Question

Verify that the n=1 wave function ${\psi }_1\left(x\right)$ of the quantum harmonic oscillator really is a solution of the Schrödinger equation. That is, show that the right and left sides of the Schrödinger equation are equal if you use the ${\psi }_1\left(x\right)$ wave function.

Short answer

$\frac{d^2{\Psi }_1\left(x\right)}{d{x}^2}=-\frac{2m}{{\hslash }^2}\left(E-\frac{1}{2}k{x}^2\right){\Psi }_1\left(x\right)$

Step-by-step solution

Step 1. Given information

The Schrödinger wave equation for a quantum harmonic oscillator is,

$\frac{d^2\Psi }{d{x}^2}=-\frac{2m}{{\hslash }^2}\left(E-\frac{1}{2}k{x}^2\right)\Psi \left(x\right)$

Here,

E= energy of the harmonic oscillator,

k= force constant

Step 2. for wave function

consider,

${\Psi }_1\left(x\right)=A_1{e}^{-\frac{x^2}{2b^2}}$

First derivative,

$\frac{d{\Psi }_1\left(x\right)}{dx}=\frac{d}{dx}\left(A_1{e}^{-\frac{x^2}{2b^2}}\right)$

$=A_1\frac{d}{dx}\left(e^{-\frac{x^2}{2b^2}}\right)$

$=-\frac{A_1}{b^2}x{e}^{-\frac{x^2}{2b^2}}$

The second derivative,

$\frac{d^2{\Psi }_1\left(x\right)}{d{x}^2}=\frac{d}{dx}\left(\frac{d{\Psi }_1\left(x\right)}{dx}\right)$

$=-\frac{A_1}{b^2}\frac{d}{dx}\left(x{e}^{\frac{x^2}{2b^2}}\right)$

$=-\frac{A_1}{b^2}{e}^{-\frac{x^2}{2b^2}}+\frac{A_1}{b^4}{x}^2{e}^{-\frac{x^2}{2b^2}}$

$=-\left(\frac{1}{b^2}-\frac{x^2}{b^4}\right)A_1{e}^{-\frac{x^2}{2b^2}}$

$=-\left(\frac{1}{b^2}-\frac{x^2}{b^4}\right){\Psi }_1\left(x\right)\left(\mathrm{As},{\Psi }_1\left(x\right)=A_1{e}^{-\frac{x^2}{2b^2}}\right)$

Step 3 Substituting  ℏmω = b in d2Ψ1(x)dx2=-1b2-x2b4Ψ1(x). 

$\frac{d^2{\Psi }_1\left(x\right)}{d{x}^2}=-\left(\frac{1}{b^2}-\frac{x^2}{b^4}\right){\Psi }_1\left(x\right)$

$=-\left(\frac{1}{{\left(\sqrt{\frac{\hslash }{m\omega }}\right)}^2}-\frac{x^2}{{\left(\sqrt{\frac{\hslash }{m\omega }}\right)}^4}\right){\Psi }_1\left(x\right)$

$=-\left(\frac{m\omega }{\hslash }-\frac{m^2{\omega }^2{x}^2}{{\hslash }^2}\right){\Psi }_1\left(x\right)$

$=-\left(\frac{m\omega }{\hslash }-\frac{m^2(k/m)x^2}{{\hslash }^2}\right){\Psi }_1\left(x\right)\left(\mathrm{As},{\omega }^2=k/m\right)$

$=-\frac{2m}{{\hslash }^2}\left(\frac{1}{2}\hslash \omega -\frac{1}{2}k{x}^2\right){\Psi }_1\left(x\right)$

The ground state energy of the harmonic oscillator is $\frac{1}{2}\hslash \omega$.

Therefore,

$\frac{d^2{\Psi }_1\left(x\right)}{d{x}^2}=-\frac{2m}{{\hslash }^2}\left(E-\frac{1}{2}k{x}^2\right){\Psi }_1\left(x\right)$