Question:

The wave function for the one dimensional harmonic oscillator with the potential energy, \(\frac{1}{2}kx^2\), is given as

\[ \Phi_0 = C\exp(-ax^2) \]

Find the constant \(C\) when the wave function is normalized. You can use the normalized

Gaussian distribution: \(\frac{1}{\sqrt{2\pi}\sigma}\int^{\infty}_{-\infty}\exp(-\frac{x^2}{2\sigma^2})dx = 1\).

Answer:

For the normalized wave function, it has to satisfy:

\[  \int^{\infty}_{-\infty}\Phi_0\Phi^*_0 dx = C^2\int^{\infty}_{-\infty}\exp(-2ax^2) dx = 1\]

where \(\Phi^*_0\) is the complex conjugate of \(\Phi_0\). This means that the density of the distribution must be 100%. Then, compare this with the Gaussian normal distribution as follows:

\[ 2a = \frac{1}{2\sigma^2}  \]

Therefore,

\[   \sigma = \frac{1}{\sqrt{4a}}\]

Plug it into the formula of Gaussian distribution.

\[ \frac{1}{\sqrt{2\pi}(1/\sqrt{4a})}\int^{\infty}_{-\infty}\exp(-\frac{x^2}{2(1/4a)})dx = 1 \]

Now, we can see

\[ C^2 = \frac{1}{\sqrt{2\pi}(1/\sqrt{4a})} = \frac{1}{\sqrt{2\pi}}\sqrt{4a} \]

\[           = \sqrt{\frac{2a}{\pi}} \]

Thus, we have

\[ C = (2a/\pi)^{1/4}\]

This idea can be associated with statistics in math.