I Does a certainty in the position imply infinite variation in speed?

Giuseppino32
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As position uncertainty decreases, the uncertainty in momentum and velocity increases, potentially approaching the speed of light. While Δv can be large, it doesn’t imply any arbitrary velocity from 0 to 𝑐; it's probabilistic and constrained.
I would like to know if this thought makes any sence or if i'm missing something

Heisenberg principle states that: ΔxΔρ ≥ ħ/2Δρ ≥ ħ /2Δx
If we consider a scenario where we increase the precision of our measurement of position, we have Δx ⇒ 0 the principle implies:
Δρ ≥ ħ/2Δx → Δρ ⇒ ∞

Assuming that any of that makes any sense, If we consider the maximum possible speed in the universe, c (the speed of light, approximately 3×10⁸m/s) the minimum possible speed as 0, and assume the mass of the particle remains constant, then the uncertainty in velocity Δv is given by: Δρ = 3.10⁸ - 0 = 3.10⁸ m/s

This illustrates that as the uncertainty in position decreases, the uncertainty in momentum—and thus velocity—can increase significantly, potentially approaching the speed of light.
Does that makes physical sense? Lets say we know where an electron e⁻ is in space so if its mass is constant 9.1093837 × 10-31 kilograms. If we find out its position does it mean it could have any value from 0 to the speed of light, in any direction, without the necessity of a force?
 
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How doe we "know" this very defined location?? Likely there are some nontrivial "forces" involved.. The Heisenberg uncertainty limit is difficult to circumvent.
 
Giuseppino32 said:
the precision of our measurement of position

##\Delta x## is not the precision of our measurement, it's standard deviation. It's sad that a lot of physicists forget about that when they talko to laypeople.


Or even physics students. Like this one professor at my alma mater, who was known for being very nitpicky and rude towards students, and making a lot of mistakes herself at the same time. But you know, she is doing experimental physics, so she does not have to remember all the theoretical details. Or even how linear function works. Oh well. Sorry for that off-topic.
 
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weirdoguy said:
##\Delta x## is not the precision of our measurement, it's standard deviation. It's sad that a lot of physicists forget about that when they talko to laypeople.


Or even physics students. Like this one professor at my alma mater, who was known for being very nitpicky and rude towards students, and making a lot of mistakes herself at the same time. But you know, she is doing experimental physics, so she does not have to remember all the theoretical details. Or even how linear function works. Oh well. Sorry for that off-topic.
Oh, thank you for this insightful response. It is a bit weird because it is almost like it still follows the exact same logic, although not necessarily they carry the same meaning, which still holds the discussion the same way. It's really sad that what motivates most people in this forum is making corrections rather than discussing the question. Not your case, though—again, thank you for this wonderful piece of information you just provided us. Keep doing the extraordinary work
 
Well, my problem is that writing longer responses in english takes me way too much time, so usually my responses are quite short. But I'm known for writing walls of texts on polish forums o0)

And with your question, what I wrote is quite an important thing, and with that in mind you should reconsider whether your question still makes sense. Sometimes such thing changes everything, so there is no point in analysing what someone wrote after wrong assumption.
 
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If we release an electron around a positively charged sphere, the initial state of electron is a linear combination of Hydrogen-like states. According to quantum mechanics, evolution of time would not change this initial state because the potential is time independent. However, classically we expect the electron to collide with the sphere. So, it seems that the quantum and classics predict different behaviours!
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