Quantum Mechanics #1

Sebuah partikel bermassa m di kotak satu dimensi , ditemukan dalam keadaan dasar

$\displaystyle \psi (x) = \sqrt{\frac{2}{a}} sin \left (\frac{\pi x}{a}\right) \tag*{}$

Tentukan $\displaystyle \Delta x \Delta p$, pada kondisi ini !

Solusi :

Gunakan $\displaystyle p = i \hbar \frac{d}{dx}$, didapat

$\displaystyle p \psi (x) = i \hbar \frac{d}{dx} \left[ \sqrt{\frac{2}{a}} sin \frac{\pi x}{a}\right] \tag*{}$

$\displaystyle = \frac{-\hbar \pi}{a} \sqrt{\frac{2}{a}} cos \left[\frac{\pi x}{a}\right] \tag*{}$

Dan

$\displaystyle p^{2} \psi (x) = i \hbar \frac{d}{dx} \left(\frac{-\hbar\pi}{a} \sqrt{\frac{2}{a}} cos \left(\frac{\pi x}{a}\right)\right)\tag*{}$

$\displaystyle = \frac{-\hbar \pi}{a} \sqrt{\frac{2}{a}} sin\left(\frac{\pi x}{a}\right) \tag*{}$

$\displaystyle \left<p\right> \rightarrow \int \psi^{*}(x)p \psi (x) \mathrm{d}x \tag*{}$

Kita temukan dalam soal dalam keadaan dasar, maka $\displaystyle \left<p\right> = 0$ , maka

$\displaystyle \left<p^{2}\right> = \int \psi^{*}(x)p^{2} \psi (x) \mathrm{d}x \tag*{}$$\displaystyle \int_{0}^{a} \sqrt{\frac{2}{a}} sin \left(\frac{\pi x}{a}\right) \frac{ \hbar^{2}\pi^{2}}{a^{2}} \sqrt{\frac{2}{a}}sin \left(\frac{\pi x}{a}\right)\mathrm{d}x$

$\displaystyle \frac{\hbar^{2}\pi^{2}}{a^{2}} \int_{0}^{a} \frac{2}{a} \left[sin\frac{\pi x}{a}\right]^{2} \mathrm{d}x\tag*{}$
$\displaystyle \frac{2\hbar^{2} \pi^{2}}{a^{3}} \int_{0}^{a} \frac{1-cos\left[\frac{2\pi x}{a}\right]}{2} \mathrm{d}x \tag*{}$

$\displaystyle \frac{\hbar^{2} \pi^{2}}{a^{3}}x|_{0}{}^{a} = \frac{\hbar^{2}\pi^{2}}{a^{2}}\tag*{}$

$\displaystyle \Delta p= \sqrt{\left<p^{2}\right>-\left<p\right>^{2}} = \sqrt{\frac{\hbar^{2}\pi^{2}}{a^{2}}} = \frac{\hbar \pi}{a}\tag*{}$

$\displaystyle \left<x\right> = \int \psi^{*} (x)x\psi(x) \mathrm{d}x\tag*{}$

$\displaystyle = \int_{0}^{a} \sqrt{\frac{2}{a}} sin \left(\frac{\pi x}{a}\right)x\sqrt{\frac{2}{a}} sin \left(\frac{\pi x}{a}\right) \mathrm{d}x$

$\displaystyle \frac{2}{a} \int_{0}^{a}x \left(sin\left(\frac{\pi x}{a}\right)\right)^{2}\mathrm{d}x\tag*{}$

$\displaystyle \frac{2}{a} \frac{a^{2}}{4} = \frac{a}{2} \tag*{}$

$\displaystyle \left<x^{2}\right> = \int \psi ^{*}(x)x^{2}\psi(x) \mathrm{d}x\tag*{}$

$\displaystyle \int_{0}^{a} \sqrt{\frac{2}{a}}x^{2} \left(sin\left(\frac{\pi x}{a}\right)\right)^{2} \mathrm{d}x\tag*{}$

$\displaystyle=\frac{a^{2}}{6}\left(2-\frac{3}{\pi^{2}}\right)\tag*{}$

$\displaystyle \Delta x = \sqrt{ \left<x^{2}\right> - \left<x\right>^{2}} \tag*{}$

$\displaystyle=\sqrt{\frac{a^{2}\left(\pi^{2}-6\right)}{12\pi^{2}}}\tag*{}$ 

$\displaystyle =\frac{a}{\pi} \sqrt{\frac{\left(\pi^{2}-6\right)}{12}} \tag*{}$

$\displaystyle\sqrt{\left(\pi^{2}-6\right)12}= 0,57\tag*{}$

And so

$\displaystyle \Delta x \Delta p \geq \frac{\hbar}{2}\tag*{}$

Referensi = Quantum Mechanics by David McMahon

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Ditulis Mohammad_fatshira — Friday, September 6, 2019 — 1 Comment — Physics