- #1
stunner5000pt
- 1,443
- 4
In the first hald of this question it was proven that
-\frac{\hbar^2}{2m} \frac{d}{dx} \left[ \phi_{m}^* \frac{d \phi_{n}}{dx} - \phi_{n} \frac{d \phi_{m}^*}{dx}\right] = (E_{m} - E_{n}) \phi_{m}^* \phi_{n}
By integrating over x and by assuming taht Phi n and Phi m are zero are x = +/- infinity show that
\int_{-infty}^{infty} \phi_{m}^*(x) \phi_{n}(x) dx = 0 if Em is not En
so for this do i simply integrate that above expression wrt x?? is it really that simple?
-\frac{\hbar^2}{2m} \frac{d}{dx} \left[ \phi_{m}^* \frac{d \phi_{n}}{dx} - \phi_{n} \frac{d \phi_{m}^*}{dx}\right] = (E_{m} - E_{n}) \phi_{m}^* \phi_{n}
By integrating over x and by assuming taht Phi n and Phi m are zero are x = +/- infinity show that
\int_{-infty}^{infty} \phi_{m}^*(x) \phi_{n}(x) dx = 0 if Em is not En
so for this do i simply integrate that above expression wrt x?? is it really that simple?
