Non-degenerate Perturbation Theory Problem : can't solve exactly. But with Unperturbed eigenvalue problem. Can solve exactly. Therefore, know and . called perturbations Copyright – Michael D. Fayer, 2007

complete, orthonormal set of states Solutions of complete, orthonormal set of states with eigenvalues and Kronecker delta Copyright – Michael D. Fayer, 2007

Substitute these series into the original eigenvalue equation Expand wavefunction and also have Have series for Substitute these series into the original eigenvalue equation Copyright – Michael D. Fayer, 2007

Sum of infinite number of terms for all powers of l equals 0. Coefficients of the individual powers of l must equal 0. zeroth order - l0 first order - l1 second order - l2 Copyright – Michael D. Fayer, 2007

First order correction also substituting Substituting this result. Want to find and . Expand Then After substitution Copyright – Michael D. Fayer, 2007

Therefore, the left side is 0. After substitution Left multiply by unless n = i, but then Therefore, the left side is 0. Copyright – Michael D. Fayer, 2007

The first order correction to the energy. We have The first order correction to the energy. Then Absorbing l into The first order correction to the energy is the expectation value of . Copyright – Michael D. Fayer, 2007

First order correction to the wavefunction Again using the equation obtained after substituting series expansions Left multiply by Equals zero unless i = j. Coefficients in expansion of ket in terms of the zeroth order kets. Copyright – Michael D. Fayer, 2007

is the bracket of with and . Therefore The prime on the sum mean j  n. zeroth order ket correction to zeroth order ket energy denominator Copyright – Michael D. Fayer, 2007

First order corrections Copyright – Michael D. Fayer, 2007

Second Order Corrections Using l2 coefficient Expanding Substituting and following same type of procedures yields l2 coefficients have been absorbed. Second order correction due to first order piece of H. Second order correction due to an additional second order piece of H. Second order correction due to first order piece of H. Second order correction due to an additional second order piece of H. Copyright – Michael D. Fayer, 2007

Energy and Ket Corrected to First and Second Order Copyright – Michael D. Fayer, 2007

Example: x3 and x4 perturbation of the Harmonic Oscillator Vibrational potential of molecules not harmonic. Approximately harmonic near potential minimum. Expand potential in power series. First additional terms in potential after x2 term are x3 and x4. Copyright – Michael D. Fayer, 2007

cubic “force constant” quartic “force constant” harmonic oscillator – know solutions zeroth order eigenvalues zeroth order eigenkets perturbation c and q are expansion coefficients like l. Copyright – Michael D. Fayer, 2007

In Dirac representation First consider cubic term. Multiply out. Many terms. None of the terms have the same number of raising and lowering operators. Copyright – Michael D. Fayer, 2007

has terms with same number of raising and lowering operators. Therefore, Using Only terms with the same number of raising and lowering operators are non-zero. There are six terms. Copyright – Michael D. Fayer, 2007

Sum of the six terms Therefore With Energy levels not equally spaced. Real molecules, levels get closer together – q is negative. Correction grows with n faster than zeroth order term decrease in level spacing. With Copyright – Michael D. Fayer, 2007

Perturbation Theory for Degenerate States and normalize and orthogonal and Degenerate, same eigenvalue, E. Any superposition of degenerate eigenstates is also an eigenstate with the same eigenvalue. Copyright – Michael D. Fayer, 2007

n linearly independent states with same eigenvalue n linearly independent states with same eigenvalue system n-fold degenerate Can form n orthonormal Can form an infinite number of sets of . Nothing unique about any one set of n degenerate eigenkets. Copyright – Michael D. Fayer, 2007

Want approximate solution to zeroth order Hamiltonian perturbation zeroth order eigenket zeroth order energy But is m-fold degenerate. Call these m eigenkets belonging to the m-fold degenerate E0 orthonormal With Copyright – Michael D. Fayer, 2007

zeroth order ket having eigenvalue, Here is the difficulty perturbed ket zeroth order ket having eigenvalue, But, is a linear combination of the We don’t know which particular linear combination it is. is the correct zeroth order ket, but we don’t know the ci. Copyright – Michael D. Fayer, 2007

Substituting the expansions for E and into To solve problem Expand E and Some superposition, but we don’t know the cj. Don’t know correct zeroth order function. Substituting the expansions for E and into and obtaining the coefficients of powers of l, gives zeroth order first order want these Copyright – Michael D. Fayer, 2007

Use projection operator To solve substitute Need Use projection operator The projection operator gives the piece of that is . Then the sum over all k gives the expansion of in terms of the . Defining Known – know perturbation piece of the Hamiltonian and the zeroth order kets. Copyright – Michael D. Fayer, 2007

Substituting this and gives this piece becomes Substituting this and gives Result of operating H0 on the zeroth order kets. Left multiplying by Copyright – Michael D. Fayer, 2007

Correction to the Energies Two cases: i  m (the degenerate states) and i > m. i  m Left hand side – sum over k equals zero unless k = i. But with i  m, The left hand side of the equation = 0. Right hand side, first term non-zero when j = i. Bracket = 1, normalization. Second term non-zero when k = i. Bracket = 1, normalization. The result is We don’t know the c’s and the . Copyright – Michael D. Fayer, 2007

is a system of m of equations for the cj’s. One equation for each index i of ci. Besides trivial solution of only get solution if the determinant of the coefficients vanish. We know the Have mth degree equation for the . Copyright – Michael D. Fayer, 2007

Get system of equations for the coefficients, cj’s. Solve mth degree equation – get the . Now have the corrections to energies. To find the correct zeroth order eigenvectors, one for each , substitute (one at a time) into system of equations. Get system of equations for the coefficients, cj’s. Know the . There are only m – 1 conditions because can multiply everything by constant. Use normalization for mth condition. Now we have the correct zeroth order functions. Copyright – Michael D. Fayer, 2007

The solutions to the mth degree equation (expanding determinant) are Therefore, to first order, the energies of the perturbed initially degenerate states are Have m different (unless some still degenerate). Copyright – Michael D. Fayer, 2007

Correction to wavefunctions Again using equation found substituting the expansions into the first order equation Left multiply by Orthogonality makes other terms zero. Normalization gives 1 for non-zero brackets. gives 1 gives 0 Therefore Normalization gives Aj = 0 for j  m. Already have part of wavefunction for j  m Copyright – Michael D. Fayer, 2007

First order degenerate perturbation theory results Correct zeroth order function. Coefficients ck determined from system of equations. Correction to zeroth order function. Copyright – Michael D. Fayer, 2007