Given the time-independent Schrödinger’s equation in one dimension $$H\psi = E\psi$$ what restrictions can we place on V(x) (inside the hamiltonian) and E to guarantee that the solutions won't have infinite norm?

My question springs from the treatment of the quantum harmonic oscillator in Griffiths. There, using the ladder operator method, we find new solutions by applying the ladder operators to existing solutions. But how can we say that the initial solution $\psi_0$ (which we use to generate new solutions) itself is normalizable (Griffiths proves that ladder operators give normalizable solutions, given that they act on a normalizable function)?

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    $\begingroup$ Related: physics.stackexchange.com/q/90101/2451 , physics.stackexchange.com/q/33668/2451 and links therein. $\endgroup$ – Qmechanic May 12 '15 at 6:35
  • $\begingroup$ @Qmechanic: I couldn't find an answer in the provided links. Also, I have edited the question to make it more specific. $\endgroup$ – Sidd May 13 '15 at 6:36
  • $\begingroup$ The edited question (v5) seems to be significantly different from previous versions. Now $\sup_{x\in\mathbb{R}} V<\infty$ is no longer assumed. $\endgroup$ – Qmechanic May 13 '15 at 8:29
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    $\begingroup$ Your question is not particularly well formulated. Not all hamiltonians have ladder operators (or, at least, ladder operators which sit inside a convenient algebra). Are you asking for conditions on $V$ such that all the eigenstates of $H$ will be square integrable? Or conditions on $V$ such that $H$ will have at least one square integrable eigenstate? Note that mixed spectra (part discrete and square-integrable, part non-integrable continuum) are very common, the easiest example being the hydrogen atom. $\endgroup$ – Emilio Pisanty May 19 '15 at 18:39
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    $\begingroup$ (In general, making your question more precise is more helpful towards getting an answer than throwing rep at it. You're free to keep it unrevised but it will make it much harder to answer. If you're specifically interested in ladder-operator methods you should make this explicit, and also present a well-formulated question which can be answered. "Eigenstates obtained by such methods" is very vague unless you're very specific about what "such methods" means.) $\endgroup$ – Emilio Pisanty May 19 '15 at 18:43

SUMMARY OF EDITED VERSION: You cannot place any conditions on $V(x)$ and $E$ that guarantee that solutions to the time-independent Schrödinger equation are normalizable, for something of a silly reason.

Initial, partial answer: If the potential is bounded below by some value $V_\text{min}$, then a solution to the time-independent Schrödinger equation with $E \leq V_\text{min}$ cannot be normalizable. Proof: suppose $\psi(x)$ is a normalized wavefunction that solves Schrödinger's equation with eigenvalue $E$. Then $$ E = \langle \psi | H | \psi \rangle = \int \left[ \frac{\hbar^2}{2m} \left| \frac{d\psi}{dx} \right|^2 + V(x) |\psi|^2 \right] dx > \int V(x) |\psi|^2 dx \geq \int V_\text{min} |\psi|^2 dx = V_\text{min}. $$ We have strict equality in the third step because $\psi$ cannot have zero derivative everywhere and still be square-integrable. (Note that this result is also a problem in Griffiths's text, with a different suggested method of solution.)


In fact, I think that it is impossible to place constraints on $V(x)$ and $E$ that guarantee a normalizable solution, for the following rather silly reason: If we view the time-independent Schrödinger equation as an ODE, then for any value of $E$ there are two linearly independent solutions to the second-order ODE $$ - \frac{\hbar^2}{2m} \frac{d^2\psi}{dx^2} + \left[ V(x) - E \right] \psi = 0. $$ So even if one square-integrable function $\psi_1(x)$ satisfies this equation, there will also be another linearly independent solution $\psi_2(x)$ that satisfies the equation as well, and this second solution will in general not be square-integrable (see below.) For example, if you try to solve the above ODE for the harmonic oscillator with $E = \hbar \omega/2$, you'll get two solutions, one of which has the usual $e^{-x^2/\sigma^2}$ behaviour (and is therefore square-integrable) and the other of which goes as $e^{x^2/\sigma^2}$ asymptotically and is therefore not square-integrable.

You might wonder whether it is possible that both $\psi_1(x)$ and $\psi_2(x)$ could both be square-integrable; but unfortunately this turns out not to be the case. To show that this can't happen, we can use logic like that of Ali Moh's answer. If the leading-order asymptotic behavior of $V(x)$ is proportional to $x^\alpha$, and the potential $V(x) \to \infty$ as $x \to \infty$, then we can write the Schrödinger equation asymptotically (after some rescaling) as $$ -\psi'' + (\beta x^\alpha - e) \psi = 0. $$ where $e$ is proportional to the energy and $\beta > 0$. If $\alpha > 0$, then the first term in the brackets will dominate, and the equation then has the approximate solution (via Mathematica) $$ \psi(x) = \left\{ \sqrt{x} I_{-1/(2+\alpha)} \left( \frac{2 \sqrt{\beta}}{2 + \alpha} x^{1 + \alpha/2} \right), \sqrt{x} I_{1/(2+\alpha)} \left( \frac{2 \sqrt{\beta}}{2 + \alpha} x^{1 + \alpha/2} \right) \right\}. $$ where $I_n(x)$ is a modified Bessel equation of the first kind. Now, if we have two solutions $\psi_1$ and $\psi_2$ that correspond to the same value of $E$ and are both square-integrable, then some linear combination of them should behave like each of these two solutions asymptotically; but both of these solutions diverge, and $\psi_1$ and $\psi_2$ both have to go to zero asymptotically to be square-integrable. Hence, one of the solutions $\psi_1$ and $\psi_2$ must not be square-integrable. (The same logic holds if $\beta < 0$; we just get ordinary Bessel functions instead, which are still not square-integrable.)

If, on the other hand, $\alpha \leq 0$, then the potential is bounded above and we recover the case discussed by Ali Moh. Again, the generic solution to the ODE will either be oscillatory or exponential, and so at most one of the linearly independent solutions to the time-dependent Schrödinger equation will be normalizable.

Beyond this, I'm not sure what exactly one could do. You might ask, "Can we put conditions on $V(x)$ and $E$ such that a normalizable solution to the Schrödinger equation exists?" But this is basically asking "Is $E$ in the spectrum of the Hamiltonian, when the Hamiltonian is acting only on the space of normalizable wavefunctions?" In other words, you're asking for the energy eigenstates and eigenvalues, which is what we're usually looking for anyway. You could always use variational-principle results to place bounds on the spectra of particular potentials; and there's the well-known result that at least one bound state ($E < 0$) exists for any attractive potential in 1D (and 2D). But I'm skeptical that a more general result exists.

  • $\begingroup$ Do you mean the time independent Schrödinger equation? Otherwise, what's $E$? $\endgroup$ – Emilio Pisanty May 19 '15 at 18:41
  • $\begingroup$ You are of course correct, and I've edited my answer accordingly. $\endgroup$ – Michael Seifert May 19 '15 at 19:11
  • $\begingroup$ No worries - typos happen. $\endgroup$ – Emilio Pisanty May 19 '15 at 19:22

Since $V(x)$ is bounded from above we have three possibilities. Either it oscillates at infinity with an upper bound, or it asymptotes to a constant $<E$ or it diverges to $-\infty$.

Since we are interested in $x\rightarrow\infty$ we may average the oscillation in the first case to the mean, and if it diverges then we concern our selves with the leading higher polynomial power (call it $\alpha$, assuming it asymptotes as a polynomial which is general enough). The result is that the behavior of the wave-function at infinity is oscillatory; trigonometric for the first two cases, and in the third case as $x^{\frac{1-\alpha/2}{2+\alpha}}J_{\frac{1}{2+\alpha}}\left(\frac{2x^{1+\alpha/2}}{2+\alpha}\right)$ which again asymptotes to an oscillatory non-decaying behavior.

All three cases are summarized by the asymptotic differential equation (where $\alpha=0$ for the first two cases) $$ x\rightarrow\infty:\qquad \left(\frac{d^2}{dx^2} + \beta x^\alpha \right)\psi(x)=0 $$

The point is that in all cases an oscillatory non-decaying asymptotic behavior of the wave-function necessarily implies that it is not normalizable.

  • $\begingroup$ My question springs from the treatment of the quantum harmonic oscillator. There, using the ladder operator method, we find new solutions, but what criteria do we use to say that the wave functions thus generated are normalizable? $\endgroup$ – Sidd May 12 '15 at 11:40
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    $\begingroup$ For the harmonic oscillator $V(x)\propto x^2$ so for any $E$ there is a certain $x_0$ such that $E<V(x) \,\,\forall\,\, |x|>x_0$ and in particular for $x\rightarrow\infty$... which is the opposite of your question. Here wave-functions are bound and normalizable. $\endgroup$ – Ali Moh May 12 '15 at 17:09
  • $\begingroup$ But if $E<V_{min}$, the states are non-normalizable. In this case, that would be $E<0$. $\endgroup$ – Sidd May 13 '15 at 12:26

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