Quantum Harmonic Oscillator Wave Function

The wave function of a quantum harmonic oscillator varies depending on the energy level of the particle being described. Using the number operator, the wave function of a ground state harmonic oscillator can be found. Repetitively applying the raising operator to the ground state wave function then allows the derivation of the general formula describing wave functions of higher energy levels.

Ground State Wave Function

The number operator returns the quantum number of a particle, so applying the number operator, $N$, to the wave function, $\psi_0(x)$, of a particle in the ground state should give:

\begin{equation}
N\psi_0(x)=0
\end{equation}
\begin{equation}
\left(\frac{\hat{p}^2}{2m\hbar\omega} + \frac{m\omega\hat{x}^2}{2\hbar} - \frac{1}{2}\right) \psi_0(x) = 0
\end{equation}

In the above equation, $\hat{p}$ and $\hat{x}$ are the one-dimensional momentum operator and the one-dimensional position operator respectively.

\begin{equation}
-\frac{\hbar}{2m\omega}\frac{d^2}{{dx}^2}\psi_0(x) + \left(\frac{m\omega x^2}{2\hbar} - \frac{1}{2}\right) \psi_0(x) = 0
\end{equation}

Since the second derivative of $\psi_0(x)$ must return the product of $\psi_0(x)$ and the sum of a constant and a term of $x^2$, it can heuristically be deduced that a possible solution is:

\begin{equation}
\psi_0(x) = Ae^{Cx^2}
\end{equation}

where $A$ and $C$ are constants yet to be determined. Substitute this possible solution for $\psi_0(x)$ into equation $(3)$ for verification and to solve for $C$.

\begin{equation}
-\frac{\hbar}{2m\omega}\left(2ACe^{Cx^2} + 4AC^2x^2e^{Cx^2}\right) + \left(\frac{m\omega x^2}{2\hbar} - \frac{1}{2}\right) Ae^{Cx^2} = 0
\end{equation}
\begin{equation}
-\frac{\hbar C}{m\omega} - \frac{1}{2} - \frac{2\hbar C^2x^2}{m\omega} + \frac{m\omega x^2}{2\hbar}= 0
\end{equation}
\begin{align}
-\frac{\hbar C}{m\omega} - \frac{1}{2} &= 0 \\
-\frac{2\hbar C^2x^2}{m\omega} + \frac{m\omega x^2}{2\hbar} &= 0
\end{align}
\begin{equation}
C = -\frac{m\omega}{2\hbar}
\end{equation}

Therefore, the solution for the ground state wave function is verified to be:

\begin{equation}
\psi_0(x) = A\exp\left(-\frac{m\omega x^2}{2\hbar}\right)
\end{equation}

To solve for the normalizing constant $A$, apply the following normalization equation:

\begin{equation}
\int_{-\infty}^\infty |\psi_0(x)|^2 dx = 1
\end{equation}
\begin{equation}
|A|^2 \int_{-\infty}^\infty \exp\left(-\frac{m\omega x^2}{\hbar}\right) dx = 1
\end{equation}

Let $u = \sqrt{m\omega/\hbar} x$.

\begin{equation}
|A|^2 \sqrt{\frac{\hbar}{m \omega}}\int_{-\infty}^\infty e^{-u^2} du = 1
\end{equation}

The definite integral in the above equation is the Gaussian integral and gives a value of $\sqrt{\pi}$.

\begin{equation}
|A|^2 \sqrt{\frac{\pi\hbar}{m \omega}} = 1
\end{equation}
\begin{equation}
A = \left(\frac{m \omega}{\pi\hbar}\right)^{1/4}
\end{equation}

Note that although $A$ can be written as being negative or even complex as long as equation $(14)$ is satisfied, normalizing constants are real and positive by convention. The ground state wave function for a harmonic oscillator is thus derived to be:

\begin{equation}
\psi_0(x) = \left(\frac{m \omega}{\pi\hbar}\right)^{1/4} \exp\left(-\frac{m\omega x^2}{2\hbar}\right)
\end{equation}

General Wave Function

Applying the raising operator, $a^\dagger$, $n$ times to the ground state wave function, $\psi_0(x)$, gives:

\begin{equation}
\left(a^\dagger\right)^n \psi_0(x) = \sqrt{n!} \psi_n(x)
\end{equation}

where $\psi_n(x)$ is the wave function with quantum number $n$.

\begin{equation}
\psi_n(x) = \frac{1}{\sqrt{n!}} \left(a^\dagger\right)^n \psi_0(x)
\end{equation}

Substituting in the raising operator and the ground state wave function allows $\psi_n(x)$ to be written as so:

\begin{equation}
\psi_n(x) = \frac{1}{\sqrt{2^n n!}} \left(\frac{m \omega}{\pi\hbar}\right)^{1/4} \left( \sqrt{\frac{m\omega}{\hbar}}\hat{x} - \frac{i}{\sqrt{\hbar m\omega}}\hat{p} \right)^n \exp\left(-\frac{m\omega x^2}{2\hbar}\right)
\end{equation}
\begin{equation}
\psi_n(x) = \frac{1}{\sqrt{2^n n!}} \left(\frac{m \omega}{\pi\hbar}\right)^{1/4} \left( \sqrt{\frac{m\omega}{\hbar}}x - \sqrt{ \frac{\hbar}{m\omega}}\frac{d}{dx} \right)^n \exp\left(-\frac{m\omega x^2}{2\hbar}\right)
\end{equation}
\begin{equation}
\psi_n(x) = \frac{1}{\sqrt{2^n n!}} \left(\frac{m \omega}{\pi\hbar}\right)^{1/4} \left( \sqrt{\frac{m\omega}{\hbar}}x - \frac{d}{d\left(\sqrt{m\omega/\hbar}x\right)} \right)^n \exp\left(-\frac{m\omega x^2}{2\hbar}\right)
\end{equation}

Thus, the wave function for a quantum harmonic oscillator of quantum number $n$ is derived to be:

\begin{equation}
\psi_n(x) = \frac{1}{\sqrt{2^n n!}} \left(\frac{m \omega}{\pi\hbar}\right)^{1/4} \exp\left(-\frac{m\omega x^2}{2\hbar}\right) H_n\left( \sqrt{\frac{m\omega}{\hbar}}x \right)
\end{equation}

where $H_n\left(\sqrt{m\omega/\hbar}x\right)$ is the $n$th-degree Hermite polynomial of $\sqrt{m\omega/\hbar}x$.