Stationary perturbation theory, non-degenerate states

Perturbed oscillator

Problem:

A one-dimensional harmonic oscillator has momentum p, mass m, and angular frequency ω.  It is subject to a perturbation U = bx⁴, where b is a suitable parameter, so that perturbation theory is applicable.
(a)  Derive an expression for a and a^† in terms of x and p, using the fact that they satisfy
[a, a^†] = 1, H = ħω (a^†a + ½).
(b)  Evaluate the first-order shift of the energy eigenvalues due to the above perturbation using creation and annihilation operators.
(c)  Does your result suggest that something "goes wrong" for large n even for small b-values?  Argue qualitatively why this may be the case.

Solution:

• Concepts:
  First order perturbation theory for non-degenerate states
• Reasoning:
  The eigenstates of the 1D harmonic oscillator are not degenerate.
• Details of the calculation:
  (a) H = p²/(2m) + mω²x²/2.
  Assume a = αx + iβp,  a^† = αx - iβp,
  [a, a^†] = 2αβħ = 1.
  H = ħω (a^†a + ½) = ħω[(αx)² + (βp)² - αβħ + ½) = ħω[(αx)² + (βp)²).
  α = √(mω/(2ħ)),  β = 1/√(2mωħ).
  (b)  x = (a + a^†)/(2α),  p = -i(a - a^†)/(2β),
  x² = (aa + a^†a^† + a^†a  + aa^†)/(4α²).
  For eigenstates |n> of H we have a^†|n> = (n + 1)^½|n + 1>, a|n> = n^½|n - 1>.
  <n|x⁴|n> = <n|(a + a^†)⁴|n>/(2α)⁴.
  <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).
  <n|x⁴|n> = (6n² + 6n + 3) ħ²/(4m²ω²) = (n² + n + ½)2ħ²/(4m²ω²).
  ΔE_n = (n² + n + ½)2bħ²/(4m²ω²).
  (c)  Even for small b ΔE_n will become large for large n, since it increases quadratically with n. First-order perturbation theory will no longer be a valid approximation.

Problem:

A charged particle of mass m and charge q is sitting in a harmonic potential U₀(x) = ½mω²x².  A weak constant electric field E is applied in the +x-direction, so that the potential is perturbed by U₁(x) = -qEx.
(a)  Show that there is no change in the energy levels to first order in E.
(b)  Calculate the second-order change in the energy levels.
(c)  This problem can be solved exactly by changing variables to x' = x - qE/(mω²).  Show that the exact energy levels agree with your results in parts (a) and (b).

The following may be used without proof:
x = (ħ/(2mω))^½(a^† + a),  a|n> = n^½|n-1>,  a^†|n> = (n+1)^½|n+1>.

Solution:

• Concepts:
  The harmonic oscillator, first and second order perturbation theory.
• Reasoning:
  We are asked to  find the first and second order energy correction due to the perturbation.

• Details of the calculation:
  (a)  Let H₀ = p²/(2m) + U₀(x).  Let {|n>} denote the orthonormal eigenbasis of H₀.
  H₀|n> = E₀ⁿ|n> = (n + ½)ħω|n>,  <n|n'> = δ_nn'.
  The eigenvalues of H₀ are not degenerate.
  Let H = H₀ + U₁.  The eigenvalues of H may be written as Eⁿ = E₀ⁿ + E₁ⁿ + E₂ⁿ + ...  . 
  E₁ⁿ = <n|U₁|n> = -qE<n|x|n> = -qE(ħ/(2mω))^½<n|(a + a^†)|n>
  = -qE(ω²2m)^-½[(nħω)^½<n|n - 1> + ((n+1)ħω)^½<n|n + 1>] = 0.
  
  (b)  E₂ⁿ = ∑_n'≠n |<n'|U₁|n>|²/(E₀ⁿ - E₀^n') = q²E²∑_n'≠n |<n'|x|n>|²/(E₀ⁿ - E₀^n').
  <n'|x|n> = (ω²2m)^-½[(nħω)^½<n'|n - 1> + ((n + 1)ħω)^½<n'|n + 1>]
  =  (ω²2m)^-½[(nħω)^½δ_n'n-1 + ((n + 1)ħω)^½δ_n'n+1].
  E₂ⁿ = q²E²(ω²2m)^-1[(nħω)/(E₀ⁿ - E₀^n-1) + ((n + 1)ħω)/(E₀ⁿ - E₀ⁿ⁺¹)
  = q²E²/(ω²2m)[n - n + 1] = -q²E²/(ω²2m).
  (c)  Let x' = x - qE/(mω²).
  ½mω²x'² = ½mω²x² - qEx + q²E²/(ω²2m)
  Then H = p²/(2m) + ½mω²x'² - q²E²/(ω²2m)
  H' = H + q²E²/(ω²2m) = p²/(2m) + ½mω²x'².
  The eigenvalues of H' are (n + ½)ħω,  the eigenvalues of H are therefore (n + ½)ħω - q²E²/(ω²2m).

Problem:

Consider a single electron moving with an anharmonic potential energy given by
U(x) = ½ kx² + k'x⁴.  Let H₀ be the Hamiltonian of the system when k' = 0.
(a)  Use perturbation theory to find the energies of the ground and excited states to first order, assuming that k' << k.
(b)  Expand the ground state to first order in terms of the eigenstates {|n>} of H₀.

Solution:

• Concepts:
  The harmonic oscillator, first  order perturbation theory for non-degenerate states.
• Reasoning:
  We are asked to find first order corrections due to a given perturbation.

• Details of the calculation:
  (a)  H = H₀ + H₁.
  The eigenvalues of the 1D Hamiltonian H₀ = p_x²/(2m) + ½kx² = p_x²/(2m) + mω²x²/2 are
  E_n⁰ = (n + ½)ħω, with  ω² = k/m.  We denote the eigenstates by {|n>}, n = 0, 1, 2, ... .
  Define a = αx + iβp_x,  a^† = αx - iβp_x,  α =√(mω/(2ħ)),  β =1/√(2mωħ).
  [a, a^†] = 2αβħ = 1.  Then
  H_x = ħω (a^†a + ½) = ħω[(αx)² + (βp_x)² - αβħ + ½) = ħω[(αx)² + (βp_x)²).
  We have x = (a + a^†)/(2α),  p_x = -i(a - a^†)/(2β).
  For eigenstates |n> of H_x we have a^†|n> = (n + 1)^½|n + 1>, a|n> = n^½|n - 1>.
  With H = H₀ + H₁, to first order the energy eigenvalues are E_n⁰ + E_n¹, with
  E_n¹= <n|H₁|n>.
  H₁ = k'x⁴.
  x² = (1/(4α²))(aa + aa^† + a^†a + a^†a^†).
  <n|x⁴|n> = (1/(16α⁴)) <n|(a + a^†)⁴|n> has 16 terms, but only those with equal numbers of a and a^† can be non-zero.  Only 6 terms survive.
  <n|aaa^†a^†|n> = (n + 1)(n + 2),  <n|a^†a^†aa|n> = n(n - 1),  <n|aa^†aa^†|n> = (n + 1)², 
  <n|a^†aa^†a|n> = n², <n|aa^†a^†a|n> = n(n + 1), <n|a^†aaa^†|n> = n(n + 1).
  E_n¹ = [k'ħ²/(4m²ω²)](6n² + 6n + 3).
  The energies to first order are E_n = (n + ½) ħω + [k'ħ²/(4m²ω²)](6n² + 6n + 3).

  (b)  The first order corrections to the eigenstates are
  |0¹> = ∑_n'≠0 [<n'|H₁|0>/(E₀⁰ - E_n'⁰)]|n'>.
  <n'|x⁴|0> has 16 terms, but only those with more a^† than a and a^† on the right can be nonzero.  Four terms survive.

  <n'|x⁴|0> = (1/(16α⁴))<n'|(aa^†a^†a^† + a^†a a^†a^† +  a^†a^†aa^† + a^†a^†a^†a^†)|0>.
  (aa^†a^†a^† + a^†a a^†a^† +  a^†a^†aa^† + a^†a^†a^†a^†)|0> = (18^½ + 8^½ + 2^½)|2> + 24^½|4>.

  |0¹> = -|2>[k'ħ²/(4m²ω²)](18^½ + 8^½ + 2^½)/(2ħω) - |4>[k'ħ²/(4m²ω²)]24^½/(4ħω)
  = -[k'ħ/(8m²ω³)](8.49 |2> + 2.45|4>).
  To first order the ground state is |0> - [k'ħ/(8m²ω³)](8.49 |2> + 2.45|4>).

Perturbed hydrogenic atoms

Problem:

Solution:

• Concepts:
  First order perturbation theory for non-degenerate states
• Reasoning:
  The ground state of the hydrogen atom with a point nucleus is non-degenerate (neglecting spin).  If the proton has a finite size, then the potential inside the proton differs from a pure Coulomb potential.  We consider this difference a small perturbation of the system with a point nucleus.

• Details of the calculation:
  The ground state is not degenerate.  The potential energy of an electron outside a uniformly charged sphere of radius r₀ and total charge q_e is U(r) = -e²/r,  r > 0 where  e² = q_e²/(4πε₀).
  Inside the sphere the potential energy is
  U(r) = [e²/(2r₀)][(r/r₀)² - 3].
  From Gauss' law we know that  E(r < r₀) = q_inside/(4πε₀r²) = q_er/(4πε₀r₀³)  (SI units).
  Φ(r) = Φ(r₀) + [q_e/(4πε₀r₀³)]∫_r^r0rdr = q_e/(4πε₀r₀) + q_e(r₀² - r²)/(8πε₀r₀³).
  U(r) = -q_eΦ(r) = [e²/(2r₀)][(r/r₀)² - 3].
  
  Let H₀ = p²/(2μ) - e²/r,  H = p²/(2μ) + [e²/(2r₀)][(r/r₀)² - 3] = H₀ + H'  (r < r₀).
  H' = e²r²/(2r₀³) - 3e²/(2r₀) + e²/r.
  E₁⁰ = <100|H'|100> = ∫₀^2πdφ∫₀^πsinθ dθ Y₀₀(θ,φ)∫₀^r0r²dr |R₁₀(r)|² H'
  ≈ ∫₀^r0r²dr |R₁₀(0)|² H',
  since r₀ << a₀.
  R₁₀(r) = (2/a₀^3/2)exp(-r/a₀),  R₁₀(0) = 2/a₀^3/2. 
  E₁⁰ = (4/a₀³)[ e²r₀⁵/(10r₀³) -  3e²r₀³/(6r₀) +  e²r₀²/2] = (2/5)(e²/a₀)(r₀/a₀)²
  = (4/5)E_I(r₀/a₀)² = (4/5)(13.6 eV)(10^-15/0.53*10^-10)² = 4*10^-9 eV.

Problem:

The correctly normalized hydrogen ground state wave function in 3D is given by
ψ₀(r) = (πa₀³)^-½exp(-r/a₀),
where a₀ = ħ²/(m_ee²) is the Bohr radius, which is numerically ~0.529 Ǻ.
(a)  Confirm that this does indeed satisfy the radial Schroedinger equation for hydrogen, and that the wave function is normalized to ∫d³r|ψ(r)|² = 1.
(b)  Two identical ions are introduced on the z-axis at locations z = +d and -d.  Assuming that the effect of each ion on the electron can be treated as a point interaction,
U_{e - ion} = ν₀ δ(r - r_ion),
calculate the change in the hydrogen atom's ground state energy using first order perturbation theory.

Solution:

• Concepts:
  The Schroedinger equation, stationary perturbation theory
• Reasoning:
  We are asked to use first order perturbation theory to calculate the perturbed atom's ground state energy.
• Details of the calculation:
  (a)  The time-independent Schroedinger equation for the hydrogen atom is
  H Φ(r) = [-(ħ²/(2μ))∇² -e2/r] Φ(r) = E Φ(r),
  where μ = m_e for infinite proton mass.. 
  Writing  Φ_nlm(r) = R_nl(r)Y_lm(θ,φ) = [u_nl(r)/r]Y_lm(θ,φ)
  we find  -[ħ²/(2μ)][(∂²/∂r²) + 2μl(l+1)/ħ²]u_lm (r) - [e²/r]u_lm0(r) = E_lmu_lm (r).
  Since Φ₁₀₀(r) = (πa₀³)^-½ exp(-r/a₀), we have u₁₀(r) = 2 a₀^-3/2 r exp(-r/a₀).
  We show that u₁₀(r) satisfies
   -[ħ²/(2μ)][∂²/∂r²] u₁₀(r)  - [e²/r] u₁₀(r) = E₁₀u₁₀(r) by simply differentiating.
  (∂²/∂r²)rexp(-r/a₀) = [-(2/a₀) + (r/a₀²)]exp(-r/a₀)
  Therefore [ħ²/(a₀μ)] - [rħ²/(2a₀²μ)] - e² = Er.
  This equation is satisfied if a₀ = ħ²/(μe²) and E = -ħ²/(2a₀²μ) = - μe⁴/(2ħ²).
  Normalization: ∫ d³r |ψ(r)|² = 4π∫₀^∞ r²dr (πa₀³)^-1 exp(-2r/a₀) = ½ ∫₀^∞ r'²dr'exp(-r') = 1.

  (b)  The ground state of the hydrogen atom is non-degenerate, we use first-order stationary perturbation theory for non-degenerate energy eigenvalues.
  The first order correction to the ground state energy is E₁¹ = <Φ₁₀₀|W|Φ₁₀₀>.

  E₁¹ = ∫_-1¹dcosθ ∫₀^∞ r²dr |R₁₀(r)|²|Y₀₀(θ,φ)|² ν₀ δ(r - d) δ(cosθ - 1)/r²
  + ∫_-1¹dcosθ ∫₀^∞ r²dr |R₁₀(r)|²|Y₀₀(θ,φ)|² ν₀ δ(r - d) δ(cosθ + 1)/r²
  = ν₀ |R₁₀(d)|²|Y₀₀(θ = 0)|² + ν₀ |R₁₀(d)|²|Y₀₀(θ = π)|² = 2ν₀ |ψ₁₀₀(d)|²
  = 2n₀exp(-2d/a₀)/(πa³).
  Alternatively:
  E₁¹ = ∫₀^∞dx∫₀^∞dy∫₀^∞dz |ψ₁₀₀ (r)|²ν₀ δ(x)δ(y) (δ(z - d) + δ(z + d)) = 2ν₀ |ψ₁₀₀(d)|².

Problem:

(a)  If the Hamiltonian for a system may be written as  H = H₀ + H₁ where H₁ is a small perturbation and H₀ψ₀ = E₀ψ₀, then the effect of H₁ gives rise to an energy correction term ΔE₀.  Deduce an expression for the energy correction term ΔE₀ if ψ₀ is non-degenerate.
(b)  The relativistic dependence of mass on velocity introduces into the energy operator for the hydrogen atom a correction term for the state
ψ₀ = (πa₀³)^-½exp(-r/a₀)
that may be written as   (-E₀²/(2mc²))[1 + 2a₀/r]².
Determine, using first order perturbation theory, the consequent level shift.

Solution:

• Concepts:
  Stationary perturbation theory.
• Reasoning:
  We are asked to derive the expression for the first order energy correction in perturbation theory.
• Details of the calculation:
  (a)  Let H = H₀ + λW.  Let |ψ₀⁰> be a non-degenerate eigenstates of H₀.
  H₀|ψ₀⁰> = E₀⁰|ψ₀⁰>.
  Assume that λW = H₁, and that λ is small.
  Let H|ψ₀> = E⁰|ψ₀>.
  Since λW is small, we assume that E⁰ and |ψ₀> can be expanded as a power series in λ.
  E⁰ = E₀⁰ + λE₁⁰ + λ²E₂⁰ + ... ,
  |ψ₀> = |ψ₀⁰> + λ|ψ₀¹> + λ²|ψ₀²> + ... .
  We may then write
  (H₀ + λW)(|ψ₀⁰> + λ|ψ₀¹> +λ²|ψ₀²> + ...)
  = (E₀⁰ + λE₁⁰ + λ²E₂⁰ + ...)(|ψ₀⁰> + λ|ψ₀¹> + λ²|ψ₀²> + ...).
  This equation is must be valid over a continuous range of λ.  Therefore we equate coefficients of equal powers of λ on both sides to obtain a series of equations that represent successively higher orders of the perturbation.
  (i)  (H₀ - E₀⁰)|ψ₀⁰> = 0
  (ii)  (H₀ - E₀⁰)|ψ₀¹> = (E₁⁰ - W)|ψ₀⁰>
  Therefore  <ψ₀⁰|(H₀ - E₀⁰)|ψ₀¹> = 0 = <ψ₀⁰|( (E₁⁰ - W)|ψ₀⁰>, or
  E₁⁰ = <ψ₀⁰|W)|ψ₀⁰> = ΔE₀.
  (b)  ΔE₀ = 4π(-E₀²/(2mc²))(πa₀³)^-1∫₀^∞r²dr exp(-2r/a₀)(1 + 4a₀/r + 4 a₀²/r²)
  = -4(E₀²/(2mc²)) ∫₀^∞x²dx exp(-2x)(1 + 4/x + 4/x²)
  = -4(E₀²/(2mc²))[¼  + 1 + 2] = -13*(E₀²/(2mc²)).

Problem:

A muonic atom is one in which an atomic electron is replaced by a muon.  The muon is 209 times more massive than the electron.
(a)  Compute the energy of the 2p - 1s muonic transition in ²⁰⁸Pb (Z = 82) under the assumption that Pb is a point nucleus.   Make reasonable assumptions and explain your assumptions.   Compare your result with the observed value of 5.8 MeV.
(b)  Use the transition-energy values computed and given in part (a) and simple scaling rules for hydrogenic atoms to give an order-of-magnitude estimate of the nuclear radius of Pb (whose actual nuclear charge radius is ~ 6 fm).
(c)  Use perturbation theory to calculate the first-order shift in the ground-state energy of an electron in hydrogenic  ²⁰⁸Pb (Z = 82) due to the finite size of this nucleus.  Assume the nucleus is a uniformly charged sphere.  Why is this not a valid approach for the muonic Pb atom?

The ground state wave function of the hydrogen atom is |1,0,0> = (πa₀³)^-½exp(-r/a₀).

Solution:

• Concepts:
  Hydrogenic atoms, perturbation theory
• Reasoning:
  We use scaling rules for hydrogenic-atom wave functions and perturbation theory to calculate the first-order shift in the ground-state energy of an electron in a hydrogenic atom due to the finite size of the nucleus.
• Details of the calculation:
  To find the eigenfunctions and eigenvalues of the Hamiltonian of a hydrogenic atom with a point nucleus we replace in the eigenfunctions of the Hamiltonian of the hydrogen atom 
  a₀ by a₀' = ħ²/(μ'Ze²) = a₀(μ/μ')(1/Z),
  and in the eigenvalues of the Hamiltonian of the hydrogen atom we replace
  E_I by E_I' = μ'Z²e⁴/(2ħ²) = E_I(μ'/μ)Z².
  (a)  Z = 82,  μ'/μ = 209,  E_n  = -82²*209*13.6 eV = -19.11 MeV/n².
  The 2p - 1s transition energy is E₂ - E₁ = 14.33 MeV.
  We neglect the presence of the electrons, because the average distance of the muon from the nucleus is 1/209 times the average distance of an electron with the same quantum numbers (n,l,m) from the nucleus.  The muon making a 2p - 1s transition therefore is barely shielded by the electrons and the effective charge is the full charge of the point nucleus.
  The calculated transition energy is ~2.5 times bigger than the observed transition energy.

  (b)  Assume the effective charge is less than the full charge of the nucleus because the most likely radial distance places the muon inside the nucleus in the 1s state.  Use the measured transition energy to find Z_eff.  Assume only the 1s energy is significantly altered.
  5.8 MeV = -Z_eff²*209*13.6 eV - Z²*209*13.6/4 eV.
  Z_eff = 61.  Z = 1.34 Z_eff.
  The most likely radial distance of the muon from the center of the nucleus is
  a₀' = a₀/(209*Z_eff) = 5.29*10^-11 m/(209*61) = 4.15*10^-15 m.
  Assume that the radius of the nucleus is proportional to Z³.
  R = (82/61)^(1/3) a₀' = 4.6 fm.  This is reasonable close to 6 fm for such a crude approximation.
  
  (c)  First order perturbation theory for non-degenerate states
  The ground state of the hydrogenic atom with a point nucleus is non-degenerate.  The potential energy of an electron outside a uniformly charged sphere of radius r₀ and total charge Zq_e is U(r) = -Ze²/r,  r > 0, with e² = q_e²/(4πε₀).
  Inside the sphere the potential energy is
  U(r) = [Ze²/(2r₀)][(r/r₀)² - 3].
  From Gauss' law we know that  E(r < r₀) = q_inside/(4πε₀r²) = Zq_er/(4πε₀r₀³)  (SI units).
  Φ(r) = Φ(r₀) + [Zq_e/(4πε₀r₀³)]∫_r^r0rdr = Zq_e/(4πε₀r₀) + Zq_e(r₀² - r²)/(8πε₀r₀³).
  U(r) = -q_eΦ(r) = [Ze²/(2r₀)][(r/r₀)² - 3].
  
  Let H₀ = p²/(2μ) - Ze²/r,  H = p²/(2μ) + [Ze²/(2r₀)][(r/r₀)² - 3] = H₀ + H'  (r < r₀).
  H' = Ze²r²/(2r₀³) - 3Ze²/(2r₀) + Ze²/r.
  E₁⁰ = <100|H'|100> = ∫₀^2πdφ∫₀^πsinθ dθ Y₀₀(θ,φ)∫₀^r0r²dr |R₁₀(r)|² H'
  ≈ ∫₀^r0r²dr |R₁₀(0)|² H',
  as long as r₀ << a.
  R₁₀(r) = (2/a^3/2)exp(-r/a),  R₁₀(0) = 2/a^3/2, with a = a₀/Z.
  E₁⁰ = (4Z/a³)[ e²r₀⁵/(10r₀³) -  3e²r₀³/(6r₀) +  e²r₀²/2] = (2/5)Z(e²/a)(r₀/a)²
  = (4/5)Z²E_I(r₀/a)²,
  since a = a₀/Z = 5.29*10^-11 m/Z = 6.45*10^-13 m and E_I = 13.6 eV.
  The ground state energy of an electron in a hydrogenic atom with Z = 82 is Z²E_I = 91.4 keV
  The first-order shift in the ground state energy due to the finite size of the nucleus with a radius r₀ = 6*10^-15 m is  
  E₁⁰ = (4/5)(91.4 keV)(6*10^-15/6.45*10^-13)² = 6.3 eV.
  
  This not a valid approach for the muonic Pb atom, because for first-order order perturbation theory to be a valid approximation the correction we need E₁⁰  << E₀⁰, and the experimental data show that this is not the case.