I'm reading this wikipedia article and I'm trying to understand the proof under "Degeneracy in One Dimension". Here's what it says:

Considering a one-dimensional quantum system in a potential $V(x)$ with degenerate states $|\psi_1\rangle$ and $|\psi_2\rangle$ corresponding to the same energy eigenvalue $E$, writing the time-independent Schrödinger equation for the system: $$-\frac{\hbar^2}{2m}\frac{ \partial^2\psi_1}{ \partial x^2} + V\psi_1 =E\psi_1$$ $$-\frac{\hbar^2}{2m}\frac{ \partial^2\psi_2}{ \partial x^2} + V\psi_2 =E\psi_2$$ Multiplying the first equation by $\psi_2$ and the second by $\psi_1$ and subtracting one from the other, we get: $$\psi_1\frac{d^2}{d x^2}\psi_2-\psi_2\frac{d^2}{d x^2}\psi_1=0$$ Integrating both sides $$\psi_1\frac{\partial \psi_2}{\partial x}-\psi_2\frac{\partial \psi_1}{\partial x}=constant$$ In case of well-defined and normalizable wave functions, the above constant vanishes, provided both the wave functions vanish at at least one point, and we find: $\psi_1(x)=c\psi_2(x)$ where $c$ is, in general, a complex constant. So, the two degenerate wave functions are not linearly independent. For bound state eigenfunctions, the wave function approaches zero in the limit $x\to\infty$ or $x\to-\infty$, so that the above constant is zero and we have no degeneracy.

Here are the parts I don't understand:

  1. How does integrating $\psi_1\frac{d^2}{d x^2}\psi_2-\psi_2\frac{d^2}{d x^2}\psi_1=0$ yield $\psi_1\frac{\partial \psi_2}{\partial x}-\psi_2\frac{\partial \psi_1}{\partial x}=constant$. Without having explicit formulas for $\psi_1(x)$ and $\psi_2(x)$, how do we just apparently pull them both out of the integral?

  2. I don't understand how the constant in the last equation vanishes just because $\psi_1$ and $\psi_2$ vanish at infinity.

  3. Even if it does vanish, how does that imply $\psi_1=c\psi_2$?


2 Answers 2


I agree with the answers to 1) and 3), but disagree with the proposed answer to 2) of buzhidao and levitopher (the same argument is used in well-known textbooks such as Shankar, p.176). It is incorrect to say that "at infinity, the function is zero". This is never true for piecewise continuous potentials: if the wave function would actually be zero from a finite point onwards, it would have to be zero everywhere because of the uniqueness property of second order ordinary differential equations (e.g. http://www.emis.de/journals/DM/v5/art5.pdf). What you actually want from a solution is for it to be square-integrable on the real line. This implies that it goes to zero at infinity but not that it attains zero.

The problem with the argument is that there exist square-integrable solutions to the Schrödinger equation where the first derivative is unbounded at infinity. An example is $\psi(x) = N \sin(x^3)/x$ for $x \geq x_0$ with $x_0$ some positive number and $N$ a suitable normalization. The derivative oscillates and is unbounded as $x \rightarrow \infty$. This is a solution to the Schrödinger equation with zero energy and potential $V(x) = 2/x^2 - 9x^4$ (setting $m=1/2, \hbar = 1$. I only mention the behavior for $x \geq x_0$ because we are interested in that regime, and my proposed $\psi$ would not be square-integrable near zero. This could be fixed by making the potential infinite for e.g. $x \leq x_0 = (2 \pi)^{1/3}$ so that $\psi(x \leq x_0) = 0$). So it is false that $\psi_1 \psi_2'$ tends to zero because $\psi_1$ and $\psi_2$ tend to zero. We have to be assured that the derivative does not mess things up by becoming infinite. To this end we should impose a constraint on the set of potentials $V$ we consider, such that this constraint ensures that $\psi'$ is bounded as $x \rightarrow \infty$ as a consequence of $\psi$ satisfying the Schrödinger equation.

In a recent article https://arxiv.org/abs/0706.1135 one refers to Messiah's quantum mechanics book (https:// archive.org/details/QuantumMechanicsVolumeI: the relevant pages are 98-106) for a proof that if $V(x) - E \geq M^2 > 0, \forall x > x_0$, for some numbers $M,x_0$, then the eigenlevel corresponding to $E$ is non-degenerate (assuming there is such an eigenlevel). He shows that in this case the eigenfunction satisfies $\psi(x) = \mathcal{O}(e^{-Mx})$ and $\psi'(x) = \mathcal{O}(e^{-Mx})$ as $x \rightarrow \infty$ which is more than enough to guarantee that buzhidao and levitopher's argument works. Note that the condition on $V$ is stronger than simply being bounded from below. Note also that this does not prove that the potential has to be bounded from below to ensure that the energy eigenlevels are non-degenerate.

  • 2
    $\begingroup$ Brilliant. Might want to send Shankar a correction. $\endgroup$ Commented Oct 11, 2020 at 22:02

(1). $\psi_1\psi_2^"-\psi_2\psi_1^"=\frac{d}{dx}(\psi_1\psi_2'-\psi_2\psi_1')=0$.

(2). At infinity, it is zero, so the constant must be 0

(3). Integrate $\psi_1\psi_2'=\psi_2\psi_1'$, you will get that

  • 6
    $\begingroup$ Just a slight rephrasing of #2: "At infinity the wave functions are zero so that left side of the last equation is identically zero. That "constant" is constant in $x$, so must be everywhere zero" $\endgroup$
    – levitopher
    Commented Sep 22, 2014 at 3:41
  • 3
    $\begingroup$ What if, e.g., we have two identical potential wells separated by an infinite barrier? It is well known that all the energy states will be doubly degenerated. Where does the proof go wrong in this case? $\endgroup$
    – user17116
    Commented Sep 22, 2014 at 5:11
  • 2
    $\begingroup$ @LBO There are two loopholes in the theorem. First, for unbounded potentials, the slope diverges at infinity and the constant in (2) is not zero. An example is the so-called "volcano potential". Secondly, if $\psi_1$ and $\psi_2$ become zero at finite $x$, the proportionality constant between them can change. If this is the case, then at least one of them needs to have a kink at the zero. The curvature diverges and this is only consistent with a singular potential to give finite $E$. An example is the double well with infinite barrier. $\endgroup$
    – Praan
    Commented Nov 13, 2015 at 9:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.