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.


  • $\begingroup$ Compute $H\psi $, compute $E\psi$ for suitable $E$. $\endgroup$
    – user223391
    Oct 16, 2017 at 14:10
  • 2
    $\begingroup$ There are currently 3 votes to close your question for "missing context or other details" These could be because you did not include the time-independent Schrödinger's equation in your post or it could be because people wanted you to show more of your own thoughts about the question. I won't presume to know what each person was thinking when they voted but you should look into improving the question. $\endgroup$ Oct 16, 2017 at 18:41
  • $\begingroup$ @TrevorGunn You are too right. I'll be better next time. Also may i ask these questions here? To me, this problem was involving mathematical calcuations. $\endgroup$
    – Maxwell
    Oct 16, 2017 at 18:57
  • $\begingroup$ @Maxwell I don't know enough about physics or about physics questions on Math.SE to give you an authoritative answer. The only reason I saw this question is from your post on Meta. I don't ever browse the physics tags or the diff eq tags so I don't know what the customs are. $\endgroup$ Oct 16, 2017 at 19:11
  • $\begingroup$ As far I've expierienced - if a question is stated in clear mathematical terms it will be answered with a short answer, often an reference either. If the math of it isn't clearly stated, or is very specific to a problem(ie not a very general question like this) it will be overlooked by the users. $\endgroup$ Oct 16, 2017 at 21:26

1 Answer 1


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.

  • $\begingroup$ I cannot work it out. If you be more clear, I'd be happy. $\endgroup$
    – Maxwell
    Oct 16, 2017 at 14:47
  • $\begingroup$ I worked it out. $\endgroup$ Oct 16, 2017 at 14:55
  • $\begingroup$ I got it now. Thanks for your great answer ;) $\endgroup$
    – Maxwell
    Oct 16, 2017 at 15:09

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