Question

A finite potential energy function U(x) allows $\mathit{\psi }\left(x\right)$ the solution of the time-independent Schrödinger equation. to penetrate the classically forbidden region. Without assuming any particular function for U(x) show that b(x) must have an inflection point at any value of x where it enters a classically forbidden region.

Short answer

The answer of given problem is$\psi \left(x\right)=\left\{\begin{array}{ll}Cex{p}^{+\alpha x}& x<0\\ \text{Asin(kL) + Bcos(kL)}& \text{0}⩽\text{x}⩽\text{L}\\ Gex{p}^{-\alpha x}& x>L\end{array}\right.$

Step-by-step solution

Wave function

The wave function $\psi \left(x\right)$

role="math" localid="1660033358067" $\psi \left(x\right)=\left\{\begin{array}{ll}Cex{p}^{+\alpha x}& x<0\\ \text{Asin(kL) + Bcos(kL)}& \text{0}⩽\text{x}⩽\text{L}\\ Gex{p}^{-\alpha x}& x>L\end{array}\right.$

Step 2:Total energy and potential energy

By definition, the inflection point is a point where the total energy is equal to zero. Total energy E is equal to potential energy U_o

Traditionally authorised and classically banned territories are separated by turning points. A turning point is a point at which the second derivative vanishes.

So, the answer is$\psi \left(x\right)=\left\{\begin{array}{ll}Cex{p}^{+\alpha x}& x<0\\ \text{Asin(kL) + Bcos(kL)}& \text{0}⩽\text{x}⩽\text{L}\\ Gex{p}^{-\alpha x}& x>L\end{array}\right.$