Schrödinger's Equation — Question 3

Question 3

Wave-function curvature identifies allowed and forbidden regions locally

A real stationary wave function is positive at a point $x_0$. Its energy is E and its potential is finite. Use $-(\hbar^2/2m)\psi\prime\prime+V\psi=E\psi$.

1. Determine the sign of curvature for $E>V(x_0)$ and for $E<V(x_0)$.

2. For an electron with $\psi\prime\prime/\psi=25.0\,\mathrm{nm}^{-2}$, infer $V-E$ using $\hbar^2/(2m_e)=.0381\,\mathrm{eV\,nm^2}$.

Question 3 - Solution

Wave-function curvature identifies allowed and forbidden regions locally

(a) Method and calculation

$$\boxed{\frac{\psi\prime\prime}{\psi}=\frac{2m}{\hbar^2}(V-E)}.$$$$\psi>0:\quad E>V\Rightarrow\psi\prime\prime<0,\qquad E<V\Rightarrow\psi\prime\prime>0.$$

(b) Interpretation and further calculation

$$V-E=\frac{\hbar^2}{2m_e}\frac{\psi\prime\prime}{\psi}.$$$$\boxed{V-E=(.0381)(25.0)\,\mathrm{eV}=.9525\,\mathrm{eV}}.$$

Physical check

The sign statement reverses where the real wave function is negative. At a node, dividing by the wave function is invalid.