I'm trying to solve the Schrödinger equation for a free particle in a helix, but I have found it difficult to understand the normalization of the wave function. Well, let $\alpha(\phi) = b\cos(\phi)\hat{i} + b\sin(\phi)\hat{j} + a\phi \hat{k}$ the parametrization of a circular helix, for $a$ and $b$ constants and $\phi \in \mathbb{R}$ an angular variable. We get the Laplacian $ \nabla^{2} = \dfrac{1}{a^{2} + b^{2}}\dfrac{d^{2}}{d\phi^{2}}.$

It follows that the Schrödinger equation is $$ -\dfrac{\hbar^{2}}{2m}\dfrac{1}{a^{2} + b^{2}}\dfrac{d^{2}\psi}{d\phi^{2}} = E\psi$$ or $$ \dfrac{d^{2}\psi}{d\phi^{2}} + k^{2}\psi = 0, $$ for $k^{2} = \dfrac{2m(a^{2} + b^{2})}{\hbar^{2}}E$. Consequently, $\psi(\phi) = Ae^{ik\phi}$, where $A$ is a normalization constant.

Looking at the geometry of the curve, I believe that the wave function must be periodic, that is, $\psi(\phi) = \psi(\phi + 2\pi)$. Thus, $k$ must be an integer: $k_{n} = n = 0, \pm 1, \pm 2, \cdots. $ With this, the possible values for energy are obtained: $$ E_{n} = \dfrac{\hbar^{2}n^{2}}{2m(a^{2} + b^{2})}.$$

My biggest problem is finding the normalization constant. If I consider $0 < \phi < 2\pi$ (because the wave function is periodic at $2\pi$), then $$ 1 = \int_{0}^{2\pi}|\psi|^{2}d\phi = |A|^{2}2\pi$$ or $A = \dfrac{1}{\sqrt{2\pi}}$.

But, in fact, $\phi$ can take on any real value in the helix. In that case, we get that $\int_{-\infty}^{\infty}|\psi|^{2}d\phi$ diverges! Which of the two cases is correct? In the first, the particle behaves similarly to a particle in a ring, in the second there are no quantized energies.

  • $\begingroup$ I've removed a number of comments that were attempting to answer the question and/or responses to them. Commenters, please keep in mind that comments should be used for suggesting improvements and requesting clarification on the question, not for answering. $\endgroup$
    – David Z
    Commented Jun 13, 2020 at 6:26

1 Answer 1


I think there are two issues here:

  • the wave function in a helix does not have to be periodic
  • the wave function does not have to be real

After one tour of the helix we do not return to the same point, so the periodicity does not apply: $$\alpha(\phi+2\pi) = \alpha(\phi) + a2\pi\hat{k}.$$ What might be the source of confusion here is distinguishing the coordinate along the helix with the rotation around its axis.

Real wave functions
Since the periodicity does not apply, we have the problem identical to the Schrödinger equation for a free particle with momentum $k$. Its solution are usually chosen as complex exponents; however, if one needs a real solution, it is always possible to solve in terms of sine and cosine waves.

  • $\begingroup$ Free particles are usually plagued with normalization problem so the box normalization (some kind of periodicity is needed but it's up to us what distance to use as the period as long as we return to the same point). $\endgroup$
    – aitfel
    Commented Jun 13, 2020 at 12:11
  • $\begingroup$ @aitfel Indeed. The point is that this helix problem is not really different from that for a free particle. $\endgroup$
    – Roger V.
    Commented Jun 13, 2020 at 12:58

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