Let's consider a confinant potential $V(x) \in C^2(\mathbb{R}, \mathbb{R})$, of the form of a harmonic oscillator ($V(x) = x^2$ for example):

Harmonic oscillator potential

Consider a classical particle which has a kinetic energy $E_i$ at $x=0$. It will forever stay between $-a$ and $a$ because of the conservation of its total energy (see picture above, where $t$ is time):

Trajectory of a classical particle in the harmonic oscillator

However, a quantum particle will behave differently. There are infinite possible wave functions for this problem, here is one of them:

A wave function for a quantum particle

You can see that there exists two points where $\psi(x) = 0$ (the orange dots). It means the particle can't possibly be there since the probability density is given by $\rho = |\psi(x)|^2$.

That fact really surprises me.

Why is the probability density of a particle equal to zero (at nodes) even though there is nothing physically preventing from it?

  • 2
    $\begingroup$ that the particle has an infinite speed near those values. Does not follow. Knowing with 100% certainty that particle is there is not the same as knowing that particle is not there. For the former we have infinite uncertainty in momentum, due to the Heisenberg principle, but for the latter - uncertainty of momentum does not apply, because particle will be somewhere else. Most probably it will be tunneling through potential wall, thus zero probability density at wall edge. $\endgroup$ Jun 2, 2022 at 11:59
  • 5
    $\begingroup$ I think you are misunderstanding what the uncertainty principle says. (the square of) that whole wavefunction you provided is the distribution for $x$ of the particle. the time derivative of the above function gives you the distribution for $p$. The uncertainty principle is a statement about the standard deviations of the two distribtions. It's unrelated to the value of either the position or momentum distribution at a single point. $\endgroup$ Jun 2, 2022 at 16:15
  • 3
    $\begingroup$ Greetings! We have noticed that you have made a large number of trivial edits to this question. This is unfair to other users, because each trivial edit to your question “bumps” it to top of the recently-active question views, where it displaces a post by someone else. Some of your edit comments suggest you are aware this an abuse of our system. Please try to limit edits to substantive improvements. To “feature” a question, you can offer a bounty. Thank you! $\endgroup$
    – rob
    Jun 6, 2022 at 1:36
  • $\begingroup$ What is a "confinant potential"? A dictionary only has confinant as a French and Catalan word. $\endgroup$ Jun 20, 2022 at 0:32

5 Answers 5


There is a fundamental misconception here.

When the wavefunction is a solution to the time-independent Schrödinger equation, the wavefunction yields a time-independent probability density: there is no "motion" of a particle and thus no notion that the particle has any speed, much less infinite speed. In fact the average momentum is $0$.

The particle does not "move" along the shape of the wavefunction (or rather the shape of its modulus squared), so it does not "travel across a node". Rather, if one measures the position of the particles using a large number of identically prepared system, one will find never find the particle in the small region of size $dx$ centered around $x_0$ where $\vert\psi(x_0)\vert^2=0$.

A good analogy would be the distribution of light resulting from the interference of two (or more) sources. The resulting intensity is a static pattern on the screen. There are small regions around the centre of dark fringes where no light will ever be recorded. This does not mean the particle of light has "infinite velocity" near the centre of the dark fringes. It simply means that light never falls in this small region.


In order to make sure I understood you correctly let me rephrase your question: why is the probability density of a particle equal to zero (at nods) even though there is nothing physically preventing from it?

This does not mean at all that the particle is accelerating/deccelerating (which indeed would lower the detection rate at specific points). Keep in mind that the probability density along an x-axis only tells (after multiplying by the volume and the total amount of measures), how many times you will find that particle at given x position (assuming you have done a sufficient amount of measures). It does not directly describe the way it moves per se, thus you cannot deduce its acceleration but there is a thing called the propagator which gives the amplitude to go from one point to another in a certain time .

Altough there is unfortunately no physical explanation behind it (yet). There are a few things to keep in mind here, which could eventually clarify some things:

  1. The whole concept of nods is only but a consequence of assuming the wave-particle duality (thus using the wave function as a more rigorous description).
  2. The only reason we are still considering wave functions is for a pragmatic reason: it correlates with empirically based data.
  3. Note that we are talking about the probability density thus a more correct way to interpret it physically is that the probability of detecting a particle in the vicinity of a given point (nods in this case) decreases to zero, but not the probability of detecting at a specific point.
  4. Logically this results from the wavelength (of the wave function in question): just like a sinusoidal wave having more nods due to its smaller wavelength.

Disclaimer: these are just my personal thoughts. I am myself a mere student of quantum mechanics.

enter image description here

  • $\begingroup$ Thanks for your answer, you understood my problem correctly (I edited my question with your answer) but when you say it is density of probability and not just probability, it still means that the probability of having the particle at the nods is $0$ (let $x_0$ be the root, just make an integral bewteen $x_0 + \epsilon$ and $x_0 - \epsilon$), and make $\epsilon$ tend to $0$ $\endgroup$ May 28, 2022 at 16:24
  • $\begingroup$ Moreover, if I understood it correctly, you tell me this is some kind of "mistake" of the theory. Is that it ? $\endgroup$ May 28, 2022 at 16:26

You can’t use the uncertainty principle to evaluate an arbitrary subset of a wavefunction. The uncertainty principle applies to the solutions of the Schrödinger equation.

Specifically, the uncertainties are

$$ \sigma_x^2 = \left<x^2\right> - \left<x\right>^2 \\ \sigma_p^2 = \left<p^2\right> - \left<p\right>^2 $$

and the uncertainty principle guarantees that

$$ \sigma_x \sigma_p \geq \hbar/2 $$

It is a useful homework problem to compute these four expectation values for arbitrary solutions to the harmonic oscillator potential and verify that the uncertainty relation is always satisfied.

As for the nodes: remember that your solutions are stationary states, superpositions of left-moving and right-moving semiclassical waves, which may exhibit destructive interference. From Wikipedia, an illustration of

  • A,B: classical oscillators
  • C—F: energy eigenstates, with real and imaginary parts varying like $e^{i\omega t}$. These have amplitude profiles $\psi^*\psi$ which are constant in time, zeros at their nodes, and zero expectation value for $\left<x\right>$ and $\left<p\right>$.
  • G: a superposition of C—F, showing non-stationary behavior.
  • H: a coherent state of the oscillator.

Harmonic Oscillator


The general solution does not have anything specific to do with the zeroes of this particular solution. We could definitely model a particle with the above normalizable wavefunction but it will depend on the nature of the problem at hand.

As we go to higher and higher excited states we see more and more arbitrary zeroes of the wavefunction. Harmonic Oscillator Eigenstates

The answer to the Q is in your Q itself, "There are infinite possible wave functions for this problem". In general a particle (a wavepacket, a general solution) is represented as a superposition of energy eigenstates. The key idea is that the harmonic oscillator eigenstates form a complete set of orthonormal states.

$\psi(x)=\sum_{p} A(p) \psi_{p}(x) $

$\psi_{n}(x)=N_{n} e^{-\beta^{2} x^{2} / 2} H_{n}(\beta x), \quad n=0,1,2,3, \ldots$

or more precisely the Hermite Polynomials form an orthogonal basis of the Hilbert space with respect to the weight function and are also complete.

$$\int_{-\infty}^{\infty} H_{m}(x) H_{n}(x) w(x) d x=0 \quad \forall m \neq n$$

$$ \int_{-\infty}^{\infty}|f(x)|^{2} w(x) d x<\infty $$

$$ \langle f, g\rangle=\int_{-\infty}^{\infty} f(x) \overline{g(x)} w(x) d x $$

It basically means any general solution can be written as a sum of these eigenstates. Now any general normalizable solution can have a zero anywhere in the real line and from completeness we know we can construct it using a linear combination of Energy eigenkets of the Harmonic oscillator. Thus a general solution to the Harmonic Oscillator problem could have a zero anywhere, not necessarily in some special points.

Hence the above zeroes are not special.

Now how we construct a solution will depend on the nature of the problem, for example if we had an Einstein Solid( a collection of N harmonic oscillators) with a hole at a particular point we could crudely model a solution as a general harmonic oscillator eigenstate with a zero at that point.

  • $\begingroup$ I am not sure of the last example as I literally made it in my mind as I wrote the answer (it'll most probably be a bad approximation to the problem). Someone could add if they know any such technique being actually used. $\endgroup$
    – Gravity_CK
    Jun 3, 2022 at 8:01

Lets say the particle is an electron in a potential. If the electron was really a classical pointparticle that is bouncing back and forth between the "potential walls", then we should come up with some sort of classical explanation for why it is not found at the nodes. Actually, we would also have to explain why it is found most often near the "crests/throughs", which is just as weird when one thinks about it. In going from classical to quantum mechanics, we cant just say that the wavefuction is an abstract mathematical thing which tells us about the probablility distribution for position measurements, but the electron is still a pointparticle which is moving back/forth until we measure it - I think this question is just another reminder that we can do quantum mechanis, but dont really understand it.


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.