Problem 19.75 Challenge

a) จากเงื่อนไขความต่อเนื่องของฟังก์ชันที่ \(x = L\) จะได้

\(0 = B(L) + C\)

\(C = -BL\)

จากเงื่อนไขความต่อเนื่องของฟังก์ชันที่\(x=\frac{L}{2}\) จะได้

\(A(\frac{L}{2})=B(\frac{L}{2})+C\)

\(AL = BL + 2C\)

\(AL = BL + 2(-BL)\)

\(AL = -BL\)

\(B = -A\)

จะได้ \(B = -A\) และ \(C = AL\)

\(_{\infty}\)

จาก\(\int |\Psi |^{2}dx=1\)

\(_{-\infty}\)

\(_{L/2L}\)

\(\int (Ax)^{2}dx +\int (Bx+C)^{2}dx=1\)

\(_{0L/2}\)

\(_{L/2L}\)

\(\int (Ax)^{2}dx +\int (-Ax+AL)^{2}dx=1\)

\(_{0L/2}\)

\(_{L/2L}\)

\(A^{2}(\int x^{2}dx +\int (L-x)^{2}dx)=1\)

\(_{0L/2}\)

\(_{L/2L}\)

\(A^{2}(\frac{x^{3}}{3}-\frac{(L - x)^{3}}{3}) = 1\)

\(_{0L/2}\)

\(A^{2}\frac{L^{3}}{12}= 1\)

\(A = \pm \sqrt{\frac{12}{L^{3}}}\)

\(\therefore A = \pm \sqrt{\frac{12}{L^{3}}}, B = \mp \sqrt{\frac{12}{L^{3}}}\) และ \(C = \pm \sqrt{\frac{12}{L}}\;\blacksquare\)

b) จากข้อ a) จะได้

\(\pm \sqrt{\frac{12}{L^{3}}}x; 0 \le  x \le \frac{L}{2}\)

\(\Psi  =\)

\(\pm \sqrt{\frac{12}{L^{3}}}(L - x);\frac{L}{2}\le  x \le  L\)

\(0; Other x\)

\(_{3L/4}\)

จาก \(ℙ = \int |\Psi |^{2}dx\)

\(_{L/4}\)

\(_{L/23L/4}\)

\(ℙ =\int \frac{12}{L^{3}}x^{2}dx +\int \frac{12}{L^{3}}(L - x)^{2}dx\)

\(_{L/4L/2}\)

\(_{L/23L/4}\)

\(ℙ =\frac{12}{L^{3}}(\int x^{2}dx +\int (L - x)^{2}dx)\)

\(_{L/4L/2}\)

\(_{L/23L/4}\)

\(ℙ =\frac{12}{L^{3}}(\frac{x^{3}}{3}-\frac{(L - x)^{3}}{3})\)

\(_{L/4L/2}\)

\(_{L/23L/4}\)

\(ℙ =\frac{12}{L^{3}}(\frac{x^{3}}{3}-\frac{(L - x)^{3}}{3})\)

\(_{L/4L/2}\)

\(ℙ=\frac{7}{8}\;\blacksquare\)

\(_{\infty}\)

c)จาก\(〈x〉=\int \Psi *f(x)\Psi dx\)

\(_{-\infty}\)

\(_{L/2L}\)

\(〈x〉=\int \frac{12}{L^{3}}x^{3}dx +\int \frac{12}{L^{3}}x(L - x)^{2}dx\)

\(_{0L/2}\)

\(_{L/2L}\)

\(〈x〉 =\frac{12}{L^{3}}( \int x^{3}dx +\int (L^{2}x-2Lx^{2}+x^{3})dx)\)

\(_{0L/2}\)

\(_{LL}\)

\(〈x〉 =\frac{12}{L^{3}}(\int x^{3}dx+\int (L^{2}x-2Lx^{2})dx)\)

\(_{0L/2}\)

\(_{LL}\)

\(〈x〉 =\frac{12}{L^{3}}(\frac{x^{4}}{4}+(\frac{L^{2}x^{2}}{2}-\frac{2Lx^{3}}{3}))\)

\(_{0L/2}\)

\(〈x〉=\frac{L}{2}\;\blacksquare\)