According to special relativity if an object is in motion relative to your frame of reference it contracts in the direction of its momentum. Thus, if a wave-particle duality, such as an electron, appeared to be moving from your frame of reference it is my understanding that said particle's wave form would appear contracted from your frame of reference.

This contraction would seemingly result in a particle's location being determinable to a greater degree of certainty with no cost to the degree of certainty with which an observer could determine the particle's momentum which would contradict the Heisenberg uncertainty principle and the limit it applies to the information an observer could collect about said particle.

So, what is the explanation for this? Is the Heisenberg uncertainty principle only applicable to particles at rest relative to the observer? Is Planck's constant relative (does it dilate proportionally with the momentum of said particle)? Or, alternatively, have I misinterpreted either special relativity or the uncertainty principle?

  • $\begingroup$ There are no single particles in relativistic theory, to begin with, so it doesn't make a whole lot of sense to try to rescue the single particle picture. The uncertainty principle still applies, in its mathematically more useful form of commutator relations for the underlying quantum fields. $\endgroup$
    – CuriousOne
    Jul 30, 2016 at 1:08
  • $\begingroup$ Why do you say that the contraction results in a particle's location being determinable to a greater degree? I don't exactly see the logic there (forgive me if this is a stupid question). $\endgroup$ Jul 30, 2016 at 1:11
  • $\begingroup$ @CuriousOne, you make it sound as if the uncertainty principle isn't applicable in all cases, which it is. $\endgroup$ Jul 30, 2016 at 1:11
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    $\begingroup$ "Wave-particle duality" is a concept with a lot of caché in the popular press that shows up almost not at all in the serious working of quantum mechanics. Because a proper view doesn't try to make things fit into those two boxes but recognizes two facts: (a) that all the world is quantum at it's root and quantum states are quantum states rather than being some classical thing; and (b) that when you probe a quantum system you drive it into the eigenstates of your probe. $\endgroup$ Jul 30, 2016 at 2:06
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    $\begingroup$ @HaruFujimura: Lots of things are still being taught in schools that educators think are up to date science. This particular thing wasn't up to date in 1929. The problem is that we can't really teach the correct concepts, so some might actually think that teaching false stuff is better than teaching none, but in reality most science teachers simply don't know better. The double slit experiment is a completely classical experiment. It tells us nothing about quantum mechanics. The first experiments that do are black body spectra and the photoelectric effect and, of course, atomic spectra. $\endgroup$
    – CuriousOne
    Jul 30, 2016 at 5:02

1 Answer 1


The uncertainty principle arises because the relationship between the position states of a particle-like system1 and the momentum states of that same system is a Fourier transformation.

Even in classical optics or electronics there exist a theorem linking the spread of a signal in the time-domain and in frequency-domain. The uncertainty principle is exactly the same math.

Notable the relationship between the two sets of states being that of a Fourier transform is not dependent on the object having any particular momentum distribution, so the Heisenberg principle is likewise insensitive to the value of a particle momentum.

1 By which I mean a quantum system that will exhibit particle-like properties if you probe it properly.

  • $\begingroup$ Unless I am mistaken everything stated above is agreed with what I am saying in my question. I was suggesting that Special Relativity would condense/contract the wave form of the particle ever so slightly if it was in motion relative to your position. $\endgroup$ Jul 30, 2016 at 2:37
  • $\begingroup$ "The length contraction is proportional to the momentum and can be calculated from the Lorentz transformation." link $\endgroup$ Jul 30, 2016 at 2:45
  • $\begingroup$ I have to admit I can't reproduce the math at this time, but I can outline an argument. To whit: If you want to apply relativity, you have to start with a relativistic quantum mechanics; that adds some surprising wrinkles, because now the operators act on a four-state. Any scalar computed from those operators is going to be a Lorentz-scalar and thus invariant on boosts. $\endgroup$ Jul 30, 2016 at 2:53
  • $\begingroup$ @ConnorMcMonigle: Why do you think "there is no cost to the degree of certainty with which an observer could determine the particle's momentum"? As dmckee said if the wave function really contracts in space, its Fourier transformation would expand in momentum space and so the uncertainty in determining momentum will increase. $\endgroup$
    – Seyed
    Jul 30, 2016 at 10:31
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    $\begingroup$ The space interval contracts by 1/γ, but the momentum interval dilates by γ, so the de Broglie relationship remains invariant. Check out the relativistic Doppler effect. $\endgroup$ Aug 1, 2016 at 16:18

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