Problem 1:

Consider a neutron (mass m = 1.67*10^-27 kg) in the laboratory that is subjected to gravity (only the vertical motion in the z direction is of interest here) and a hard wall at z = 0.

(a)  What is the Hamiltonian that governs this system?
(b)  Make an estimate for the wave length of the quantum mechanical ground state of a gravitationally bound neutron.
(c)  Compute the action S(E) = ∮dz p(z) and quantize it to find the quantized energies.
(d)  How do energies scale with the principal quantum number?

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 q = z and k² = (2m/ħ²)(E - U(z)).
• Details of the calculation:
  (a)  H = T + U = p²/(2m) + U(z) = -(ħ²/(2m))∂²/∂z²  + U(z).
  U(z) = mgz for z ≥ 0, U(z) = ∞ for z < 0.
  
  (b)  One way to estimate λ is to start with the deBroglie wavelength λ = h/p. 
  Then T = h²/(2mλ²).  Assume that z_max is on the order of λ and write E = h²/(2mλ²) + mgλ.
  The ground state has the lowest energy.  Differentiate with respect to λ to find E_min.
  -h²/(mλ³) + mg = 0,  λ = (h²/(m²g))^1/3 = h^2/3/(m^2/3g^1/3) ~ 25 μm.
  The estimate for the ground state energy then becomes E_min ~ (3/2)h^2/3m^1/3g^2/3.
  
  (c)  The WKB approximation requires that ∫_cylcepdz = ∫_cylceħk(z)dz = (n - ¼)h or ∫_z1^z2 p(z)dz = (n - ¼)(h/2)
  if the potential has one vertical wall.
  Here n = 1, 2, ..., ∫_cylce denotes an integral over one complete cycle of the classical motion, z₁ and z₂ are the classical turning points, and k² = (2m/ħ²)(E - U(z)).
  For this problem ∫₀^zmax p(z)dz = (n - ¼)h/2, and k² = (2m/ħ²)(E - mgz).
  We need (2m)^½∫₀^zmax (E - mgz)^½dz = (n - ¼)πħ, with mgz_max = E,  z_max = E/(mg).
  Therefore (2m)^½∫₀^E/(mg) (E - mgz)^½dz = (n - ¼)πħ.
  This integral evaluate to E^3/2 = (3/4)(n - ¼)hm^½g/2^½. 
  E ~ 0.655 (n - ¼)^2/3 h^2/3m^1/3g^2/3 = 0.655*(n - ¼)^2/3*4.31*10^-31 J.
  
  (d)  For large n, E is proportional to n^2/3.

Problem 2:

Submit this problem on Canvas as Assignment 5.  If you used an AI as a Socratic tutor, submit a copy of your session leading to your solution and reflect on your session.  If you did not need any help or worked with another student, explain your reasoning, do not just write down formulas. 

A particle is moving in a one dimensional potential,
U(x) = ∞  for x < 0 and x > π,
U(x) = 0  for 0 < x < π/2,
U(x) = U₀ for π/2 < x < π.
(Units:  Assume all quantities are expressed in terms of consistent small length and mass units.)

(a)  The variational method can be used to find an upper bound for the ground-state energy of the particle.
For the trial function Ψ = N(2/π)^½(sin(x) + a sin(2x)), find the optimal value of the variational parameter a.
(b)  For U₀ = 0.1*ħ²/(2m), estimate the ground state energy of the particle in terms of ħ²/(2m).
(c)  Find the wave function and energy of the ground state for the cases where U₀ --> 0 and U₀ --> ∞.

Solution:

• Concepts:
  The variational method
• Reasoning:
  To find the optimal value of the variational parameter a, we have to minimize <ψ|H|ψ> with respect to a.
• Details of the calculation:
  (a)  Normalize the trial wave function:
  Ψ = N(Ψ₁ + aΨ₂), where Ψ₁ and Ψ₂ are normalized eigenfunctions of the infinite well with width π.
  E_n = n²π²ħ²/(2mπ²) = n²ħ²/(2m)
  N²(1 + a²) = 1,  N = (1/(1 + a²))^½.
  <H> = <ψ|H|ψ> = -(ħ²/(2m))∫₀^πψ*(∂²/∂x²)ψ dx + U₀∫_{π /2}^π ψ*ψ dx.
  <H> = N²E₀ + N²4a²E₀.
  + U₀[N²∫_π/2^πψ₁*ψ₁dx + N²a²∫_π/2^πψ₂*ψ₂ dx + 2N²a(2/π)∫_π/2^πsin(x) sin(2x) dx].
  Here E₀ = ħ²/(2m).
  <H> = E₀(1 + 4a²)/(1 + a²) + U₀*[0.5(1 + a²)/(1 + a²) - (8/(3π))a/(1 + a²)]
  = [U₀/(1 + a²)](x + ya + za²), with x = E₀/U₀ + 1/2, y = -8/(3π), z = 4E₀/U₀ + 1/2.
  d<H>/da = 0.
  [-2a/(1 + a²)][x + ya + za²)] + (y + 2za) = 0.
  [-2ax - 2a²y - 2za³)] + (y + 2za) + (a²y + 2za³)= 0.
  [-2ax - a²y)] + (y + 2za) = 0.
  a² + 2(x/y - z/y) a - 1 = 0.
  a = -(x/y - z/y) - [(x/y - z/y)² + 1]^1/2.
  a = -1.125 π E₀/U₀ + ((1.125 π E₀/U₀)² + 1)^½.
  (b)  For U₀ = 0.1*E₀ we have a = -11.25 π + ((11.25 π)² + 1)^½ = +0.01414,
  <H>_min = E₀*1.0494.
  Note:  The suggested form of Ψ is only an acceptable wave function as long as <H>_min > U₀.  For <H>_min < U₀ the wave function needs to exponentially decay in the region π/2 < x < π.
  (c)  As U₀ --> 0  the variational parameter a approaches 0 and N --> 1.
  The ground state wave function and energy will be that of the infinite well of width π.
  Ψ = (2/π)^½sin(x),  E = ħ²/(2m).
  As U₀ --> ∞  the variational parameter approaches 1 and N --> 1/√2.
  But Ψ = (1/π)^½(sin(x) + sin(2x)),  0 < x < π, Ψ = 0 elsewhere is not an acceptable trial wave function for the ground state, and the variational method will not yield an approximate ground state energy.
  The ground state wave function and energy as  U₀ --> ∞ will be that of the infinite well of width π/2.  
  Ψ = (4/π)^½sin(2x),  0 < x < π/2, Ψ = 0 elsewhere.  E = 2ħ²/m.

Problem 3:

A particle of mass m is bound in a 3D radially symmetric potential well which is weakly anharmonic, U(r) = ½mω²r² + εr⁴.
(a)  The ground state energy can be written as a power series E₀ = (3/2)ħω + O(ε) + O(ε²) + ... .   Use first order perturbation theory to determine the O(ε) term exactly.
(b)  The first order correction to the ground state wave function will mix the unperturbed Φ₀⁰(r) with a limited set of Φ₀ⁿ(r) of unperturbed basis states.  Which basis states are mixed at O(ε) in the ground state corrected to first order?  
Specify these using the Cartesian (n_x,n_y,n_z) labels for the harmonic oscillator.

[Note:  H₀ = H_0x + H_0y + H_0z.  H_0x = ħω (a^†a + ½),  a = αx + iβp,  a^† = αx - iβp, α = √(mω/(2ħ)),  β = 1/√(2mωħ).
For eigenstates |n> of H_0x we have a^†|n> = (n + 1)^½|n + 1>, a|n> = n^½|n - 1>.]

Solution:

• Concepts:
  First order perturbation theory for non-degenerate states
• Reasoning:
  The ground state of the 3D isotropic harmonic oscillator is not degenerate.
• Details of the calculation:
  (a) H₀ = p²/(2m) + ½mω²(x² + y² + z²).
  U = ε(x⁴ + y⁴ + z⁴ + 2x²y² + 2x²z² + 2y²z²).
  ΔE = <0_x,0_y,0_z|U|0_x,0_y,0_z> = first order energy correction to O(ε).
  <0_x,0_y,0_z|x⁴|0_x,0_y,0_z> = <0_x|x⁴|0_x> = <0|x⁴|0>.
  <0_x,0_y,0_z|x²y²|0_x,0_y,0_z> = <0_x|x²|0_x><0_y|y²|0_y> = <0|x²|0><0|y²|0>.
  Therefore, from symmetry, ΔE = 3ε<0|x⁴|0> + 6ε<0|x²|0>².
  
  x = (a + a^†)/(2α).
  <0|x²|0> = <0|(a + a^†)²|0>/(2α)².  <0|x⁴|0> = <0|(a + a^†)⁴|0>/(2α)⁴.
  x² = (aa + a^†a^† + a^†a  + aa^†)/(4α²).
  <0|x²|0> = <0|aa + aa^† + a^†a + a^†a^†|0>(2α)² = <0|aa^†|0>(2α)² = 1/(2α)².
  
  <n|(a + a^† )⁴|n> has 16 terms, but only those with equal numbers of a and a^† will be non-zero.  Only six terms survive.
  <n|(a + a^†)⁴|n> = <n|(aaa^†a^† +  aa^†aa^† + aa^†a^†a +a^†aaa^† + a^†aa^†a + a^†a^†aa)|n>.
  <n|(a + a^† )⁴|n> = (n + 1)(n + 2) + (n + 1)² + 2n(n + 1) + n² + n(n - 1).
  <0|x⁴|0> = 3/(2α)⁴.
  ΔE = 3ε<0|x⁴|0>/(2α)⁴ + 6ε<0|x²|0>²(2α)⁴ = 15ε/(2α)⁴ = 15εħ²/(4m²ω²).
  
  We can also evaluate ΔE using the ground state wave function for 3D radially symmetric harmonic oscillator potential.
  The normalized ground state wave function of the 1-dimensional harmonic oscillator is
  Φ₀(x) = (mω/(πħ))^¼exp(-½mωx²/ħ).  The ground state wave function for the 3D therefore is
  Φ₀(r) = (mω/(πħ))^3/4exp(-½mωr²/ħ).
  ΔE = ε(mω/(πħ))^3/4 4π ∫r⁶dr exp(-mωr²/ħ).   ∫x²ⁿdx exp(-ax²) = [1/(2ⁿ⁺¹aⁿ)][1*3*5*...*(2n-1)](π/a)^1/2.
  ΔE =15εħ²/(4m²ω²).
  
  (b)  The first correction to a non-degenerate eigenvector |0,0,0> is
  |ψ₀¹> = ∑_nx,ny,nz≠0 b_nx,ny,nz|Φ_nx,ny,nz> .
  b_nx,ny,nz = <n_x,n_y,n_z|U|0_x,0_y,0_z>/(E₀ - E_nx,ny,nz).
  <0|x²|n_x,n_y,n_z> = 0 unless n_x = 0 or 2, and n_y = n_z = 0.
  <0|x⁴|n_x,n_y,n_z> = 0 unless n_x = 0, 2 or 4 and n_y = n_z = 0.
  Basis states are mixed at O(ε) in the ground state corrected to first order therefore are the ones shown in the table below.
  
  

  nx	ny	nz	_	_	_	nx	ny	nz
  0	0	0				2	2	0
  2	0	0				2	0	2
  0	2	0				0	2	2
  0	0	2
  4	0	0
  0	4	0
  0	0	4

  [Aside:  Different eigenbasis for H₀ exist, for example {|n, l, m>}, with eigenvalues E_nl = (3/2 + 2(n - 1) +  l)ħω, n = 1, 2, ... , l = 0, 1, 2, ... .
  Using this basis, states that are mixed at O(ε) in the ground state corrected to first order are
  |n, l, m> = |1, 0, 0>,  |2, 0, 0>, and |3, 0, 0>.
  These are states with the same energy eigenvalues and l = 0.]