# Particle in one-dimensional box

1. Nov 12, 2008

### kasse

By separation of variables, I have found that

$$\frac{-\hbar}{2mg(x)}\frac{d^{2}g(x)}{dx^{2}} = \frac{i}{h(t)}\frac{d h(t)}{dt}$$

Both sides there have to equal the same constant. But why is this constant the total energy of the particle?

2. Nov 12, 2008

### nicksauce

The idea is that you have that Hamiltonian $$H = \frac{p^2}{2m} = -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial^2x}$$. The energy of the particle are eigenvalues of the Hamiltonian, so $$H\psi=E\psi$$, so $$\frac{-\hbar^2}{2m}\frac{\partial^2}{\partial^2x}\psi=E\psi$$.

(And in the way you wrote it, the constant won't be the energy of the particle, but rather the energy divided by hbar)