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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

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

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

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Sum of infinite number of terms for all powers of equals 0. Coefficients of the individual powers of must equal 0. zeroth order - 0 first order - 1 second order - 2 Copyright – Michael D. Fayer, 2007

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

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

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

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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

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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

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First order corrections Copyright – Michael D. Fayer, 2007

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Second Order Corrections Using 2 coefficient Expanding Substituting and following same type of procedures yields 2 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

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

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Example: x 3 and x 4 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 x 2 term are x 3 and x 4. Copyright – Michael D. Fayer, 2007

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

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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

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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

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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. Copyright – Michael D. Fayer, 2007

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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

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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, 2007

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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 E 0 orthonormal With Copyright – Michael D. Fayer, 2007

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Here is the difficulty perturbed ket zeroth order ket having eigenvalue, But, is a linear combination of the We dont know which particular linear combination it is. is the correct zeroth order ket, but we dont know the c i. Copyright – Michael D. Fayer, 2007

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To solve problem Expand E and Some superposition, but we dont know the c j. Dont know correct zeroth order function. Substituting the expansions for E and into and obtaining the coefficients of powers of, gives zeroth order first order want these Copyright – Michael D. Fayer, 2007

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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

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this piece becomes Left multiplying by Substituting this and gives Result of operating H 0 on the zeroth order kets. Copyright – Michael D. Fayer, 2007

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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 dont know the cs and the. Copyright – Michael D. Fayer, 2007

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

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Solve m th 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, c j s. There are only m – 1 conditions because can multiply everything by constant. Use normalization for m th condition. Now we have the correct zeroth order functions. Know the. Copyright – Michael D. Fayer, 2007

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The solutions to the m th 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

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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 A j = 0 for j m. Already have part of wavefunction for j m Copyright – Michael D. Fayer, 2007

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

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