I was reading Introduction to Quantum Mechanics by David Griffiths and I am in Chapter 2, page 45. I know that since the solutions from Schrödinger equation cannot be normalized for a free particle. This must imply that the wavefunction of free particle must be non separable i.e.,
$$\Psi(x,t)\neq \psi(x)\phi(t)$$
because the solution which we get from solving the Schrödinger equation in this case is
$$\psi(x) = Ae^{ikx}+Be^{-ikx}$$
cannot be normalized. Thus, this is not a valid solution.
But Griffiths writes that
The general solution to the time-dependent Schrodinger equation is still a linear
combination of separable solutions (only this time it's an integral over the continuous variable $k$, instead of a sum over the discrete index)
But how is this possible, because we already know that we cannot use separation of variables to solve the Schrödinger equation?
Best Answer
The assumption that separability and normalizability are somehow linked is wrong. It can be directly shown to be wrong because we can use the separability ansatz, use $\hat{H}\vert\psi\rangle=E\vert\psi\rangle$, and get a complete set of solutions and they agree with our separability ansatz. So, you can say that the success of the process justifies the ingredient ansatz.
Now, coming to your core confusion. An energy eigenstate of a free particle with energy $E$ is given by $e^{-iEt}(Ae^{ikx}+Be^{-ikx})$ where $E=\frac{k^2}{2m}$. You can check that this solution indeed satisfies the Schrodinger equation. This is not normalizable but it is perfectly separable.
Finally, it should be noted that non-normalizable wavefunctions, indeed, cannot describe a physical particle/system. However, they are useful because they arise as eigenfunctions of position and momentum operators and they form a complete basis for all the normalizable wavefunctions. So, the non-normalizable wavefunctions do not themselves live in the Hilbert space but they provide a basis for the Hilbert space. And thus, solving the Schrodinger equation in this basis is equivalent to solving it for all wavefunctions.
For example, a Gaussian wavepacket of the form $\psi(x,0)=Ae^{-x^2/\Delta}$ can be seen as the linear combination $\psi(x,0)=\int \frac{dp}{2\pi}\ e^{ipx}\tilde\psi(p)$ where $\tilde\psi(p)$ is the Fourier transform of $\psi(x,0)$. Now, we can trivially calculate $\psi(x,t)=\int \frac{dp}{2\pi}\ e^{-ip^2t/2m}e^{ipx}\tilde\psi(p)$. Of course, I simply mentioned the Gaussian wavefunction as an example, the same formulae and the procedure can be used for any initial wavefunction.