บทที่ 19 · ฟิสิกส์อะตอมและควอนตัมฟิสิกส์
Problem 19.73 Challenge
หัวข้อ: กลศาสตร์ควอนตัม (Quantum Mechanics)
\(〈x〉 = 0 \;\blacksquare\)
ดูวิธีทำ
\(_{\infty}\)
a)จาก\(\int |\Psi |^{2}dx=1\)
\(_{-\infty}\)
\(_{L/4}\)
\(\int A^{2}\cos ^{2}(\frac{2\pi x}{L})dx = 1\)
\(_{-L/4}\)
\(_{L/4}\)
\(\frac{A^{2}}{2}\int (1 + \cos (\frac{4\pi x}{L}))dx = 1\)
\(_{-L/4}\)
\(_{L/4}\)
\(\frac{A^{2}}{2}(x +\frac{L}{4\pi}\sin (\frac{4\pi x}{L}))= 1\)
\(_{-L/4}\)
\(\frac{A^{2}}{2}(\frac{L}{2})=1\)
\(A =\frac{2}{L}\;\blacksquare\)
\(_{L/8}\)
b)จาก\(ℙ=\int |\Psi |^{2}dx\)
\(_{0}\)
\(_{L/8}\)
\(ℙ=\int A^{2}\cos ^{2}(\frac{2\pi x}{L})dx\)
\(_{0}\)
\(_{L/8}\)
\(ℙ =\frac{4}{2L}\int (1 + \cos (\frac{4\pi x}{L}))dx\)
\(_{0}\)
\(_{L/8}\)
\(ℙ=\frac{2}{L}(x+\frac{L}{4\pi}\sin (\frac{4\pi x}{L}))\)
\(_{0}\)
\(ℙ=\frac{1}{4}+\frac{1}{2\pi}= 0.41\)
\(_{\infty}\)
c)จาก\(〈x〉=\int \Psi *f(x)\Psi dx\)
\(_{-\infty}\)
\(_{L/4}\)
\(〈x〉 =\int (\frac{2}{L}\cos (\frac{2\pi x}{L}))x(\frac{2}{L}\cos (\frac{2\pi x}{L}))dx\)
\(_{-L/4}\)
\(_{L/4}\)
\(〈x〉 =\frac{4}{L}\int xcos^{2}(\frac{2\pi x}{L})dx\)
\(_{-L/4}\)
\(_{L/4}\)
\(〈x〉 =\frac{4}{2L}\int x(1 + \cos (\frac{4\pi x}{L}))dx\)
\(_{-L/4}\)
\(_{L/4}\)
\(〈x〉=\frac{2}{L}(\frac{x^{2}}{2}+ (x\frac{L}{4\pi}\sin (\frac{4\pi x}{L}) - \int \frac{L}{4\pi}\sin (\frac{4\pi x}{L})dx))\)
\(_{-L/4}\)
\(_{L/4}\)
\(〈x〉=\frac{2}{L}(\frac{x^{2}}{2}+\frac{Lx}{4\pi}\sin (\frac{4\pi x}{L}) + (\frac{L}{4\pi})^{2}\cos (\frac{4\pi x}{L}))\)
\(_{-L/4}\)
\(〈x〉 = 0 \;\blacksquare\)