- #1

CAF123

Gold Member

- 2,950

- 88

## Homework Statement

Denote the ground state and the first excited state of a 1D quantum system by ##\psi_{0}## and ##\psi_{1}##. If it is given that $$\psi_{1}(x) = x \psi_{0}(x)\,\,\,\text{and}\,\,V(0)=0$$ find the potential V(x).

## Homework Equations

TISE

## The Attempt at a Solution

If I sub ##\psi_1## into the TISE I get ##-\frac{\hbar^2}{2m} \psi_1'' + V(x)\psi_1 = E_1 \psi_1##(1). I can replace ##\psi_1## with ##x\psi_o## to give after differentiation, ##-\frac{\hbar^2}{2m} \left(2\psi_o' + x\psi_o''\right) + V(x)x\psi_o = E_1x\psi_o##. (2)

But I also know that ##-\frac{\hbar^2}{2m} \psi_0'' + V(x)\psi_0 = E_2\psi_0## Rearrange (2) yields $$-\frac{\hbar^2}{m}\psi_o' + x\left(-\frac{\hbar^2}{2m}\psi_o'' + V(x)\psi_o)\right) = E_1x\psi_o.$$The coefficient of x is precisely ##E_2\psi_o## so I end up with a diff eq for ##\psi_o##: $$\psi_o' = -\frac{m(E_2 - E_1)}{\hbar^2}x\psi_0 \Rightarrow \psi_o = K \exp\left(-\frac{m(E_1-E_2)}{\hbar^2} \frac{x^2}{2}\right)$$, K a constant.

So it then easy to get ##\psi_1## by multiplying ##\psi_o## by x. Subbing ##\psi_1## back into (1) and differentiating twice I get that $$V(x) = \frac{m}{2}\frac{(E_1-E_2)^2}{\hbar^2}x^2$$, neglecting the constant terms. Does this seem okay?

Last edited: