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9.18. The differential equation tells us that we want a function whose derivative is equal to the function itself times a constant and p :\frac{d}{d p} \phi_{0}(\pi)=-\frac{1}{\mu \omega \eta} \pi \phi_{0}(\pi)We know that the derivative of the exponential function e^{p} is itself, so to get the extra factor of p we need a p^{2} in the exponent. To get the multiplicative factor correct, the function must be e^{-\pi^{2} / 2 \mu \omega \eta}. Now normalize:1=\int_{-\infty}^{\infty}\left|\phi_{0}(\pi)\right|^{2} \delta \pi=\int_{-\infty}^{\infty}\left|A \varepsilon^{-\pi^{2} / 2 \mu \omega \eta}\right|^{2} \delta \pi=|A|^{2} \int_{-\infty}^{\infty} \varepsilon^{-\pi^{2} / \mu \omega \eta} \delta \xiFrom Eq. (F.22) we get the integral, giving1=|A|^{2} 2 \frac{\sqrt{\mu \omega \eta}}{2} \sqrt{\pi}=|A|^{2} \sqrt{\pi \mu \omega \eta}Choosing the normalization constant to be real and positive gives ...