Question

Suppose that ${\mathrm{\psi }}_1\left(\mathrm{x}\right)$and ${\psi }_2\left(x\right)$are both solutions to the Schrödinger equation for the same potential energy $\mathrm{U}\left(\mathrm{x}\right)$. Prove that the superposition $\mathrm{\psi }\left(\mathrm{x}\right)={\mathrm{A\psi }}_1\left(\mathrm{x}\right)+{\mathrm{B\psi }}_2\left(\mathrm{x}\right)$is also a solution to the Schrödinger equation.

Short answer

The prove is done.

Step-by-step solution

Given information 

We have given, two wave function for the same potential.

We have to prove that the superposition will also be the solution.

Simplify

We know the Schrodinger equation is,

$-\frac{\hslash }{2\mathrm{m}}\frac{{\mathrm{d}}^2\mathrm{\psi }\left(\mathrm{x}\right)}{{\mathrm{dx}}^2}+{\mathrm{U}}_0\mathrm{\psi }\left(\mathrm{x}\right)=\mathrm{E\psi }\left(\mathrm{x}\right)$

So we can write,

$$\begin{gathered} -\frac{\hslash }{2\mathrm{m}}\frac{{\mathrm{d}}^2{\mathrm{\psi }}_1\left(\mathrm{x}\right)}{{\mathrm{dx}}^2}+{\mathrm{U}}_0{\mathrm{\psi }}_1\left(\mathrm{x}\right)={\mathrm{E}}_1{\mathrm{\psi }}_1\left(\mathrm{x}\right)..............\left(1\right) \\ -\frac{\hslash }{2\mathrm{m}}\frac{{\mathrm{d}}^2{\mathrm{\psi }}_2\left(\mathrm{x}\right)}{{\mathrm{dx}}^2}+{\mathrm{U}}_0{\mathrm{\psi }}_2\left(\mathrm{x}\right)={\mathrm{E}}_2{\mathrm{\psi }}_2\left(\mathrm{x}\right)..................\left(2\right) \end{gathered}$$

Now, add them as,$\mathrm{A}\times \left(1\right)+\mathrm{B}\times \left(2\right)$

Then,

$$\begin{gathered} \mathrm{A}(-\frac{\hslash }{2\mathrm{m}}\frac{{\mathrm{d}}^2{\mathrm{\psi }}_1\left(\mathrm{x}\right)}{{\mathrm{dx}}^2}+{\mathrm{U}}_0{\mathrm{\psi }}_1(\mathrm{x}\left)\right)+\mathrm{B}(-\frac{\hslash }{2\mathrm{m}}\frac{{\mathrm{d}}^2{\mathrm{\psi }}_2\left(\mathrm{x}\right)}{{\mathrm{dx}}^2}+{\mathrm{U}}_0{\mathrm{\psi }}_2(\mathrm{x}\left)\right)=\mathrm{i}\hslash \frac{\mathrm{d}}{\mathrm{dt}}{\mathrm{\psi }}_1\left(\mathrm{x}\right)+\mathrm{i}\hslash \frac{{\mathrm{d\psi }}_2\left(\mathrm{x}\right)}{\mathrm{dt}} \\ -\frac{\hslash }{2\mathrm{m}}\frac{{\mathrm{d}}^2\left({\mathrm{A\psi }}_1\right(\mathrm{x})+{\mathrm{B\psi }}_2(\mathrm{x}\left)\right)}{{\mathrm{dx}}^2}+{\mathrm{U}}_0\left({\mathrm{A\psi }}_1\right(\mathrm{x})+{\mathrm{B\psi }}_2(\mathrm{x}\left)\right)=\mathrm{i}\hslash \frac{\mathrm{d}\left({\mathrm{A\psi }}_1\right(\mathrm{x})+{\mathrm{B\psi }}_2(\mathrm{x})}{\mathrm{dt}} \\ -\frac{\hslash }{2\mathrm{m}}\frac{{\mathrm{d}}^2\mathrm{\psi }\left(\mathrm{x}\right)}{{\mathrm{dx}}^2}+{\mathrm{U}}_0\mathrm{\psi }\left(\mathrm{x}\right)=\mathrm{i}\hslash \frac{\mathrm{d\psi }}{\mathrm{dt}} \end{gathered}$$

hence proved.