Normalization of a wave function

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. Note that 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 $\Phi_0$. Compare this with the Gaussian normal distribution.
$$2a = \frac{1}{2\sigma^2}  \\    \sigma = \frac{1}{\sqrt{4a}}$$
Plug it in the formula of Gaussian distribution.
$$\frac{1}{\sqrt{2\pi}(1/\sqrt{4a})}\int^{\infty}_{-\infty}\exp(-\frac{x^2}{2(1/4a)})dx = 1$$
We can see
$$C^2 = \frac{1}{\sqrt{2\pi}(1/\sqrt{4a})} = \frac{1}{\sqrt{2\pi}}\sqrt{4a} \\            = \sqrt{\frac{2a}{\pi}}$$
Therefore,
$$C = (2a/\pi)^{1/4}$$

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