Constant Potential

Goal

We have a particle in an environment with constant Potential $V(x)=k$.

Solution

The associated Hamiltonian is given by $H=\frac{P^2}{2m}+V(x)=\frac{-{\hslash }^2}{2m}\frac{{\partial }^2}{\partial x^2}+k$. The associated Time-independent Schrodinger equation is given by

$$\begin{array}{rll}E|\psi ⟩& =H|\psi ⟩& \text{Expanding the Hamiltonian}\\ & =\left(\frac{-{\hslash }^2}{2m}\frac{{\partial }^2}{\partial x^2}+k\right)|\psi ⟩& \\ & =\frac{-{\hslash }^2}{2m}\frac{{\partial }^2}{\partial x^2}|\psi ⟩+k|\psi ⟩& \\ & \Updownarrow & \\ (E-k)|\psi ⟩& =\frac{-{\hslash }^2}{2m}\frac{{\partial }^2}{\partial x^2}|\psi ⟩& \end{array}$$

This is a standard Differential Equation, whose solution can be written in the position basis as:

$$\psi (x)=\{\begin{array}{ll}A\exp \left(\frac{\sqrt{2m(E-k)}}{\hslash }x\right)+B\exp \left(-\frac{\sqrt{2m(E-k)}}{\hslash }x\right),& \frac{-{\hslash }^2}{2m(E-k)}\ge 0\\ Ax+B,& E=k\\ A\cos \left(\frac{\sqrt{2m(E-k)}}{\hslash }x\right)+B\sin \left(\frac{\sqrt{2m(E-k)}}{\hslash }x\right),& \frac{-{\hslash }^2}{2m(E-k)}\ge 0\end{array}$$

Simplifying the conditions by comparing the values of $E$ and $k$ directly, we get

$$\psi (x)=\{\begin{array}{ll}A\exp \left(\frac{\sqrt{2m(k-E)}}{\hslash }x\right)+B\exp \left(-\frac{\sqrt{2m(k-E)}}{\hslash }x\right),& E\le k\\ Ax+B,& E=k\\ A\cos \left(\frac{\sqrt{2m(E-k)}}{\hslash }x\right)+B\sin \left(\frac{\sqrt{2m(E-k)}}{\hslash }x\right),& E\ge k\end{array}$$