- #1
gfd43tg
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Homework Statement
Homework Equations
The Attempt at a Solution
a) For this part, I know for distinguishable particles, the expectation value of the square distance
$$\langle (x_{1}^{2} - x_{2}^{2}) \rangle = \langle x^{2} \rangle_{2} + \langle x^{2} \rangle_{3} - 2 \langle x \rangle_{2} \langle x \rangle_{3}$$
This just gets really ugly...
$$\langle x^{2} \rangle_{2} = \int \psi_{2}^{*}x^{2}\psi_{2} dx = \frac {2}{a} \int x^{2} sin^{2}\Big (\frac {2 \pi}{a} x \Big ) dx $$
Using the equation given in the problem (is that even right? It seems suspicious that it is independent of ##n##)
$$\langle x^{2} \rangle_{2} = \frac {2}{a} \Big [ \frac {\pi^{3}}{6} - \frac {\pi}{4(\frac {2 \pi}{a})^{2}} \Big ] $$
$$\langle x^{2} \rangle_{2} = \frac {2}{a} \Big [ \frac {\pi^{3}}{6} - \frac {a^{2}}{16 \pi} \Big ] $$
$$ = \frac {\pi^{3}}{3a} - \frac {a}{8 \pi} $$
Similarly for ##\langle x^{2} \rangle_{3}##
$$\langle x^{2} \rangle_{3} = \frac {2}{a} \Big [ \frac {\pi^{3}}{6} - \frac {\pi}{4(\frac {3 \pi}{a})^{2}} \Big ]$$
$$\langle x^{2} \rangle_{3} = \frac {2}{a} \Big [ \frac {\pi^{3}}{6} - \frac {a^{2}}{36 \pi} \Big ] $$
$$=\frac {\pi^{3}}{3a} - \frac {a}{18 \pi}$$
For ##\langle x \rangle_{2}##
$$\langle x \rangle_{2} = \int \psi_{2}^{*}x \psi_{2} dx$$
$$\langle x \rangle_{2} = \frac {2}{a} \int x sin^{2}(\frac {2 \pi}{a} x) dx $$
From the equation given in the problem statement
$$\langle x \rangle_{2} = \frac {2}{a} \frac {\pi^{2}}{4} = \frac {\pi^{2}}{2a} $$
The same goes for ##\langle x \rangle_{3}##
$$\langle x \rangle_{3} = \frac {2}{a} \int x sin^{2}(\frac {3 \pi}{a} x) dx$$
$$\langle x \rangle_{3} = \frac {2}{a} \frac {\pi^{2}}{4} = \frac {\pi^{2}}{2a}$$
$$\langle (x_{1}^{2} - x_{2}^{2}) \rangle = \langle x^{2} \rangle_{2} + \langle x^{2} \rangle_{3} - 2 \langle x \rangle_{2} \langle x \rangle_{3}$$
$$\langle (x_{1}^{2} - x_{2}^{2}) \rangle = \Big (\frac {\pi^{3}}{3a} - \frac {a}{8 \pi} \Big ) + \Big (\frac {\pi^{3}}{3a} - \frac {a}{18 \pi} \Big ) - 2 \Big (\frac {\pi^{2}}{2a} \Big ) \Big ( \frac {\pi^{2}}{2a} \Big ) $$
$$\langle (x_{1}^{2} - x_{2}^{2}) \rangle = \frac {2 \pi^{3}}{3a} - \frac {13a}{72 \pi} - \frac {\pi^{4}}{2a^{2}} $$
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(b) For this problem, I am basically looking at this webpage
http://galileo.phys.virginia.edu/classes/252/symmetry/Symmetry.html
symmetric:
$$\psi_{23} (x_{1},x_{2}) = A sin \Big (\frac {2 \pi}{a} x_{1} \Big ) cos \Big (\frac {3 \pi}{a} x_{2} \Big ) $$
Not sure how to get A, but anyways
For symmetric
$$ \psi_{23}^{S} = \frac {1}{\sqrt{2}} \Big [ sin \Big (\frac {2 \pi}{a} x_{1} \Big ) cos \Big (\frac {3 \pi}{a} x_{2} \Big ) +sin \Big (\frac {3 \pi}{a} x_{1} \Big ) cos \Big (\frac {2 \pi}{a} x_{2} \Big ) \Big ]$$
For antisymmetric
$$ \psi_{23}^{A} = \frac {1}{\sqrt{2}} \Big [ sin \Big (\frac {2 \pi}{a} x_{1} \Big ) cos \Big (\frac {3 \pi}{a} x_{2} \Big ) - sin \Big (\frac {3 \pi}{a} x_{1} \Big ) cos \Big (\frac {2 \pi}{a} x_{2} \Big ) \Big ] $$
This is slightly different from my notes, which had two sine terms instead of a cosine term, and I am not sure if that is because this time it has electrons. I am just so confused...
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