บทที่ 19 · ฟิสิกส์อะตอมและควอนตัมฟิสิกส์
Problem 19.84 Challenge
หัวข้อ: สมการชเรอดิงเงอร์ (The Schrödinger Equation)
\(\therefore E =\frac{\alpha \hbar ^{2}}{m}\) และ \(U(x) = \frac{2\alpha ^{2}\hbar ^{2}x^{2}}{m}\;\blacksquare\)
ดูวิธีทำ
จาก \((-\frac{\hbar ^{2}}{2m}\frac{\partial ^{2}}{\partial x^{2}}+ U(x))\Psi (x) = E\Psi (x)\)
\(-\frac{\hbar ^{2}}{2m}\frac{\partial ^{2}\Psi (x)}{\partial x^{2}}+ U(x)\Psi (x) = E\Psi (x)\)
\(_{-\alpha x^{2}}\)
\(-\frac{\hbar ^{2}}{2m}\frac{\partial ^{2}}{\partial x^{2}}(Ae) + U(x)\Psi (x) = E\Psi (x)\)
\(_{-\alpha x^{2}}\)
\(-A\frac{\hbar ^{2}}{2m}\frac{\partial}{\partial x}(-2\alpha xe) + U(x)\Psi (x) = E\Psi (x)\)
\(_{-\alpha x^{2}}_{-\alpha x^{2}}\)
\(-A\frac{\hbar ^{2}}{2m}(4\alpha ^{2}x^{2}e- 2\alpha e) + U(x)\Psi (x) = E\Psi (x)\)
\(_{-\alpha x^{2}}\)
\(-\frac{\hbar ^{2}}{2m}(4\alpha ^{2}x^{2} - 2\alpha )(Ae) + U(x)\Psi (x) = E\Psi (x)\)
\(-\frac{\hbar ^{2}}{2m}(4\alpha ^{2}x^{2} - 2\alpha )\Psi (x) + U(x)\Psi (x) = E\Psi (x)\)
\(-\frac{\hbar ^{2}}{2m}(4\alpha ^{2}x^{2} - 2\alpha ) + U(x) = E\)
\(E - U(x) =\frac{\alpha \hbar ^{2}}{m}-\frac{2\alpha ^{2}\hbar ^{2}x^{2}}{m}\)
\(\therefore E =\frac{\alpha \hbar ^{2}}{m}\) และ \(U(x) = \frac{2\alpha ^{2}\hbar ^{2}x^{2}}{m}\;\blacksquare\)