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

[jaumzaum]

Is it possible for an electron in the 1s orbital of an hydrogen atom to be indefinitely far from the nucleus in a given instant?

From the Schrodinger equation we can see that the radial probability is NEVER zero, so it would be possible to see an electron in the moon, for example.

But if I understood correctly, all orbitals have an specific energy associated to it. In the cas of the 1s Hydrogen orbital is -13,6eV. The average potential energy of an electron in that orbital is -2 *13,6 eV and the average kinetic energy is +13,6eV.

But consider an electron in the moon. The potential electric energy of that electron would be practically zero. So, for the total energy to be -13,6 eV, the kinetic energy would have to be -13,6eV, which is impossible.

What is wrong with this reasoning?

 

[anuttarasammyak]

1s state of hydrogen is ruined by position measurement of electron. You will find it around $\mathbf{r}$ with probability amplitude of $|\psi(\mathbf{r})|^2$ even on the moon, but thus found electron does not belong to 1s state anymore.

 

[jaumzaum]

@anuttarasammyak Why is that? I would suppose it has something to do with Heisenberg Uncertanity principle, but could you clarify your statements a little bit more?

 

[Nugatory]

Is it possible for an electron in the 1s orbital of an hydrogen atom to be indefinitely far from the nucleus in a given instant?

Your wording is imprecise in a way that doesn't matter with classical physics, but does with quantum mechanics. In quantum mechanics it makes no sense to talk about an electron being anywhere, whether near the nucleus or very far away. So instead we have to ask the more narrow question "Is it possible for an electron in the 1s orbital of an hydrogen atom to be detected indefinitely far away from the nucleus?" and the answer is "yes". If by "far away" we mean something like the distance to the moon, then that probability is extraordinarily small, but it's not zero so not impossible.

But consider an electron in the moon. The potential electric energy of that electron would be practically zero. So, for the total energy to be -13,6 eV, the kinetic energy would have to be -13,6eV, which is impossible.
What is wrong with this reasoning?

Measuring the position of the electron requires an exchange of energy between the electron and the measuring apparatus. The energy of the entire system consisting of the nucleus, the electron, and the measuring apparatus is conserved; in the case you're describing we might say that the apparatus has lost 13.6 ev, the electron is at zero, and the energy books still balance. (But be aware that this is a semi-classical explanation, something that you will have to unlearn when you study QM seriously).

 

[jaumzaum]

Your wording is imprecise in a way that doesn't matter with classical physics, but does with quantum mechanics. In quantum mechanics it makes no sense to talk about an electron being anywhere, whether near the nucleus or very far away. So instead we have to ask the more narrow question "Is it possible for an electron in the 1s orbital of an hydrogen atom to be detected indefinitely far away from the nucleus?" and the answer is "yes". If by "far away" we mean something like the distance to the moon, then that probability is extraordinarily small, but it's not zero so not impossible.

Measuring the position of the electron requires an exchange of energy between the electron and the measuring apparatus. The energy of the entire system consisting of the nucleus, the electron, and the measuring apparatus is conserved; in the case you're describing we might say that the apparatus has lost 13.6 ev, the electron is at zero, and the energy books still balance. (But be aware that this is a semi-classical explanation, something that you will have to unlearn when you study QM seriously).

Thanks for the answer @Nugatory.
You said it's possible for the electron of the 1s orbital to be detected on the moon, but you also said that for that to be true we first need to transfer energy from some apparatus to the electron. So I would say that "naturally" the electron cannot go to the moon, only if we try to measure its position. But then you could argue that it does not matter where the electron is if we can't measure and therefore are not sure of its position. But is "does not matter" the same as "does not occur"? At that given instant, we measured the position of the electron and we found it was in the moon. What about if we haven't measured anything? Where would the electron be? What would be the answer to that question?

Would it be like: "We cannot be sure, but it would be somewhere, and could indeed be in the moon" or more like: "It would be nowhere"

Also, why you said that In quantum mechanics it makes no sense to talk about an electron being anywhere, whether near the nucleus or very far away ?

Thank you

 

[PeterDonis]

What about if we haven't measured anything? Where would the electron be?

If you haven't measured the electron's position, asking where it is is meaningless.

Would it be like: "We cannot be sure, but it would be somewhere, and could indeed be in the moon" or more like: "It would be nowhere"

Neither. It would be like "asking where the electron is is a meaningless question". "Meaningless" in the same sense as, for example, asking what color the number three is.

 

[HomogenousCow]

In an empty universe save for an atomic nuclei and one electron, yes it is possible to find the electron anywhere in the universe. However in practice this is meaningless, the probability density of the wavefunction even a few bohrs away from the nuclei is orders of magnitude below unity.

 

[Nugatory]

What about if we haven't measured anything? Where would the electron be? What would be the answer to that question?
Would it be like: "We cannot be sure, but it would be somewhere, and could indeed be in the moon" or more like: "It would be nowhere"

It’s not somewhere but we don’t know where. It has no position. It’s like asking where my lap is when I’m standing up, or where my fist is when my hand is open.

For a more stark example, consider a photon after it has passed through a horizontal polarizing filter: Quantum mechanics says that if we measure the polarization of that photon there is a 100% certainty that it will be horizontal. It is natural to think this means that the photon is horizontally polarized whether we measure it or not (and Einstein made pretty much that argument back in 1935). However, it turns out that there are subtle differences between “The particle is horizontally polarized” and “The particle has no polarization unless and until we measure it; if we do the result of the measurement will be horizontal polarization”. These differences can be detected in experiments; the experiments have been done; and they confirm the quantum mechanical model. For more about this, you can google for “Bell’s Theorem”

 

[PeroK]

Is it possible for an electron in the 1s orbital of an hydrogen atom to be indefinitely far from the nucleus in a given instant?

From the Schrodinger equation we can see that the radial probability is NEVER zero, so it would be possible to see an electron in the moon, for example.

One exercise might be for you to describe broadly an experiment where the outcome was "the position of the electron in a hydrogen atom was measured and found to be on the moon". What sort of apparatus would you need and how would you conduct the experiment?

 

[vanhees71]

If you haven't measured the electron's position, asking where it is is meaningless.

It's not meaningless, but the only way to answer the question is to measure it.

If you say the electron is "prepared" as an electron in a hydrogen atom in its ground state, all you can say is the probability to find this electron at a given place when you try to detect it at this place. This probability is $\mathrm{d}^3 r |\psi(\vec{r})|^2$. This probability goes exponentially to zero as a function of the distance from the hydrogen atom's location. Thus there's non-zero probability to detect it anywhere, including at far distances from the atom.

Of course when measuring the electron's position, i.e., detecting it in some small region defined by the position of the detector, you'll usually destroy the hydrogen atom, such that the electron is in another state after the measurement.

 

[PeterDonis]

It's not meaningless, but the only way to answer the question is to measure it.

If the only way to answer the question is to measure it, then asking the question without measuring it is meaningless.

If you say the electron is "prepared" as an electron in a hydrogen atom in its ground state, all you can say is the probability to find this electron at a given place when you try to detect it at this place.

Yes, when you try to detect it at this place. But if you are not trying to detect it anywhere, you can't say it has a position at all. The interpretation of the squared modulus of the spatial wave function as the probability of the electron having a particular position is only meaningful if you are going to try to detect its position. The fact that the OP was trying to apply this interpretation in a scenario where no position measurement is going to be made is the root of the problem the OP is having with his reasoning.

 

[Dadface]

I think some of the answers need clarifying:

If you haven't measured the electron's position, asking where it is is meaningless

Does that mean if you have measured where it is, it's meaningful to ask where it is?

The interpretation of the squared modulus of the spatial wave function as the probability of the electron having a particular position is only meaningful if you are going to try to detect its position.

So it seems that the interpretation is meaningful to Bill, who intends to detect the position but not to Ben who doesn't intend to detect the position.
If Bill changes his mind and decides not to attempt detecting the position does that previous meaningful interpretation become meaningless?

 

[Nugatory]

Does that mean if you have measured where it is, it's meaningful to ask where it is?

No, but it is meaningful to talk about the position you found when you measured it. But that's where it was then, when you measured it, not where it is now, some time after we've measured it.

So it seems that the interpretation is meaningful to Bill, who intends to detect the position but not to Ben who doesn't intend to detect the position. If Bill changes his mind and decides not to attempt detecting the position does that previous meaningful interpretation become meaningless?

The squared modulus is the probability of getting a partcular result if we measure. That's meaningful, just as it is meaningful to say "if I toss I toss an honest coin it ha a 50% probability of landing heads-up"; that's true whether anyone wants to toss the coin or not. In the case of the coin it make no sense to talk about whether it has landed heads or tails before we toss it. That doesn't surprise anyone; the heads/tails result doesn't exist yet, just as my lap doesn't exist until I sit down.

What is surprising to people who first encounter quantum mechanics is that the position of a quantum particle works the same way. We can calculate the probability of the particle having a particular position when and if we measure the position (just as I can calculate the probability of getting a particular result from the toss of a coin or the roll of a die) but that probability is all there is until we make the measurement (just as the coin isn't heads or tails unless or until we toss it).

 

[PeterDonis]

So it seems that the interpretation is meaningful to Bill, who intends to detect the position but not to Ben who doesn't intend to detect the position.

It's not a matter of intent. It's a matter of what measurement you are actually analyzing. Either you're analyzing a position measurement, or you're not. The analysis of a particular measurement is a perfectly objective piece of math that doesn't depend on anyone's intentions.

 

[Dadface]

No, but it is meaningful to talk about the position you found when you measured it. But that's where it was then, when you measured it, not where it is now, some time after we've measured it.
The squared modulus is the probability of getting a partcular result if we measure. That's meaningful, just as it is meaningful to say "if I toss I toss an honest coin it ha a 50% probability of landing heads-up"; that's true whether anyone wants to toss the coin or not. In the case of the coin it make no sense to talk about whether it has landed heads or tails before we toss it. That doesn't surprise anyone; the heads/tails result doesn't exist yet, just as my lap doesn't exist until I sit down.

What is surprising to people who first encounter quantum mechanics is that the position of a quantum particle works the same way. We can calculate the probability of the particle having a particular position when and if we measure the position (just as I can calculate the probability of getting a particular result from the toss of a coin or the roll of a die) but that probability is all there is until we make the measurement (just as the coin isn't heads or tails unless or until we toss it).

Thank you but I already have a basic idea of what the Born rule is about. I was trying to point out that some of the comments made were lacking in clarity and my questions were attempts to show how some people might interpret those comments. From your response it seems that my comments in post 12 are lacking in clarity. I hope I have compensated for that in my notes here.
I think, perhaps, your second sentence above, which I have highlighted in bold, is lacking in clarity. What do you mean by "where it is now"? It seems that you are saying or implying that it is meaningful to describe that particles can have a position, where it is now, when not being measured. But I know what you really mean. I think.

 

[Dadface]

It's not a matter of intent. It's a matter of what measurement you are actually analyzing. Either you're analyzing a position measurement, or you're not. The analysis of a particular measurement is a perfectly objective piece of math that doesn't depend on anyone's intentions.

Of course. I was trying to point out that the way you expressed your comments laid them open to strange interpretations. Take another look at what you wrote.

 

[ZapperZ]

From the Schrodinger equation we can see that the radial probability is NEVER zero, so it would be possible to see an electron in the moon, for example.

We get this type of question very often, i.e. people seem to not know when something is highly unlikely to occur. If you look at statistical physics, there is a non-zero probability that a vase that has been broken into a thousand pieces, will reassemble itself back into the original vase when you throw it onto the floor.

So when was the last time you saw that happening? In fact, in reality, we take it that such a thing is practically impossible to occur!

Now, when you calculate the forces acting on, say, a bowling ball, do you also account for all the gravitational forces due to the people surrounding it, and include the gravitational forces from the moon, from the stars int he alpha centauri clusters, etc... to be able to accurately describe the motion of the bowling ball on its way to the pins? Do you think these forces are necessary in calculating the stability of a bridge that we often build here on earth? How many structure catastrophes have been attributed to us not including the gravitational forces from other stars around us?

So now, at what point do you think the wavefunction for the 1s orbital in the hydrogen atom is no longer realistically significant that is might as well be considered to be non-existent? There is a difference between something to be theoretically non-zero versus something to be realistically irrelevant, and you need to understand the difference here. Otherwise, every single physical phenomenon that you see around you are all "approximations" that did not include an infinite other effects.

Please note that astrologers incorrectly think that stars and planets that seem to align in the sky have effects on human activities here on earth. They forget that these stars and planets are in a 3D space, and that a star that may look to be behind another star can be extremely far away from the one in front of it. In fact, there are many cases where a star on the opposite side of the sky may be closer to the first star than the one directly behind it. So if you consider gravitational forces alone, an alignment of stars in the sky means nothing. Once again, there is a lack of realistic and quantitative understanding of the situation here.

Zz.

 

[PeterDonis]

I was trying to point out that the way you expressed your comments laid them open to strange interpretations.

I'll leave it to the OP of this thread to ask questions if he thinks he needs clarification. You appear to understand what is meant.

 

[jaumzaum]

We get this type of question very often, i.e. people seem to not know when something is highly unlikely to occur. If you look at statistical physics, there is a non-zero probability that a vase that has been broken into a thousand pieces, will reassemble itself back into the original vase when you throw it onto the floor.

Thanks @ZapperZ, but my goal here was not to analyze if a particular event is probable or not, I think some people got me wrong. The moon thing was only an example, I probably should have gave another one. My main question here is why the Kinetic energy seems to be negative there, and in any other place where the potential energy > -13,6eV, as all these places are allowed to occur.

As some people said, if I try to find the electron in the moon I will need to make an experiment in which I would need to give him at least 13,6eV so that I could detect his position there. As moon is making people say it's not realistic, let's consider a more realistic situation:

Consider an Hydrogen atom, and an experiment that tries to find the position of the electron in this Hydrogen atom. The Hydrogen atom is in a Vacuum and is the only atom in that recipient. Given that the Borh radius is equal to $R=5.291 \times 10^{ -11}m$, consider that we could find the electron at 4R from the nucleus. The potential energy associated to that position is $-27,2eV/4=-6,8eV$. If we had conservation of energy (i.e. no work being made on the electron by the experiment) we would say that the kinetic energy would also be -6,8eV. We know negative kinetic energy is impossible. So, by what I understood from @Nugatory, to detect this electron, we NEED to spend at least 6,8eV of work. Is this correct?

My main issue is with the Schrodinger equations. I don't know how these equations are proved, but do they account for this strange phenomenon? The fact that we actually need to spend energy to dettect the position of the electron, because if it doesn't, any distance that is greater than $2 R$ would be impossible.

Also, in the experiment I mentioned, what ensure me that the electron was 4R? If I need to gave the electron some energy to detect his position, it could be in a different place and use that energy to "move" to a new place, so that the previous location was different.

I'm having trouble to understand this.
Can you guys help me?

 

[PeroK]

Thanks @ZapperZ, but my goal here was not to analyze if a particular event is probable or not, I think some people got me wrong. The moon thing was only an example, I probably should have gave another one. My main question here is why the Kinetic energy seems to be negative there, and in any other place where the potential energy > -13,6eV, as all these places are allowed to occur.

As some people said, if I try to find the electron in the moon I will need to make an experiment in which I would need to give him at least 13,6eV so that I could detect his position there. As moon is making people say it's not realistic, let's consider a more realistic situation:

Consider an Hydrogen atom, and an experiment that tries to find the position of the electron in this Hydrogen atom. The Hydrogen atom is in a Vacuum and is the only atom in that recipient. Given that the Borh radius is equal to $R=5.291 \times 10^{ -11}m$, consider that we could find the electron at 4R from the nucleus. The potential energy associated to that position is $-27,2eV/4=-6,8eV$. If we had conservation of energy (i.e. no work being made on the electron by the experiment) we would say that the kinetic energy would also be -6,8eV. We know negative kinetic energy is impossible. So, by what I understood from @Nugatory, to detect this electron, we NEED to spend at least 6,8eV of work. Is this correct?

My main issue is with the Schrodinger equations. I don't know how these equations are proved, but do they account for this strange phenomenon? The fact that we actually need to spend energy to dettect the position of the electron, because if it doesn't, any distance that is greater than $2 R$ would be impossible.

Also, in the experiment I mentioned, what ensure me that the electron was 4R? If I need to gave the electron some energy to detect his position, it could be in a different place and use that energy to "move" to a new place, so that the previous location was different.

I'm having trouble to understand this.
Can you guys help me?

In general, the expected value of the potential energy in a hydrogen atom is twice the energy level and the expected value of the KE is equal in magnitude to the energy level. E.g. for the ground state we have:
$$E = -13.6eV, \ \ \langle PE \rangle = -27.2eV, \ \ \langle KE \rangle = +13.6eV$$
But, energy and position are incompatible observables. If you measure the position of the electron you destroy the previous energy eigenstate (ground state in this case). There's no sense in which the electron "really was" at that position while the atom was in in the ground state. According to orthodox QM, the measured value of position required a position measurement which inevitably destroyed the ground state.

You can never say that a) the electron is at radius $r$ and b) the atom is in the ground state. If the atom is in the ground state, then the electron does not have a well-define position. And, if you measure the position of the electron, then the atom is no longer in an energy eigentsate.

 

[Dadface]

Jaumzaun, Because we can deal with differences in potential energy only, not absolute values, we choose two things:

1. A particle separation
2. A value of potential energy for the chosen separation.

Any choices we make would be somewhat arbitrary but the most useful and logical choices, which are widely accepted, are infinity for separation and zero for potential energy.

The potential energy at infinity has its maximum value and is zero by convention only. If the particles move to the Bohr separation the potential energy reduces by 27.2eV approximately, half of which is radiated to the surroundings.

 

[Nugatory]

Any choices we make would be somewhat arbitrary...

That is true, but how we make that arbitrary choice should neither affect the kinetic energy nor allow the negative kinetic energy that appears in @jaumzaun's original post and restatement in #19.

This thread has several explanations of this apparent paradox. Although it's not obvious just from reading them, they're different ways of saying the same thing: an interaction is required to produce a position measurement; that interaction necessarily changes the state; the negative kinetic energy is the result of erroneously ignoring that state change.

 

[Dadface]

jaumzaun calculated the kinetic energy to be negative because he thought that the potential energy at infinity really is zero instead of being at a maximum value. When he gets used to the potential energy convention he should calculate that when the atom moves to a ground separation at the Bohr radius the potential energy becomes - 27.2 eV half of this energy being radiated away and the other half being retained by the atom. As usual energy is conserved.

 

[jaumzaum]

jaumzaun calculated the kinetic energy to be negative because he thought that the potential energy at infinity really is zero instead of being at a maximum value. When he gets used to the potential energy convention he should calculate that when the atom moves to a ground separation at the Bohr radius the potential energy becomes - 27.2 eV half of this energy being radiated away and the other half being retained by the atom. As usual energy is conserved.

I don't think I messed up with the frames. I stick with my frame of reference, in which I consider the potential energy to be zero at infinity. Considering the same frame of reference, without changing it, I know that at Bohr radius the potential energy would be -27,2eV and the kinetic energy +13,6 eV. The total energy would be -13,6 eV, and that's no problem with that being negative, I know it's based on the frame of reference, the kinetic energy is the one that cannot be negative because it doesn't depend on any inertial reference frame. When I move the electron, keeping it in the 1s orbital, to 4R, the potential energy, considering the same frame of reference, becames -6,8eV. By conservation of energy the kinetic energy must be -6,8eV for the sum to be -13,6eV.

If you consider the potential energy at infinity to any other value, say +100eV, then the potential energy at Bohr radius would be 72,8eV, the kinetic energy would remain 13,6eV and the total energy would be 86,4 eV. At 4R, the potential energy would be 93,2eV and the kinetic energy should be -6,8eV for the sum to be 86,4eV. Kinetic energy doesn't depend of the frame of reference chosen (as it shouldn't), and would be negative in both scenarios in the problem.

My question actually is why is this paradox happening. But I think, as @Nugatory said, the measurement is changing the previous state of the atom, and hat is giving the electron some amount of energy

 

[PeterDonis]

jaumzaun calculated the kinetic energy to be negative because he thought that the potential energy at infinity really is zero instead of being at a maximum value.

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.

The mistake made in the OP was to implicitly assume that the unbound state and the bound state have the same total energy. They don't. The unbound state (if we assume that the proton and electron are at rest relative to each other) has energy 13.6 eV higher than the bound state.

When I move the electron, keeping it in the 1s orbital, to 4R, the potential energy, considering the same frame of reference, becames -6,8eV. By conservation of energy the kinetic energy must be -6,8eV for the sum to be -13,6eV.

Wrong. By "moving the electron" (what you are really describing is changing the energy level the electron is in, which is not the same as "moving" it, but we don't need to delve into all those complications right now) you added energy to the system. So the system's total energy is now -3.4 eV (potential energy -6.8 eV, kinetic energy +3.4 eV) instead of -13.6 eV.

Similarly, if you removed the electron from the atom altogether, ending up with an unbound proton and electron at rest relative to each other, you would have to add 13.6 eV of energy to the system, so its total energy would be zero (zero potential plus zero kinetic).

 

[jaumzaum]

Wrong. By "moving the electron" (what you are really describing is changing the energy level the electron is in, which is not the same as "moving" it, but we don't need to delve into all those complications right now) you added energy to the system. So the system's total energy is now -3.4 eV (potential energy -6.8 eV, kinetic energy +3.4 eV) instead of -13.6 eV.

Sorry, I didn't use the right words. 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.

 

[PeroK]

Sorry, I didn't use the right words. 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.

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.

 

[ZapperZ]

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.

I've been thinking of the same thing and have been scratching my head at all this. Look at the radial part of the wavefunction, even for the 1s orbital, r=4R is INCLUDED in that "spread"! The electron distribution includes that distance from the nucleus.

So this whole scenario makes no sense.

Zz.

 

[PeterDonis]

The situation I was describing was a bound electron in the position x=R going to the position x=4R by itself

This is meaningless if you are not measuring the position. If the electron is in the ground state energy level, the only thing you can say about it is that its energy is -13.6eV. You cannot say where it is, and you cannot talk about it "moving". You can only meaningfully talk about its energy.

This is possible by my understanding right?

No. See above.

it seems to be possible by Schrodinger equations

Why?

 

[Dadface]

I don't think I messed up with the frames. I stick with my frame of reference, in which I consider the potential energy to be zero at infinity. Considering the same frame of reference, without changing it, I know that at Bohr radius the potential energy would be -27,2eV and the kinetic energy +13,6 eV. The total energy would be -13,6 eV, and that's no problem with that being negative, I know it's based on the frame of reference, the kinetic energy is the one that cannot be negative because it doesn't depend on any inertial reference frame. When I move the electron, keeping it in the 1s orbital, to 4R, the potential energy, considering the same frame of reference, becames -6,8eV. By conservation of energy the kinetic energy must be -6,8eV for the sum to be -13,6eV.

If you consider the potential energy at infinity to any other value, say +100eV, then the potential energy at Bohr radius would be 72,8eV, the kinetic energy would remain 13,6eV and the total energy would be 86,4 eV. At 4R, the potential energy would be 93,2eV and the kinetic energy should be -6,8eV for the sum to be 86,4eV. Kinetic energy doesn't depend of the frame of reference chosen (as it shouldn't), and would be negative in both scenarios in the problem.

My question actually is why is this paradox happening. But I think, as @Nugatory said, the measurement is changing the previous state of the atom, and hat is giving the electron some amount of energy

Use a different convention, let the potential energy at infinity be equal to X. We can write energy equations for the particles at infinity and at the ground state.

Infinity P.E. = X , K.E = 0
Bohr radius P.E. = X -27.2 ,K.E = +13.6, Energy lost = 13.6

Whatever convention is used energy is conserved. I think a problem is that when some people look at the convention of setting X equal to zero they think there is no potential energy at all. In this example the maximum at an infinite separation.

 

[PeterDonis]

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.

 

[Stephen Tashi]

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?

 

[PeterDonis]

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.

 

[Dadface]

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.

 

[PeterDonis]

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.