Identical particles in 1D infinite well

[E92M3]

Homework Statement

I need to find the mean square distance between the 2 particles. Before I can do that, I need the expectation of $$x_1^2$$ and $$x_2^2$$ , then $$x_1x_2$$. I an on the first part and got stuck.

Homework Equations

$$<X_1^2>=\int_0^L \int_0^L x_1^2 |\psi_{n,m}(x_1,x_2)|^2 dx_1dx_2$$
where:
$$\psi_{n,m}(x_1,x_2)=\frac{1}{\sqrt{2}}[\psi_n(x_1)\psi_m(x_2)+\psi_n(x_2)\psi_m(x_1)]$$
This is used because the spin state is antisymmetric.
$$\psi_n(x_1)=\sqrt{\frac{2}{L}}\sin (\frac{n \pi}{L}x_1)$$
$$\psi_m(x_1)=\sqrt{\frac{2}{L}}\sin (\frac{m \pi}{L}x_1)$$
$$\psi_n(x_2)=\sqrt{\frac{2}{L}}\sin (\frac{n \pi}{L}x_2)$$
$$\psi_m(x_2)=\sqrt{\frac{2}{L}}\sin (\frac{m \pi}{L}x_2)$$

The Attempt at a Solution

I play around a bit and got this:

$$\int_0^L \int_0^L x_1^2 |\psi_{n,m}(x_1,x_2)|^2 dx_1dx_2$$
$$=\frac{1}{2}[\int_0^L \psi_m(x_2)^2 \int_0^L x_1^2 \psi_n(x_1)^2dx_1dx_2+2\int_0^L \psi_n(x_2)\psi_m(x_2) \int_0^L x_1^2 \psi_n(x_1)$$$$\psi_m(x_1)dx_1dx_2+\int_0^L \psi_n(x_2)^2 \int_0^L x_1^2 \psi_m(x_1)^2dx_1dx_2]$$
$$=\frac{1}{2}[L^2(\frac{1}{3}-\frac{1}{2n^2\pi^2})+\frac{8}{L^2}\int_0^L \sin (\frac{n \pi}{L}x_2)\sin (\frac{m \pi}{L}x_2)dx_2\int_0^L x_1^2$$$$\sin (\frac{n \pi}{L}x_1)\sin (\frac{m \pi}{L}x_1)dx_1+L^2(\frac{1}{3}-\frac{1}{2m^2\pi^2})]$$
$$=\frac{1}{2}[L^2(\frac{2}{3}-\frac{1}{2n^2\pi^2}-\frac{1}{2m^2\pi^2})+\frac{8}{L^2}\int_0^L \sin (\frac{n \pi}{L}x_2)\sin (\frac{m \pi}{L}x_2)dx_2\int_0^L x_1^2$$$$\sin (\frac{n \pi}{L}x_1)\sin (\frac{m \pi}{L}x_1)dx_1]$$

Am I correct so far? I am stuck from this point. Thanks for the help!

 

[Jhero]

Yes, your approach is correct so far. To continue, you can use the orthogonality property of the sine functions to simplify the integral:

\int_0^L \sin (\frac{n \pi}{L}x)\sin (\frac{m \pi}{L}x)dx=\begin{cases} 0 & \text{if } n\neq m \\ \frac{L}{2} & \text{if } n=m \end{cases}

Using this property, you can simplify the integral and solve for the expectation value of x_1^2. Similarly, you can use the orthogonality property to solve for the expectation value of x_2^2. Once you have both values, you can use the formula for mean square distance to calculate the final result.