Problem 1:

Consider the one-dimensional harmonic oscillator with potential V(x) = mω²x²/2.  Use the variational principle to find the ground state energy employing a Gaussian wave function
ψ(x) = (b/π)^¼exp(-bx²/2),
which is already normalized to 1.  "b" is a variational parameter that you must optimize.
(a)  Find the kinetic energy <T> in this state.
(b)  Find the potential energy <V> in this state.
(c)  Optimize "b".
(d)  Find the optimal variational energy.  How does the result compare with the exact result for a 1D harmonic oscillator?

Solution:

• Concepts:
  The variational method
• Reasoning:
  The Hamiltonian is H = T + V = (-ħ²/(2m))(∂²/∂x²) + mω²x²/2.  The variational method often yields a very good estimate for the ground state energy of a system.
• Details of the calculation:
  (a)  <T> = (b/π)^½∫_-∞^∞ exp(-bx²/2)(-ħ²/(2m))(∂²/∂x²))exp(-bx²)dx
  = 2(b/π)^½(-ħ²/(2m))∫₀^∞ [-b + b²x²]exp(-bx²)dx
  = 2(b/π)^½[(-ħ²/(2m))b^½∫₀^∞ exp(-x²)dx - 2(b/π)^½[(b²ħ²/(2m))b^3/2∫₀^∞ x²exp(-x²)dx.
  ∫₀^∞ exp(-x²)dx = π^½/2.  ∫₀^∞  x²exp(-x²)dx = π^½/4.
  <T> = 2(b/π)^½[(-ħ²/(2m))b^½(π^½/2) - 2(b/π)^½[(b²ħ²/(2m))b^3/2(π^½/4).
  <T> = ħ²b/(4m).
  
  (b)  <V> = mω²(b/π)^½∫_-0^∞ x²exp(-bx²)dx = mω²(b/π)^½b^-3/2∫_-0^∞ x²exp(x²)dx
  = mω²(b/π)^½b^-3/2π^½/4.
  <V> = mω²/(4b).
  <H> = ħ²b/(4m) + mω²/(4b).
  
  (c)  d<H>/db = 0  -->  ħ²/(4m) - mω²/(4b²) = 0,  b = mω/ħ. 
  
  (d)  <E> = ħ²b/(4m) + mω²/(4b) = ħω/4 + ħω/4  = ħω/2, the exact ground-state energy.

Problem 2:

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.
  <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 3:

A particle of mass m is constrained to move in an infinitely deep, one-dimensional square well extending from 0 to +a. 
If this particle is under the influence of a delta-function perturbation H' = -αδ(x - a/2), where α is a constant.
(a) (i) Find the first-order correction to the allowed energies.
     (ii) Explain why the energies are not perturbed for even n.
(b)  Find the first three nonzero correction terms in the first-order expansion of the ground state wave function.

Solution:

• Concepts:
  First order perturbation theory for non-degenerate states, the infinite square well
• Reasoning:
  The energy level of the 1D infinite square well are non-degenerate.
• Details of the calculation:
  (a)  H = H₀ + H'.
  (i)  H₀|Φ_n> = E₀ⁿ|Φ_n>.   Φ_n(x) = (2/a)^½sin(nπx/a), E₀ⁿ= n²π²ħ²/(2ma²),  n = 1, 2, 3, ... .
  First order corrections to the allowed energies:
  E₁ⁿ = <Φ_n|H'|Φ_n> = (2α/a)∫₀^a dx sin²(nπx/a) δ(x - a/2) = (2α/a) sin²(nπ/2).
  E₁ⁿ = 0 for even n.  E₁ⁿ = (2α/a) for odd n.
  (ii)  The energies are not perturbed for even n because the probability of finding the particle at the position of the delta function is zero for even n.  The particle therefore does not "see" the perturbation.
  
  (b)  First-order perturbation theory yields for the ket
  |ψ₁> = |Φ₁> + ∑_n>1<Φ_n|H'|Φ₁>/(E₀¹ - E₀ⁿ) |Φ_n> ,
  <Φ_n|H'|Φ₁> = (2α/a)∫₀^a dx sin(πx/a)sin(nπx/a) δ(x - a/2) = (2α/a)sin(π/2)sin(nπ/2).
  <Φ_n|H'|Φ₁> = 0 for even n,
  <Φ_n|H'|Φ₁> = (2α/a) for odd n such that (n + 1)/2 is odd, (5, 9, 13, ...).
  <Φ_n|H'|Φ₁> = -(2α/a) for odd n such that (n + 1)/2 is even, (3, 7, 11, ...).
  E₀¹ - E₀ⁿ = -(n² - 1)π²ħ²/(2ma²). 
  To first order, the ground-state wave function is
  |ψ₁> = |Φ₁> + [4maα/(8π²ħ²)]|Φ₃> - [4maα/(24π²ħ²)]|Φ₅> + [4maα/(48π²ħ²)]|Φ₇> + ....

Problem 4:

A particle of mass m in an infinitely deep square well,
U = 0 for 0 < x < a,  0 < y < a,  U = infinite elsewhere,
is subject to a perturbation H' = -Bxy.
The positive constant B has units energy/length².
(a)  Calculate the zeroth order energy levels and energy eigenfunctions.
(b)  Determine the effects of the perturbation on the two lowest zeroth order energy levels.
Is there splitting of either of these levels?

Solution:

• Concepts:
  Stationary perturbation theory for non-degenerate and for degenerate states
• Reasoning:
  The ground state of a two-dimensional infinite potential square well is not degenerate.  The first excited state is two-fold degenerate.
• Details of the calculation:
  (a)  H = H₀ + H'.
  The eigenfunctions of H₀ are Φ_nl(x,y) = (2/a)sin(nπx/a)sin(lπy/a), with eigenvalues
  E = (n² + l²)π²ħ²/(2ma²).
  (b)  H' = -Bxy.
  For the ground state n = l = 1.  The  ground state is not degenerate.
  The unperturbed ground state energy is E₀⁰ = π²ħ²/(ma²).  The first order perturbation correction is
  E₁⁰ = <Φ₁₁|H'|Φ₁₁> = -(4B/a²)∫₀^asin²(πx/a) x dx ∫₀^asin²(πy/ a) y dy = -Ba²/4 .
  E₁ = ħ²/(ma²) - Ba²/4 to first order.
  
  b)  The first excited state is two-fold degenerate, E₁⁰ = 5π²ħ²/(ma²).
  We can have n = 2, l = 1, or n = 1, l = 2.
  We have to diagonalize the matrix of H' in the subspace spanned by these two degenerate states.

	<Φ12|H'|Φ12> - λ	<Φ12|H'|Φ21>
	<Φ21|H'|Φ12>	<Φ21|H'|Φ21> - λ

  = 0.

 

 

<Φ₁₂|H'|Φ₁₂> = -(4B/a²)∫₀^asin²(πx/a) x dx ∫₀^asin²(2πy/a) y dy = -Ba²/4 = -C
<Φ₂₁|H'|Φ₂₁> = -(4B/a²)∫₀^asin²(2πx/a) x dx ∫₀^asin²(πy/a) y dy  = -Ba²/4 = -C
<Φ₁₂|H'|Φ₂₁> = -(4B/a²)∫₀^asin(πx/a) sin(2πx/a) x dx ∫₀^asin(πy/a) sin(2πy/a) y dy

sin(x) sin(2x) = 2sin²(x) cos(x) = 2(1 - cos²(x)) cos(x)
= 2cos(x) - 2 cos³(x) = 2cos(x) - ½cos(3x) - 3/2 cos(x) = ½(cos(x) - cos(3x)).
∫cos(x) x dx = cos(x) + x sin(x).

<Φ₁₂|H'|Φ₂₁> = -Ba²(256/(81π⁴)) = -bC = <Φ₂₁|H'|Φ₁₂>.
C = Ba²/4, b = 1024/(81π⁴) = 0.13.
det(H' - λI) = 0.  (-C - λ)² - (bC)² = 0,  C + λ = ±bC,  λ = -C ± bC.
λ = -C + bC:  Φ = 2^-½ [Φ₁₂ - Φ₂₁],  E₂⁺ = 5π²ħ²/(2ma²) - C + bC.
λ = -C - bC:  Φ = 2^-½ [Φ₁₂ + Φ₂₁],  E₂^- = 5π²ħ²/(2ma²) - C - bC.
The perturbation removes the degeneracy.  There is splitting.