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If a person is sitting on a chair his momentum is zero and his uncertainty in position should be infinite. But we can obviously position him at most within few chair lengths.

What am I missing? Do we have to invoke earth's motion, motion of the galaxy etc. to resolve the issue?

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    $\begingroup$ I wonder if the quantum phenomena can still be observed in such a large scale system... $\endgroup$ – K_inverse Nov 12 '18 at 1:16
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    $\begingroup$ @K_inverse Yes, they can. But as soon as you try that, you'd realize neither the momentum nor the position is perfectly localized, so the premise of the question is false - you decidedly don't have "zero momentum" when sitting on a chair. $\endgroup$ – Luaan Nov 12 '18 at 11:22
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    $\begingroup$ If it is a rocking chair you don't have zero uncertainty in the position even at macroscopic level :-) $\endgroup$ – Francesco Nov 12 '18 at 11:48
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    $\begingroup$ You're confusing the momentum with the uncertainty in momentum. $\endgroup$ – mkrieger1 Nov 12 '18 at 13:54
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    $\begingroup$ Since I unexpectedly gained huge momentum very unexpectedly while sitting on an IKEA chair, I have never again felt any certainty about my position in the universe. $\endgroup$ – Pavel Nov 12 '18 at 19:44
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If a person is sitting on a chair his momentum is zero...

How close to zero?

The uncertainty principle says that if $\Delta x$ is the uncertainty in position and $\Delta p$ is the uncertainty in momentum, then $\Delta x\,\Delta p\sim \hbar$. So, consider an object with the mass of a person, say $M = 70\ \mathrm{kg}$. Suppose the uncertainty in this object's position is roughly the size of a proton, say $\Delta x = 10^{-15}\ \mathrm m$. The uncertainty principle says that the uncertainty in momentum must be $$ \Delta p\sim\frac{\hbar}{\Delta x}\approx\frac{1 \times 10^{-34}\ \mathrm m^2\ \mathrm{kg/s}}{10^{-15}\ \mathrm m}\approx 1\times 10^{-19}\ \mathrm{m\ kg/s}, $$ so the uncertainty in the object's velocity is $$ \Delta v=\frac{\Delta p}{M}\approx \frac{\approx 1\times 10^{-19}\ \mathrm{m\ kg/s}}{70\ \mathrm{kg}}\sim 1\times 10^{-21}\ \mathrm{m/s}. $$ In other words, the uncertainty in the person's velocity would be roughly one proton-radius per month.

This shows that the uncertainties in a person's position and momentum can both be zero as far as we can ever hope to tell, and this is not at all in conflict with the uncertainty principle.

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    $\begingroup$ Reminds me that science is closing in on producing quantum effects on the macro scale. We're a ways off from a human, but a 120 carbon atom bucky ball? Smashed. (A coherent beam of humans would be an interesting engineering challenge...) $\endgroup$ – Draco18s Nov 12 '18 at 15:09
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    $\begingroup$ @Draco18s Isn't that a marching column? $\endgroup$ – Pilchard123 Nov 12 '18 at 15:57
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    $\begingroup$ @Pilchard123 Yeah, but they don't maintain their momentum uncertainties when walking through double doors. $\endgroup$ – Draco18s Nov 12 '18 at 16:20
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    $\begingroup$ +1 Great job taking a joke/troll question, applying correct physics, and ending with "zero as far as we can ever hope to tell". $\endgroup$ – AnoE Nov 12 '18 at 23:03
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    $\begingroup$ @AnoE - I wouldn't say that's a joke/troll question. It's interest to grasp the basics of uncertainty principle. In fact, basic physics textbooks examples are not far away from that question. $\endgroup$ – Pere Nov 13 '18 at 10:33
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If we pretend that person is a quantum mechanical particle of mass $m=75$ kg and we localize him in a box of length $L=1$ m, then the resulting uncertainty in his velocity would be about one Planck length per second. Are you sure you know his velocity to within one Planck length per second?

Applying quantum mechanical principles to classical systems is always a recipe for disaster, but this underlying point is a good one - in macroscopic systems, the uncertainty principle implies fundamental uncertainties which are so small as to be completely meaningless from an observational point of view. If you were moving at a planck length per second for a hundred quadrillion years, you'd be about halfway across a hydrogen atom.

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    $\begingroup$ I tried doing the experiment you suggest in your last sentence but the damned hydrogen atom wouldn't sit still and I gave up after a couple of weeks. $\endgroup$ – David Richerby Nov 12 '18 at 17:50
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    $\begingroup$ @DavidRicherby I'd be more concerned if you had convinced a hydrogen atom to stay still $\endgroup$ – SGR Nov 13 '18 at 13:34

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