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| classes:2009:fall:phys4101.001:lec_notes_1209 [2009/12/14 16:08] – youmans | classes:2009:fall:phys4101.001:lec_notes_1209 [2009/12/14 19:23] (current) – x500_santi026 |
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| **Main class wiki page: [[home]]** | **Main class wiki page: [[home]]** |
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| **Non-degenerate Perturbation Theory** | In this lecture we continued in our exploration of non-degenerate perturbation theory with a specific example from the textbook (problem 6.3) and went on to discuss the case of two-fold degenerate perturbations. |
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| <math>E_n^{(1)}=<\psi_n^{(0)|H'|\psi_n^{(0)}> </math> | ** I. Non-Degenerate Perturbation Theory ** |
| <math>|\psi_n^{(0)}>=\sum_{n\neq m}\frac{|\psi_m^{(0)}><\psi_m^{(0)|H'|\psi_n^{(0)}>}{E_n^{0}-E_m^{0}}</math> | |
| <math>E_n^{(2)}=<\psi_n^{(0)|H'|\psi_n^{(0)}>=\sum_{n\neq m}\frac{|<\psi_m^{(0)}|H'|\psi_n^{(0)}>|^2}{E_n^{0}-E_m^{0}}</math> | |
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| **Non-Degenerate Example** | Summarizing from Monday's lecture, we know that in order to approximate a solution to the Schrödinger equation for a specified potential, the eigenvalues (energies) and eigenfunctions (wavefunctions) are rewritten as power series so nth order corrections may be obtained. The 1st order correction to the energy is the expectation value of the perturbation <math>H'</math> in the non-perturbed state <math>\psi_n^{(0)}</math>:\\ |
| | <math>E_n^{(1)}=<\psi_n^{(0)}|H'|\psi_n^{(0)}> </math>\\ |
| | The corresponding wavefunction correction is:\\ |
| | <math>|\psi_n^{(1)}>=\sum_{n\neq m}\frac{|\psi_m^{(0)}><\psi_m^{(0)}|H'|\psi_n^{(0)}>}{E_n^{0}-E_m^{0}}</math>\\ |
| | The 2nd order correction to the energy yields a similar result:\\ |
| | <math>E_n^{(2)}=<\psi_n^{(0)}|H'|\psi_n^{(0)}>=\sum_{n\neq m}\frac{|<\psi_m^{(0)}|H'|\psi_n^{(0)}>|^2}{E_n^{0}-E_m^{0}}</math>\\ |
| | Note that even the 1st order correction to the wavefunction is "notoriously poor" as Griffiths states, so the 2nd order correction is not useful and therefore not shown. With these formulae we can examine the effects of small changes in the idealized, standard potentials that have already been established in previous chapters. |
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| | **II. Non-Degenerate Example ** |
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| This example problem is related to problem 6.3 in the text. | This example problem is related to problem 6.3 in the text. |
| <math>E_{1st}^{(1)}=\int\int|\psi_{1st}^{(0)}(x_{1},x_{2})|^{2}V(x_{1},x_{2})dx_{1}dx_{2}=-2V_{0}</math>\\ | <math>E_{1st}^{(1)}=\int\int|\psi_{1st}^{(0)}(x_{1},x_{2})|^{2}V(x_{1},x_{2})dx_{1}dx_{2}=-2V_{0}</math>\\ |
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| **Degenerate Case** | **III. Degenerate Perturbation Theory** |
| 2-fold degeneracy | |
| By definition, <math>H_{0}|\psi_a^{0}>=E^{0}|\psi_a^{0}></math>\\ | |
| <math>H_{0}|\psi_b^{0}>=E^{0}|\psi_b^{0}></math>\\ | |
| For <math>E^{1}</math>, we only need <math>|\psi_a^{0}></math> and <math>|\psi_b^{0}></math>. | |
| <math>H|/psi>=(E^{0}+E^{1})|/ps | |
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| | When the situation calls for multiple, distinct states with the same energy, we must reformulate the above process to include the complications that arise. If we deal with the simpler case of two-fold degeneracy first, the generalization to higher orders follows easily. |
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| | To begin, the two distinct states are defined to be\\ |
| | <math>H^0|\psi_a^{0}>=E^0|\psi_a^{0}></math> and <math>H^0|\psi_b^{0}>=E^0|\psi_b^{0}></math>.\\ |
| | Note that the Hamiltonian is <math>H=H^0+H'</math>. For the 1st order correction of the energy <math>E^{(1)}</math>, we can represent <math>|\psi_a^{(0)}></math> and <math>|\psi_b^{(0)}></math> as a linear combination such that <math>\psi</math> is still an eigenstate of the unperturbed Hamiltonian:\\ |
| | <math>H|\psi>=(E^{0}+E^{1})|\psi>\Rightarrow|\psi>=\alpha|\psi_a^{0}>+\beta|\psi_b^{0}></math>\\ |
| | We can write this in matrix form:\\ |
| | <math>\begin{pmatrix} H_{aa} H_{ab}\\H_{ba} H_{bb}\end{pmatrix} \begin{pmatrix} \alpha\\\beta\end{pmatrix}=\begin{pmatrix} E^{(0)}+E^{(1)}\end{pmatrix} \begin{pmatrix} \alpha\\beta\end{pmatrix}</math>\\ |
| | The matrix elements of this Hamiltonian follow the form: <math>H_{ij}=<\psi_i^{0}|H^0+H'|\psi_j^{0}></math>. Separating the unperturbed and perturbed portions, the above matrix equation becomes:\\ |
| | <math>\begin{pmatrix} E^0+H'_{aa} H'_{ab}\\H'_{ba} E^0+H'_{bb}\end{pmatrix} \begin{pmatrix} \alpha\\\beta\end{pmatrix}=\begin{pmatrix} E^0+E^1\end{pmatrix} \begin{pmatrix} \alpha\\\beta\end{pmatrix}</math>\\ |
| | The unperturbed energy can be eliminated to yield an equation that provides the 1st order correction for the energy:\\ |
| | <math>\begin{pmatrix} H'_{aa} H'_{ab}\\H'_{ba} H'_{bb}\end{pmatrix} \begin{pmatrix} \alpha\\\beta\end{pmatrix}=\begin{pmatrix} E^1\end{pmatrix} \begin{pmatrix} \alpha\\\beta\end{pmatrix}</math>. |
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