Question

Obtain the expectational value of the potential energy v(x) = 1/2 mw^2x^2 of the one dimensional harmonic oscillator in the first excited state Ψ1(x) = 2(a/2√2)^1/2 axe^-a^2x^2/2

Answer 1

Answer on Question#55008 - Physics - Mechanics - Relativity

Obtain the expectation value of the potential energy $V(x) = \frac{1}{2} mw^2 x^2$ of the one dimensional harmonic oscillator in the first excited state $\Psi_{1}(x) = 2\left(\frac{a}{2\sqrt{2}}\right)^{\frac{1}{2}} axe^{-\frac{a^2x^2}{2}}$.

Solution:

The expectation value is given by

$$\begin{aligned} \langle V(x) \rangle_1 &= \int_{-\infty}^{\infty} \Psi_1(x)V(x)\Psi_1^*(x)dx = \int_{-\infty}^{\infty} 4\frac{a}{2\sqrt{2}}a^2x^2e^{-a^2x^2} \frac{1}{2}mw^2x^2dx = \\ &= \int_{-\infty}^{\infty} \frac{mw^2a^3}{\sqrt{2}}x^4e^{-a^2x^2}dx = \frac{mw^2}{\sqrt{2}a^2} \int_{-\infty}^{\infty} (ax)^4e^{-(ax)^2} d(ax) = |y = ax| = \\ &= \frac{mw^2}{\sqrt{2}a^2} \int_{-\infty}^{\infty} y^4e^{-y^2}dy = \frac{mw^2}{\sqrt{2}a^2} \frac{3\sqrt{\pi}}{4} = \frac{3}{4}\sqrt{\frac{\pi}{2}} \frac{mw^2}{a^2} \end{aligned}$$

Answer: $\frac{3}{4}\sqrt{\frac{\pi}{2}}\frac{mw^2}{a^2}$.

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