I need to prove: For variable separable solutions to the Schrodinger equation to be normalizable, the separation constant must be real.
This can be proven as follows: \begin{align} \int_{-\infty}^\infty |\Psi(x,t)|^2 dx & = \int_{-\infty}^\infty |\psi(x)\phi(t)|^2 dx\\ & = \int_{-\infty}^\infty |\psi(x)e^{-ikt/\hbar}|^2 dx\\ & = \int_{-\infty}^\infty |\psi(x)e^{-i(k_R + ik_I)t/\hbar}|^2 dx\\ & = e^{2k_It/\hbar}\int_{-\infty}^\infty |\psi(x)|^2 dx\\ \end{align} The second factor (integral) is independent of $t$ and therefore for the product to be a finite non-zero constant, the first factor should also be independent of time, which is possible iff $k_I = 0$ (i.e., iff $k \in \mathbb{R}$). (QED)
(Here $k = k_R + i k_I$ (where $k_R$ and $k_I$ are real numbers) is the separation constant and exp(constant of integration) from $\phi(t)$ has been absorbed in $\psi(x)$. Also, please note that I am using $k$ instead of $E$ to denote the separation constant.)
BUT I have not been successful in trying to prove the same thing using Equation 1.26, Section 1.4, Introduction to Quantum Mechanics (2nd Ed) (by David Griffiths). The equation I am referring to, is:
$\frac{d}{dt} \int_{-\infty}^\infty |\Psi(x,t)|^2 dx = \frac{i\hbar}{2m}(\Psi^*\frac{\partial\Psi}{\partial x}-\Psi\frac{\partial\Psi^*}{\partial x})|_{-\infty}^{\infty}$ (if the potential energy function is real)
If I replace $\Psi(x,t)$ by $\psi(x)\phi(t)$ in the above equation, I get:
$\frac{d}{dt} \int_{-\infty}^\infty |\Psi(x,t)|^2 dx = \frac{i\hbar}{2m} |\phi(t)|^2 (\psi^*\frac{\partial\psi}{\partial x}-\psi\frac{\partial\psi^*}{\partial x})|_{-\infty}^{\infty} = \frac{i\hbar}{2m} e^{2k_It/\hbar} (\psi^*\frac{\partial\psi}{\partial x}-\psi\frac{\partial\psi^*}{\partial x})|_{-\infty}^{\infty}$
Now, shouldn't $\psi$ go to zero as $x$ goes to $\pm\infty$, for the wave-function to be normalizable? (Griffiths says, just below equation 1.26, that "... $\Psi(x,t)$ must go to zero as $x$ goes to $\pm\infty$ - otherwise, the wave-function would not be normalizable." Here, wouldn't this mean that $\psi(x)$ must go to zero as $x$ goes to $\pm\infty$?) If it does go to zero, then the RHS would be zero irrespective of whether $|\phi(t)|^2$ is a constant or not, i.e., irrespective of whether the separation constant is real or not.
Am I missing something here?
Could it be possible that $\psi(x)$ does not go to zero when $x$ goes to $\pm\infty$ for $k \in \mathbb{C \setminus R}$?