Why doesn't the electron stick on the proton in the L=0 states?

In summary, the conversation discusses the L=0 states in Hydrogen atom and the maximum probability for the electron to exist at r=0, the location of the proton. It is logical for the probability density to be suppressed near the origin and peak at a distance from the origin. The conversation also touches on the uncertainty principle and its role in preventing the electron from falling on the proton, as well as the four forces and their relationship to the uncertainty principle.
  • #1
Abu Abdallah
26
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In The L=0 states in Hydrogen atom, the electron has a maximum probablity to exist at r=0 ( at the location of the proton ) Is this logical? What prevents the electron then from falling on the proton? There is no angular momentum to prevent that !
 
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  • #2
Please read our FAQ in the General Physics forum.

Zz.
 
  • #3
Abu Abdallah said:
In The L=0 states in Hydrogen atom, the electron has a maximum probablity to exist at r=0 ( at the location of the proton ) Is this logical? What prevents the electron then from falling on the proton? There is no angular momentum to prevent that !

The probability density is [itex] \vert \psi(\vec r,t) \vert^2 dV [/itex], not [itex] \vert \psi(\vec r,t) \vert^2 [/itex]. Because of the r^2 in the volume element dV you actually find that the probaility density is suppressed near the origin! The probability density actually peaks at soem distance from the orgin (to find the actual distance, just find the value of r for which [itex] \vert \psi(\vec r,t) \vert^2 r^2 [/itex] is maximum).

Patrick
 
  • #4
Abu Abdallah said:
In The L=0 states in Hydrogen atom, the electron has a maximum probablity to exist at r=0 ( at the location of the proton ) Is this logical? What prevents the electron then from falling on the proton? There is no angular momentum to prevent that !
Patrick is right: the probability density peaks at the Bohr radius.

What prevents the fuzzy position of the electron relative to the proton to get less fuzzy is the fuzziness of the electron's momentum relative to the proton. If a fuzzy position propagates (in time if not in space) with a sharp momentum, the measure of its fuzziness (e.g., the standard deviation of the corresponding probability distribution) remains unchanged. If it propagates with a fuzzy momentum, it becomes fuzzier. As a result you have a stable equilibrium between two tendencies of the internal relative position of atomic hydrogen:
  • the tendency to become less fuzzy due to the electric attraction between proton and electron, and
  • the tendency to become more fuzzy due to the fuzziness of the momentum mandated by the uncertainty principle (which is a mistranslation of the German original Unschärfe Prinzip, the correct translation of which is "fuzziness principle").
 
  • #5
nrqed said:
The probability density is [itex] \vert \psi(\vec r,t) \vert^2 dV [/itex], not [itex] \vert \psi(\vec r,t) \vert^2 [/itex]. Because of the r^2 in the volume element dV you actually find that the probaility density is suppressed near the origin!
Patrick

nrqed, calculate the probabiliyt density with respect to another frame (whose origin not at the proton) and you will find that [itex] \vert \psi(\vec r,t) \vert^2 [/itex] has a maximum at the proton while the volume element is not zero there.


Koantum, I know the force that attracts the electron to the proton : it's
Coulomb's force. But what is this force that tries to impose the
Fuzziness principle ? which one of the four force is it?
 
  • #6
Abu Abdallah said:
nrqed, calculate the probabiliyt density with respect to another frame (whose origin not at the proton) and you will find that [itex] \vert \psi(\vec r,t) \vert^2 [/itex] has a maximum at the proton while the volume element is not zero there.

Have you done this?

Zz.
 
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  • #7
Abu Abdallah said:
nrqed, calculate the probabiliyt density with respect to another frame (whose origin not at the proton) and you will find that [itex] \vert \psi(\vec r,t) \vert^2 [/itex] has a maximum at the proton while the volume element is not zero there.

I'd like to see your calculation for this.

Notice that the full Schrodinger equation involves both the proton and the electron kinetic energies. What one does is the standard treatment is to decouple the equation in the center of mass frame from the overall constant velocity translation of the center of mass frame (this is pretty much the same as in classical mechanics). The meaning of the Schrodinger equation for hydrogen that is usually given in textbooks is only the part hat corresponds to the equation in the center of mass frame (and the mass that is there should really be the reduced mass, not the electron mass). So, by definition, if you change frame, this equation will not change, you will just be adding momentum to the center of mass.
See any textbook, for example Griffiths (Intro to quantum mechanics, section 5.1 and especially Problem 5.1 which shows all the steps)
 
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  • #8
Abu Abdallah said:
Koantum, I know the force that attracts the electron to the proton : it's Coulomb's force. But what is this force that tries to impose the Fuzziness principle ? which one of the four force is it?
None of the four forces. You know the first force from classical physics. You don’t know the second because it's purely quantum. Besides, it's not a force but a tendency that every relative position has: to grow fuzzier unless counterbalanced by an attractive force that gives rise to the opposite tendency. Nothing needs to act on a fuzzy relative position to make it grow fuzzier. This tendency, encapsulated by the uncertainty principle (to stick to the conventional name) is part and parcel of the quantum behavior of matter.
 

FAQ: Why doesn't the electron stick on the proton in the L=0 states?

Why doesn't the electron stick to the proton in the L=0 state?

In the L=0 state, the electron and proton are in a state of equilibrium where they are attracted to each other by the electromagnetic force. However, the electron also experiences repulsion from the proton due to its own negative charge. This balance between attraction and repulsion prevents the electron from sticking onto the proton.

Is there a specific reason for the electron not sticking to the proton in the L=0 state?

The lack of electron sticking to the proton in the L=0 state is a result of the quantum mechanical properties of the electron and the proton. The electron behaves as a wave and can only exist in certain discrete energy levels, while the proton has a fixed positive charge. These properties do not allow for the electron to stick onto the proton in the L=0 state.

Can the electron stick to the proton in any other states?

Yes, in higher energy states the electron can stick to the proton due to the increased attraction between the two particles. In these states, the electron's wave function overlaps with the proton's nucleus, allowing for a stable binding between the two particles.

How does the L=0 state differ from other states in terms of electron sticking to the proton?

In the L=0 state, the electron is in the lowest energy level and is not able to stick to the proton due to the balance of forces mentioned earlier. In higher energy states, the electron is able to stick to the proton as it has more energy and can overcome the repulsion from the proton's positive charge.

Can the electron ever stick to the proton in the L=0 state?

No, the electron cannot stick to the proton in the L=0 state as it would require a change in the quantum mechanical properties of the electron and the proton. This is not possible as these properties are fundamental to the nature of these particles and cannot be changed without external forces acting on them.

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