Recent content by Ziezi

You are right, I've omitted a crucial part: Inside the box, the wavefunction is: \begin{equation} \frac{\hbar^2}{2m} \frac{d^2 \psi(x)}{dx^2} = E \psi(x) \iff \frac{d^2 \psi(x)}{dx^2} = k^2 \psi(x) \end{equation} where ##k = \frac{\sqrt{2mE}}{\hbar}##. Outside the box...

That is not a bad idea, however, equation (12) is useful assuming that the values of ##\xi## are valid. The second value of ##\xi## produces an energy, ##E = 331.98##, making it a bit ambiguous given that the potential is only ##U_0 = 40##. Most probably there are mistakes in both, analytical...

The analytical solutions are: \begin{equation} \psi(x) = \begin{cases} Ce^{\alpha x}, \text{if } x < -\frac{L}{2}\\ Asin(kx) + Bcos(kx), \text{if } -\frac{L}{2} \leq x \leq \frac{L}{2}\\ Fe^{-\alpha x} , \text{if } x > \frac{L}{2} \end{cases} \end{equation}...