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suppose we have a potential that's independent of time $V(x,t) = V(x)$ so in Schrödinger equation we get: $$i\hbar \frac{\partial \Psi (x,t)}{\partial t}=-\frac{\hbar^2}{2m}\frac{\partial^2 \Psi (x,t)}{\partial x^2}+V(x)\Psi(x,t)$$ since the LHS involves a variation of $\Psi$ with $x$ and the LHS involves a variation of $\Psi$ with $t$ can we say that a basis for the solutions will be with the form $\Psi(x,t)=\psi(x)T(t)$?

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I'm actually quite surprised at this question: yes, yes you can. In fact, it's extremely famous, and is called the time-independent Schrodinger Equation, or (perhaps more appropriately) the Energy Eigenvalue equation, since it reduces to:

$$\hat{H} \psi_n(x) = E_n \psi_n(x),$$

which is the eigenvalue equation for the Hamiltonian perator ${\hat{p}^2}/{2m} + V(\hat{x})$.

Note that we cannot say that all solutions of the equation can be written in so-called variable separable form, only a special class of solutions. However, the general solution can be obtained by a linear combination of such separable solutions.

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    $\begingroup$ can you explain why exactly we can say that? i understand why it benefits us but why we we can say that since each side has a derivative with respect to one variable why we can say that there will be a solution of the form $\Psi= \psi (x)T(t)$ $\endgroup$ – Elad Elmakias Nov 16 '20 at 15:46
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    $\begingroup$ That's a good question! This is a standard method that is used to solve many partial differential equations, and has had a lot of success with some equations. The main reason we use it is precisely because it works! :) "Guessing and checking" is a perfectly acceptable method to solve differential equations! However, I don't believe there is a set of conditions for when it will work. This question is more mathematical, and there are some very good answers to similar questions over at Mathematics. For example: Why separation of variables works in PDEs?. $\endgroup$ – Philip Nov 16 '20 at 15:53
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Yes, see for example this small set of notes that outlines how the Schrodinger equation comes apart into two separate equations: $$-\frac{\hbar^2}{2m}\frac{d^2\psi(x)}{dx^2}+V(x)\psi(x)=E\psi(x) \tag{1},$$ and $$i\hbar\frac{d\phi(t)}{dt}=E\phi(t) \tag{2}.$$

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Yes. There is a very standard proof for that solution. Just put that form of the wavefunction into the TDSE and you will get the time independent Schrodinger equation along with the time component of a stationary state. Of course, variable separable solutions are not the only solutions of the Schrodinger's equation in case of a time independent potential, but these solutions are extremely special because they give rise to stationary states. Try doing it with a time dependent potential and you will find that the mathematics doesn't allow you to write such solutions.

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It is not possible to write ALL solutions of the Time-Dependent Schrödinger Equation as $$\Psi(x,t)=\psi(x)T(t)$$

However there are some solutions which are separable in that way. Those solutions are

$$\Psi(x,t)=\psi(x)e^{-itE/\hbar}$$

where $\psi(x)$ is a solution of the Time-Independent Schrödinger Equation with energy $E$. These "special" separable solutions are exactly the states which have definite energy, e.g. when you measure their energy there is a 100% probability of getting $E$. Realistic states tend not to be in perfect energy eigenstates so in that sense these separable solutions are rare. Nonetheless they are incredibly essential in QM since you can build any solution out of them.

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