Problem 1:

Consider a particle of mass m in a one-dimensional potential energy well
U(x) = ½kx² for x > 0 and U(x) = ∞ for x < 0.  What is the ground-state energy?

Solution:

• Concepts:
  The eigenfunctions of the 1D harmonic oscillator
• Reasoning:
  The odd eigenfunctions of the harmonic oscillator with U(x) = ½ kx² = ½ mω²x² for all x are eigenfunctions of H for this potential which satisfy the boundary conditions.
• Details of the calculation:
  The eigenvalues therefore are E = (n + ½)ħω, n = odd,
  or rewriting, E = (2n + 3/2)ħω, n = 0, 1, 2, ... .
  The ground state energy is (3/2)ħω = (3/2)(k/m)^½.

Problem 2: 

Follow these steps to estimate the total energy of a helium atom.

(a)  What would be the total energy of a helium atom in its ground state in the approximation that you ignore completely the electrostatic force between the two electrons?

(b)  Interpret the sign of your answer in part (a).  What does it mean for the stability of the He atom?  How good of an estimate do you think this is?  Does it over or underestimate the energy?  Why?

(c)  Now consider the correction to the potential energy due to the Coulomb interaction between the electrons.  Assume that the electrons are classical particles in the first Bohr orbit (but remember Z = 2 for the helium nucleus).  The two electrons will always stay on opposite sides of the orbit from each other to minimize their energy.  What is the potential energy due to their interaction under this assumption?  Combine this number with your answer to part (a) to obtain the total energy of the two electrons.  Compare your result to the observed value of -79.0 eV. 

Solution:

• Concepts:
  Hydrogenic atoms, scaling rules
• Reasoning:
  E_n = -13.6 eV (μ'/μ)Z²/n²,  here μ' = μ = m_e.
• Details of the calculation:
  (a)  Ignoring the electron-electron interaction, we have two electrons with energy -E_I' = -E_I(μ'/μ)Z², where E_I is the ionization energy of hydrogen, E_I = 13.6 eV. 
  The Bohr radius for each electron is a₀' = a₀(μ/μ')(1/Z). 
  We can let μ' = μ = m_e and so the total energy of a helium atom in its ground state would be
  -2*4*13.6 eV = -108.8 eV and a₀' = a₀/2 for each electron.
  The He atom would be stable.  The assumption of no electron-electron interaction makes the total energy more negative and the He atom would be harder to ionize.
  
  (b)  The electron-electron interaction raises the potential energy by
  [1/(4πε₀)]q_e²/(2a₀') = 9*10⁹(1.6*10^-19)²/(0.529*10^-10) J = 27.2 eV.
  This would raise the total energy of the He atom to - 81.6 eV,
  much closer to the measured value of -79 eV.

Problem 3:

A particle of mass m in three dimensions is subjected to the radial potential  V(r) =[ħ²κ/(2m)]δ(r - R).  Here R > 0 and κ > 0 are parameters.  In what follows, we consider s-waves only. 
(a)  Compute the wave functions that are positive-energy solutions.
Hint: Parametrize the outside wave function such that its asymptotic (i.e. for r >> R) form is
ψ(r) = sin(kr + δ(k))/r  and its energy is  E = ħ²k²/(2m).  Derive a formula for the phase shift δ(k).
Hint:  Use cot(α + β)=(cotα cotβ - 1)/(cotα + cotβ).

(b)  Take the limit κ --> ∞  and compute the phase shift.  What physical system is this? 
Hint:  Look also at the inside (i.e. for r < R) wave function.

(c)  Assume and attractive potential, i.e.  κ < 0, and make an ansatz for a bound-state wave function with negative energy  E = -ħ²γ²/(2m).    Under what conditions (on R and κ) will the system have a bound state?

Solution:

• Concepts:
  Spherical δ-function potential
• Reasoning:
  In a central potential Vr) the wave function ψ_klm(r,θ,φ) = R_kl(r)Y_lm(θ,φ) = [u_kl(r)/r]Y_lm(θ,φ) is a product of a radial function R_kl(r) and the spherical harmonic Y_lm(θ,φ).  The differential equation for u_kl(r) is
  [-(ħ²/(2m))(∂²/∂r²) + ħ²l(l + 1)/(2mr²) + V(r)]u_kl(r) = E_klu_kl(r).
  To calculate δ(k) for this problem we must solve the radial equation for l = 0.
• Details of the calculation:
  (a)  (∂²/∂r² + k² - 2mV(r)/ħ²)u_k0(r) = 0,   k² = 2mE/ħ².
  Region 1:  V(r) = 0, u¹_k0(r) = C₁sin(kr).
  Region 2:  V(r) = 0, u²_k0(r) = C₂sin(kr + δ).
  Boundary conditions at a point where U is discontinuous:
  (i)  At a finite step the boundary conditions are that Φ(r) and (∂/∂r)Φ(r) are continuous.
  (ii)  At an infinite step (∂/∂r)Φ_ε(r)|_ε-->0 is discontinuous, but it has a finite discontinuity. 
  Therefore Φ_ε(r) remains continuous as ε --> 0.
  
  ∂Φ_2ε(R + ε)/∂r - ∂Φ_1ε(R- ε)/∂r = -(2m/ħ²)∫_R-ε^R+ε (E - (ħ²κ/(2m)) δ(r - R)) Φ(r) dr = κ Φ(R).
  
  u¹_k0(R) = u²_k0(%), C₁sin(kR) = C₂sin(kR + δ).
  ∂u²_k0(r)/∂r|_R+ε - ∂u¹_k0(r)/∂r|_R-ε = κ u(R).
  kC₁cos(kR) + κC₁sin(kR) = kC₂cos(kR + δ).
  kcot(kR) + κ = kcot(kR + δ).
  
  Using cot(kR + δ) = (cot(kR)cotδ - 1)/(cot(kR) + cot δ).
  cot(kR) + κ/k = (cot(kR) cot(δ - 1)/(cot(kR) + cot δ).
  cot²(kR) + (κ/k)cot(kR)  + 1 = -(κ/k)cot δ .
  -cot(δ(k)) = (k/κ) cot²(kR) + cot(kR) + (k/κ).
  
  (b)  As κ --> ∞  cot(δ(k)) = -cot(kR),   δ(k)) = -kR.
  The system is a spherical potential well with an infinitely hard boundary.
  The inside wave function is ψ_n00(r,θ,φ) = (1/(2πR)^½)sin(nπr/R)/r.
  Outside we have the l = 0 partial wave for scattering from a perfectly rigid sphere of radius R.
  
  (c)  For a bound state E is negative,  ρ² = -2mE/ħ² = γ².  (Let γ be a positive number.)
  (∂²/∂r² - γ²)u_k0(r) = 0 in regions 1 and 2.
  Solutions that satisfy boundary conditions at zero and infinity are
  Region 1:  u¹_k0(r) = C₁sinh(γr). 
  Region 2:  u²_k0(r) = C₂ exp(-γr)
  
  u¹_k0(R) = u²_k0(R), C₁sinh(γR) = C₂exp(-γR).
  ∂u²_k0(r)/∂r|_R+ε - ∂u¹_k0(r)/∂r|_R-ε = κ u(R).
  -γC₁cosh(γR) - γC₂cexp(-γR)  = -|κ|C₁sinh(-γR).
  
  γcosh(γR) + γsinh(γR))  = |κ|sinh(-γR).
  γexp(γR) = |κ|sinh(-γR) = |κ|(exp(γR) - exp(-γR))/2.
  γ = |κ/2|(1 - exp(-2γR)).
  γ, κ, and R are all positive numbers, so a solution for γ always exists.

Problem 4:

Submit this problem on Canvas as Assignment 3.  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 coherent state |λ> for a simple harmonic oscillator with frequency ω is an eigenstate of the lowering operator.  
a|λ> = λ|λ>.   (Note that λ can be a complex number.)
A coherent state is not an eigenstate of the Hamiltonian H, but can be expanded in terms of the eigenstates of H, |λ> = ∑₀^∞b_n|n>.
(a)  Show that, up to an overall normalization, a coherent state can be expressed as
|λ> = exp(λa^†)|0>.   Here a^† is the raising operator and |0> is the ground state.
Hint:  Evaluate a|λ> = a∑₀^∞b_n|n> and show that for a coherent state the relationship between b_n+1 and b_n is
b_n+1 = λb_n/√(n+1), or b_n = λⁿb₀/√(n!).

(b)  Start with a coherent state |λ₀> at time t = 0.  a|λ₀> = λ₀|λ₀>.
Evaluate |λ(t)> = U(t,0)|λ₀>.  Show that |λ(t)> is also a coherent state.  Show that the eigenvalue is λ(t) = λ₀exp(-iωt).

(c)  Calculate time dependent expectation values of coordinate and momentum for a simple harmonic oscillator in a coherent state.  Show that these quantities evolve as the classical coordinate and momentum of a harmonic oscillator.  Start with a coherent state at time t = 0 taking λ₀ to be a real number.

Useful relations:
a|n> = √n|n-1>,  a^†|n> = √(n+1)|n+1>,  x = (ħ/(2mω))^½<a^† + a>,  p = i(mωħ/2)^½<a^† - a>

Solution:

• Concepts:
  Evolution of coherent states
• Reasoning:
  A coherent state is a linear superposition of eigenstates of the harmonic oscillator with a specific relationship between the expansion coefficients.
• Details of the calculation:
  (a)  |λ> = ∑₀^∞b_n|n>, 
  a|λ> = ∑₀^∞b_na|n> =  ∑₁^∞b_n√n|n-1> = ∑₀^∞b_n+1√(n+1)|n> = λ∑₀^∞b_n|n>.
  b_n+1 = λb_n/√(n+1),  b_n = λⁿb₀/√(n!).
  |λ> = b₀∑₀^∞[λⁿ/√(n!)]|n>.  But |n> =   [(a^†)ⁿ/√(n!)] |0>.
  Therefore |λ> = b₀∑₀^∞[λⁿ(a^†)ⁿ/n!)|0> = b₀ exp(λa^†)|0>.
  
  (b) Let  |λ> = ∑₀^∞[λ₀ⁿ/√(n!)]|n>, i.e. let b₀ = 1.
  |λ(t)> = U(t,0)|λ₀> = exp(-iHt/ħ)|λ₀> = exp(-iHt/ħ) ∑₀^∞[λ₀ⁿ/√(n!)]|n>
  = ∑₀^∞[λ₀ⁿ/√(n!)] exp(-i(n + ½)ωt)|n> = exp(-iωt/2)∑₀^∞[λ₀ⁿ/√(n!)] exp(-iωt)ⁿ|n>
  = exp(-iωt/2) ∑₀^∞[λ(t)ⁿ/√(n!)]|n>, with λ(t) = λ₀exp(-iωt).
  |λ(t)> is an eigenstate of a as shown in part (a) with eigenvalue λ(t).
  
  (c)  <x(t)> = (ħ/(2mω))^½<a^† + a>(t) = (ħ/(2mω))^½(<aλ(t)|λ(t)> + <λ(t)|aλ(t)>)
   = (ħ/(2mω))^½(λ(t)* + λ(t)) = (2ħ/(mω))^½Re(λ₀exp(-iωt)) = (2ħ/(mω))^½λ₀cos(ωt).
  <x(t)> oscillates with frequency ω like the classical coordinate x.
  <p(t)> = i(mωħ/2)^½<a^† - a>(t) = (mωħ/2)^½(<aλ(t)|λ(t)> - <λ(t)|aλ(t)>)
   = i(mωħ/2)^½(λ(t)* - λ(t)) = (2mωħ)^½Im(λ₀exp(-iωt)) = -(2mωħ)^½λ₀sin(ωt).
  <p(t)> oscillates with frequency ω like the classical coordinate p.