Perturbation Theory is a method of approximation not limited to the ground state of a system. It is useful when we know a solution to a similar problem which we can use as a first approximation to the problem that must be solved. Let’s say we want to solve the Schrödinger Eq.

H y _N = E _{Ny N},

where N=state of the system; N=0, ground state; N=1, first excited state, etc.

Let’s say we know the solution to a similar problem

H⁰ y _N⁰ = E _N^0y _N⁰,

where H⁰ is very similar to H such that

H = H⁰ + H¹,

where H¹is small. Assume that:

H⁰ is Hermitian

Its eigenfunctions, y _N⁰,are orthonormal.

We can use the y _N⁰ to find the y _N.

One-dimensional Anharmonic Oscillator has

H = -h²/(2m) {d²/dx²} + k x²/2 + c x³ + d x⁴

One-dimensional Harmonic Oscillator has

H⁰ = -h²/(2m) {d²/dx²} + k x²/2

If c & d are small, then the eigenfunctions & eigenvalues of the anharmonic oscillator must be closely related to those of the harmonic oscillator--which we know. But how can we use this information? We can expand the wavefunction & energy of the anharmonic oscillator in a Taylor series around the wavefunction & energy that we know.

Define the system with H⁰ as the unperturbed system & the with H as the perturbed system.

H = H⁰ + H¹

H¹ = H - H⁰ = c x³ + d x⁴ for the anharmonic oscillator

Introduce the parameter l , the perturbation parameter. Essentially, l is used for "bookkeeping". The following is the use of perturbation theory for nondegenerate energy levels (degenerate levels will not be covered due to lack of time).

Let

y _N⁰ = wavefunction of the unperturbed nondegenerate level N

E_N⁰ = energy of the unperturbed nondegenerate level N

y _N⁰ ® y _N & E_N⁰ ® E_N when the perturbation is applied.

We want to solve

H y _N = (H⁰ + l H¹) y _N = E_{Ny N}

Since H depends on l , E_N & y _N will too.

Expand y _N in a Taylor Series around y _N⁰ (i.e. y _N evaluated at l = 0, or y _{Nô l =0}):

y _N = y _{Nô l =0} + (¶ y _N/¶ l )ô l _{=0× l} + (¶ ^2y _N/¶ l ²)ô l _{=0× l} ²/2! +

(¶ ^3y _N/¶ l ³)ô l _{=0× l} ³/3! + ....

= y _N⁰ + (¶ y _N/¶ l )ô l _{=0× l} + (¶ ^2y _N/¶ l ²)ô l _{=0× l} ²/2! +

(¶ ^3y _N/¶ l ³)ô l _{=0× l} ³/3! + ....

Also expand E_N in a Taylor Series around E_N⁰

E_N = E_{Nô l =0} + (¶ E_N/¶ l )ô l _{=0× l} + (¶ ²E_N/¶ l ²)ô l _{=0× l} ²/2! +

(¶ ³E_N/¶ l ³)ô l _{=0× l} ³/3! + ....

= E_N⁰ + (¶ E_N/¶ l )ô l _{=0× l} + (¶ ²E_N/¶ l ²)ô l _{=0× l} ²/2! +

(¶ ³E_N/¶ l ³)ô l _{=0× l} ³/3! + ....

Define y _N^k = (¶ ^ky _N/¶ l ^k)ô l ₌₀/k!

E_N^k = (¶ ^kE_N/¶ l ^k)ô l ₌₀/k!

Then

y _N = y _N⁰ + l y _N¹ + l ² y _N² + l ³ y _N³ + ...

E_N =E_N⁰ + l E_N¹ + l ² E_N² + l ³ E_N³ + ...

Substitute these expansions into

H y _N = (H⁰ + l H¹) y _N = E_{Ny N ®}

(H⁰ + l H¹) (y _N⁰ + l y _N¹ + l ² y _N² + l ³ y _N³ + ...)

= (E_N⁰ + l E_N¹ + l ² E_N² + ...) (y _N⁰ + l y _N¹ + l ² y _N²+ ...)

Collect terms in l :

H^0y _N⁰ + l (H^0y _N¹ + H^1y _N⁰) + l ²(H^0y _N² + H^1y _N¹) + ...

= E_N^0y _N⁰ + l (E_N^1y _N⁰ + E_N^0y _N¹)

+ l ²(E_N^0y _N² + E_N^1y _N¹ + E_N^2y _N⁰) + ...

If the left side equals the right side, then the left & right sides must be equal for each power of l . This gives the Perturbation Eq.:

l ⁰: H^0y _N⁰ = E_N^0y _N⁰ (1)

l ¹: H^0y _N¹ + H^1y _N⁰ = E_N^1y _N⁰ + E_N^0y _N¹

(H⁰ - E_N⁰) y _N¹ = (E_N¹ - H¹) y _N⁰ (2)

l ²: H^0y _N² + H^1y _N¹ = E_N^0y _N² + E_N^1y _N¹ + E_N^2y _N⁰

(H⁰ - E_N⁰) y _N² = E_N^2y _N⁰ + (E_N¹ - H¹) y _N¹ (3)

etc.

Eq. (1) is the Schrödinger Eq. for the unperturbed problem. We already know the solutions y _N⁰ & E_N⁰. We can use these to solve Eq. (2) for E_N¹, the first order correction to the energy, & y _N¹, the first order correction to the wavefunction. At this point, we will have improved our approximation to E_N:

E_{N »} E_N⁰ + E_N¹, l =1

y _{N »} y _N⁰ + y _N¹, l =1

Solve Eq. (2):

(H⁰ - E_N⁰) y _N¹ = (E_N¹ - H¹) y _N⁰

Since H⁰ is Hermitian, then the y _N⁰ are a complete set. So we can expand the unknown y _N¹ in terms of the y _N⁰. Let

y _N¹ = S a_j y _j⁰

Then (H⁰ - E_N⁰) S a_j y _j⁰ = (E_N¹ - H¹) y _N⁰

= S a_j (H⁰ - E_N⁰) y _j⁰

j

= S a_j H^0y _j⁰ - S a_jE_N^0y _j⁰

j j

= S a_j E_j^0y _j⁰ - S a_jE_N^0y _j⁰

j j

= S a_j (E_j ⁰ - E_N⁰) y _j⁰ = (E_N¹ - H¹) y _N⁰

j

We want to calculate E_N¹. Multiply both sides by (y _m⁰)^* & integrate over the spatial coordinates:

ò (y _m⁰)^{* S} a_j (E_j ⁰ - E_N⁰) y _j⁰ dt

j

= ò (y _m⁰)^* (E_N¹ - H¹) y _N⁰dt

S a_j (E_j ⁰ - E_N⁰) ò (y _m⁰)^{* y} _j⁰ dt

j

S a_j (E_j ⁰ - E_N⁰)d _mj = E_N^{1 d} _mN - ò (y _m⁰)^* H¹ y _N⁰dt

j

a_m (E_m ⁰ - E_N⁰) = E_N^{1 d} _mN - H_mN¹

0 = E_N¹ - H_NN¹

E_N¹ = H_NN¹ (Note: Don’t need y _N¹ to get E_N¹)

a_m (E_m ⁰ - E_N⁰) = - H_mN¹

a_m = - H_mN¹/(E_m ⁰ - E_N⁰) = H_mN¹/(E_N ⁰ - E_m⁰)

Since y _N¹ = S a_j y _j⁰

from the above we know the form of all the a_j’s except a_N. We can set a_N = 0. So

y _N¹ = S [H_jN¹/(E_N ⁰ - E_j⁰)] y _j⁰

And

y _{N »} y _N⁰ + y _N¹, l =1

» y _N⁰ + S [H_jN¹/(E_N ⁰ - E_j⁰)] y _j⁰

Expand y _N² in terms of the y _N⁰

y _N² = S b_j y _j⁰

Then (H⁰ - E_N⁰) y _N² = E_N^2y _N⁰ + (E_N¹ - H¹) y _N¹

becomes

(H⁰ - E_N⁰) S b_j y _j⁰ = E_N^2y _N⁰ + (E_N¹ - H¹) y _N¹

= S b_j (H^0y _j⁰ - E_N^0y _j⁰)

= S b_j (E_j ^0y _j⁰ - E_N^0y _j⁰)

= S b_j (E_j ⁰ - E_N⁰) y _j⁰

Multiply both sides by (y _m⁰)^* & integrate over the spatial coordinates:

ò (y _m⁰)^{* S} b_j (E_j ⁰ - E_N⁰) y _j⁰ dt

Can we simplify E_N² ? We must evaluate ò (y _N⁰)^* y _N¹ dt . But

y _N¹ = S a_j y _j⁰

and a_j = H_jN¹/(E_N ⁰ - E_j⁰), a_N = 0.

The S below stands for the sum over j not equal to N:

So ò (y _N⁰)^* y _N¹ dt = ò (y _N⁰)^{* S} a_j y _j⁰ dt

= S a_{j ò} (y _N⁰)^* y _j⁰ dt

= S a_{j d Nj}

= 0 since j cannot be equal to N.

Therefore

E_N² = ò (y _N⁰)^* H¹ y _N¹ dt

Using the definition of y _N¹

E_N² = ò (y _N⁰)^* H¹ S a_j y _j⁰

= S a_j ò (y _N⁰)^* H¹ y _j⁰

= S a_j H_Nj¹

= S {H_jN¹/(E_N ⁰ - E_j⁰)} H_Nj¹

j not equal to N

Since H¹ is Hermitian, H_Nj¹ = (H _jN¹)^*. So

E_N² = S H_jN¹(H _jN¹)^*/(E_N ⁰ - E_j⁰)

j not equal to N

= S ½ H_jN^{1½ 2}/(E_N ⁰ - E_j⁰)

j not equal to N

So through second order in the energy,

E_N = E_N ⁰ + H_NN¹ + S ½ H_jN^{1½ 2}/(E_N ⁰ - E_j⁰)

j not equal to N

We could continue in the same method as above (Rayleigh- Schrödinger Perturbation Theory) to higher order of perturbation. But how do you know when to stop? Does the perturbation series converge? Expect ½ E_N^0½ > ½ E_N^1½ > ½ E_N^2½ > ...

What does the perturbation procedure actually do?

Since y _N¹ is expanded in terms of y _j ⁰ for k not equal to N, the effect of the procedure is to add into first order those states from the complete set that were left out of zeroth order. But they aren’t added in with equal weight to that of y _N ⁰ (i.e. 1), but rather with weight a_j. Since

a_j µ 1/(E_N ⁰ - E_j⁰)

the terms which have E_j⁰ close to E_N ⁰ will cause 1/(E_N ⁰ - E_j⁰) to be large & therefore result in a larger contribution from the states close to N.

Practical Considerations:

(1) E_N¹ is easy to evaluate since only y _N⁰ is used in

H_NN¹ = ò (y _N⁰)^* H¹ y _N⁰ dt

(2) E_N² is difficult to evaluate because you would need the complete set of y _j⁰ (except j=N). This is an infinite number of terms! If you have a Hamiltonian that supports both discrete & continuous states, as with the hydrogen atom, then you must include not only the sum over all discrete states, but the integral over all the continuous states, as well. This is impossible to do exactly. Instead you could use a combination variation-perturbation treatment.

He has 2 protons & 2 electrons. The nucleus has a charge of +2e & each electron has a charge of -e. Electrons are indistinguishable particles, but we will number them to keep track of them. Electron #1 is at distance r₁ from the nucleus; Electron #2 is at distance r₂ from the nucleus. The distance between the two elecrons is r₁₂. Let us set up the problem to treat a general two-electron ion by using a nuclear charge of +Ze. Then

H = - h²/(2m) Ñ ₁² - Z(e’) ²/ r₁ - h²/(2m) Ñ ₂² - Z(e’) ²/ r₂+(e’) ²/ r₁₂

In spherical polar coordinates

y = y (r₁, q ₁, f ₁, r₂, q ₂, f ₂)

r₁₂ = [(x₁ - x₂) ² + (y₁ - y₂) ² + (z₁ - z₂) ²]^1/2

Let H₁⁰ be the hydrogen-like Hamiltonian for electron #1; Let H₂⁰ be the hydrogen-like Hamiltonian for electron #2:

H₁⁰ = - h²/(2m) Ñ ₁² - Z(e’) ²/ r₁; E₁ = -(Z²/N₁²)(e’) ²/(2a₀)

H₂⁰ = - h²/(2m) Ñ ₂² - Z(e’) ²/ r₂; E₂ = -(Z²/N₂²)(e’) ²/(2a₀)

where

H₁⁰ F₁ (r₁, q ₁, f ₁) = E₁ F₁ (r₁, q ₁, f ₁)

H₂⁰ F₂ (r₂, q ₂, f ₂) = E₂ F₂ (r₂, q ₂, f ₂)

and F₁ & F₂ are the hydrogen-like wavefunctions & E₁ & E₂ are the hydrogen-like energies.

Define H⁰ = H₁⁰ + H₂⁰.

If the Hamiltonian is a sum of two Hamiltonians, then the wavefunction is a product of their eigenfunctions:

y ⁰ (r₁, q ₁, f ₁, r₂, q ₂, f ₂) = F₁ (r₁, q ₁, f ₁) F₂ (r₂, q ₂, f ₂)

Then

H^{0y 0} = (H₁⁰ + H₂⁰) F₁ F₂

= F₂ H₁⁰ F₁ + F₁H₂⁰ F₂

= F₂ E₁ F₁ + F₁E₂ F₂

= (E₁ + E₂) F₁ F₂

= E^{0y 0}, E⁰ = E₁ + E₂ = -(Ze’) ² /(2a₀)(1 /N₁² + 1 /N₂²)

The assumption in this choice of H⁰ is that the two electrons exert no repulsive force on each other (i.e. they operate as independent particles). This is not physically correct, but can be used as a first approximation.

In the ground state of helium, both electrons are in the 1s orbit, but have opposite spins. Indicate the electronic configuration by 1s ². So the zeroth-order approximation to the ground state wavefunction can be given

= -[(Ze’) ² /(2a₀)](2); (e’) ² /(2a₀) = 13.6 eV

= - (4) (2) (13.6eV) for He

= -108.8eV for He.

How does this compare to the actual ground state energy of He?

He has E = 0 when both electrons have been ionized, So

E_true = -(IP¹ + IP²); IP = ionization potential

= - (24.6 eV + 54.4 eV) = -79.0 eV

So -108.8 eV is off by 38%! This is not unexpected because the perturbation

H¹ = H - H⁰ = (e’) ²/ r₁₂

is not necessarily small. When r₁ is close to r₂, (e’) ²/ r₁₂ is large.

Let’s see what improvement we can get by calculating E¹:

× sin q ₁ sin q ₂ dr₁ dr₂ dq ₁ dq ₂ df ₁ df ₂

1/r₁₂ can be expanded in terms of r₁ & r₂ & the Spherical Harmonics, Y_l^m(q ,f ):

1/r₁₂ = S S [4p /(2l+1)] (r_<^l/ r_>^l+1) [Y_l^m(q ₁,f ₁)]* [Y_l^m(q ₂,f ₂)],

where r_< is the smaller of r₁ & r₂ & r_> is the larger of r₁ & r₂. Skim through the details of how the integral is done (p. 230-231). The result is

E¹ = (5Z/8) (e’)²/a₀

then the energy is -74.8 eV

Choice (II): Try making the trial function more flexible by replacing the atomic number, Z, by a variational parameter, r . This allows for the screening of the nuclear charge by the other electrons. (Full nuclear charge corresponds to r =2; full screening, r =2-1=1)

Optimize I with respect to r , the effective nuclear charge.

I = <f ½ H½ f >

Choose f = f₁f₂

and [-h²/(2m) Ñ ₁² - r (e’) ²/r₁]f₁ = -r ²(e’) ²/(2a₀) f₁

[-h²/(2m) Ñ ₂² - r (e’) ²/r₂]f₂ = -r ²(e’) ²/(2a₀) f₂

We can simplify the calculation if we write H to contain the Hamiltonians for these eigenfunctions. We must add & subtract terms in r so that we still have the He atom Hamiltonian.

H = - h²/(2m) Ñ ₁² - Z(e’) ²/ r₁ - h²/(2m) Ñ ₂² - Z(e’) ²/ r₂+(e’) ²/ r₁₂

= [-h²/(2m) Ñ ₁² - r (e’) ²/r₁] + [-h²/(2m) Ñ