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TDSE: >>>> $i\hbar \frac{\partial\psi(x,t)}{\partial t}=-\frac{\hbar}{2m}\frac{\partial^2\psi(x,t)}{\partial t^2}+V(x)\psi(x,t)$ navi eta funksia latzura >>>> $\psi(x,t)=T(t)u(x)$ Plugging ze valuz ma ra >>>> $i\hbar \frac{(T(t)u(x))}{\partial t}=-\frac{\hbar}{2m}\frac{\partial^2(T(t)u(x))}{\partial t^2}+V(x)T(t)u(x)$ Yallak lechachev! >>>> $u(x)\cdot i\hbar \frac{\partial T(t)}{\partial t}=T(t) \cdot(-\frac{\hbar^2}{2m})\cdot\frac{\partial ^2u(x)}{\partial x^2}+V(x)T(t)u(x)$ Dividing bai $T(t)u(x)$ ma ra >>>>$\frac{1}{T(t)}\cdot i\hbar\frac{\partial T(t)}{\partial t}=\frac{1}{u(x)}(-\frac{\hbar^2}{2m})\frac{\partial^2u(x)}{\partial x^2}+V(x)$ And we did it! ילק אכשף צך רק להשוות כל צד לקבוע נתחיל עם הטיים מה רה >>>>$\frac{1}{T(t)}\cdot i\hbar\frac{\partial T(t)}{\partial t}=E$
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Thu, 11 Jun 2020 14:35 GMT