__Perturbed oscillator__

A one-dimensional harmonic oscillator has momentum p, mass m, and angular
frequency ω. It is subject to a perturbation U = bx^{4}, 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 = p^{2}/(2m) + mω^{2}x^{2}/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),

x^{2 }= (aa + a^{†}a^{†}+ a^{†}a + aa^{†})/(4α^{2}).

For eigenstates |n> of H we have a^{†}|n> = (n + 1)^{½}|n + 1>, a|n> = n^{½}|n - 1>.

<n|x^{4}|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) + n^{2}+ n(n - 1).

<n|x^{4}|n> = (6n^{2 }+ 6n + 3) ħ^{2}/(4m^{2}ω^{2}) = (n^{2}+ n + ½)2ħ^{2}/(4m^{2}ω^{2}).

ΔE_{n}= (n^{2}+ n + ½)2bħ^{2}/(4m^{2}ω^{2}).

(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.

A charged particle of mass m and charge q is sitting in a harmonic potential
U_{0}(x) = ½mω^{2}x^{2}. A weak constant
electric field E is applied in the +x-direction, so that the potential is
perturbed by U_{1}(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/(mω^{2}).
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 H_{0}= p^{2}/(2m) + U_{0}(x). Let {|n>} denote the orthonormal eigenbasis of H_{0}.

H_{0}|n> = E_{0}^{n}|n> = (n + ½)ħω|n>, <n|n'> = δ_{nn'}.

The eigenvalues of H_{0}are not degenerate.

Let H = H_{0 }+ U_{1}. The eigenvalues of H may be written as E^{n }= E_{0}^{n }+ E_{1}^{n }+ E_{2}^{n }+ ... .

E_{1}^{n }= <n|U_{1}|n> = -qE<n|x|n> = -qE(ħ/(2mω))^{½}<n|(a + a^{†})|n>

= -qE(ω^{2}2m)^{-½}[(nħω)^{½}<n|n - 1> + ((n+1)ħω)^{½}<n|n + 1>] = 0.

(b) E_{2}^{n}= ∑_{n'≠n }|<n'|U_{1}|n>|^{2}/(E_{0}^{n }- E_{0}^{n'}) = q^{2}E^{2}∑_{n'≠n }|<n'|x|n>|^{2}/(E_{0}^{n }- E_{0}^{n'}).

<n'|x|n> = (ω^{2}2m)^{-½}[(nħω)^{½}<n'|n - 1> + ((n + 1)ħω)^{½}<n'|n + 1>]

= (ω^{2}2m)^{-½}[(nħω)^{½}δ_{n'n-1}+ ((n + 1)ħω)^{½}δ_{n'n+1}].

E_{2}^{n}= q^{2}E^{2}(ω^{2}2m)^{-1}[(nħω)/(E_{0}^{n }- E_{0}^{n-1}) + ((n + 1)ħω)/(E_{0}^{n }- E_{0}^{n+1})

= q^{2}E^{2}/(ω^{2}2m)[n - n + 1] = -q^{2}E^{2}/(ω^{2}2m).

(c) Let x' = x - qE/(mω^{2}).

½mω^{2}x'^{2}= ½mω^{2}x^{2}- qEx + q^{2}E^{2}/(ω^{2}2m)

Then H = p^{2}/(2m) + ½mω^{2}x'^{2}- q^{2}E^{2}/(ω^{2}2m)

H' = H + q^{2}E^{2}/(ω^{2}2m) = p^{2}/(2m) + ½mω^{2}x'^{2}.

The eigenvalues of H' are (n + ½)ħω, the eigenvalues of H are therefore (n + ½)ħω - q^{2}E^{2}/(ω^{2}2m).

Consider a single electron moving with an anharmonic potential energy given
by

U(x) = ½ kx^{2} + k’x^{4}. Let H_{0} 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
H_{0}.

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 = H_{0}+ H_{1}.

The eigenvalues of the 1D Hamiltonian H_{0}= p_{x}^{2}/(2m) + ½kx^{2}= p_{x}^{2}/(2m) + mω^{2}x^{2}/2 are

E_{n}^{0}= (n + ½)ħω, with ω^{2}= 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 = H_{0}+ H_{1}, to first order the energy eigenvalues are E_{n}^{0}+ E_{n}^{1}, with

E_{n}^{1}= <n|H_{1}|n>.

H_{1}= k’x^{4}.^{ }x^{2}= (1/(4α^{2}))(aa + aa^{†}+ a^{†}a + a^{†}a^{†}).

<n|x^{4}|n> = (1/(16α^{4})) <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> = n^{2}, <n|aa^{†}a^{†}a|n> = n(n + 1), <n|a^{†}aaa^{†}|n> = n(n + 1).

E_{n}^{1}= [k’ħ^{2}/(4m^{2}ω^{2})](6n^{2}+ 6n + 3).

The energies to first order are E_{n}= (n + ½) ħω + [k’ħ^{2}/(4m^{2}ω^{2})](6n^{2}+ 6n + 3).(b) The first order corrections to the eigenstates are

|0^{1}> = ∑_{n’≠0 }[<n’|H_{1}|0>/(E_{0}^{0 }– E_{n’}^{0})]|n’>.

<n’|x^{4}|0> has 16 terms, but only those with more a^{†}than a and a^{†}on the right can be nonzero. Four terms survive.<n’|x

^{4}|0> = (1/(16α^{4}))<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>.|0

^{1}> = -|2>[k’ħ^{2}/(4m^{2}ω^{2})](18^{½}+ 8^{½}+ 2^{½})/(2ħω) - |4>[k’ħ^{2}/(4m^{2}ω^{2})]24^{½}/(4ħω)

= -[k’ħ/(8m^{2}ω^{3})](8.49 |2> + 2.45|4>).

To first order the ground state is |0> - [k’ħ/(8m^{2}ω^{3})](8.49 |2> + 2.45|4>).

__Perturbed hydrogenic atoms__

|1,0,0> = (πa

and the Bohr constant is a

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 r_{0}and total charge q_{e}is U(r) = -e^{2}/r, r > 0 where e^{2}= q_{e}^{2}/(4πε_{0}).

Inside the sphere the potential energy is

U(r) = [e^{2}/(2r_{0})][(r/r_{0})^{2}- 3].

From Gauss’ law we know that E(r < r_{0}) = q_{inside}/(4πε_{0}r^{2}) = q_{e}r/(4πε_{0}r_{0}^{3}) (SI units).

Φ(r) = Φ(r_{0}) + [q_{e}/(4πε_{0}r_{0}^{3})]∫_{r}^{r0}rdr = q_{e}/(4πε_{0}r_{0}) + q_{e}(r_{0}^{2}- r^{2})/(8πε_{0}r_{0}^{3}).

U(r) = -q_{e}Φ(r) = [e^{2}/(2r_{0})][(r/r_{0})^{2}- 3].

Let H_{0}= p^{2}/(2μ) - e^{2}/r, H = p^{2}/(2μ) + [e^{2}/(2r_{0})][(r/r_{0})^{2}- 3] = H_{0}+ H' (r < r_{0}).

H' = e^{2}r^{2}/(2r_{0}^{3}) - 3e^{2}/(2r_{0}) + e^{2}/r.

E_{1}^{0}= <100|H'|100> = ∫_{0}^{2π}dφ∫_{0}^{π}sinθ dθ Y_{00}(θ,φ)∫_{0}^{r0}r^{2}dr |R_{10}(r)|^{2}H'

≈ ∫_{0}^{r0}r^{2}dr |R_{10}(0)|^{2}H',

since r_{0}<< a_{0}.

R_{10}(r) = (2/a_{0}^{3/2})exp(-r/a_{0}), R_{10}(0) = 2/a_{0}^{3/2}.

E_{1}^{0}= (4/a_{0}^{3})[ e^{2}r_{0}^{5}/(10r_{0}^{3}) - 3e^{2}r_{0}^{3}/(6r_{0}) + e^{2}r_{0}^{2}/2] = (2/5)(e^{2}/a_{0})(r_{0}/a_{0})^{2}

= (4/5)E_{I}(r_{0}/a_{0})^{2}= (4/5)(13.6 eV)(10^{-15}/0.53*10^{-10})^{2}= 4*10^{-9}eV.

The correctly normalized hydrogen ground state wave
function
in 3D is given by

ψ_{0}(r)
= (πa_{0}^{3})^{-½}exp(-r/a_{0}),

where a_{0} =
ħ^{2}/(m_{e}e^{2})
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 ∫d^{3}r|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}
= ν_{0 }
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**) = [-(ħ^{2}/(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 -[ħ^{2}/(2μ)][(∂^{2}/∂r^{2}) + 2μl(l+1)/ħ^{2}]u_{lm}(r) – [e^{2}/r]u_{lm0}(r) = E_{lm}u_{lm}(r).

Since Φ_{100}(r) = (πa_{0}^{3})^{-½ }exp(-r/a_{0}), we have u_{10}(r) = 2 a_{0}^{-3/2 }r exp(-r/a_{0}).

We show that u_{10}(r) satisfies

-[ħ^{2}/(2μ)][∂^{2}/∂r^{2}] u_{10}(r) – [e^{2}/r] u_{10}(r) = E_{10}u_{10}(r) by simply differentiating.

(∂^{2}/∂r^{2})rexp(-r/a_{0}) = [-(2/a_{0}) + (r/a_{0}^{2})]exp(-r/a_{0})

Therefore [ħ^{2}/(a_{0}μ)] - [rħ^{2}/(2a_{0}^{2}μ)] – e^{2}= Er.

This equation is satisfied if a_{0}= ħ^{2}/(μe^{2}) and E = -ħ^{2}/(2a_{0}^{2}μ) = - μe^{4}/(2ħ^{2}).

Normalization: ∫ d^{3}r |ψ(r)|^{2}= 4π∫_{0}^{∞}r^{2}dr (πa_{0}^{3})^{-1 }exp(-2r/a_{0}) = ½ ∫_{0}^{∞}r'^{2}dr'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 E_{1}^{1}= <Φ_{100}|W|Φ_{100}>.E

_{1}^{1}= ∫_{-1}^{1}dcosθ ∫_{0}^{∞}r^{2}dr |R_{10}(r)|^{2}|Y_{00}(θ,φ)|^{2}ν_{0 }δ(r - d) δ(cosθ - 1)/r^{2}

+ ∫_{-1}^{1}dcosθ ∫_{0}^{∞}r^{2}dr |R_{10}(r)|^{2}|Y_{00}(θ,φ)|^{2}ν_{0 }δ(r - d) δ(cosθ + 1)/r^{2 }= ν_{0}|R_{10}(d)|^{2}|Y_{00}(θ = 0)|^{2}+ ν_{0}|R_{10}(d)|^{2}|Y_{00}(θ = π)|^{2}= 2ν_{0}|ψ_{100}(d)|^{2}

= 2n_{0}exp(-2d/a_{0})/(πa^{3}).

Alternatively:

E_{1}^{1}= ∫_{0}^{∞}dx∫_{0}^{∞}dy∫_{0}^{∞}dz |ψ_{100}(**r**)|^{2}ν_{0 }δ(x)δ(y) (δ(z - d) + δ(z + d)) = 2ν_{0}|ψ_{100}(d)|^{2}.

(a) If the Hamiltonian for a system may be written as H =
H_{0} + H_{1} where H_{1} is a small perturbation and H_{0}ψ_{0}
= E_{0}ψ_{0},
then the effect of H_{1} gives rise to an energy correction term
ΔE_{0}. Deduce an expression
for the energy correction term ΔE_{0
}if ψ_{0}
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

ψ_{0} = (πa_{0}^{3})^{-½}exp(-r/a_{0})

that
may be written as (-E_{0}^{2}/(2mc^{2}))[1
+ 2a_{0}/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 = H_{0 }+ λW. Let |ψ_{0}^{0}> be a non-degenerate eigenstates of H_{0}.

H_{0}|ψ_{0}^{0}> = E_{0}^{0}|ψ_{0}^{0}>.

Assume that λW = H_{1}, and that λ is small.

Let H|ψ_{0}> = E^{0}|ψ_{0}>.

Since λW is small, we assume that E^{0}and |ψ_{0}> can be expanded as a power series in λ.

E^{0 }= E_{0}^{0 }+ λE_{1}^{0 }+ λ^{2}E_{2}^{0 }+ ... ,

|ψ_{0}> = |ψ_{0}^{0}> + λ|ψ_{0}^{1}> + λ^{2}|ψ_{0}^{2}> + ... .

We may then write

(H_{0}+ λW)(|ψ_{0}^{0}> + λ|ψ_{0}^{1}> +λ^{2}|ψ_{0}^{2}> + ...)

= (E_{0}^{0 }+ λE_{1}^{0 }+ λ^{2}E_{2}^{0 }+ ...)(|ψ_{0}^{0}> + λ|ψ_{0}^{1}> + λ^{2}|ψ_{0}^{2}> + ...).

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) (H_{0 }- E_{0}^{0})|ψ_{0}^{0}> = 0

(ii) (H_{0 }- E_{0}^{0})|ψ_{0}^{1}> = (E_{1}^{0 }- W)|ψ_{0}^{0}>

Therefore <ψ_{0}^{0}|(H_{0 }- E_{0}^{0})|ψ_{0}^{1}> = 0 = <ψ_{0}^{0}|( (E_{1}^{0 }- W)|ψ_{0}^{0}>, or

E_{1}^{0}= <ψ_{0}^{0}|W)|ψ_{0}^{0}> = ΔE_{0}.

(b) ΔE_{0}= 4π(-E_{0}^{2}/(2mc^{2}))(πa_{0}^{3})^{-1}∫_{0}^{∞}r^{2}dr exp(-2r/a_{0})(1 + 4a_{0}/r + 4 a_{0}^{2}/r^{2})

= -4(E_{0}^{2}/(2mc^{2})) ∫_{0}^{∞}x^{2}dx exp(-2x)(1 + 4/x + 4/x^{2})

= -4(E_{0}^{2}/(2mc^{2}))[¼ + 1 + 2] = -13*(E_{0}^{2}/(2mc^{2})).

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

(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

The ground state wave function of the hydrogen atom is |1,0,0> = (πa_{0}^{3})^{-½}exp(-r/a_{0}).

Solution:

- Concepts:

Hydrogenic atoms, perturbation theory - Reasoning:

We are used to use scaling rules for hydrogenic-atom wave functions and to 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

a_{0}by a_{0}' = ħ^{2}/(μ'Ze^{2}) = a_{0}(μ/μ')(1/Z),

and in the eigenvalues of the Hamiltonian of the hydrogen atom we replace

E_{I}by E_{I}' = μ'Z^{2}e^{4}/(2ħ^{2}) = E_{I}(μ'/μ)Z^{2}.

(a) Z = 82, μ'/μ = 209, E_{n }= -82^{2}*209*13.6 eV = -19.11 MeV/n^{2}.

The 2p – 1s transition energy is E_{2}– E_{1}= 14.33 MeV.

We neglect the presence of the electrons, because the average distance of the muon from the nucleus is ½09 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_{eff}^{2}*209*13.6 eV - Z^{2}*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

a_{0}' = a_{0}/(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 Z^{3}.

R = (82/61)^{⅓}a_{0}' = 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 r_{0}and total charge Zq_{e}is U(r) = -Ze^{2}/r, r > 0, with e^{2}= q_{e}^{2}/(4πε_{0}).

Inside the sphere the potential energy is

U(r) = [Ze^{2}/(2r_{0})][(r/r_{0})^{2}- 3].

From Gauss’ law we know that E(r < r_{0}) = q_{inside}/(4πε_{0}r^{2}) = Zq_{e}r/(4πε_{0}r_{0}^{3}) (SI units).

Φ(r) = Φ(r_{0}) + [Zq_{e}/(4πε_{0}r_{0}^{3})]∫_{r}^{r0}rdr = Zq_{e}/(4πε_{0}r_{0}) + Zq_{e}(r_{0}^{2}- r^{2})/(8πε_{0}r_{0}^{3}).

U(r) = -q_{e}Φ(r) = [Ze^{2}/(2r_{0})][(r/r_{0})^{2}- 3].

Let H_{0}= p^{2}/(2μ) - Ze^{2}/r, H = p^{2}/(2μ) + [Ze^{2}/(2r_{0})][(r/r_{0})^{2}- 3] = H_{0}+ H' (r < r_{0}).

H' = Ze^{2}r^{2}/(2r_{0}^{3}) - 3Ze^{2}/(2r_{0}) + Ze^{2}/r.

E_{1}^{0}= <100|H'|100> = ∫_{0}^{2π}dφ∫_{0}^{π}sinθ dθ Y_{00}(θ,φ)∫_{0}^{r0}r^{2}dr |R_{10}(r)|^{2}H'

≈ ∫_{0}^{r0}r^{2}dr |R_{10}(0)|^{2}H',

as long as r_{0}<< a.

R_{10}(r) = (2/a^{3/2})exp(-r/a), R_{10}(0) = 2/a^{3/2}, with a = a_{0}/Z.

E_{1}^{0}= (4Z/a^{3})[ e^{2}r_{0}^{5}/(10r_{0}^{3}) - 3e^{2}r_{0}^{3}/(6r_{0}) + e^{2}r_{0}^{2}/2] = (2/5)Z(e^{2}/a)(r_{0}/a)^{2}

= (4/5)Z^{2}E_{I}(r_{0}/a)^{2.}

since a = a_{0}/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 Z^{2}E_{I}= 91.4 keV

The first-order shift in the ground state energy due to the finite size of the nucleus with a radius r_{0}= 6*10^{-15}m is

E_{1}^{0}= (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 E_{1}^{0}<< E_{0}^{0}, and the experimental data show that this is not the case.