Question related to proof of: For variable separable solutions to the Schrodinger equation to be normalizable, the separation constant must be real

Question

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 $k \in \mathbb{C \setminus R}$?

Answer 1

The tricky thing is that what your last equation is actually proving is that the sufficient condition for the normalization to be constant is for the net probability current through the boundaries (at $\pm\infty$) to be zero.

Now, strictly speaking, even if $k_I\neq0$ the wave function is still normalizable - doing the normalization will just cancel the factor of $\exp(k_It)$, removing it from your wave function. Say your original un-normalized wave function is $\Psi$ define $$\Psi_N = \frac{\Psi}{\sqrt{\int |\Psi|^2 \, \mathrm{d}x}} \tag1$$ and watch it vanish.

That's just a nitpick of terminology, though. The real problem is not that $\Psi$ isn't normalizable, it's that it fails an unstated boundary condition: $\Psi$ has to be finite for $t\rightarrow \pm \infty$. That boundary condition is less strict than requiring that $\Psi$ be normalized (note: not normalizable, normalized) but it is sufficient to do the job.

There is, of course, another route you can take. You can focus on the $x$ equation instead of the $t$ one: $$-\frac{\hbar^2}{2m} \psi'' + V(x)\psi = k \psi. \tag2$$ Take the complex conjugate of (2) and call it (3). Multiply (2) by $\psi^*$ and (3) by $\psi$ then integrate both sides over all $x$. Now subtract the pairs of equations, and manipulate the integrals using integration by parts to leave only surface terms (i.e. something of the form $[\ldots]_{x=-\infty}^\infty$). The right hand side will be $(k - k^*) \int |\psi|^2 \, \mathrm{d}x$. Normalizability requires the left hand side of the equation to vanish. The right hand side can only vanish if $k=k^*$, q.e.d.

This last version is an example of how to prove that a Hermitian operator's eigenvalues are real.