บทที่ 19 · ฟิสิกส์อะตอมและควอนตัมฟิสิกส์
Problem 19.83 Challenge
หัวข้อ: สมการชเรอดิงเงอร์ (The Schrödinger Equation)
\(\therefore U(x) =\frac{\hbar ^{2}\alpha}{m}(2\alpha x^{2} - 3) \;\blacksquare\)
ดูวิธีทำ
จาก \((-\frac{\hbar ^{2}}{2m}\frac{\partial ^{2}}{\partial x^{2}}+ U(x))\Psi (x) = E\Psi (x)\)
\(-\frac{\hbar ^{2}}{2m}\frac{\partial ^{2}\Psi (x)}{\partial x^{2}}+ U(x)\Psi (x) = E\Psi (x)\)
\(_{-\alpha x^{2}}\)
\(-\frac{\hbar ^{2}}{2m}\frac{\partial ^{2}}{\partial x^{2}}(Axe) + U(x)\Psi (x) = E\Psi (x)\)
\(_{-\alpha x^{2}}_{-\alpha x^{2}}\)
\(-A\frac{\hbar ^{2}}{2m}(\frac{\partial}{\partial x}(-2\alpha x^{2}e) +\frac{\partial}{\partial x}e) + U(x)\Psi (x) = E\Psi (x)\)
\(_{-\alpha x^{2}}_{-\alpha x^{2}}_{-\alpha x^{2}}\)
\(-A\frac{\hbar ^{2}}{2m}(4\alpha ^{2}x^{3}e- 4\alpha xe- 2\alpha xe) + U(x)\Psi (x) = E\Psi (x)\)
\(_{-\alpha x^{2}}\)
\(-\frac{\hbar ^{2}\alpha}{m}(2\alpha x^{2} - 3)(Axe) + U(x)\Psi (x) = E\Psi (x)\)
\(-\frac{\hbar ^{2}\alpha}{m}(2\alpha x^{2} - 3)\Psi (x) + U(x)\Psi (x) = E\Psi (x)\)
\(-\frac{\hbar ^{2}\alpha}{m}(2\alpha x^{2} - 3) + U(x) = E\)
\(-\frac{\hbar ^{2}\alpha}{m}(2\alpha x^{2} - 3) + U(x) = 0\)
\(\therefore U(x) =\frac{\hbar ^{2}\alpha}{m}(2\alpha x^{2} - 3) \;\blacksquare\)