on-degenerate Perturbation Theory Suppose one wants to solve the eigenvalue problem ĤΦ = Φ where µ =,1,2,, E µ µ µ and where Ĥ can be written as the sum of two terms, ˆ ˆ ˆ ˆ ˆ ˆ H = H + ( H H ) = H + V and where one knows the eigenfunctions and eigenvalues of Ĥ Ĥ Φ = E Φ µ µ µ The difference between the two Hamiltonians, V ˆ is called the perturbation and to the extent that V ˆ is (in some sense) small relative to Ĥ we expect the eigenfunctions and eigenvalues of Ĥ to be similar to those of Ĥ. The perturbationv ˆ could be the result of putting the original system in an electric or magnetic field. For example a hydrogen atom in a constant electric field F along the z axis has the perturbation ˆV = ezf cos(θ) where θ is the angle between the position vector to the electron an the electric field F. In this instance the energy and eigenfunctions of the perturbed Hamiltonian depend parametrically on the electric field. As illustrated in the figure below, the energy and the wavefunction of the various unperturbed states varies with field strength. -6-7 state 3-8 Energy -9-1 -11 state 1 state 2-12 unperturbed Field Strength perturbed J. F. Harrison Michigan State University 12/12/217 1
Sometimes the form of the perturbation is subtle. For example we might be interested in the effect of the size of the nucleus on the energy of a one-electron atom. When we ignore the size of the nucleus we write the electrostatic potential experienced by the electron as Ze 2 4πε r where r <. When we take the size of the nucleus into account the potential becomes Ze 2 4πε R for r R and Ze2 4πε r for R r < where R is the radius of the nucleus and the perturbation then becomes ˆV = Ze2 1 4πε r 1 R for r R and for R r < In this instance there isn t an external parameter like an electric field. Both types of perturbations may be dealt with using Rayleigh Schrodinger perturbation theory, which we will now develop. We begin by introducing the Hamiltonian ˆ ˆ ˆ H( λ) = H + λv when the perturbation is due to an external field λ is the field strength. If there is no external field Ĥ(λ) is a fictitious Hamiltonian that is equal to the Hamiltonian of interest, Ĥ when λ = 1 and to the Hamiltonian Ĥ, whose eigenvalues and eigenfunctions we know, when λ =. λ is often called the ordering parameter. Lets suppose that we are interested in the effect of the perturbation on the Φ state of the unperturbed system so the eigenvalue problem for H ˆ ( λ ) is Ĥ(λ)Φ (λ) = E (λ)φ (λ) where as λ Φ (λ) & E (λ) Φ & E. We assume that both Φ (λ) & E (λ) have a Maclaurin series expansion in λ and write and Φ (λ) = =! λ d Φ dλ (λ) λ= = λ Φ ( ) = J. F. Harrison Michigan State University 12/12/217 2
E (λ) = = d E (λ)! dλ λ λ= = λ E ( ) Inserting these expansions into the Schrodinger equation gives = ( Ĥ + λv ˆ ) λ ( ) Φ = λ Q+P ( E P) (Q) Φ = P= Q= The sum λ Q+P ( E P) (Q) Φ may be transformed to a more convenient form as follows. P= Q= Q 4 3 2 P+Q=4 1 P+Q=1 1 2 3 4 P If we let = P + Q then we can cover all points in the PQ plane by summing along the diagonals λ Q+P ( E P) (Q) Φ = λ ( E P) ( P) Φ = λ ( E P) ( P) Φ P= Q= From which we obtain = P= = P= λ Ĥ Φ + (1 δ ) Vˆ Φ 1 E P P Φ P=1 = = and since we want this to be true for arbitrary λ we set the coefficient of zero and obtain the Rayleigh-Schrodinger equations. Ĥ Φ + (1 δ ) Vˆ Φ 1 P=1 The first equation in this sequence is E P P Φ =, for =,1,2,, λ equal to J. F. Harrison Michigan State University 12/12/217 3
Ĥ Φ = E Φ and is the unperturbed problem for which we know all of the eigenvalues and eigenfunctions. The next, or first order equation, is Ĥ Φ + ˆ V Φ = E Φ + E Φ or Ĥ ( E )Φ + Vˆ ( E )Φ = Keep in mind that the unknowns are the first order correction to the energy and wavefunction, E & Φ. To find E we multiply both sides of the above equation by the complex conjugate of the unperturbed solution, Φ and integrate. This gives us Φ Ĥ E Φ + Φ ˆ V E 1 Φ = and because Ĥ is Hermitian Φ Ĥ E Φ = and the first order correction to the energy is simply the average of the perturbing potential over the unperturbed wavefunction. Φ ˆ V Φ = E ote that this result depends on the eigenvalue E being non-degenerate. If for example the unperturbed function Φ 1 (orthogonal to Φ ) also has the eigenvalue E and we formed Φ 1 Ĥ E Φ + Φ 1 ˆ V E 1 Φ = we now have Φ 1 ˆ V Φ = J. F. Harrison Michigan State University 12/12/217 4
which may not be true. This possibility is addressed in our discussion of degenerate perturbation theory and we will continue this discussion assuming the state we are considering is non-degenerate. We will leave the determination of Φ for the time being and consider the second order equation Ĥ ( E )Φ + Vˆ ( E )Φ = E Φ To isolate E we once again multiply both sides by the complex conjugate of the unperturbed solution and integrate. And because Ĥ is Hermitian this results in Φ ˆ V E Φ = E To deal with Φ Φ, the coefficient of E, we need to consider the normalization of the perturbed solution, Φ (λ). Rather than selecting the conventional normalization Φ (λ) Φ (λ) = 1 we choose the more convenient intermediate normalization in which the overlap between the perturbed and unperturbed wavefunctions is taken to be 1, i.e., Φ Φ (λ) = 1= Φ Φ + Φ Φ λ + Φ Φ λ 2 + Φ Φ (3) λ 3 + Of course after we obtain Φ (λ) we may renormalize it to 1. Intermediate normalization is convenient because it requires that all of the corrections Φ ( ), are orthogonal to the unperturbed wavefunction Φ Φ Φ ( ) = δ Returning to the second order equation we see that Φ Φ order correction to the energy is = and so the second Φ ˆ V Φ = E J. F. Harrison Michigan State University 12/12/217 5
and as yet we do not know Φ. It is interesting that E is always negative. We may show this as follows. E = Φ ˆ V Φ = Φ ˆ V E Φ = ( ˆ V E )Φ Φ From the first order equation So Vˆ ( E )Φ = Ĥ ( E )Φ E = Φ Ĥ E Φ But from the variation principle Φ Ĥ Φ E, and E Lets continue and determine E (3). Writing the third order equation Ĥ ( E )Φ (3) + Vˆ ( E )Φ = E (3) Φ + E Φ multiplying by the complex conjugate of Φ and integrating we obtain Φ ˆ V E Φ = Φ ˆ V Φ = E (3) We may write this in terms of Φ as follows. As we have seen from the first order equation we have ( Vˆ E )Φ = (Ĥ E )Φ and since Φ ˆ V E Φ = E (3) = ( ˆ V E )Φ Φ we have J. F. Harrison Michigan State University 12/12/217 6
E (3) = Φ Ĥ E Φ Using the second order equation this becomes E (3) = Φ ˆ V E Φ and the third order correction to the energy is determined by the first order correction to the wavefunction. This is an example of the 2 + 1 rule, which states that corrections to th the wavefunction through order determines the corrections to the energy through ( 2 + 1) th order. Let s now find Φ, the first order correction to the wavefunction. Occasionally one can solve the differential equation directly for Φ and we illustrate this approach in the examples. Most of the time however this is not possible and a more general approach is called for. We will assume that the function Φ can be expanded in the eigenfunctions of the unperturbed Hamiltonian, Ĥ so that Φ = Φ ν ν C ν where we exclude the unperturbed state Φ from the summation because Φ Φ = To determine the coefficients C ν we insert the expansion into the first order equation, multiply through by the complex conjugate of one of the unperturbed eigenfunctions, say Φ η and integrate. This results in C η = Φ η ˆ V Φ E E η and therefore Φ = ν Φ ν ˆ V Φ Φ ν E E ν Using this expression the second and third order corrections to the energy become J. F. Harrison Michigan State University 12/12/217 7
Φ ˆ E V Φ ν Φ ˆ ν V Φ Φ ˆ V Φ 2 ν = E = E ν E = E ν ν ν ν V ν 2 E ν where V ν = Φ ˆ V Φ ν & E ν = E E ν and in a similar fashion E (3) V = ν V νσ V σ V V ν ν σ E ν E σ ν 2 E 2 ν J. F. Harrison Michigan State University 12/12/217 8