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7_Perturbation_Theory

# 7_Perturbation_Theory - VII Time Independent Perturbation...

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VII. Time Independent Perturbation Theory Gri ths Chapter 6 1

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A. Nondegenerate Perturbation Theory Suppose we have solved the time independent Schrödinger equation for some potential. 1. Formulation H 0 | n 0 = E 0 n | n 0 Therefore we know the eigenvalues E n 0 and the orthonormal eigenfunctions n 0 | m 0 = δ nm . Now we perturb the potential slightly. What are the new eigenvalues and eigenfunctions? 2
new Hamiltonian In general, we won’t be able to solve the new Schrödinger equation, but we can use the old eigenvalues and eigenfunctions to get an approximat e solution. H = H 0 + H 1 Next, we assume the new eigenvalues and eigenfunctions can be expanded about the old ones. | n = | n 0 + | n 1 + | n 2 + · · · E n = E 0 n + E 1 n + E 2 n + · · · The superscripts gives the power of matrix element of H 1 that we expect the term to come out proportional to. H | n = E n | n A. Nondegenerate Perturbation Theory Let’s write the new Hamiltonian, corresponding to the perturbed potential, in terms of the old Hamiltonian. 3

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H = H 0 + H 1 | n = | n 0 + | n 1 + | n 2 + · · · E n = E 0 n + E 1 n + E 2 n + · · · Substitute everything in... H | n = E n | n 2. Zeroth Order Theory H 0 | n 0 = E 0 n | n 0 We recover the unperturbed system 3. First Order Theory E 1 n = n 0 | H 1 | n 0 | n 1 = m = n m 0 | H 1 | n 0 ( E 0 n E 0 m ) | m 0 1st order in the matrix element of H 1 , as expected. The first order correction to the energy is the expectation value of the perturbation in the unperturbed state. See board VII.A.3 A. Nondegenerate Perturbation Theory 4
4. An Example Gri ths Problem 6.1: Suppose we put a delta function bump in the centre of an infinite square well: H 1 = αδ ( x a/ 2) where α is a constant. ( a ) Find the first order correction to the allowed energies. Explain why the energies are not perturbed for even n. ( b ) Find the first three non - zero terms in the expansion of the correction to the ground state wave function. Solution: See board VII.A.4 A. Nondegenerate Perturbation Theory 5

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5. Second Order Energies The second order energies are found, using the same method, to be E 2 n = m = n | m 0 | H 1 | n 0 | 2 E 0 n E 0 m Most books stop here ( Shankar, Gri ths, for example ) ignoring the 2nd order wave functions. We’ll stop here too. For those interested, Libo ff gives the 2nd order wave functions. 6 A. Nondegenerate Perturbation Theory See board VII.A.5
6. Another Example Gri ths Problem 6.4: Suppose we put a delta function bump in the centre of an infinite square well: H 1 = αδ ( x a/ 2) where α is a constant. Solution: See board VII.A.6 Find the second order correction to the energies. 7 A. Nondegenerate Perturbation Theory

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B. Degenerate Perturbation Theory We found the first order perturbation correction for the wave function was 1. The Problem | n 1 = m = n m 0 | H 1 | n 0 ( E 0 n E 0 m ) | m 0 If two di ff eren t states, |n 0 and |m 0 have the sam e energy, E n 0 = E m 0 , then this expression blows up.
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