Calculate (x) ,(x^2) , (p) , (p^2) , σx and σp, for the nth stationary state of the infinite square well. Check that the uncertainty principle is satisfied. Which state comes closest to the uncertainty limit?

Question

Calculate [latex] \left\langle x\right\rangle ,\left\langle x^2\right\rangle , \left\langle p\right\rangle , \left\langle p^2\right\rangle , σ_x [/latex] and [latex] σ_p [/latex] , for the nth stationary state of the infinite square well. Check that the uncertainty principle is satisfied. Which state comes closest to the uncertainty limit?.

Accepted Answer

[latex] \langle x\rangle=\int x|\psi|^{2} d x=\frac{2}{a} \int_{0}^{a} x \sin ^{2}\left(\frac{n \pi}{a} x\right) d x [/latex]. Let [latex] y \equiv \frac{n \pi}{a} x [/latex], so [latex] dx = \frac{a}{nπ}dy [/latex]; y : 0→ nπ.

[latex] =\frac{2}{a}\left(\frac{a}{n \pi}\right)^{2} \int_{0}^{n \pi} y \sin ^{2} y d y=\left.\frac{2 a}{n^{2} \pi^{2}}\left[\frac{y^{2}}{4}-\frac{y \sin 2 y}{4}-\frac{\cos 2 y}{8}\right]\right|_{0} ^{n \pi} [/latex]

[latex] =\frac{2 a}{n^{2} \pi^{2}}\left[\frac{n^{2} \pi^{2}}{4}-\frac{\cos 2 n \pi}{8}+\frac{1}{8}\right]= \frac{a}{2} [/latex] (Independent of n.)

[latex] \left\langle x^{2}\right\rangle=\frac{2}{a} \int_{0}^{a} x^{2} \sin ^{2}\left(\frac{n \pi}{a} x\right) d x=\frac{2}{a}\left(\frac{a}{n \pi}\right)^{3} \int_{0}^{n \pi} y^{2} \sin ^{2} y d y [/latex]

[latex] =\frac{2 a^{2}}{(n \pi)^{3}}\left[\frac{y^{3}}{6}-\left(\frac{y^{2}}{4}-\frac{1}{8}\right) \sin 2 y-\frac{y \cos 2 y}{4}\right]_{0}^{n \pi} [/latex]

[latex] =\frac{2 a^{2}}{(n \pi)^{3}}\left[\frac{(n \pi)^{3}}{6}-\frac{n \pi \cos (2 n \pi)}{4}\right] = a^{2}\left[\frac{1}{3}-\frac{1}{2(n \pi)^{2}}\right] [/latex].

[latex] \langle p\rangle=m \frac{d\langle x\rangle}{d t}=0[/latex]

(Note : Eq. 1.33 is much faster than Eq.1.35:)

[latex] \left\langle p^{2}\right\rangle=\int \psi_{n}^{*}\left(\frac{\hbar}{i} \frac{d}{d x}\right)^{2} \psi_{n} d x=-\hbar^{2} \int \psi_{n}^{*}\left(\frac{d^{2} \psi_{n}}{d x^{2}}\right) d x [/latex]

[latex] =\left(-\hbar^{2}\right)\left(-\frac{2 m E_{n}}{\hbar^{2}}\right) \int \psi_{n}^{*} \psi_{n} d x=2 m E_{n} = \left(\frac{n \pi \hbar}{a}\right)^{2} [/latex].

[latex] \sigma_{x}^{2}=\left\langle x^{2}\right\rangle-\langle x\rangle^{2}=a^{2}\left(\frac{1}{3}-\frac{1}{2(n \pi)^{2}}-\frac{1}{4}\right)=\frac{a^{2}}{4}\left(\frac{1}{3}-\frac{2}{(n \pi)^{2}}\right) [/latex]; [latex] \sigma_{x}=\frac{a}{2} \sqrt{\frac{1}{3}-\frac{2}{(n \pi)^{2}}} [/latex].

[latex] \sigma_{p}^{2}=\left\langle p^{2}\right\rangle-\langle p\rangle^{2}=\left(\frac{n \pi \hbar}{a}\right)^{2} [/latex] ; [latex] \sigma_{p}=\frac{n \pi \hbar}{a} [/latex].

∴ [latex] \sigma_{x} \sigma_{p} = \frac{\hbar}{2} \sqrt{\frac{(n \pi)^{2}}{3}-2} [/latex].

The product [latex] \sigma_{x} \sigma_{p} [/latex] is smallest for n = 1; in that case, [latex] \sigma_{x} \sigma_{p}=\frac{\hbar}{2} \sqrt{\frac{\pi^{2}}{3}-2}=(1.136) \hbar / 2>\hbar / 2 [/latex].

Question 2.4: Calculate (x) ,(x^2) , (p) , (p^2) , σx and σp, for the nth ...

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