I Can an electron in the 1s orbital be indefinitely far from the nucleus?

Click For Summary
An electron in the 1s orbital of a hydrogen atom can theoretically be detected at any distance, including far from the nucleus, due to the non-zero radial probability density described by the Schrödinger equation. However, this detection requires a measurement that involves an energy exchange, which alters the electron's state. Without measurement, discussing the electron's position is considered meaningless in quantum mechanics, as it does not have a defined location until observed. The probability of finding the electron at a significant distance, such as the moon, is extremely low but not zero. Ultimately, the interpretation of the electron's position is contingent upon the act of measurement, which collapses its wave function to a specific state.
  • #31
jaumzaum said:
The situation I was describing was a bound electron in the position x=R going to the position x=4R by itself, while still bounded to the atom and in the 1s orbital with that characteristic energy of -13,6eV, no external work being made.

Even if we put aside all the valid issues that have been raised with this in terms of QM, and just consider a classical electron orbiting a classical proton with the given energy, and rule out the electron radiating any energy away (which classically it would, it would not remain in a stable orbit but spiral into the nucleus while emitting radiation), we still need to be clear about how classical orbits work.

If the electron's total orbital energy is -13.6eV, that's its total energy at all positions in its orbit. If its orbit is elliptical, so that its position is not always x=R, its total energy is still -13.6eV at all positions in its orbit. That means that it is impossible for the classical orbit to reach any position where the potential energy would be greater than or equal to -13.6eV, no matter how elliptical it gets, since, as you correctly believe, it is impossible for its orbital kinetic energy to be zero or negative in the classical model. That means that even the most elongated possible elliptical orbit would have to have its greatest distance from the proton less than the distance at which the potential energy is -13.6eV. In particular, it would be impossible for a classical elliptical orbit to reach x=4R, since the potential energy there is -6.8eV, which is greater than -13.6eV.

So another way of illustrating the issue you are having is that you are trying to combine two inconsistent views of the electron. You are trying to think of it like a classical particle with a position that is moving in some kind of orbit; but you are also trying to make its orbit have properties that a classical orbit simply cannot have. The "negative kinetic energy" at x=4R is simply a manifestation of this underlying problem. The way to resolve it is to stop trying to combine two inconsistent views of the electron: i.e., to understand that the QM model of the electron is simply incompatible with any way of thinking of the electron in a hydrogen atom as a classical particle "orbiting" the proton. That simply doesn't work.
 
Physics news on Phys.org
  • #32
If we detected an electron on the moon, how would we know which atom it "belonged to"? Does answering the question in OP require setting up a situation where we somehow know we are detecting an electron that belongs to the the atom in question and not to an atom that's part of the moon?
 
  • Like
Likes PeroK
  • #33
Dadface said:
Use a different convention, let the potential energy at infinity be equal to X

The potential energy zero point, as has already been said, has nothing to do with any of the issues being discussed in this thread. Only differences in potential energy matter for this discussion.
 
  • #34
PeterDonis said:
No, the zero point of potential energy makes no difference at all to the calculation, since the difference in potential energy between the unbound state and the bound state is the only thing that appears in the calculation.
Of course the choice of convention makes no difference to the calculations. I never said it did. But it can be confusing whatever convention is used. Some people may look at the chosen values and think they are absolute values.
 
  • #35
Dadface said:
Some people may look at the chosen values and think they are absolute values.

I don't think the OP is doing that since he has already agreed, in the post that you quoted, that changing the zero point of potential energy does not change anything in his calculations.
 
  • #36
PeterDonis said:
I don't think the OP is doing that since he has already agreed, in the post that you quoted, that changing the zero point of potential energy does not change anything in his calculations.
Yes I had another look at his post and I see what you mean. It seems that he forgot to take into account the fact that during a transition to the ground state the ionisation energy is radiated to the surroundings.
 
  • #37
Stephen Tashi said:
If we detected an electron on the moon, how would we know which atom it "belonged to"? Does answering the question in OP require setting up a situation where we somehow know we are detecting an electron that belongs to the the atom in question and not to an atom that's part of the moon?
I think a good reply to your question has been given by ZapperZ in post 17.
 
  • #38
Dadface said:
It seems that he forgot to take into account the fact that during a transition to the ground state the ionisation energy is radiated to the surroundings.

And that, in the reverse process, you have to add that energy back, yes.
 
  • #39
Dadface said:
I think a good reply to your question has been given by ZapperZ in post 17.

Post #17 contrasts emprical facts with theoretical probabilities. How does this answer the question of determining which atom an electron "belongs to". Are we saying that, empirically, a detected electron can be assigned to one particular atom?
 
  • #40
Stephen Tashi said:
Are we saying that, empirically, a detected electron can be assigned to one particular atom?

I don't think this is the case in general. In some particular situations it might be, for example, if we have a single atom confined in a trap. But nobody is going to build a single-atom trap the size of the Earth-Moon distance.
 
  • #41
Dadface said:
Does that mean if you have measured where it is, it's meaningful to ask where it is?

That is to some extent interpretation dependent. From the formalism Peter is of course correct. But let's take say Consistent History's (CH) interpretation. It says if you measure it then it has that position. But after the measurement its wavefunction will then start to spread. Again the formalism is silent on it's position, but the CH interpretation says it has a position somewhere between the areas of the wavefunction that are not zero. Over time my view of QM has changed quite a bit and my view now is like Peters - but interpretations have various takes on the issue. It's part of the reason Wienberg doesn't really like any interpretation. Maybe we just have to accept reality is just a mathematical description, and that's all you can really say, especially in areas beyond our direct experience from which our everyday view of the world is formed. As quantum effects leak into that world because of advancing technology expect to see some really weird things eg a well known one is the behaviour of liquid Helium.

Thanks
Bill
 
  • #42
bhobba said:
That is to some extent interpretation dependent.

While this is true, if we are going to go down that rabbit hole, it should be in a separate thread in the interpretations forum. :wink:

For this thread and this forum, "the formalism" is what we should use.
 
  • Like
Likes bhobba
  • #43
jaumzaum said:
The situation I was describing was a bound electron in the position x=R going to the position x=4R by itself, while still bounded to the atom and in the 1s orbital with that characteristic energy of -13,6eV, no external work being made. This is possible by my understanding right? At least it seems to be possible by Schrodinger equations if I'm not misunderstanding it.
You are misunderstanding it. The Schrodinger equation gives us the probability of finding the electron at a given position, but that doesn't imply that that it "goes to" that position from somewhere else - not even if we had previously measured the position and gotten a different result.

Let's say that we measure the position at time ##T_1## and find the electron at ##x=R##. Terrific... we can say that electron was at that position at the time of the measurement. But what about a moment later? We don't have a position measurement from a moment later so we cannot say that the electron is still at that position a moment later - no measurement, no position. Instead we use the Schrodinger equation to calculate the wave function of an electron at time ##t=T_1+\Delta{t}## given that it was at ##x=R## at time ##T_1##; we'll get a probability distribution that says it's likely to be fairly close to ##R## but continues to spread out over time.

So we measure it again, at time ##T_2##, and this time we find it at ##x=4R##. That doesn't mean that it moved from ##x=R## to ##x=4R## during the time interval between ##T_1## and ##T_2##; that interpretation would imply that it had a definite position between measurements, and it doesn't. Instead, the first measurement took the electron from the state "no definite position" to the state "at position ##R## at time ##T_1##"; and the second measurement took it from the state "no definite position" to the state "at position ##4R## at time ##T_2##". Both measurements necessarily changed the energy of the electron (although the total energy of the system including the measuring device is conserved).
 
Last edited:
  • Like
Likes Vanadium 50, PeterDonis and bhobba
  • #44
PeroK said:
This does not make sense in QM. If the electron is in a bound state it has no well-defined position. It cannot "go from A to B"; it cannot "be at A" and it cannot "move to B".

None of these classical descriptions make any sense in terms of the SDE, which describes the wavefunction.
An energy eigenstate describes a static configuration, i.e., nothing moves at all. It also implies that only conserved quantities, whose representing self-adjoint operators commute with the Hamiltonian, can have determined values. All others (among them position and momentum in the case of interacting particles) are indetermined and you can only give probabilities for finding a certain possible value when measuring one of these observables (Born's rule).
 

Similar threads

  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 18 ·
Replies
18
Views
2K
  • · Replies 21 ·
Replies
21
Views
2K
  • · Replies 47 ·
2
Replies
47
Views
4K
  • · Replies 36 ·
2
Replies
36
Views
7K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 36 ·
2
Replies
36
Views
5K