Question

2096) Two noninteracting particles 1 and 2, each of mass m, are in a 1-D infinite square well ol width a. If one is in the state V'in and the other in the state (n! /), calculate C(xI-x), assuming (a) (6%) they are distinguishable particles, (b) (7%) Ihey are identical bosons, and (c) (796) Ihey arc identical fermions. 4.

Answer 1

$\langle(x_1-x_2)^2\rangle=\langle x_1^2\rangle+\langle x_2^2\rangle-2\langle x_1x_2\rangle=\langle x_1^2\rangle+\langle x_2^2\rangle-2\langle x_1\rangle \langle x_2\rangle$

because for 2 independent(non-interacting) variables A and B, $\langle AB\rangle=\langle A\rangle\langle B\rangle$

Also, single particle wavefunctions are

$\psi_n=\sqrt{\frac{2}{a}}\sin\left ( \frac{n\pi x}{a} \right )$

Now, lets begin!

(a) If particles are distinguishable (which is a non-realistic situation for quantum particles, but anyway) there is no condition on the overall wavefunctions, hence we have simply

$\Psi_{nl}(x_1,x_2)=\psi_n(x_1)\psi_l(x_2)=\frac{2}{a}\sin\left ( \frac{n\pi x_1}{a} \right )\sin\left ( \frac{l\pi x_2}{a} \right )$

Note: Because we can label the particle(possible for distinguishable), I have explicitly chosen "1" in n-th state and "2" in l-th state.

$\langle(x_1-x_2)^2\rangle=\iint x_1^2\Psi^2dx_1dx_2+\iint x_2^2\Psi^2dx_1dx_2-2\iint x_1x_2\Psi^2dx_1dx_2$

Some simplification can be done without actually expanding

$\langle(x_1-x_2)^2\rangle=\int x_1^2\psi_n^2\,dx_1+\int x_2^2\psi_l^2\,dx_2-2\int x_1\psi_n^2\,dx_1\int x_2\psi_l^2\,dx_2$

$\langle(x_1-x_2)^2\rangle=a^2\left ( \frac{1}{3}-\frac{1}{2n^2\pi^2} \right )+a^2\left ( \frac{1}{3}-\frac{1}{2l^2\pi^2} \right )-2\frac{a}{2}\cdot\frac{a}{2}$

$\langle(x_1-x_2)^2\rangle=a^2\left ( \frac{1}{6}-\frac{1}{2n^2\pi^2}-\frac{1}{2l^2\pi^2} \right )$

(b,c) Can't label them anymore.

$\Psi_{nl}(x_1,x_2)=\frac{1}{\sqrt2}\left [\psi_n(x_1)\psi_l(x_2)\pm\psi_l(x_1)\psi_n(x_2) \right ]$

+ for Bosons and - for Fermions. On doing the calculation you will get the same answer as (a)