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, 2017 1 1 1

complete, orthonormal set of ket vectors Solutions of complete, orthonormal set of ket vectors with eigenvalues and Kronecker delta Copyright – Michael D. Fayer, 2017 2 2 2

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, 2017 3 3 3

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, 2017 4 4 4

First order correction also substituting Substituting this result. Want to find and . Expand Then After substitution Copyright – Michael D. Fayer, 2017 5 5 5

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, 2017 6 6 6

The first order correction to the energy. We have The first order correction to the energy. (Expectation value of in zeroth order state ) number, kets normalized, and transposing, Then Absorbing l into The first order correction to the energy is the expectation value of . Copyright – Michael D. Fayer, 2017 7 7 7

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, 2017 8 8 8

is the bracket of with and . Therefore The prime on the sum mean j  n. zeroth order ket energy denominator correction to zeroth order ket Copyright – Michael D. Fayer, 2017 9 9 9

First order corrections Copyright – Michael D. Fayer, 2017 10 10 10

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, 2017 11 11 11

Energy and Ket Corrected to First and Second Order Copyright – Michael D. Fayer, 2017 12 12 12

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, 2017 13 13 13

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, 2017 14 14 14

In Dirac representation First consider cubic term. Multiply out. Many terms. None of the terms have the same number of raising and lowering operators. (At second order will not be zero.) Copyright – Michael D. Fayer, 2017 15 15 15

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, 2017 16 16 16

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, 2017 17 17 17

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, 2017 18 18 18

n linearly independent states with same eigenvalue n linearly independent states with same eigenvalue system n-fold degenerate Can form an infinite number of sets of . Nothing unique about any one set of n degenerate eigenkets. Can form n orthonormal Copyright – Michael D. Fayer, 2017 19 19 19

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 E10 orthonormal With Copyright – Michael D. Fayer, 2017 20 20 20

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. The correct zero order ket depends on the nature of the perturbation. p states of the H atom in external magnetic field – p1, p0, p-1 electric field – px, pz, py Copyright – Michael D. Fayer, 2017 21 21 21

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, 2017 22 22 22

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, 2017 23 23 23

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, 2017 24 24 24

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, 2017 25 25 25

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, 2017 26 26 26

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, 2017 27 27 27

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, 2017 28 28 28

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, 2017 29 29 29

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, 2017 30 30 30