Quantum Harmonic Oscillator

Setup

The Schrodinger Equation is given by $H={\sum }_{i=0}^n\frac{P_i}{2m}+\frac{1}{2}m{\omega }_i{X}_i^2$, as described in the Quantum Harmonic Oscillator. We impose no boundary conditions other than getting a Wave function that can be normalised

Solution

Since we have a Separable Hamiltonian, we can consider each dimension independently. As such, we can focus on the simpler Hamiltonian $H=\frac{P}{2m}+\frac{1}{2}m\omega X$. The associated Time-independent Schrodinger equation is given by

$$\begin{array}{rll}E\psi & =H\psi & \text{Expanding the Hamiltonian Operator}\\ & =-\frac{{\hslash }^2}{2m}\frac{{\partial }^2}{\partial x^2}\psi +\frac{1}{2}m\omega x^2\psi & \text{Using the ansatz ψ(x)=e−kx2f(x), hoping to cancel the x2 term}\\ & =-\frac{{\hslash }^2}{2m}[4k^2{x}^{2}f(x)+2(-2kx)f^{\prime }(x)+f^{\prime \prime }(x)]e^{-k{x}^2}+\frac{1}{2}m\omega x^{2}f(x)e^{-k{x}^2}& \text{Removing the e−kx2 term from both sides}\\ & \Downarrow & \\ Ef(x)& =-\frac{{\hslash }^2}{2m}[4k^2{x}^{2}f(x)-4kx{f}^{\prime }(x)+f^{\prime \prime }(x)]+\frac{1}{2}m\omega x^{2}f(x)& \text{Choosing k=2imℏω to cancel the x2 term, via −2mℏ24k2=21mω}\\ & =-\frac{{\hslash }^2}{2m}[-4kx{f}^{\prime }(x)+f^{\prime \prime }(x)]& \end{array}$$

With the above we get the “simpler” equation $\frac{2mE}{{\hslash }^2}f(x)=4kx{f}^{\prime }(x)-f^{\prime \prime }(x)$.

• Figure out how to solve this 2nd order differential equation. We are supposed to get the energy levels $E=(n+\frac{1}{2})\hslash \omega ,n\in \mathbb{N}$.