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Non-Degenerate Example

This example problem is related to problem 6.3 in the text.

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.

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>

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>

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>

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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