I Quantum Mechanics and Electrodynamics/Electrostatics

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Hi all, I have a question relating to the title above.

The uncertainty relation tells us that an electron that is localised (in terms of its PDF) is space has a large uncertainty in momentum space. However in classical electrostatics/dynamics we seem to make attempts to do things like approximating the magnetic field caused by an electron moving at a given velocity from A to B.

Isn't this a little wonky since we are assuming that we know position and momentum at the same time? If so, is there a reason why classical theory still holds up pretty well (slight understatement)?

Thanks in advance!
 

Vanadium 50

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Thanks for your response.

Because h is small.
Do you mind elaborating a little on what this means? I'm afraid I don't follow.
 

Vanadium 50

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If h were zero, QM would look like classical physics.
 
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Isn't this a little wonky since we are assuming that we know position and momentum at the same time? If so, is there a reason why classical theory still holds up pretty well (slight understatement)?
As @Vanadium 50 said above, the reason it works is because h is small in terms of the scale where classical theory holds up well. Recall that the uncertainty principle does not merely tell us that we cannot know position and momentum at the same time, but it also puts specific constraints on our knowledge of position and momentum. Specifically: ##\sigma_x \sigma_p \ge \hbar/2##.

So if we have an electron whose velocity we prepare to an accuracy of 1 m/s then we simultaneously prepare its position to an accuracy of 58 microns. If it is a proton with the same velocity then we can also prepare its position to an accuracy of 32 nanometers. On classical scales that works OK.
 
472
9
As @Vanadium 50 said above, the reason it works is because h is small in terms of the scale where classical theory holds up well. Recall that the uncertainty principle does not merely tell us that we cannot know position and momentum at the same time, but it also puts specific constraints on our knowledge of position and momentum. Specifically: ##\sigma_x \sigma_p \ge \hbar/2##.

So if we have an electron whose velocity we prepare to an accuracy of 1 m/s then we simultaneously prepare its position to an accuracy of 58 microns. If it is a proton with the same velocity then we can also prepare its position to an accuracy of 32 nanometers. On classical scales that works OK.
Thanks a lot!
 

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