Stationary perturbation theory is concerned with finding the changes in the discrete energy levels and the changes in the corresponding energy eigenfunctions of a system, when the Hamiltonian of a system is changed by a small amount.  Let H=H₀+H’=H₀+lW.  H₀ is the unperturbed Hamiltonian whose eigenvalues E₀^p and eigenstates |fⁱ_p> are known. Let {|fⁱ_p>}denote an orthonormal eigenbasis of H₀, H₀|fⁱ_p>= E₀^p|fⁱ_p>.  Here i denotes the degeneracy.  Assume that the matrix elements of H’ in the eigenbasis of H₀ are small compared to the matrix elements of H₀.

<fⁱ_p|H'|fⁱ_p>   <<    <fⁱ_p|H₀|fⁱ_p>=E₀^p.

We write H’=lW with l<<1 and <fⁱ_p|W|fⁱ_p>»E₀^p.

• Example:

  

  H=H₀+H’ is the Hamiltonian of a perturbed one-dimensional harmonic oscillator.

We are looking for the eigenvalues E(l) and the eigenstates |y(l)> of H(l)=H₀+lW.

H(l)|y_p>=E^p(l)|y_p> or, keeping the dependence on l in mind without specifically writing it down, H|y_p>=E^p|y_p>.

Since lW is small, we assume that E and |y> can be expanded as a power series in l.

E^p=E₀^p+lE₁^p+l²E₂^p+... ,
|y_p>=|y_p⁰>+l|y_p¹>+l²|y_p²>+... .

We may then write

(H₀+lW)(|y_p⁰>+l|y_p¹>+l²|y_p²>+...)

= (E₀^p+lE₁^p+l²E₂^p+...)(|y_p⁰>+l|y_p¹>+l²|y_p²>+...).

This equation is must be valid over a continuous range of l.  Therefore we equate coefficients of equal powers of l on both sides to obtain a series of equations that represent successively higher orders of the perturbation.

• (H₀-E₀^p)|y_p⁰>=0
• (H₀-E₀^p)|y_p¹>=(E₁^p-W)|y_p⁰>
• (H₀-E₀^p)|y_p²>=(E₁^p-W)|y_p¹>+E₂^p|y_p⁰>
• (H₀-E₀^p)|y_p³>=(E₁^p-W)|y_p²>+E₂^p|y_p¹>+E₃^p|y_p⁰>
• ...

(H₀-E₀^p)|y_p⁰>=0 implies that |y_p⁰> is a linear combination of unperturbed eigenfuctionns |fⁱ_p> with the corresponding eigenvalue E₀.  We choose <y_p⁰|y_p⁰>=1. |y_p^s> is not uniquely defined.  We can add an arbitrary multiple of |y_p⁰> to each |y_p^s> without affecting the left hand side of the above equations.  Most often this multiple is chosen so that <y_p⁰|y_p^s>=0.  The perturbed ket is then not normalized.

We then have

0                            =                       

To calculate the energy to sth order, we only need to know the state vector to order s-1.

First-order perturbation theory for non-degenerate levels

Consider a particular non-degenerate eigenvalue E₀ⁿ of H₀. H₀|f_n>=E₀ⁿ|f_n>.  The other eigenvalues of H may or may not be degenerate.  We have |y_n⁰>=|f_n> and E₁ⁿ=<f_n|W|f_n>.

The first-order energy correction therefore is lE₁ⁿ=<f_n|H’|f_n>.

We have Eⁿ=E₀ⁿ+<f_n|H’|f_n>+O(l²).

• Example:

   

  

  Eⁿ=E₀ⁿ +O(l²). To first order, the eigenvalues of H are equal to the eigenvalues of the unperturbed Hamiltonian H₀.

First-order eigenvector corrections:

(H₀-E₀^p)|y_p¹> =( E₁^p-W)|y_p⁰> = (E₁^p-W)|f_p>. |y_p⁰> is an eigenstate of the unperturbed Hamiltonian.  We may expand

in terms of the basis vectors |fⁱ_p'>.  In the expansion b_p=0 because <y_p⁰|y_pⁱ>=0.

Multiply from the left by <fⁱ_p''|.

Therefore

• Example:

   

  

  

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Second-order perturbation theory for non-degenerate levels

Second-order energy corrections:

Since we have found the expression for the state vector to first order, we can now find the expression for the energy to second order.

• Example:

   

  

   

Let H=H₀+H’=H₀+lW.  In practice, after having derived the perturbation expansion, we often set l=1 and let H’=W be small.

Problems:

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^-13cm. The ground state wave function of the hydrogen atom is and the Bohr constant is a₀=0.53´10^-10m.

• Solution:
  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

  Inside the sphere the potential energy is 

  From Gauss’ law we know that  (SI units).

  

  

  Let

  

  

  

  since r₀ << a₀,

  

  

  =4´10^-9eV.

(a)   Calculate the first-order energy shift of the nth eigenvalue due to the perturbation.

(b)  Write out the first three non vanishing terms for the first-order perturbation expansion of the ground state in terms of the unperturbed eigenfunctions of the infinite well.

(c)  Calculate the second-order energy shift for the ground state.

• Solution:

  H =H₀+H’.

  

  The eigenvalues of H₀ are not degenerate.

   

  • (a) 

    

  • (b)  First-order perturbation theory yields

    

    

    

      if n is odd.

    

    

    

  • (c) 

    second-order energy shift for the ground state.

    The second-order energy shift of the ground state is always negative.