Bacat
- 149
- 1
Homework Statement
I am actually looking for the expectation of x for the wavefunction that is \Psi (x) = \sqrt{\frac{2}{L}}Sin(\frac{\pi x}{L}) for 0 < x < L.
To do this I need to find the solution to this integral:
f = \int_0 ^L \! \Psi^* x \Psi \, dx = \int_0^L \! x*(\sqrt{\frac{2}{L}}Sin(\frac{\pi x}{L}))^2 \, dx
The Attempt at a Solution
Using Mathematica, I know that the answer to the integral is \frac{L^2}{4}.
However, when I attempt the solution by integration by parts, I get 0. Help!
f = \frac{2}{L} \int_0^L \! xSin^2(\frac{\pi x}{L}) \, dx = \frac{2}{L} \left[ uv - \int v du \right]
Let u=x, du=dx, dv=Sin^2(\frac{\pi x}{L}) dx
Then v=\int dv = \int_0^L Sin^2(\frac{\pi x}{L}) dx = \frac{1}{2} \int_0^L (1-Cos(\frac{\pi x}{L}) dx)
v = \frac{L}{2} - Sin^2(\pi) = \frac{L}{2}
Then f = \frac{2}{L}\left[uv - \int v du\right] = \frac{2}{L}\left[x\frac{L}{2}\right|_{0}^{L} - \int_0^L \frac{L}{2} dx\right] = \frac{2}{L}\left[ \frac{L^2}{2} - \frac{L^2}{2}\right] = 0.
Where have I made an error?