For a electron orbiting around the nucleus, why must it be in phase?

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SUMMARY

The discussion centers on the necessity for an electron orbiting a nucleus to be in phase, as described by the equation nλ = 2πr. This condition ensures that the electron's wave function forms standing waves, preventing destructive interference that would lead to zero probability amplitude and thus the non-existence of the electron. The conversation highlights the transition from classical to quantum mechanics, emphasizing the role of wave-particle duality and the implications of the Heisenberg Uncertainty Principle in understanding electron behavior in atomic orbitals.

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Why nλ= 2∏r?
this is the explanation from website ,
Now, one complete orbit has the same length as the circumference of the circle that the orbit traces out, so is given by 2πr. This distance must be equal to an integral number of wavelengths. (If it were not, the wave traced out by the particle on its first orbit would be out of phase with those traced out on all subsequent orbits.)

Over an infinite number of orbits, the out of phase waves would cancel each other out to zero amplitude, which implies that the particle cannot exist under such circumstances. Only if the waves traced out on all orbits overlap exactly, i.e. if the orbit is an integral number of wavelengths, is the situation a satisfactory one under which the particle can exist.
My question is , 1)electron is moving like a wave when it is orbiting the nucleus, correct?
2) why out of phase will cancel out each other? What is being cancelled? Waves of electron? In the orbit, they are not radiating , they only orbiting around the nucleus, what is going to be cancel?
Thank you.
 
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The explanation you're describing is from "old quantum theory" which was never really more than a set of rules of thumb. Getting the energy levels of hydrogen with this method is generally understand to be an example of getting the right answer for the wrong reason. So you're questions don't really have good answers because the scenario you're describing isn't really what's going on. Nonetheless, if you want to understand the reasoning of the early quantum founders when they developed this argument, it is essentially:
1.) Yes, the electron's wavelike behaviour means that while orbiting the nucleus it is like a wave traveling in a loop.
2.) What is being canceled is the electron's probability amplitude (as physicists later came to understand the quantum wavefunction). It doesn't really matter, though, what the de Broglie wave represents as long as you accept that it is necessarily associated with an electron traveling at a particular velocity. It's the same principle as strings on a guitar: when you pluck the string, only standing wave patterns are possible since both ends are fixed. The same goes for closed loops of string: they can only vibrate in standing wave patterns. This a general property of waves and is true for both vibrations on strings and de Broglie waves—whatever they may be. Hence, the reasoning was that if the electron is a wave traveling around the nucleus then only standing waves—discrete frequencies corresponding to discrete energy levels—are possible.

I should point out that even this isn't quite the same argument Bohr and company made—they were thinking initially in terms of quantizing angular momentum, not energy. But the reasoning is similar.

And it really should be emphasized that this reasoning, though compelling, really isn't correct. It doesn't capture the real nature of atomic orbitals due to taking the idea of orbiting electrons a bit too literally. Also, as you alluded to, one would expect an electron orbiting a nucleus to radiate energy, according to classical electrodynamics. This was another inconsistency of old quantum mechanics that wouldn't be properly fixed until Feynman and his colleagues put the finishing touches on quantum electrodynamics.
 
This a general property of waves and is true for both vibrations on strings and de Broglie waves—whatever they may be.

To clarify, standing waves are a property of de Broglie waves of bound particles. Only standing probability waves are allowed for a bound particle...like an electron orbital or one in a lattice like a metal.
 
Thank you guys
LastOneStanding said:
2.) What is being canceled is the electron's probability amplitude (as physicists later came to understand the quantum wavefunction). It doesn't really matter, though, what the de Broglie wave represents as long as you accept that it is necessarily associated with an electron traveling at a particular velocity.

Can you explain more please?
the electron orbiting around the nucleus is described by the wave function, if they have discrete orbits then how are they going to overlap and cancel?
 
It isn't the wavefunctions of different electrons interfering with each other, it's the wavefunction of a single electron interfering with itself as it loops around in a circle. It's exactly the same principle as standing waves in an air column or on a string with its ends fixed. You might find it helpful to review these topics.
 
Here is some nice introductory detail about my post #3

Modern conceptions and connections to the Heisenberg Uncertainty Principle
Immediately after Heisenberg discovered his uncertainty relation,[17] it was noted by Bohr that the existence of any sort of wave packet implies uncertainty in the wave frequency and wavelength, since a spread of frequencies is needed to create the packet itself.[18] In quantum mechanics, where all particle momenta are associated with waves, it is the formation of such a wave packet which localizes the wave, and thus the particle, in space. In states where a quantum mechanical particle is bound, it must be localized as a wave packet, and the existence of the packet and its minimum size implies a spread and minimal value in particle wavelength, and thus also momentum and energy.

http://en.wikipedia.org/wiki/Atomic_orbital

You may also want to read about particle confinement in a 'potential well' or 'box'...Wiki must have something...the idea is that an electron in free space has a probability density amplitude [wave] extending all over the place...it might be anywhere... BUT stick that electron in a [confinement] box and now the wave [de Broglie wave if you like] is stuck inside the box...It has zero chance of being found outside the box, so it's wavelength can be no larger than the box. Whereas before confinement it's wave consisted of all/any frequencies, when confined it's wavelengths are now only those of the box..and harmonics...it's 'localized'...a 'wavepacket' in Wiki terminology above.

same concept for a bound orbital electron...
 
"Over an infinite number of orbits, the out of phase waves would cancel each other out to zero amplitude, which implies that the particle cannot exist under such circumstances. Only if the waves traced out on all orbits overlap exactly, i.e. if the orbit is an integral number of wavelengths, is the situation a satisfactory one under which the particle can exist..."

This is starting to sound suspiciously like "Wheeler-Feynman Emitter-Absorber" Theory. 'N that's very interesting. W-F worked in an opaque universe. It had a following and then dropped off the radar screen. It is always an intriguing stop along the way.

Like Dirac's Sea. Even when it was "wrong", it was right. An area of research was found where it worked, in Cold Physics (See SciAm, Helium3 Superfluids, June, 1990, for ex.).

Very interesting.

CW
 
Thank you guys.
 

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