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Excerpt from my textbook

It is impossible to determine simultaneously, the exact position and exact momentum (or velocity) of an electron.

It rules out the existence of definite trajectories of electrons and other similar particles.

Ok, so we cannot know the exact position and velocity of an electron at any instant. But how does this conclude that electrons don't follow definite paths.

Isn't the fact that we need light to reflect off something in order to see it, merely a limitation on our part, a limitation of human eye?

To observe an electron, we need to illuminate it with "light" or electromagnetic radiation. The "light" used must have a wavelength smaller than the dimensions of an electron. The high momentum photons of such light would change the energy of electrons by collisions. We would be able to calculate the position of the electron but we would know very little about its velocity after collision.

This might seem like a silly question, in fact I don't know much about Quantum mechanics. My textbook goes over these topics vaguely.

It would be helpful if the answer would be in simple words and not in terms of mathematical equations.

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  • $\begingroup$ There are other, more fundamental reasons why there are no such things as definite trajectory. You are dealing with probability, where will the particle pop up next. A basic quide to QfT will explain this better than I can, if you are interested. Not my d/v, I hasten to add $\endgroup$ – user163104 Aug 7 '17 at 19:41
  • $\begingroup$ Your text book is misleading, that is why you got the D/v, so it's not you personally, imo. $\endgroup$ – user163104 Aug 7 '17 at 19:52
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    $\begingroup$ @Countto10 it's really not necessary to get into QFT to understand why we don't think of electron trajectories as classical paths (i.e. like you would for a macroscopic ball). $\endgroup$ – DanielSank Aug 7 '17 at 20:11
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    $\begingroup$ You haven't said what you mean by "trajectory". But if what you mean is a function $f$ that maps times to positions, such that at time $t$ the electron is in position $f(t)$ (meaning an eigenstate of the position operator with eigenvalue $f(t)$), then this is certainly impossible for several reasons, starting with the fact that the position operator has no eigenstates. $\endgroup$ – WillO Aug 7 '17 at 20:22
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    $\begingroup$ First off, forget completely that you can think of a particle in any way like a very small football, so there is no direct line between points A and B. In QM, you don't deal with particle creation and annihilation, the Schrodinger equation is not enough to explain this. But in Quantum Field Theory, the fundamental "thing" is a field, from which particles can appear and disappear $\endgroup$ – user163104 Aug 7 '17 at 20:35
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There is a formulation of quantum mechanics called the de-Broglie-Bohm theory, in which particles move on definite trajectories. These trajectories are the solutions $\mathbf{Q}_k$ of the so-called guiding equation $$ \frac{d\mathbf{Q}_k}{dt} = \Im \frac{\psi^\dagger \nabla_k \psi}{|\psi|^2}, $$ where $\psi$ is the solution of the usual Schrödinger equation.

This answers your question with no. The uncertainty relation can be derived in the de-Broglie-Bohm theory for the outcomes of measurements. This is where it becomes relevant: Although the theory features exact trajectories of electrons, they cannot be measured to arbitrary accuracy. But this is, as pointed out e.g. in many papers of John Bell, not a shortcoming of the theory but something that should be expected in a theory of small things: "To admit things not visible to the gross creatures we are is, in my opinion, to show a decent humility, and not just a lamentable addiction to metaphysics."

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It has been shown that Dirac's bispinor equation can be expressed, in an equivalent tensor form, as a constrained Yang-Mills equation in the limit of an infinitely large coupling constant. It was also shown that the free tensor Dirac equation is a completely integrable Hamiltonian system with Lie algebra type Poisson brackets, from which Fermi quantization can be derived directly without using bispinors. The Yang-Mills equation for a finite coupling constant is investigated. Given the following expressions {Φ1(x), G} = −∇ · Gv, {Φ2(x), G} = −AGρ − ∇ · ρ0ρ −1 (∇ × v) × Gv + ∇ · ρ0ρ −1∇sGs −∇ · ρ0ρ −1 [∇GB] · B + ∇ · ρ0ρ −1∇ · [BGB] , we deduce various contributions to the Dirac bracket (7): ZZ d 3xd3x ′ {F, Φ1(x)}C −1 11 (x, x ′ ){Φ1(x ′ ), G} = − Z d 3x∇ · FvA −1∇ · (ρ 2 0ρ −1 (∇ × v) × ∇A−1∇ · Gv), ZZ d 3xd3x ′ {F, Φ1(x)}C −1 12 (x, x ′ ){Φ2(x ′ ), G} = Z d 3x∇ · Fv

Gρ + A −1∇ · (ρ0ρ −1 (∇ × v) × Gv) −A−1∇ · (ρ0ρ −1∇sGs) + A −1∇ · (ρ0ρ −1 [∇GB] · B) − A−1∇ · ρ0ρ −1∇ · [BGB] , ZZ d 3xd3x ′ {F, Φ2(x)}C −1 21 (x, x ′ ){Φ1(x ′ ), G} = − Z d 3x

Fρ + A −1∇ · (ρ0ρ −1 (∇ × v) × Fv) −A−1∇ · (ρ0ρ −1∇sFs) + A −1∇ · (ρ0ρ −1 [∇FB] · B) − A−1∇ · ρ0ρ −1∇ · [BFB] ∇ · Gv. The Dirac bracket now reads {F, G}∗ = Z d 3x  ρ −1 (∇ × v) ·  Fˆ v × Gˆ v − ρ −1∇s ·  FsGˆ v − Fˆ vGs −  ρ −1Fˆ v · [∇GB] − ρ −1Gˆ v · [∇FB] · B − B ·  [∇  ρ −1Fˆ v ] · GB − [∇  ρ −1Gˆ v ] · FB i (A.2), where Fˆ v = PA ·Fv = Fv −ρ0∇

A−1It is shown that the nonlinear Yang-Mills equation has exact plane wave solutions in one-to-one correspondence with the plane wave solutions of Dirac's bispinor equation. The theory of nonlinear dispersive waves is applied to establish the existence of wave packets.THE TRUE UNCERTAINTY LIES HERE.

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  • $\begingroup$ Quoting from the question: "It would be helpful if the answer would be in simple words and not in terms of mathematical equations. " $\endgroup$ – D. Halsey May 2 at 17:23
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Yes it does. There is a common misunderstanding of the uncertainty principle as our own lack of knowledge. When you read poor descriptions like "it is impossible to determine the momentum and position at the same time", you may interpret "impossible" as a limitation of our knowledge or tools. It is not. The uncertainty principle means that the electron DOES NOT HAVE the exact position and exact momentum at the same time. No matter how great the tools are, you cannot measure what is not there to measure.

Why? Simple. Particles interact with each other as particles, but between interactions they fly as waves. For example, if you send an electron through a screen with two slits, it will pass through both at the same time. Look up the double slit experiment for more information.

This is called particle/wave dualism that describes the nature of our reality. Particles are not microscopic "balls". Particles are waves that only interact with each other like "balls" when they meet. Quantum mechanics is also is known as wave mechanics, as it describes particles as wave functions with quantum properties.

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    $\begingroup$ whether it rules out definite trajectories is more interpretational. Example: De Broglie-Bohm theory, where particles can have classical trajectories, and you still get a theory that produces all standard QM results. Further, you can't really say a particle is a wave. All you can say is that theories that model it as a wave, produce good results. $\endgroup$ – CDCM Aug 8 '17 at 7:05
  • $\begingroup$ @CDCM: I appreciate your comment, but I respectfully disagree. The uncertainty principle is more fundamental that quantum mechanics or any its interpretation, because it reflects the symmetry of nature. "Interpretation" is a nice word, but what it really means is "understanding". I've stated mine and you are welcome to state yours, but saying that "this is interpretational" doesn't really mean much, because, if one has different understandings, he has none. On the nature of the electron, I agree, we don't know what it is, but if it walks like a duck and quacks like a duck, then, you know... :) $\endgroup$ – safesphere Aug 8 '17 at 7:33
  • $\begingroup$ I agree the uncertainty principle is a fundamental thing, but my point is it doesn't prohibit definite trajectories. You can recover all of quantum mechanics, while keeping classical trajectories. (I admit I'm somewhat playing the devil's advocate, in that I think pilot wave theory is just an item of interest). The theory is essentially unfalsifiable, as it makes no distinct predictions different from QM. But it does mean you can't say HUP rules out definite trajectories. $\endgroup$ – CDCM Aug 8 '17 at 9:49
  • $\begingroup$ @CDCM: In the de-Broglie-Bohm theory, trajectories exist, but are unknown, while the particle still obeys the wave function. So this theory is but an attempt to explain the world upside down. Sure, such an attempt may be mathematically consistent, just stand on your head and describe what you see. It breaks the Occam's razor principle, as inconsistent even with the special relativity without the foliation of space or some other extension. So saying that trajectories exist and depends on the wave function, but are unknown is like saying that Earth is the center of the universe. Sure, why not. $\endgroup$ – safesphere Aug 8 '17 at 17:09
  • $\begingroup$ I don't necessarily disagree with most of those critiques of pilot wave theory, but it does definitively allow classical trajectories to exist, which was in fact the original question. I'm not saying they exist, merely that HUP doesn't rule out the possibility they could exist, which you must admit, is interesting. I disagree more with the latter half of your comment. It's essentially an under-developed theory. It's unfair to critique it on being in an inelegant form. $\endgroup$ – CDCM Aug 8 '17 at 17:31

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