Problem 1:

Find the ground state energy of the He atom using the variational method.

Useful information:
ψ_1s(r) = (1/πa₀³)^½exp(-r/a₀) for the hydrogen atom.
<ψ_1s|(1/r)|ψ_1s> = 1/a₀.
<ψ_1s|∇²|ψ_1s> = -1/a₀².
∫∫d³r d³r'|ψ_1s(r)|²|ψ_1s(r')|²(1/|r - r'|) = 5/(8a₀).

Solution:

• Concepts:
  The variational method
• Reasoning:
  We are asked to find the ground state energy of the He atom using the variational method.
• Details of the calculation:
  H = P₁²/(2m) + P₂²/(2m) - 2e²/r₁ - 2e²/r₂ + e²/r₁₂ = H₁ + H₂ + V₁₂.
  e² = q_e²/(4πε₀) in SI units.
  Let us use as a trial function  ψ(r₁,r₂) = (1/πa³)exp(-(r₁ + r₂)/a).
  We are guided by the ground state wave function of hydrogen, ψ(r) = (1/πa₀³)^½exp(-r/a₀).

  The adjustable parameter in our trial function is a = a₀/Z'. 
  ψ(r₁,r₂) is a product function, ψ(r₁,r₂) = Φ₁(r₁)Φ₂(r₂).
  Φ₁(r₁) = (1/πa³)^½exp(-r₁/a),  Φ₂(r₂) = (1/πa³)^½exp(-r₂/a).
  <ψ|H|ψ> = <Φ₁|H₁|Φ₁> + <Φ₂|H₂|Φ₂> + <ψ|V₁₂|ψ>.

  Φ₁(r₁) is the wave function of an electron in a hydrogenic atom with charge Z'.
  <Φ₁|p²₁/(2m)|Φ₁> = kinetic energy of an electron in a hydrogenic atom with charge Z'.
  <Φ₁|Z'e²/r|Φ₁> = potential energy of an electron in a hydrogenic atom with charge Z'.
  Therefore <Φ₁|H₁|Φ₁> = kinetic energy of an electron in a hydrogenic atom with charge Z', plus 2/Z' times the potential energy of an electron in a hydrogenic atom with charge Z'.
  For an electron in the ground state of a hydrogenic atom with charge Z' we have
  E_total = -Z'²E_I,  2<T> = -<V>, (virial theorem)  E_total = T + V = -<T>.
  Therefore <T> = Z'²E_I,  <V> = -2Z'²E_I.
  Using our trial wave function we therefore have <Φ₁|H₁|Φ₁> = <Φ₂|H₂|Φ₂> = Z'²E_I - 4Z'E_I, and
  <ψ|H|ψ> =  2Z'²E_I - 8Z'E_I + <ψ|V₁₂|ψ>.
  <ψ|V₁₂|ψ> = (e²/(π²a⁶)) ∫∫d³r₁d³r₂ exp(-2(r₁ + r₂)/a)/|r₁ - r₂|
  = (5/4)e²/(2a) = (5/4)Z'E_I,  where E_I = e²/(2a₀).**
  <H> = 2E_I(Z'² - 4Z' + (5/8)Z').
  d<H>/dZ' = 0 --> 2Z' - 4 + 5/8 = 0.  Z' = 2 - 5/16 = 1.688.
  <H> = -2.85*2E_I = -77.5 eV.
  The experimental value for the ground state energy of He is -78.8 eV
  The variational method often yields a very good estimate for the ground state energy of a system.  But an arbitrarily chosen trial ket can give a good approximation to the ground state energy and still be very different from the true eigenket.  So one must be very careful when using wave functions obtained by the variational method to calculate other physical quantities of the system.
  
  **How to evaluate the last integral if it is not given:
  We can recognize that the same integral can represent twice the electrostatic energy of a spherically symmetric charge distributions with charge density -q_eρ(r) = [1/(πa³)]exp(-2r/a). 
  For such a charge density the potential a r is
  Φ(r) = [-1/(4πε₀)][∫_v' dV' q_eρ(r')/|r - r'|
  The electrostatic energy of such a distributions is
  U = ½∫ρ(r)Φ(r)d³r = (e²/2)∫∫ρ(r)(ρ(r')/|r-r'|)d³rd³r'.
  Therefore U = (e²/2)∫∫d³r d³r'|ψ_1s(r)|²|ψ_1s(r')|²(1/|r - r'|) .
  
  The electrostatic energy of the charge distribution can also be found by evaluating a different integral.
  The electric field produced by the distribution is found from Gauss' law.
  E(r) = E(r)(r/r),  E(r)4πr² = Q_inside/ε₀ in SI units.
  The self energy of the distribution is given by U = (ε₀/2 )∫₀^∞ E²(r)4πr²dr.
  Here Q_inside = -4πq_e∫₀^rρ(r')r'²dr' = (q_e/2)[2 - e^-x(x² + 2x + 2)],
  with x = 2r/a,
  U = (e²/8)∫₀^∞ [2 - e^-x(x² + 2x + 2)]²dr/r² = (e²/(4a))∫₀^∞ (2 - e^-x(x² + 2x + 2))²dx/x²,
  U = (e²/(4a)) 5/4.
  <ψ|V₁₂|ψ> = (5/4)(e²/(2a)).

  We can also evaluate the integral directly.  We write
  1/|r₁ - r₂| = (4π/(2l+1))∑_l=0^∞∑_m=-l^l [r_<^l/r_>^l+1]Y^*_lm(θ₁,φ₁)Y_lm(θ₂,φ₂).

  To carry out the angular integration we then use
  ∫₀^πsinθ dθ∫₀^2πdφ Y_lm(θ,φ) = (4π)^½ ∫₀^πsinθ dθ∫₀^2πdφ Y₀₀⁸(θ,φ)Y_lm(θ,φ)  = (4π)^½δ_0lδ_0m.
  
  The integrals for the radial integration come straight out of common integral tables.

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.
  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:

Consider a particle of mass m placed in an infinite two-dimensional potential well of width a.
U(x,y) = 0 if 0 < x < a and 0 < y < a, U(x,y) = ∞ everywhere else.
The particle is also subject to a perturbation W described by
W(x,y) = W₀ for 0 < x < a/2 and 0 < y < a/2, W(x,y) = 0 everywhere else.
(a)  Calculate, to first order in W₀, the perturbed energy of the ground state.
(b)  Calculate, to first order in W₀, the perturbed energy of the first excited state.
Give the corresponding wave functions to 0th order in W₀.

Solution:

• Concepts:
  Stationary perturbation theory for non-degenerate and for degenerate states
• Reasoning:
  The ground state of two-dimensional infinite potential square well is not degenerate.  The first excited state is two-fold degenerate.
• Details of the calculation:
  (a)  H = H₀ + W.
  The eigenfunctions of H₀ are Φ_nl(x,y) = (2/a)sin(nπx/a)sin(lπy/a), with eigenvalues
  E = (n² + l²)π²ħ²/(2ma²).
  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> = (4W₀/a²)∫₀^a/2sin²(πx/a) dx ∫₀^a/2sin²(πy/a) dy = W₀/4.
  E₁ = π²ħ²/(ma²) + W₀/4 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 W in the subspace spanned by these two degenerate states.

	<Φ12|W|Φ12> - λ	<Φ12|W|Φ21>
	<Φ21|W|Φ12>	<Φ21|W|Φ21> - λ

  = 0.

 

 

<Φ₁₂|W|Φ₁₂> = (4W₀/a²)∫₀^a/2sin²(πx/a) dx ∫₀^a/2sin²(2πy/a) dy = W₀/4.
<Φ₂₁|W|Φ₂₁> = (4W₀/a²)∫₀^a/2sin²(2πx/a) dx ∫₀^a/2sin²(πy/a) dy  = W₀/4.
<Φ₁₂|W|Φ₂₁> = (4W₀/a²)∫₀^a/2sin(πx/a) sin(2πx/a) dx ∫₀^a/2sin(πy/a) sin(2πy/a) dy
= (4W₀/π²)([-cos(x)/2 - cos(3x)/6]|₀^π/2)² = (4W₀/π²)(½ + 1/6)² = (16W₀/(9π²)).
<Φ₂₁|W|Φ₁₂> = (16W₀/(9π²)).

(W₀/4 - λ)² - (16W₀/(9π²))² = 0,  λ = W₀/4 ± (16W₀/(9π²)).
E₁^1a = W₀/4 + (16W₀/(9π²)),   E₁^1b = W₀/4 - (16W₀/(9π²)) are the first order perturbation correction.
|Φ_1a> = (1/√2)(|Φ₁₂> + |Φ₂₁>),  |Φ_1b> = (1/√2)(|Φ₁₂> - |Φ₂₁>) are the corresponding wave functions to 0th order in W₀.
The perturbation removes the degeneracy.  There is splitting.

Problem 4:

The spin Hamiltonian for an electron (a spin-½ particle), in an external magnetic field is given by
H = -γS∙B,
where the gyromagnetic ratio γ = -q_e/m, q_e being the magnitude of the charge of the electron.
Suppose that the magnetic field consists of two components along the z and y axes, respectively, i.e. let B = B₀k + B₂j.  Let us consider the case that B₂ << B₀.  Using perturbation theory, evaluate the possible energies of the electron to second order in the ratio B₂/B₀.

Solution:

• Concepts:
  Spin ½, stationary perturbation theory
• Reasoning:
  Let us write H as H₀ + U ,where H₀ = ω₀S_z and U = ω₂S_y, where we have set ω₀ = q_eB₀/m
  and ω₂ = q_eB₂/m, respectively.  Since B₂ << B₀ we have U << H₀ and perturbation theory is an appropriate approximation method.
• Details of the calculation:
  The eigenvectors of H₀ are the spin-up and spin-down kets |+> and |->, respectively. 
  The unperturbed energy eigenvalues are E₀⁺ = ħω₀/2 and E₀^- = -ħω₀/2, respectively.
  In the basis spanned by |+> and |-> the matrix representation for U is given by
  <+|U|+> = <-|U|-> = 0,   <+|U|-> = -iħω₂/2,  <-|U|+> = iħω₂/2.
  The first-order shifts of the energy eigenvalues are zero, since clearly the diagonal matrix elements of U in the unperturbed basis vanish.
  The second order shifts are given by
  ΔE⁺ = |<-|U|+>|²/(E₀⁺ - E₀^-) = (ħω₂/2)²/(ħω₀/2 - (-ħω₀/2)) = (ħω₀/4)(ω₂/ω₀)²
  and
  ΔE^- = |<+|U|->|²/(E₀^- - E₀⁺) = (ħω₂/2)²/(-ħω₀/2 - ħω₀/2) = -(ħω₀/4)(ω₂/ω₀)².

Problem 5:

Suppose the potential energy between an electron and a proton had a term U₀(a₀ /r)² in addition to usual electrostatic potential energy -e²/r, where e² = q_e²/(4πε₀).
To the first order in U₀, where U₀ = 0.01 eV, by how much would the ground state energy of the hydrogen atom be changed?

Solution:

• Concepts:
  First order perturbation theory for non-degenerate states
• Reasoning:
  The ground state of the hydrogen atom is non-degenerate (neglecting spin).  We add the perturbation H' = U₀(a₀ /r)².
• Details of the calculation:
  The ground state wave function of the hydrogen atom is ψ(r) = (πa₀³)^-½exp(-r/a₀).
  The first order correction to the ground state energy is
  E₀¹ = [4πU₀a₀²/(πa₀³)] ∫₀^∞dr r² exp(-2r/a₀) (1/r²)  = 2U₀.
  The ground state energy is shifted up by 0.02 eV.