A particle can be set off in a certain direction by giving them momentum. Momentum is a vector, so the particle heads off in a specific direction. But the wave function of the particle allows it to obtain other momentum values, which would steer the particle on a different path. How then can we "shoot" electrons and other particles in straight lines? How can they maintain their momentum in the face of quantum uncertainty?


The uncertainty principle's restriction on the minimum spreads in position and momentum is really small. An electron can be confined to a region in both position and momentum space that's extremely small compared to anything human-sized, but still have more than enough spread to obey the uncertainty principle.

To give you an idea of how small this is, $\hbar=1.05457173 × 10^{-34} J\cdot s$, so an electron with a standard deviation in it's position of 10 micrometers ($10^{-5}m)$ has a minimum uncertainty in it's velocity of about $6 m/s$. Electrons in any sort of beam are usually travelling at some appreciable fraction of the speed of light ($3 \times 10^{8} m/s)$, so this uncertainty is tiny.

  • $\begingroup$ Thanks, Dan, this is a very clear answer. So am I correct in understanding that the photons coming from a distant star have only a tiny variation in their velocity, so that they can be considered "linear" for practical purposes? $\endgroup$ – Koen Van Damme May 27 '13 at 14:40
  • $\begingroup$ Photons do no have mass, and are following a geodesic $\endgroup$ – anna v May 27 '13 at 17:40
  • $\begingroup$ @KoenVanDamme: They don't have to have any variation in their velocity, at least not because of the position-momentum uncertainty principle. They always travel at the speed of light, and their momentum is determined by their wavelength. $\endgroup$ – Dan May 30 '13 at 2:05
  • $\begingroup$ @KoenVanDamme: Also, photons are not billiard balls. They're quantum objects that don't necessarily obey classical laws. $\endgroup$ – Dan May 30 '13 at 2:08

Dan's answer is quite nice but I think the answer to this really depends on the interpretation one chooses to use.

The straightforward answer I think is that you probably are confusing collimation with the creation of electrons.

In classical mechanics one can independently specify the $\vec{x}$ and $\vec{p}$, very much independently for the reason that Dan describes in his answer. To illustrate this suppose you created an $e^-$ with thermal ionization then it's momentum and the position will have distribution which will be given by the wavefunction of the electron. These thermal electrons are then collimated into a beam of electrons which you can say have reasonably sharp peak in k-space. Although it seems that one can make beam as thin as possible and violating transverse uncertainty principle, but after a while when the opening size becomes of the order of $\lambda$ of electron the beam start diffracting, hence saving the uncertainty principle from violating.

But you might ask what about the longitudinal condition? Well that one actually is already saved as the position of particle a priori is unknown.

I think you should think about a better and more clearer alleged absurdity claimed by quantum theory (BTW this is not exactly absurd since this has been verified by experiments e.g. Davisson and Germer). It is now pretty much a handy rule not to think of the particles traveling in the straight line. Then how would you explain the tracks seen in the cloud chamber which seemingly violates the uncertainty principle. This problem seems to go away when one realise that it is ensamble to which this uncertainty principle apply not to one single measurement.

PS: I think the last couple of lines might interpreted differently in many worlds and consistent histories. For professionals and experts, I am just a student yet so do correct me if you find any part wrong.

  • 2
    $\begingroup$ cloud chamber tracks do not violate the uncertainty principle. At most the Delta(x) dimensions are in microns and the measurement momentum errors from the curvature in the magnetic field of the order of keV. $\endgroup$ – anna v May 27 '13 at 17:37
  • $\begingroup$ @annav I did not mean to claim that Quantum theory violates the most holy principle. I simply was emphasizing that it is slippery floor one is walking on. As in the underlying theory trajectory has no meaning at all. But one does see the electrons with tracks which are classical in nature i.e. (in the quantum language) one can say the electrons with different trajectory actually has different wave-fn. I think it is resolved by using decoherence. Thanks for asking because it helped me. Kindly reply with any criticism. :) $\endgroup$ – The Imp May 30 '13 at 10:40
  • $\begingroup$ HUP is a fundamental postulate of QM. The mathematical formalism of QM includes it in the value of the commutators en.wikipedia.org/wiki/Uncertainty_principle there is no conflict. Uncertainty always exists in our measurements, called measurement error. A straight line drawn within two points is straight within errors and the track considered a line within errors. QM tells us that as we go to smaller and smaller scales there is an inherent "error" when trying to measure simultaneously position and momentum that cannot be clarified further by better measurements. $\endgroup$ – anna v May 30 '13 at 12:34

The uncertainties on $x$ an $p$ do not imply any randomness of the trajectory until they are measured. If you don't measure anything, the particle wave function is completely determined and goes in a straight line to infinity, with a given probability distribution of $x$ and $p$. When you measure something, then the wave function is modified and may not go in the same average direction.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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