Answer to Question #152933 in Quantum Mechanics for adnan

Question

A wave function is given to be 𝜓(𝑥) = 𝑁( 𝑥 /𝑥0 )^ 𝑛𝑒 −𝑥/𝑥0, 𝑁, 𝑛, 𝑥0 𝑎𝑟𝑒 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡𝑠 From Schrodinger wave equation find potential V(x) and the energy E for which 𝜓(𝑥) is an Eigen function.

Answer 1

Let's write Schrodinger equation applied to $\psi(x)$ :

$-\frac{\hbar^2}{2m} \psi'' +V\psi = E\psi$

First calculate $\psi''$ :

$\psi''=N e^{-x/x_0} (\frac{n(n-1)}{x_0^2} (\frac{x}{x_0})^{n-2}-2\frac{n}{x_0^2}(\frac{x}{x_0})^{n-1}+\frac{1}{x_0^2}(\frac{x}{x_0})^n)$

$\psi''=Ne^{-x/x_0} (\frac{x}{x_0})^n (\frac{1}{x_0^2}-\frac{2n}{xx_0}+\frac{n(n-1)}{x^2}) = \psi (\frac{1}{x_0^2}-\frac{2n}{xx_0}+\frac{n(n-1)}{x^2})$

Now we substitue this expression into the Schrodinger equation :

$(\frac{-\hbar^2}{2m}(\frac{1}{x_0^2}-\frac{2n}{xx_0}+\frac{n(n-1)}{x^2})+V)\psi = E\psi$

As $\psi\neq 0$ everywhere except for $x=0$, we can simplify by $\psi$ and find

$\frac{-\hbar^2}{2m}(\frac{1}{x_0^2}-\frac{2n}{xx_0}+\frac{n(n-1)}{x^2})+V = E$

$V=E+\frac{\hbar^2}{2m}(\frac{1}{x_0^2}-\frac{2n}{xx_0}+\frac{n(n-1)}{x^2})$

As $V\to 0$ when $|x|\to+\infty$ , we find that

$E=-\frac{\hbar^2}{2mx_0^2}$

$V=\frac{\hbar^2}{2m}(-\frac{2n}{xx_0}+\frac{n(n-1)}{x^2})$