# Field lines of electrons in an atomic orbital

sophiecentaur
Science Advisor
Gold Member
2020 Award
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
Staff Emeritus
Science Advisor
Education Advisor
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
Science Advisor
Gold Member
2020 Award
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.

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 wanna 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 ?

Last edited:
Vanadium 50
Staff Emeritus
Science Advisor
Education Advisor
The integral is over all space.

mfb
Mentor
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.

sophiecentaur said:
.... 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.

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.

1 person
Vanadium 50
Staff Emeritus
Science Advisor
Education Advisor
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
Science Advisor
Gold Member
2020 Award
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?

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 ?

I don't think there should be any polarization because of the spherical symmetry inherent in the problem. The time-averaged field should have no angular dependence.

I think in order to get a time-averaged field you may want to integrate away the angular components.

Vanadium 50
Staff Emeritus
Science Advisor
Education Advisor
S-orbitals have spherically symmetric fields. The other orbitals do have polarizations, yes.

Cool. Makes sense now I think. Thanks again :)