### 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 (aa + 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 (aa + 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 + aa + aa  + aa)/(4α2).
For eigenstates |n> of H we have a|n> = (n + 1)½|n + 1>, a|n> = n½|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|(aaaa+  aaaa + aaaa +aaaa + aaaa + aaaa)|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).
ΔEn = (n2 + n + ½)2bħ2/(4m2ω2).
(c)  Even for small b ΔEn 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> = n½|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'> = δnn'.
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)[(nħω)½<n|n - 1> + ((n+1)ħω)½<n|n + 1>] = 0.

(b)  E2n = ∑n'≠n |<n'|U1|n>|2/(E0n - E0n') = q2E2n'≠n |<n'|x|n>|2/(E0n - E0n').
<n'|x|n> = (ω22m)[(nħω)½<n'|n - 1> + ((n + 1)ħω)½<n'|n + 1>]
=  (ω22m)[(nħω)½δn'n-1 + ((n + 1)ħω)½δn'n+1].
E2n = q2E222m)-1[(nħω)/(E0n - E0n-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 = px2/(2m) + ½kx2 = px2/(2m) + mω2x2/2 are
En0 = (n + ½)ħω, with  ω2 = k/m.  We denote the eigenstates by {|n>}, n = 0, 1, 2, ... .
Define a = αx + iβpx,  a = αx - iβpx,  α =√(mω/(2ħ)),  β =1/√(2mωħ).
[a, a] = 2αβħ = 1.  Then
Hx = ħω (aa + ½) = ħω[(αx)2 + (βpx)2 - αβħ + ½) = ħω[(αx)2 + (βpx)2).
We have x = (a + a)/(2α),  px = -i(a - a)/(2β).
For eigenstates |n> of Hx we have a|n> = (n + 1)½|n + 1>, a|n> = n½|n - 1>.
With H = H0 + H1, to first order the energy eigenvalues are En0 + En1, with
En1= <n|H1|n>.
H1 = k'x4.
x2 = (1/(4α2))(aa + aa + aa + aa).
<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|aaaa|n> = (n + 1)(n + 2),  <n|aaaa|n> = n(n - 1),  <n|aaaa|n> = (n + 1)2
<n|aaaa|n> = n2, <n|aaaa|n> = n(n + 1), <n|aaaa|n> = n(n + 1).
En1 = [k'ħ2/(4m2ω2)](6n2 + 6n + 3).
The energies to first order are En = (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 - En'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'|(aaaa + aa aa +  aaaa + aaaa)|0>.
(aaaa + aa aa +  aaaa + aaaa)|0> = (18½ + 8½ + 2½)|2> + 24½|4>.

|01> = -|2>[k'ħ2/(4m2ω2)](18½ + 8½ + 2½)/(2ħω) - |4>[k'ħ2/(4m2ω2)]24½/(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:

Calculate the first-order shift in the ground state of the hydrogen atom caused by the finite size of the proton.  Assume the proton is a uniformly charged sphere of radius r = 10-13 cm.  The ground state wave function of the hydrogen atom is
|1,0,0> = (πa03)exp(-r/a0)
and the Bohr constant is a0 = 0.53*10-10 m.

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 qe is U(r) = -e2/r,  r > 0 where  e2 = qe2/(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) = qinside/(4πε0r2) = qer/(4πε0r03)  (SI units).
Φ(r) = Φ(r0) + [qe/(4πε0r03)]∫rr0rdr = qe/(4πε0r0) + qe(r02 - r2)/(8πε0r03).
U(r) = -qeΦ(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> = ∫0dφ∫0πsinθ dθ Y00(θ,φ)∫0r0r2dr |R10(r)|2 H'
≈ ∫0r0r2dr |R10(0)|2 H',
since r0 << a0.
R10(r) = (2/a03/2)exp(-r/a0),  R10(0) = 2/a03/2
E10 = (4/a03)[ e2r05/(10r03) -  3e2r03/(6r0) +  e2r02/2] = (2/5)(e2/a0)(r0/a0)2
= (4/5)EI(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)exp(-r/a0),
where a0 = ħ2/(mee2) 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,
Ue - ion = ν0 δ(r - rion),
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 = me for infinite proton mass..
Writing  Φnlm(r) = Rnl(r)Ylm(θ,φ) = [unl(r)/r]Ylm(θ,φ)
we find  -[ħ2/(2μ)][(∂2/∂r2) + 2μl(l+1)/ħ2]ulm (r) - [e2/r]ulm0(r) = Elmulm (r).
Since Φ100(r) = (πa03)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 = 2ν0100(d)|2
= 2n0exp(-2d/a0)/(πa3).
Alternatively:
E11 = ∫0dx∫0dy∫0dz |ψ100 (r)|2ν0 δ(x)δ(y) (δ(z - d) + δ(z + d)) = 2ν0100(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)exp(-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.
H000> = E0000>.
Assume that λW = H1, and that λ is small.
Let H|ψ0> = E00>.
Since λW is small, we assume that E0 and |ψ0> can be expanded as a power series in λ.
E0 = E00 + λE10 + λ2E20 + ... ,
0> = |ψ00> + λ|ψ01> + λ202> + ... .
We may then write
(H0 + λW)(|ψ00> + λ|ψ01> +λ202> + ...)
= (E00 + λE10 + λ2E20 + ...)(|ψ00> + λ|ψ01> + λ202> + ...).
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)-10r2dr exp(-2r/a0)(1 + 4a0/r + 4 a02/r2)
= -4(E02/(2mc2)) ∫0x2dx 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)exp(-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
EI by EI' = μ'Z2e4/(2ħ2) = EI(μ'/μ)Z2.
(a)  Z = 82,  μ'/μ = 209,  En  = -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 Zeff.  Assume only the 1s energy is significantly altered.
5.8 MeV = -Zeff2*209*13.6 eV - Z2*209*13.6/4 eV.
Zeff = 61.  Z = 1.34 Zeff.
The most likely radial distance of the muon from the center of the nucleus is
a0' = a0/(209*Zeff) = 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 Zqe is U(r) = -Ze2/r,  r > 0, with e2 = qe2/(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) = qinside/(4πε0r2) = Zqer/(4πε0r03)  (SI units).
Φ(r) = Φ(r0) + [Zqe/(4πε0r03)]∫rr0rdr = Zqe/(4πε0r0) + Zqe(r02 - r2)/(8πε0r03).
U(r) = -qeΦ(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> = ∫0dφ∫0πsinθ dθ Y00(θ,φ)∫0r0r2dr |R10(r)|2 H'
≈ ∫0r0r2dr |R10(0)|2 H',
as long as r0 << a.
R10(r) = (2/a3/2)exp(-r/a),  R10(0) = 2/a3/2, with a = a0/Z.
E10 = (4Z/a3)[ e2r05/(10r03) -  3e2r03/(6r0) +  e2r02/2] = (2/5)Z(e2/a)(r0/a)2
= (4/5)Z2EI(r0/a)2,
since a = a0/Z = 5.29*10-11 m/Z = 6.45*10-13 m and EI = 13.6 eV.
The ground state energy of an electron in a hydrogenic atom with Z = 82 is Z2EI = 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.