MODULE 13 Time-independent Perturbation Theory Let us suppose that we have a system of interest for which the Schrödinger equation is We know that we can find functions (  that will satisfy this equality if we first know the form of the hamiltonian, ^H. The perturbation method starts up by looking for another system for which we know the solution and which bears a reasonably close relationship to the system we want to solve. In the case of the anharmonic oscillator we can use the harmonic oscillator as the known related system. In the case of a molecule in an electric field we can use the field- free molecule as our model.

MODULE 13 The way we approach the problem is to propose that the hamiltonian for the system of interest is the sum of the hamiltonian for the related system plus a perturbation term, viz., where Ĥ (0) is the hamiltonian for the related, known system and Ĥ (1) is the perturbation. The Schrödinger equation for the known system is

MODULE 13 In Module 9 we saw that any general function can be expanded as a superposition of the set of eigenfunctions of another operator. [Remember that a straight line can be approximated over a defined interval as a superposition of exponentials (eigenfunctions of the derivative operator) or as a superposition of sine functions (eigenfunctions of the second derivative operator).] Both operators that we are dealing with are hamiltonians, but one of somewhat different from to the other. Thus, what we seek to do is to expand the functions after perturbation as superpositions of those prior to perturbation. The known, unperturbed hamiltonian,Ĥ (0), has an infinite set of orthonormal eigenfunctions as defined by where m = 1,2,...., and the E m are the corresponding eigenvalues.

MODULE 13 Our superposition is then defined by where the a m are unknown coefficients The Schrödinger equation that we need is obtained by substituting for  into the Schrödinger equation In order to keep things simple at first we shall restrict ourselves to a superposition of just two of the known (orthonormal) eigenstates of Ĥ (0) which we label |1> and |2>.

The corresponding wavefunctions are  1 (0) and  2 (0), with associated energy eigenvalues, E 1 and E 2. MODULE 13

These 2 simultaneous equations can be written as a determinant of the coefficients a 1 and a 2

MODULE 13 where E 1 and E 2 are the expectation values of the H 11 and H 22 elements, respectively, and  2 = H 12.H 21 = H 12 2 (Hermitian). Thus H 12 is identifiable with the perturbation term in the hamiltonian, and  can be recognized as the energy arising out of the perturbation. When  = 0, E + = E 1 and E - = E 2, the two unperturbed energies of the model. Rearranging above equation under conditions where the perturbation (  ) is very small compared to the level separation,  E = E 2 -E 1

MODULE 13 Then and the quantity under the square root sign can be expanded according to the series which can be truncated at the second term when x<<<1. These two solutions converge to the exact energies for the pure states when (2  /  E) 2 <<<1

MODULE 13 This figure and the next show particular situations

MODULE 13 D

We can make some general observations about the results of the procedure outlined above:  When the perturbation is greater than zero, the lower of the levels is lowered and the upper is raised.  The closer the two pure levels are, the greater is the effect of the perturbation.  The stronger the perturbation, the greater is the proportional effect. Generally speaking the perturbation acts to drive the levels apart; they tend to avoid each other. This is an example of the non-crossing rule..

MODULE 13 The form of the wavefunctions  + and  - can be obtained by successively putting E = E+ and E = E- into the simultaneous equations and solving for the coefficients. It is useful to express the coefficients as trigonometric functions

MODULE 13 In the degenerate model system (E 1 = E 2 ), tan2  = ∞, or  =  /4. In this case the perturbed wavefunctions are or, each perturbed state is a 50% mixture of the two model states. In contrast, for a perturbation that mixes two states that are widely separated in energy, we see that And since sin z ~ z, and cos z ~ 1, it follows that the new wavefunctions are similar to the model functions with a small amount of contamination, the one from the other

MODULE 13 Many Level Systems The superposition of two sine curves does not make a good approximation to a straight line except over a very small interval. It is equally as clear that we can do better if we take more terms in the superposition. So it is with perturbation methods. Moving on from the two-state model we can use the complete orthonormal set of eigenfunctions of our ideal function,  (0)

MODULE 13 We shall be calculating the perturbed form of the I0> state of energy E 0, but this is not necessarily the ground state of the system. We also need the following identities which represent the various orders of perturbation of the unperturbed function where the first term on the RHS of the last is the energy of the unperturbed state, the second term is the first order correction to the energy and the third term is the second order correction, and so on. We need to solve the unknown Schrödinger equation

MODULE 13 We need to solve the unknown Schrödinger equation and to do so we substitute the above expressions for H,  now expanding  as a linear combination of states, and proceed to solve. The first solution we obtain is the first order correction to the energy. we can think of this as being the average value of the first order perturbation over the unperturbed wavefunction.

MODULE 13 An example: the anharmonic oscillator. We can write the hamiltonian for the anharmonic situation as the anharmonic terms are the third and fourth on the RHS and are treated as the perturbation. Then where v = 0,1,2,...is a quantum number, H v is a Hermite polynomial, and

MODULE 13 We are now in position to evaluate the first order correction to the energy using the equation focusing on the v = 0 state for which the polynomial = 1, the wavefunction becomes The first integral on the RHS is zero because the integrand is an odd function (x 3 ). The second integral has an area that is twice the area of the integral from zero to infinity, hence

MODULE 13 and the total v = 0 energy is The first order correction to the wavefunction After a little mathematical manipulation, we can obtain the wavefunction of our general system corrected to the first order. the prime on the summation sign tells us not to include k = 0 in the sum. The quantity in the braces is the mixing coefficient.

MODULE 13 Note the resemblance to the two-level system but now there is a summation over k states. The wavefunction for the unknown state is similar to  (0), but distorted by contributions from other states. The extent of the distortion depends on the sum of the magnitudes of the mixing coefficients. A given state makes no contribution if the matrix element ( ) is zero (e.g. symmetry reasons). The larger is E 0 - E k, the less that state will contribute, for a given value of the matrix element. Read about correcting to the second order at the end of Module 13