A Problem on Schrödinger's Equation Let $\Psi(x,t) = F(x)\cdot G(t)$ be a solution of the time-dependent S.E. Then, how do we show that $F(x)$ satisfies the time-independent Schrödinger's Equation ?
I know that the time-dependent Schrödinger's Equation is given by:
$$i\hbar\frac{\partial \Psi}{\partial t} = -\frac{\hbar ^2}{2m}\frac{\partial ^2 \Psi}{\partial x^2} + V\Psi \tag {1}$$
Any help is appreciated.
Regards!
 A: This is a standard Separation of Variables (SoV) question.
The Schrodingers equation reads: $i \frac{\partial}{\partial t}\psi = H \psi$, while the time-independent variant is $E = H\psi$. Let's now assume that the solution is a separable function ie we can write $$\psi(x,t) = f(x) g(t)$$ 
for some $f(x)$ and $g(t)$. 
Let us now plug in the SoV expression for $\psi$.
$$i \frac{\partial f(x)g(t)}{\partial t} = - \frac{\partial^2 f(x)g(t)}{\partial x^2} + Vf(x) g(t)$$
The goal is now to split this into LHS that only dependeds on $t$ and RHS only depending on $x$. Work out the derivatives and divide the whole expression by $f(x)g(t)$. This results in:
$$i \frac{1}{g(t)}\frac{\partial g(t)}{\partial t} = \frac{1}{f(x)}Hf(x)$$
Since we now have a function of $t$ and a function of $x$ being equal they must be constant. Let's call that constant $E$.
$$i \frac{1}{g(t)}\frac{\partial g(t)}{\partial t} = \frac{1}{f(x)}Hf(x) = E$$
(This is how one actually derives the solutions to simple problems like particle in a box)
Lets now focus on: $$\frac{1}{f(x)}Hf(x) = E.$$ Multiply both sides by $f(x)g(t)$ to get:
$$Hf(x)g(t) = E f(x)g(t)$$
and thus we have shown that if $\psi = f(x)g(t)$ solves the time dependent problem then it solves the time independent one as well.
