Discussion
The wavefunction of a 1-D harmonic oscillator between \(x = +\infty\) and \(x = -\infty\) is given by \(\psi(x) = N(2x^{2}-1)e^{-x^{2}/2}\). The value of N that normalizes the function \(\psi(x)\) is
(Given: \[ \begin{aligned} &\int_{-\infty}^{+\infty} x^{2n}e^{-x^{2}}dx \\ &= \frac{1\cdot 3\cdot 5\cdots(2n-1)}{2^{n}}\sqrt{\pi} \end{aligned} \])
Answer: C
The normalisation constant is \(N = \left(\dfrac{1}{2\sqrt{\pi}}\right)^{1/2}\).
Normalisation requires \(\int \psi^{2}\,dx = 1\). Squaring the function gives \(N^{2}(4x^{4} - 4x^{2} + 1)e^{-x^{2}}\). Using the given integral with n = 2, 1 and 0:
\[ \begin{aligned} \int x^{4}e^{-x^{2}}dx &= \tfrac{3}{4}\sqrt{\pi} \\ \int x^{2}e^{-x^{2}}dx &= \tfrac{1}{2}\sqrt{\pi} \\ \int e^{-x^{2}}dx &= \sqrt{\pi} \end{aligned} \]
\[ \begin{aligned} N^{2}\sqrt{\pi}\,(4 \cdot \tfrac{3}{4} - 4 \cdot \tfrac{1}{2} + 1) &= 1 \\ 2\sqrt{\pi}\,N^{2} &= 1 \\ N &= (2\sqrt{\pi})^{-1/2} \end{aligned} \]
The 8 in the first option is what the Hermite form 4x2 - 2, twice this polynomial, would give. 4 is what dropping the cross term -4x2 gives (3 + 1), and 3 is the x4 term alone.
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