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| classes:2009:fall:phys4101.001:lec_notes_1209 [2009/11/23 22:16] – created yk | classes:2009:fall:phys4101.001:lec_notes_1209 [2009/12/14 19:23] (current) – x500_santi026 |
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| ===== Dec 09 (Wed) ===== | ===== Dec 09 (Wed) ===== |
| ** Responsible party: ice IX, Jake22 ** | ** Responsible party: ice IX, Jake22, chavez ** |
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| **To go back to the lecture note list, click [[lec_notes]]**\\ | **To go back to the lecture note list, click [[lec_notes]]**\\ |
| **Main class wiki page: [[home]]** | **Main class wiki page: [[home]]** |
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| Please try to include the following | 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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| * main points understood, and expand them - what is your understanding of what the points were. | ** I. Non-Degenerate Perturbation Theory ** |
| * expand these points by including many of the details the class discussed. | |
| * main points which are not clear. - describe what you have understood and what the remain questions surrounding the point(s). | |
| * Other classmates can step in and clarify the points, and expand them. | |
| * How the main points fit with the big picture of QM. Or what is not clear about how today's points fit in in a big picture. | |
| * wonderful tricks which were used in the lecture.\\ | |
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| | 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. |
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| | We are given the potential <math>V(x_{1},x_{2})=-a V_{0} \delta(x_{1}-x_{2})</math> for two identical bosons initially in an infinite square well. Our strategy for attacking this problem will be to first find the zeroth order wavefunctions and eigenvalues for the ground and first excited states. From there, we will calculate the first order corrections due to our perturbed potential. |
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| | Recall for a single particle in an infinite square well:\\ |
| | <math>E_{n}=\frac{n^{2}\pi^{2}\hbar^{2}}{2m a^{2}}</math>, n=1,2,...\\ |
| | <math>\psi_{n}=\sqrt{\frac{2}{a}}sin(\frac{n\pi x}{a})</math>, <math>(0 \lt x \lt a)</math>\\ |
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| | For the ground state of our two boson system(n=1 for each particle) we have:\\ |
| | <math>E_{g}^{(0)}=\frac{\pi^{2}\hbar^{2}}{2m a^{2}}+\frac{\pi^{2}\hbar^{2}}{2m a^{2}}=\frac{\pi^{2}\hbar^{2}}{m a^{2}}</math>\\ |
| | <math>\psi_{g}^{(0)}(x_{1},x_{2})=\psi_{1}(x_{1})\psi_{1}(x_{2})</math>, so that\\ |
| | <math>\int|\psi_{g}(x_{1},x_{2})|^{2}dx_{1}dx_{2} = \int|\psi_{1}(x_{1})|^{2}dx_{1}+\int|\psi_{1}(x_{2})|^{2}dx_{2}</math>\\ |
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| | For the first excited state we have n=1 for one particle and n=2 for the other. We must also recall that the total wavefunction of bosons must be symmetric under the exchange of any two particle labels.\\ |
| | <math>E_{1st}^{(0)}=\frac{\pi^{2}\hbar^{2}}{2m a^{2}}+\frac{4\pi^{2}\hbar^{2}}{2m a^{2}}=\frac{5\pi^{2}\hbar^{2}}{2 m a^{2}}</math>\\ |
| | <math>\psi_{1st}^{(0)}(x_{1},x_{2})=\frac{1}{\sqrt{2}}[\psi_{1}(x_{1})\psi_{2}(x_{2})+\psi_{1}(x_{2})\psi_{2}(x_{1})]</math>\\ |
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| | We are now able to calculate the first order corrections to the energy eigenvalues:\\ |
| | <math>E_{g}^{(1)}=\int\int|\psi_{g}^{(0)}(x_{1},x_{2})|^{2}V(x_{1},x_{2})dx_{1}dx_{2}=-aV_{0}\int_{0}^{a}\int_{0}^{a}sin^{2}(\frac{\pi x_{1}}{a})sin^{2}(\frac{\pi x_{2}}{a})\delta(x_{1}-x_{2})dx_{1}dx_{2}\\ |
| | =-aV_{0}\int_{0}^{a}sin^{2}(\frac{\pi x_{1}}{a})dx_{1}\int_{0}^{a}sin^{2}(\frac{\pi x_{2}}{a})\delta(x_{1}-x_{2})dx_{2}=-aV_{0}(\frac{2}{a})^{2}\int_{0}^{a}sin^{4}(\frac{\pi x_{1}}{a})dx_{1}=-\frac{3}{2}V_{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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| | **III. Degenerate Perturbation Theory** |
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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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