What is the relationship between energy and position in quantum mechanics?

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In quantum mechanics, the hydrogen atom's 1s orbital has a defined energy of 13.6 eV, yet the electron's position is probabilistic, allowing it to be found anywhere from the nucleus outward. This apparent contradiction is resolved by understanding that while position is uncertain, the electron's momentum varies correspondingly, maintaining a precise energy value. The probability density decreases with distance from the nucleus but never reaches zero, allowing for phenomena like quantum tunneling into classically forbidden regions. The time-energy uncertainty principle indicates that the electron cannot remain outside these barriers indefinitely. Overall, the relationship between energy and position is governed by the wavefunction's spread in both position and momentum space, ensuring energy remains well-defined.
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In the hydrogen atom the 1s orbital has a clearly defined energy of 13.6 eV, but the probability density and radial probability density says you are liable to find the 1s electron anywhere from the nucleus on out.

How does this exact energy value jive with this variable position?
 
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does the law really say that the electron can be found in this range?
 
The energy of the atom

can be exact with the position not being exact because the canonical conjugate to the the position, the momentum, also varies. To put it very crudely, the variation in the momentum compensates for the variation in position, allowing the energy to have an exact value.

That's about as good an answer as I can come up with under the circumstances, but when you learn a little more about wavefunctions and operators and such you'll understand it better.
 
Originally posted by Sacroiliac
In the hydrogen atom the 1s orbital has a clearly defined energy of 13.6 eV, but the probability density and radial probability density says you are liable to find the 1s electron anywhere from the nucleus on out.

How does this exact energy value jive with this variable position?

Yes. That's quite correct. The probability density decreases as you go further from the nucleus of the atom but it never becomes zero. The electron can tunnel into the potential barrier, i.e. into the classically forbidden regions. The farther it travels into such a region the shorter the time it stays there. The time-energy uncertainty principle implies that the ectron will not stay outside the barrier for a time equal to or greater than the time it takes to measure an energy violation.

Pete
 


Originally posted by Tyger
can be exact with the position not being exact because the canonical conjugate to the the position, the momentum, also varies. To put it very crudely, the variation in the momentum compensates for the variation in position, allowing the energy to have an exact value.


Are you saying that as the particle gets closer to the nucleus its potential energy decreases but because it is more localized the HUP causes its momentum to increase and exactly offset the decrease in Potential energy?

PMB I didn't understand your post. Sorry.
 


Originally posted by Sacroiliac
Are you saying that as the particle gets closer to the nucleus its potential energy decreases but because it is more localized the HUP causes its momentum to increase and exactly offset the decrease in Potential energy?

I did say that was a very crude description, but the basic idea is correct. You see, we can't really say the electron has a specific momentum or position because its wavefunction is spread out in space around the proton. But while it is spread out in both coordinate space and momentum space the phase of its amplitude changes with time at a constant rate, so its energy is well defined.
 
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