This might be a slightly naive question, and if so I apologize, but I am currently a little confused as to why the Heisenberg Uncertainty principle should apply to particles, i.e. our system (say an electron) after we observe it and collapse it’s wave function.

From what I understand, the Heisenberg Uncertainty principle just comes from the fact that momentum is the Fourier transform of position (wave number technically I think, but all the same since momentum is related to wavelength which is related to wave number). The more localized one is, the less localized the other will be because ‘localized’ things require a larger distribution of frequencies to localize them.

Nonetheless, it seems as those this should only hold, if our object is treated as a wave, but if we treat it like a particle, it feels like this should just go away. Even if you represent a particle like a wave by using something like the Dirac delta function or whatnot, you would get essentially an infinite number of corresponding wave numbers, in other words total uncertainty on the momentum which seem strange if we think of things like particles classically. It just feels like in order for Heisenberg to hold, things always need to be ‘wave-like’ in some sense. I apologize for the long winded question, but any help would be appreciated.

Edit: Thank you all for your responses. I think my confusion has been cleared up.

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    $\begingroup$ A photon is neither a classical particle nor a classical wave. The same is true for an electron. This may help - How can a red light photon be different from a blue light photon? $\endgroup$
    – mmesser314
    Commented Feb 3, 2023 at 4:42
  • $\begingroup$ "if our object is treated as a wave, but if we treat it like a particle, it feels like this should just go away" - you can't make a property of an object go away by simply choosing how you'll treat it. If you want to treat it one way or the other, then you're making the assumption that, in that particular context, the aspects you are ignoring have an insignificant effect for what you're trying to calculate/predict. That assumption may not always be valid $\endgroup$ Commented Feb 6, 2023 at 6:40

4 Answers 4


I understand your confusion. It is due to an old-fashioned way of introducing Uncertainty relations based on wave formalism, dating back to Heisenberg, but probably quite misleading.

Quantum mechanics (QM) does not say that particles are waves. That was de Broglie's original point of view, but today is untenable. Particle dynamics may be described using waves. But this is not the same as saying particles are waves. There are many reasons for that. I mention a couple of them: quantum wavefunctions for more than one particle are not functions of a single space point; in measurements, nobody ever measured a fraction of charge, spin, or any other property of the particle like it would happen if the physical properties would have been spread over an extended field.

QM is a probabilistic theory from which we can extract consequences on the statistical behavior of many measurements on equally prepared systems. However, in most cases, the outcome of an individual measurement is a random variable. Moreover, QM can be formulated differently, and wavefunctions in a Hilbert space are just one of the possibilities. The real issue is the calculation of probabilities.

The actual content of the Heisenberg relations is captured by the Robertson-Schrödinger theorem: $\Delta x \Delta p_x \geq \frac{\hbar}{2}$ is a statement about the variances of the random variables corresponding to independently measured position and momentum in an ensemble of equally prepared particles. As such, it is neither a statement about the measure of both momentum and position of a single particle nor an effect of the interaction with a measurement device.

Limits for combined measurements on the same system exist, but it is a different story, and there are strong indications that such limits differ from the usual Robertson-Schrödinger result.

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    $\begingroup$ This is an extraordinarily important answer, I think this answer should be pinned to every highschool introduction of QM books $\endgroup$
    – Babu
    Commented Feb 4, 2023 at 5:06

Nonetheless, it seems as those this should only hold, if our object is treated as a wave, but if we treat it like a particle, it feels like this should just go away.

I think the sentence above is the crux of the problem. One has to distinguish the physical reality, and the way we describe it (which exists only in our minds.) The object is neither a wave, nor a particle, but something that has the properties of both. Depending on the situation, it might exhibit properties more akin to those of a wave or to those of a particle, and in some cases we can treat it as one or the other. But it won't change its behavior just because we are treating it differently. If it behaves as a wave, but we describe it as a particle, we will be just describing it incorrectly.

See also Wave-particle duality:

Wave–particle duality is the concept in quantum mechanics that every particle or quantum entity may be described as either a particle or a wave. It expresses the inability of the classical concepts "particle" or "wave" to fully describe the behaviour of quantum-scale objects.

(emphasis mine)


The particle is always described by its wave function. The difference between what we call a particle and a wave is just a matter of how localised the wave function is.

As a general rule, when a particle is travelling freely and not interacting with anything, its wave function is well approximated by an infinite plane wave. However, when the particle is interacting with other particles, the interaction usually changes the wave function into a more localised state, and this is when we use the term particle.

So for example, in a cathode ray tube the electrons have a plane wave like wave function when they are travelling from the filament to the screen, and they evolve to a more localised particle like wave function when they interact with the phosphor. But in both cases, we can calculate the standard deviation of the position and momentum from the wave function. And in both cases the standard deviations obey the Heisenberg uncertainty principle.


I am answering this part of your question :

Nonetheless, it seems as those this should only hold, if our object is treated as a wave, but if we treat it like a particle, it feels like this should just go away.

Are you aware of what the wavefunction is that is assigned to particles and their interactions?

It is the solution of the pertinent quantum mechanical differential equation for the problem with the boundary conditions at hand. It has been experimentally determined that the postulates of quantum mechanics describe and predict the observations and measurements at the level of particles.

The wavefunction postulate assigns a wavefunction $Ψ$ to the system under observation/measurement where $Ψ^*Ψ$ is the probability of observation of the system , what is waving is not the particle, but how probable it is to see it at (x,y,z,t). One needs an accumulation of events to test the wave nature, $Ψ$ , which controls the probability.

See my answer here on how the probability is built up in a double slit single electron experiment.

Mathematically it can be shown that the wave nature of particles is observed in the probability distributions.

A single electron in a vacuum at uniform velocity cannot be modeled with a plane wave, it is modeled with a wave-packet, see here,

  • $\begingroup$ Ψ∗Ψ is the probability You mean the probability density? $\endgroup$
    – Gert
    Commented Feb 3, 2023 at 14:40
  • $\begingroup$ @Gert the link I forgot to attach and have given now, is about probability. $\endgroup$
    – anna v
    Commented Feb 3, 2023 at 15:36
  • $\begingroup$ @Gert I found this about probability density: aapt.scitation.org/doi/10.1119/1.5092484 . probability density has units, whereas the postulate is about the probability definition. $\endgroup$
    – anna v
    Commented Feb 4, 2023 at 18:02
  • $\begingroup$ That link to Hyperphysics is wrong IMO. The probability at $x,t$ is always zero (normalised or not) $\endgroup$
    – Gert
    Commented Feb 4, 2023 at 19:26

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