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While deriving the Time Independent Schrödinger equation we assume that the wave function is composed of the two separate functions of time and space. And since we do not have any information regarding the wave function, then how is it correct to assume this?

P.S- I am still a beginner in Quantum mechanics so it would be great if you will elaborate your answer.

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The Time Dependent Schrödinger Equation (TDE) is a linear, second order partial differential equation (PDE) with variables time $t$ and position $\mathbf{r}$.

When trying to obtain the Time Independent Schrödinger Equation (TISE) we apply a well-known method known as separation of variables (SoV). This method is not inherent to quantum mechanics, wave functions or the Schrödinger Equation, it just works for that type of PDE.

Using the method of SoV we assume that the wave function can be written as:

$$\psi(\mathbf{r},t)=\Psi(\mathbf{r})\varphi(t)$$

Insertion of this assumption into the TDSE then yields two ordinary differential equations, one for $\Psi(\mathbf{r})$ and one for $\varphi(t)$.

Similar SoV methods are used for the classical wave function, the heat equation and the diffusion equation (among others).

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  • $\begingroup$ I suppose it would help the OP to mention that the point is that more generic solutions can be formed via a linear combination of such solutions. For example, a solution of the TDSE could be $e^{-iE_1t}\psi_{E_1}(x) + e^{-iE_2t}\psi_{E_2}(x)$ but it can be formed by the linear superposition of $e^{-iE_1t}\psi_{E_1}(x)$ and $e^{-iE_2t}\psi_{E_2}(x)$, both of which are of the form $\psi(x)\phi(t)$. $\endgroup$
    – user87745
    Commented Mar 25, 2021 at 21:17
  • $\begingroup$ Yes, I agree that using this technique we get an ODE which makes the calculations much easier. But the point is not all functions are variable separable and without having any information about the wave function beforehand, how is it correct to presume wave function is separable? $\endgroup$
    – Devesh Vaish
    Commented Mar 26, 2021 at 6:44
  • $\begingroup$ @DeveshVaish We know that SoV works here because it works on ALL linear PDEs. It's just a mathematical 'trick' that splits linear PDEs into ODEs. We don't have to have 'any information about the wave can apply SoV beforehand', we just need to know (as we do) that the TDSE is a linear, 2nd order PDE, because then we know we can apply SoV. $\endgroup$
    – Gert
    Commented Mar 26, 2021 at 15:04
  • $\begingroup$ @DeveshVaish E.g. when solving the Heat equation (Fourier) we also don't know what will be the result but because $\frac{\partial T}{\partial t}=\alpha \nabla^2 T$ is linear, then clearly SoV becomes a first guess ($T(x,y,z,t)=\Theta(t)G(x,y,z)$). $\endgroup$
    – Gert
    Commented Mar 26, 2021 at 15:12

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