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

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  • $\begingroup$ @ZeroTheHero The part you are referring to, is probably the one with $k_I$. The exponent is a real number. I have already removed the imaginary part. $\endgroup$ – A B Nov 19 '19 at 13:45
  • $\begingroup$ @ZeroTheHero. Please check it again. I have added the clarification. $\endgroup$ – A B Nov 19 '19 at 13:48
  • $\begingroup$ $|exp(-ikt)|^2=1$ . Either "k" is real or imaginary. $\endgroup$ – baponkar Nov 19 '19 at 14:58
  • $\begingroup$ @baponkar I am assuming that $k$ is a general complex number $k_R+ik_I$ where $k_R$ and $k_I$ are real. So, $-ik = -i(k_R+ik_I) = -ik_R+k_I$. $\endgroup$ – A B Nov 19 '19 at 15:13
  • $\begingroup$ $exp(-ikt)=exp(-ik_R t).exp(k_I t)$;$exp(ikt)=exp(ik_R t).exp(-k_I t)$.so |exp(-ik t)|^2=1.just algebra $\endgroup$ – baponkar Nov 19 '19 at 15:19
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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.

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  • $\begingroup$ Thank you for responding to my question. May I ask you another question: how exactly is normalizability defined? (And what exactly is a normalizable wavefunction?) Also, this is actually related to a problem in Introduction to Quantum Mechanics (2nd Ed) (by David Griffiths) - Problem 2.1 asks the reader to prove that "For normalizable solutions, the separation constant must be real." $\endgroup$ – A B Nov 19 '19 at 16:53
  • $\begingroup$ @AB Applying equation (1) is called normalizing a wave function. So, a normalizable wave function is one that we can apply (1) to without any divide by zero of infinity problems. The word 'normalizable' literally means, "Able to be normalized". I suspect that Griffiths is just being sloppy with his usage. $\endgroup$ – Sean E. Lake Nov 19 '19 at 17:05
  • $\begingroup$ 1) Would it be more appropriate to say "... the sufficient condition for the probability that the particle exists (or the probability of finding the particle between $x \to -\infty$ and $x \to +\infty$) to be constant is for the net probability current through the boundaries to be zero." Also, isn't this condition necessary as well? And 2) Why is '$\Psi$ has to be finite for $t \to \pm \infty$' a boundary condition? What physical restriction does it convey? Is it related to ensuring the finite-ness of $|\Psi|^2$ for all $x$ and $t$? $\endgroup$ – A B Nov 20 '19 at 6:24
  • $\begingroup$ And 3) Can I say "The separation constant must real so that $\Psi$ satisfies the boundary condition that '$\Psi$ has to be finite for $t \to \pm \infty$' or so that $\Psi$ is normalizable for all values of $t$ (as $\Psi_N$ takes the form of $\infty/\infty$ and $0/0$ for $t \to \pm \infty$ (or in the other order, depending on whether $k_I$ is positive or negative))" $\endgroup$ – A B Nov 20 '19 at 6:25
  • $\begingroup$ @AB "Can I say..." You can say anything, but you'll want to discuss with your TA and/or professor what they'll accept. 1) Be careful, there. IIRC, the full logic is that for $H$ to be Hermitian the probability flux on the boundaries has to vanish, and $H$ being hermitian implies the real constant. If you satisfy that condition and $H$ isn't Hermitian, no dice. $\endgroup$ – Sean E. Lake Nov 20 '19 at 6:57

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