx),  ψ₂(x) = C(x - 2a),  A sin(ka) = -Ca,  kAcos(ka) = C, tan(ka) = -ka.
  We have π/2 < ka < π for the smallest non-zero value of ka,  ka = (π/2)(1 + ε).
  Use a calculator to find ε ~ 0.3.
  k = (π/(2a))(1 + ε), U₀ = ħ²k²/(2m).

  

  (b)  In region 1 ψ(x) = A sin(kx).  The wavelength λ = 2π/k = 4a/(1 + ε) is somewhat shorter than 4a.  In region 2 the wave function decreases linearly to zero.
  
  Wave function ψ(x):                        Probability density: