Field lines of electrons in an atomic orbital

[xortdsc]

Hello,
I have a rather conceptual question I couldn't really find an answer to yet.
The electric field lines of an isolated resting electron would simply point from everywhere in space towards the position the electron is located in, with a length inversely proportional to the squared distance to that electron position.
What I'm wondering now is how electric field lines would look like for an electron in an hydrogen atom. So if one would subtract the influence of the positive proton on the actual hydrogen field lines, would the (electric) field lines of an electron in an 1s orbital still have that same shape as an isolated electron (resting at the position of the nucleus) would have ? How about the 2s orbital ? Does the length of the field lines go to zero at the radial node ? How about p-orbitals ? Where do the field lines even point to in this case ? Or is the electric field simply the integral of the isolated electron field lines and the electron density distribution of the orbital (so the field lines do not go to zero at the nodes of the orbital) ? I'm struggling with picturing this conceptually. Or is my question ill-posed ? Maybe someone could help me out here ?
Thanks in advance :)

 

[mfb]

It is problematic to mix classical concepts (field lines) with quantum mechanics (electron orbitals). For some setups, you can treat the orbital as a charge distribution and calculate field lines just based on that. For distances large compared to the orbital you get the same (independent of the type), close to the orbitals you get something different. Close to the nucleus, the field (from the electron alone) would drop to zero, indeed.

 

[dauto]

Forget about field lines. Learn about the electromagnetic field

 

[xortdsc]

Forget about field lines. Learn about the electromagnetic field

Well, but isn't the electromagnetic field a set of 2 3d-vector-fields (the E and B fields) ? so just considering the E field the vectors must point in some direction and have some magnitude. That's what I mean by "electric field lines" - maybe that was confusing. So maybe it makes more sense to say: How does the E-field look like for those orbitals considering the electron alone ?

 

[xortdsc]

It is problematic to mix classical concepts (field lines) with quantum mechanics (electron orbitals). For some setups, you can treat the orbital as a charge distribution and calculate field lines just based on that. For distances large compared to the orbital you get the same (independent of the type), close to the orbitals you get something different. Close to the nucleus, the field (from the electron alone) would drop to zero, indeed.

so there is no way to compute a electro-magnetic (near) field from those orbitals ?

 

[sophiecentaur]

so there is no way to compute a electro-magnetic (near) field from those orbitals ?

That would be difficult and not a valid concept when you think that the electron is not actually 'anywhere' in particular when it's in a bound state. It behaves like a standing wave in many respects so how could you assign it a position and direction in order to treat it the same as an electron in free space, as in a CRT?
You have to break free from the concrete thinking of classical Physics.

PS They are called "orbitals" and not 'orbits', for a good reason. Electrons are not in orbit.

 

[xortdsc]

That would be difficult and not a valid concept when you think that the electron is not actually 'anywhere' in particular when it's in a bound state. It behaves like a standing wave in many respects so how could you assign it a position and direction in order to treat it the same as an electron in free space, as in a CRT?
You have to break free from the concrete thinking of classical Physics.

PS They are called "orbitals" and not 'orbits', for a good reason. Electrons are not in orbit.

I'm aware that they're spread out over the whole orbital and not just an orbiting point, but i thought it should still be possible to compute a concrete electrical field. Maybe it could apply that the resulting electrical field would be sort of the convolution of the electron density function of the orbital with the electric field of a stationary electron (the inverse square law). So in effect it would be like chopping up the electron's charge into infinitesimal parts, distribute them over the orbital accourding to its electron density function and compute the electrical field for each of these (infinite) parts of charge and accumulate over all of them.
That was just an idea, though. Is this so far off conceptually ? Sounds fairly reasonable, I think. :)

@mfb: after re-reading your post, it sounds like you suggest something similar (i think i misunderstood at first - so don't mind my first reply to your post ;) but maybe you can approve/disapprove if the way I explained in this post would yield what you meant by "distributing the charge"), but can you clarify conceptually for which setups it would work and in which cases it'll break down ?

Thanks for all the input so far :)

 

[sophiecentaur]

I'm aware that they're spread out over the whole orbital and not just an orbiting point, but i thought it should still be possible to compute a concrete electrical field. Maybe it could apply that the resulting electrical field would be sort of the convolution of the electron density function of the orbital with the electric field of a stationary electron (the inverse square law). So in effect it would be like chopping up the electron's charge into infinitesimal parts, distribute them over the orbital accourding to its electron density function and compute the electrical field for each of these (infinite) parts of charge and accumulate over all of them.
That was just an idea, though. Is this so far off conceptually ? Sounds fairly reasonable, I think. :)

@mfb: after re-reading your post, it sounds like you suggest something similar (i think i misunderstood at first - so don't mind my first reply to your post ;) but maybe you can approve/disapprove if the way I explained in this post would yield what you meant by "distributing the charge"), but can you clarify conceptually for which setups it would work and in which cases it'll break down ?

Thanks for all the input so far :)

But you would need to do a vectorial addition of all those constituent fields. This would have to assume that they were all in every spot at the same time? (Are they?) I can't see how any calculation you might do would be likely to yield a meaningful result. You seem to be after a classical description of something that is just not classical. Why not make the leap and accept that QM rules in these circumstances? And, whilst you are about it, what about the field due to the Nucleus? The electrons are only doing what they do in atoms because of the presence of the other charges in the atom.

 

[mfb]

The concept of the quantum defects treats all electrons (apart from the one where the energy is calculated) as effective contribution to the potential the electron sees, and as far as I know the Hartree-Fock method uses a similar approach for each electron.

it sounds like you suggest something similar ([...] but maybe you can approve/disapprove if the way I explained in this post would yield what you meant by "distributing the charge")

I don't understand your suggestion.

but can you clarify conceptually for which setups it would work and in which cases it'll break down ?

I don't know, but maybe the two links above can help.

 

[xortdsc]

But you would need to do a vectorial addition of all those constituent fields. This would have to assume that they were all in every spot at the same time? (Are they?) I can't see how any calculation you might do would be likely to yield a meaningful result.

Well, yes there is a vectorial addition implicitly. I'd mean to convolve the electric field of a single free/isolated electron (3-vector field) with the normalized electron density function of the orbital (scalar field whose integral is unity so total charge will be conserved) which would yield a new convolved 3-vector field which is ought to be the electric field of the electron in the orbital.
And yes, it would assume the electron is everywhere in that orbital at the same time, which I think is how it is supposed to be interpreted (no "moving electron" but a "stretched out electron" instead). At least that what I keep reading everywhere :)

The concept of the quantum defects treats all electrons (apart from the one where the energy is calculated) as effective contribution to the potential the electron sees, and as far as I know the Hartree-Fock method uses a similar approach for each electron.
I don't know, but maybe the two links above can help.

This seems to consider multi-electron atoms which are much more complicated. I'm just talking about a single electron hydrogen atom.

 

[mfb]

I'm just talking about a single electron hydrogen atom.

You are talking about the electric field of a single electron. Where is the point in knowing that, if you don't want to consider its effect on other particles? Other electrons in the atom are those particles...

 

[sophiecentaur]

xortdsc
The only time that the field around an electron is calculable (or meaningful) is when it is isolated. In the presence of just a single proton (bound in a Hydrogen atom) it (if you wanted to consider it) will be severely modified and asymmetrical.
In a similar, very classical vein, you must have seen the magnetic field of a bar magnet and then seen how it changes when another magnet is brought close to it. In that situation it's the field of both magnets that is seen and not just that of the original magnet.
Why not just accept that it's an invalid notion with little or no relevance and not any more knowable than the actual 'position' of the electron at any time.

 

[xortdsc]

I'm sorry to insist that something sounds wrong to me when you say it is impossible or wouldn't make any sense.
Mainly because the fact that the electric field of multiple (isolated) electrons can be computed easily by super-positioning their individual fields - so I don't really see how you can say that you can compute the field for a single magnet, but not for two - at least for charges this works nicely and I think so it should for magnets.
And for another example which is closer to my question: Consider molecular dynamics simulations; They are very accurate in the resulting dynamics so there must be some way to compute the charge distribution (and field) of those molecules to compute their interaction. How could this be possible if even for the simplest possible case, a hydrogen atom in the ground state, it would be impossible ? They seem to be able to somehow solve a much more complicated problem where I'm only interested in a single hydrogen atom.

 

[sophiecentaur]

There is just one electron BUT where is it? You could say it is in random places around the nucleus. How does that help you to achieve anything? In fact what is it all supposed to mean?

 

[xortdsc]

regardless of whether the electron pops up in random places very quickly or is in a stable distributed state, the resulting time-averaged electrical field it produces should still be identical and stable, i'd think. it's this (averaged) field i'd be interested in (as it gives rise to interactions).

 

[sophiecentaur]

regardless of whether the electron pops up in random places very quickly or is in a stable distributed state, the resulting time-averaged electrical field it produces should still be identical and stable, i'd think. it's this (averaged) field i'd be interested in (as it gives rise to interactions).

Are you suggesting that the various 'bits' of the electron are located around the atom in a fixed pattern? Surely the whole point of QM is that such things are totally random in time and space.

I really can't see how such a random set of vectors, from each of the constituent bits, each one varying in time (according to a random presence or absence of a charge in anyone particular pace), could combine to produce a stable field. I think it's accepted that there is a sort of fuzzy statistical distribution of Potential around an atom or molecule - which is what the polarity of molecules is based on. Field, otoh, is a different matter.
But this model is really far too classical to be workable. You start with QM, to get your charge distribution (is that even a valid step?) and then revert back into classical for working out a field. It doesn't seem valid to me.

 

[xortdsc]

no, I'm not assuming a fixed pattern. only a distribution (which is sort of the averaged presence of the electron for each point in space). so the basis of the calculation should be this averaged presence, disregarding the random nature of the actual process.

 

[sophiecentaur]

no, I'm not assuming a fixed pattern. only a distribution (which is sort of the averaged presence of the electron for each point in space). so the basis of the calculation should be this averaged presence, disregarding the random nature of the actual process.

There's your problem. If you disregard the random nature then your resulting numbers would have no connection with reality. And what magnitude and direction would you give the resultant of a random set of vectors? If you don't take the direction and amplitude of each vector into account then you will get nonsense. If you really believe in this, I suggest you sit down and try to draw out a simple version of what you would actually do. Start with a simple model of two charge centres with a range of different amplitudes, varying in time. Add the vectors at different times and see what you get. Then do the same with three charges - and so on. The result will tend to the field of a single charge (the sum) at a distance (i.e. the same as an isolated electron). But don't you really need the Proton charge there, too? The field, then would go to zero at infinity.
I still think the Potential distribution would be a more meaningful thing to go for. For a start, it would be easier / possible to plot.

 

[mfb]

Quantum mechanics is not random, it is a deterministic theory. The interpretations of measurements can be, but if we consider processes where those measurements do not happen everything is certainly deterministic.

 

[sophiecentaur]

Quantum mechanics is not random, it is a deterministic theory. The interpretations of measurements can be, but if we consider processes where those measurements do not happen everything is certainly deterministic.

OK So where does this take the problem of adding vectors? Does it make the idea more valid - or less?
My brain hurts with QM but it hurts even more in trying to apply classical ideas in the context of QM.

 

[Dale]

Hmm, I don't know about the idea that the field is undefined or anything. The field is a measurable quantity, so there should be some operator that you can apply to the wave function to get the distribution for any measurement of the field. Of course, I would assume that the wavefunction isn't usually in an eigenstate of the field, so the best you could do would be to get an expected value.

 

[mfb]

OK So where does this take the problem of adding vectors? Does it make the idea more valid - or less?
My brain hurts with QM but it hurts even more in trying to apply classical ideas in the context of QM.

Where is the problem? For each point in phase-space, you calculate the contribution to the fields, and then integrate over the whole phase-space, with the squared amplitude as weight.

 

[sophiecentaur]

Where is the problem? For each point in phase-space, you calculate the contribution to the fields, and then integrate over the whole phase-space, with the squared amplitude as weight.

Does that not require you to be sure that the charge is actually in existence all the time? This is not how I understood the probability density of the bound electron; I thought there was a random element. What is the modern opinion on this?

 

[mfb]

The electron and its charge exist all the time.

 

[xortdsc]

Hmm, I don't know about the idea that the field is undefined or anything. The field is a measurable quantity, so there should be some operator that you can apply to the wave function to get the distribution for any measurement of the field. Of course, I would assume that the wavefunction isn't usually in an eigenstate of the field, so the best you could do would be to get an expected value.

That was exactly my way of reasoning. It is measurable so it should be computable, except the theory is incomplete on a fundamental level. :)
The expected (idealized) value would be good enough for me.

Where is the problem? For each point in phase-space, you calculate the contribution to the fields, and then integrate over the whole phase-space, with the squared amplitude as weight.

That's what I thought, but after further thinking about it there really IS a problem (merely a computational one): For a traditional point-like charge the field goes to infinity at the position of the charge, independent of its magnitude (which is possibly not 100% correct, but all we have in the classical theory). So, if one would use the charge density as a source and apply that traditional law to the partial charges it would yield infinity everywhere. So for this reason it may really be impossible. :/

But isn't there a better way of doing it ? Is QM really that incomplete when it comes to the EM-fields it gives rise to ?

 

[sophiecentaur]

The electron and its charge exist all the time.

... in all places at all times? I don't think you mean that. But, if there is a probability density function of the position of a putative electron in the bound state (which seems quite reasonable) then how can this be used to produce a map of the expected field around the atom. I can't work out how you could give a (vector) value of the field at any location or display it. I suppose it would be reasonable to produce a map of most probable or mean values. Has this been done? How could it go further than that? But is it a valid thing to take a the QM model and then expect to use it to predict the Field, which, I think, is a classical quantity.
Could you convince me about this? My mind is open.

 

[Vanadium 50]

Given an atom in an energy eigenstate, the field is static. ('But what if it isn't?' is beyond the scope of this course, the next course, and the one after that) If the field were changing, there would be radiation, and that takes energy, and then the atom is not in that energy eigenstate any more.

So what's the field? Superposition tells us it's the field of the proton, e/r^2 plus the field of the electron, which QM tells us. It works out to

$$\frac{+e}{r^3}\vec{r}-e\int \frac{\psi^* \psi d\vec{r}}{r^3}$$

Once one has the field, one is free to draw field lines, but as has been pointed out, this is not helpful. Note that while, in an energy eigenstate the electron does not have a well-defined position, the electric field is well defined.

 

[sophiecentaur]

Given an atom in an energy eigenstate, the field is static. ('But what if it isn't?' is beyond the scope of this course, the next course, and the one after that) If the field were changing, there would be radiation, and that takes energy, and then the atom is not in that energy eigenstate any more.

So what's the field? Superposition tells us it's the field of the proton, e/r^2 plus the field of the electron, which QM tells us. It works out to

$$\frac{+e}{r^3}\vec{r}-e\int \frac{\psi^* \psi d\vec{r}}{r^3}$$

Once one has the field, one is free to draw field lines, but as has been pointed out, this is not helpful. Note that while, in an energy eigenstate the electron does not have a well-defined position, the electric field is well defined.

I like that argument. It makes sense.

 

[xortdsc]

Given an atom in an energy eigenstate, the field is static. ('But what if it isn't?' is beyond the scope of this course, the next course, and the one after that) If the field were changing, there would be radiation, and that takes energy, and then the atom is not in that energy eigenstate any more.

So what's the field? Superposition tells us it's the field of the proton, e/r^2 plus the field of the electron, which QM tells us. It works out to

$$\frac{+e}{r^3}\vec{r}-e\int \frac{\psi^* \psi d\vec{r}}{r^3}$$

Once one has the field, one is free to draw field lines, but as has been pointed out, this is not helpful. Note that while, in an energy eigenstate the electron does not have a well-defined position, the electric field is well defined.

Thank you very much. This seems to go in the direction I'm interested in. :)
As I'm not a mathematician nor a physisist, I just want to make sure I interpret the formula correctly:
$$r$$ is the 3d position of interest, where the origin of the coordinate system is at the nucleus (the e+)
$$\psi^* \psi$$ just boils down to the square of the wave-function of the orbital of interest, as I'm only interested in real orbitals.

The integral is just the (indefinite) integral of the (squared) wave-function evaluated from 0 to r ?
Is that right ?

 

[Vanadium 50]

The integral is over all space.

 

[mfb]

That's what I thought, but after further thinking about it there really IS a problem (merely a computational one): For a traditional point-like charge the field goes to infinity at the position of the charge, independent of its magnitude (which is possibly not 100% correct, but all we have in the classical theory). So, if one would use the charge density as a source and apply that traditional law to the partial charges it would yield infinity everywhere. So for this reason it may really be impossible. :/

That's fine, the integral is still well-defined - even if you would add the magnitudes, the result would be finite, and if you add the directions this region will cancel nearly completely.

... in all places at all times?

That's a matter of definition and interpretation, but it is not what I meant.

But, if there is a probability density function

It is a probability only if you do measurements. I prefer the term "wave function" here. You can simply treat the electron as a wave and use it as classical object to get a meaningful (!) result.

then how can this be used to produce a map of the expected field around the atom

It is not an expectation value.

 

[dipole]

Wow, I'm really surprised at the amount of nonsense going around in the earlier parts of this thread...

Given an atom in an energy eigenstate, the field is static. ('But what if it isn't?' is beyond the scope of this course, the next course, and the one after that) If the field were changing, there would be radiation, and that takes energy, and then the atom is not in that energy eigenstate any more.

So what's the field? Superposition tells us it's the field of the proton, e/r^2 plus the field of the electron, which QM tells us. It works out to

$$\frac{+e}{r^3}\vec{r}-e\int \frac{\psi^* \psi d\vec{r}}{r^3}$$

Once one has the field, one is free to draw field lines, but as has been pointed out, this is not helpful. Note that while, in an energy eigenstate the electron does not have a well-defined position, the electric field is well defined.

That's exactly how I'd calculate the field, although this statement is incorrect,

The integral is over all space.

The integral is indeed from zero to $r$, as the OP originally surmised. Integrating over all space you're removing the position dependence of the field.

It's not clear to me though that on short time-scales there wouldn't be fluctuations in the field, because although the electron is in a stationary state the "standing wave" wave function $\Psi(\vec{r},t)$ is still vibrating about the nucleus. However, I totally agree with Vanadium's calculation for, at the very least, the time-averaged field.

 

[Vanadium 50]

Integrating over all space you're removing the position dependence of the field.

I was sloppy, as I used r as both a position variable and as a variable of integration. If you integrate over "r-integration" (maybe it should be r-prime) you will be left with a function of r. I was also sloppy in that psi itself is a function of r.

 

[sophiecentaur]

Wow, I'm really surprised at the amount of nonsense going around in the earlier parts of this thread...


.

Where were you when we needed you then?

 

[xortdsc]

Thank you guys, this was helpful.
So as I quickly did some calculations it seems like for S-orbitals the electric field is always pointing away from the nucleus and decays to zero at infinity (though the falloff is fairly steep and not constant but wavey due to the S-orbital nodes (for n>1)).
For other orbital-shapes there seems to be a slight "polarization". So let's say P-orbitals have slight negative electric charge along the direction of the dumbbell and equal positive charge in the other directions. This is a very subtle and rather short-range effect of course.
Would you say that's conceptually correct ?