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For the derivation from time-depdent Schrodinger equation to time-indepdent Schrodinger equation, if the Hamiltonian is time-indepdent, we assume the spatial and temporal variables in the wave-function are separable.

$\psi(x,t)=\varphi(x)T(t)$

Is this an assumption or truth? If it's an assumption, what is the assumption in details?

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  • $\begingroup$ Neither nor. It is a general technique to solve PDE's: en.wikipedia.org/wiki/… $\endgroup$ – AccidentalFourierTransform Jul 29 '18 at 0:16
  • $\begingroup$ thanks. Is there an entanglement between spatial and temporal function if the Hamiltonian is time-indepdent? $\endgroup$ – kinder chen Jul 29 '18 at 0:42
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I'm not sure if this is what you're looking for, but (as AccidentalFourierTransform points out), the technique is called separation of variables.

The question is, how do we know all the possible solutions to the Schrodinger equation are of this form (separable) when the Hamiltonian is time-independent?

The way to check, which I understand is standard, is to look at all the possible general solutions you get out of the spatial and temporal ODEs post-separation, and see if their products $φ_i(x)T_j(t)$ form a basis over $L^2[(0,T)\times \mathbb{R}]$. This would mean you can match any function in $L^2[(0,T)\times \mathbb{R}]$ almost everywhere with linear combinations of those products, including all the solutions to the original equation.

Luckily, there are entire theoretical programs that were dedicated to precisely this, and you'll find that such theorems indicate the affirmative for the Schrodinger equation (that all solutions for the Schrodinger equation are separable when the Hamiltonian is time-independent).

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    $\begingroup$ In short, when you're working in rectangular coordinates you can use separation of variables because of Fourier transforms and Fourier series. In many other coordinate systems, there are other special functions that serve as orthonormal bases. $\endgroup$ – DanielSank Jul 29 '18 at 4:11
  • $\begingroup$ Thanks. Are there any interaction between the spatial and temporal part? Or it only works for the eighen-function? $\endgroup$ – kinder chen Jul 30 '18 at 21:22
  • $\begingroup$ I’m not sure if this is what you’re looking for @kinderchan, but you may see that the spatial and temporal families of functions you get from the separation of variables process will be linked by a scalar parameter $\lambda$. This almost always comes from the fact that the spectrum of solutions to each ODE is linked from the separation of variables process. This is one mathematical way of how the de Broglie relations pop up in QM, which I suspect would be of great interest to you. $\endgroup$ – aghostinthefigures Aug 6 '18 at 21:31

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