Heisenberg trouble

As to wat I have, learnt, it can only be applied to moving objects!!

For an object at rest, ∂x and ∂p are both 0, as both its position and velocity are known with perfect precision!!!

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Something more..

the problem with ∆x.
The mass of a macroscopic entity is much larger than that of a quantum entity, and by the same thermodynamic requirement(Zero kelvin) it is certain that it is always moving, at least at the quantum level. This makes its de Broglie’s wavelength, and hence ∆x, vanishingly small.It may be worth arguing that if a classical entity is sitting still, its de Broglie’s wavelength should be the average wavelength of all quantum entities belong to it. But even if this case is true the de Broglie’s wavelength of the classical entity still ends up being much smaller than its representative length scale

When we measure a quantum entity, a positional error gives the feeling that the particle (macroxcopic)has moved from one position to the other On the other hand, the inability to measure the position of the macroscopic particle exactly (due to imperfect tools)will at most give the feeling that the particle(macroscopic) is vibrating. In realistic cases, this “vibrating motion” is too small for the classical scale. This is why we don’t see the effect of positional uncertainty in the macroscopic world.

Contributing a bit of what I ahve learnt over 3 years...[1]

Heisenberg Uncertainity principle is specifically designed for all microscopic and macroscopic particles.

the contradiciton::
consider a stationary block ..it has zero velcocity ,hence zero momentum =>Δp=0
so from heisenberg's principle we arrive at conclusion that Δx=∞ which is definitely not true becoz we know the positon of block.

thsi contradiciton arises becoz we are assuming velocity to eb zero which is not at all true....every macroscopic body contains clouds of molecules/particles which are having some velcoity.....which ultimately give zero average velocity ,,which we interpret to be simply velocity...

Observation from bit diff angle
sinec measuremment tools are not 100% perfect
so Δx.Δp≥h/2π
=>Δx.Δp=h/2π+K
=>Δx.Δp=M M=constant
now if we say that Δx=0,we are fooling ourselves only becoz we very well know that Δx cannot be less than λ
where λ=de broglie wavelength
=>Δp≠∞

Also if we say that Δp=0,it will be wrong becoz Δp cannot eb less than p.
the only way for ∆p to be zero is when p=0 but if p=0 the velocity of the particle(microscopic or macrosopic) =0
=>it cannot exist in the physical sense and have zero mass, otherwise it would move at the speed of light and hence possess momentum OR If the velocity is zero, the particle either does not exist or it sits still in 3D space . But we know that v can never be exactly 0 because the absolute temperature would have to be 0 K, which is not realistic. However, the case of very small momentum is real. It implies that λ could be very large, and this is why sometimes the quantum scale is not “microscopic”

hey i got...delV is the error n not the vel.if delV is 0 then it means that there is no uncertainity bt we cnt say that it is @ rest since delV is 0...

yes akhilesh is ri8...thus in macroscopic particles HUP has very less significance..bt i hav a q.wat does delX or delP signify if its value is infinite?dat is we cnt kno evr where the object is??

Heisenberg's uncertainity principle can be applied to only microscopic particles and not macroscopic objects such as cars.

consider the following situation if the principle is applied to a particle of say about a miligram(10^-6 kg) then Δx.Δv is very small equal to 10^-28 m²/s Imagine how small it would be for a moving car weighing 1000kg or more.

But when an object is at rest, its position is clearly known, so how can it be said that uncertainity in position is ∞ [7][7]

OK..... according to the formula, wat U say is true, but wat abt the experimental value??

hey kalyan delX is nt xero bt it is infinite coz delV is 0.take it to R.H.S. n 1/0 makes ur R.H.S.zero.so here it is infinite nt zero for a staionary object...

well sir......i was wrong...........again msp tells HUP can be applied to stationary objects too......then are all options in the question correct ?

Someone please see this quest.......my explanation : Ans. A. electron

a moving microscopic particle

But sir, can heisenberg uncertainty be applied to a moving car ???