12
$\begingroup$

I was studying about the arrangements of orbitals in an atom and saw a simulation of the arrangement and that some area of a smaller orbital such as a 1s is contained inside a bigger orbital of 2s (a similar also happens in bonding as well when the orbitals overlap).

So, I have a question regarding this that if the orbitals have some of their areas overlap, then doesn't a case arrive in which there is a slight probability of finding an electron of the 1s orbital and the 2s orbital in the same position in space? As the orbitals are just regions around the nucleus where the electron has the highest probability of being founded, then surely this case should arrive(even with extremely low probability) that the 1s orbital electron and the 2s orbital electron occupy the same position?

I found some answers that it is not possible due to the "Pauli's Exclusion Principle" but it just states that two electrons cannot have the "same quantum numbers", there is no mention of "position" anywhere. (in our case also the two electrons in question have different quantum numbers, so do they need to have different positions?)

Also, if they do exist in the same position, what are the consequences of this happening? Do they just immediately repel each other or something else?

I have read some answers regarding some similar questions but no luck in finding a satisfactory answer.

$\endgroup$
2
  • 1
    $\begingroup$ For clarification, the "Pauli Principle" (requirement for antisymmetry with respsect to exchange of electrons) has two consequences - one is that two electrons cannot have the same quantum numbers and the other is that they can't be at the same position. The latter is sometimes called "Pauli repulsion" rather than "Pauli exclusion", but I don't think there's consistency in the terminology. $\endgroup$
    – Andrew
    Jun 12, 2023 at 11:09
  • $\begingroup$ Does the Question assume that the electrons occupy merely the same place in space, or does space-time matter? $\endgroup$ Jun 21, 2023 at 19:39

6 Answers 6

12
$\begingroup$

It makes no sense to think of orbitals as representations of one electron. The n- electron field is an antisymmetric function of n variables, so nothing, especially 3d-space, belongs to only one electron. The configuration space is 3n-dimensional as in classical mechanics, one 3d-coordinated space for a labeled system of n elementary electrons.

What makes more sense is to span the Hilbert space of wave functions eg. for the Schrödinger H-Atom in an orthogonal series of 3d-wave functions as eigenvectors of the Hamiltonian.

Then the n-particle functions can be constructed in the tensor product of the orthogonal basis, one factor for each electron.

From these gereral tensor product as the basis of all functions, the Pauli principle selects the sub-tensor product, antisymmetric under exchange of any two coordinate lables.

Of course, this priciple forbides two identical one-particle states in the product.

But the projecton to the common 3d physical space does not forbide overlaps.

Overlaps are total normal, all the basis functions live together, but their scalar products has to vanish in order to be used in the Pauli principle basis.

$\endgroup$
3
  • 2
    $\begingroup$ There should be many more downvotes, because its an scientific text by a scientist. $\endgroup$
    – Roland F
    Jun 11, 2023 at 12:53
  • 1
    $\begingroup$ Although I couldn't understand all of it, this is a really good answer and thanks for answering and sharing your knowledge, no idea who downvoted the answer. $\endgroup$ Jun 11, 2023 at 13:22
  • 3
    $\begingroup$ @RolandF: Yes, answers that require expertise in the subject to understand are not useful here. This is a Q&A forum, not a scientific journal. You seem to understand that, so why not answer in a way that's more accessible? $\endgroup$ Jun 12, 2023 at 6:11
7
$\begingroup$

Maybe the answer depends on what you mean by "place."

So-called "quantum numbers," in effect, define a "place" within some well defined system. E.g., the traditional four quantum numbers from chemistry define the place of an electron within the system that consists of a single atomic nucleus and the electrons that are bound to it. No two electrons can have the same quantum numbers—no two can occupy the same "place"—within that system. But, the electrons belonging to some other, isolated atom, then you are discussing a completely different system. No connection between them.

If you want to understand the behavior of electrons in molecular orbitals (e.g., electrons participating in π-bonds in a benzene ring,) or valence electrons in a macroscopic chunk of metal, or free electrons in a particle accelerator, etc. then you are dealing with a more complicated system, and you need something other than those four familiar quantum numbers from chemistry to describe an electron's "position."

I wish I could say more, but I'm really just a spectator here. I know nothing about how physicists talk about the quantum states of electrons in those more complex systems.

$\endgroup$
6
$\begingroup$

The probability of getting the position of two electrons with the same spin component $s_z$, at ${\bf r}_1$ and ${\bf r}_2$ positions, is encoded in the density probability $$ p({\bf r}_1,{\bf r}_2) = \left|\psi({\bf r}_1,{\bf r}_2)\right|^2. $$ where $\psi({\bf r}_1,{\bf r}_2)$ is the spatial part of the two-electrons wavefunction. Since the total (spatial+spin) wavefunction must be antisymmetric with respect to the exchange of particles, and taking into account that the spin part of the wavefunction of two equal spin particles is symmetric, the spatial part must be an antisymmetric function: $$ \psi({\bf r}_1,{\bf r}_2) = -\psi({\bf r}_2,{\bf r}_1). $$ Such antisymmetry implies that the probability density for the two electrons at the same position must vanish.

Notice, however, a subtle point related to the fact that the modulus squared of the wavefunction is a density probability and not directly a probability. Indeed, for any continuous probability distribution of a quantity $x$, the probability of finding precisely one value of $x$ is always zero, even in connection with a non-zero probability density, because the probability is obtained from the density probability by integrating it over a finite volume, and a point is always a zero-measurement set.

Therefore, the rigorous statement in the case of two electrons with the same spin is that the probability of finding the two electrons in the same small volume $\Delta V$ vanishes faster than $\Delta V$ when $\Delta V$ goes to zero.

Note added after reading the edited question

Of course, the general statement about the antisymmetry of the two- or many-electron wavefunction can be made specific to the case of the so-called one-electron approximations without changes to the general result. Indeed, in Slater determinant wavefunctions, electronic coordinates and states play the role of rows and columns (or vice-versa) of the matrix. The determinant vanishes if two states are the same or two coordinates coincide. Thus, it should be clear that it is not directly the overlap in the same region of two one-particle states which makes the wavefunction zero but the request that two electrons are localized at the same point.

$\endgroup$
5
  • $\begingroup$ Quick follow-up: Does the full N-body Hamiltonian operator also contain a "$1/\left|r_1-r_2\right|$"-type term from electron-electron interactions (with the proper quantization), which would mean that the probability of observing a small $|r_1-r_2|$ in any state present in nature is even smaller than what just the anti-symmetry gives? (My classically-wired mind would say so, but it's been 10 years since I actually did any quantum physics so I'm not sure how these things actually work. I'm just surprised why electron-electron potentials aren't often mentioned when this question comes up.) $\endgroup$
    – JiK
    Jun 12, 2023 at 13:36
  • $\begingroup$ @JiK The Coulomb repulsion certainly increases somewhat the volume near $r=0$ where the pair correlation of two up-up or down-down electrons is close to zero. However, the effect depends on density, being large at low densities, and small at large densities (remember that the ideal quantum behavior corresponds to the limit of very high density). At metallic densities, the up-down pair correlation is not zero, even in the presence of a Coulomb repulsion. $\endgroup$ Jun 12, 2023 at 15:17
  • $\begingroup$ "the spin part of the wavefunction of two equal spin particles is symmetric" — wait what? Don't singlet states exist? $|0, 0 \rangle = |\uparrow, \downarrow \rangle - |\downarrow, \uparrow \rangle$ and all that? $\endgroup$ Jun 12, 2023 at 19:17
  • 1
    $\begingroup$ @MichaelSeifert "equal spin == equal z-projection of the spin." There is only one state: $ | \uparrow \uparrow >$, which is symmetric. $\endgroup$ Jun 13, 2023 at 3:52
  • 2
    $\begingroup$ Hmm... I suppose that's correct if that's what you mean by "equal spin"; I thought you meant "two spin-$\frac12$ particles". I would argue, though that it's misleading to only discuss that case and not discuss the singlet case. In fact, if two electrons are in a singlet state, then the opposite effect occurs: the probability of finding the particles near each other is enhanced rather than reduced. $\endgroup$ Jun 13, 2023 at 11:32
4
$\begingroup$

You can start by a one body problem : H atom where all the orbitals are orthogonal so $1s$ and $2s$ are orthogonal. The solutions of SE equation are always orthogonal. For a many electronic atom this orthogonality holds but you will never have a complete analytic description of the radial part, the angular part does not depend on the energy and is often invariant.

You can check easily that $\left<1s|2s\right> = 0$ for H atom. This means even if your you plot the density of the radial part, they overlap, the total WF of $1s$ and $2s$ do not overlap. This statement has nothing to do with the Pauli principle for a many electronic system.

The intuitive way to understand this principle is to say that if two electrons are in the same position without talking about the intrinsic angular momentum, you will have an infinite coulomb energy $+\frac{1}{r\rightarrow 0} \rightarrow \infty$. The overlap of two WF can only appear if there is another contribution $-\frac{1}{r\rightarrow 0} $ called exchange energy strictly related to the angular momentum an consequently the symmetry. If three electrons are together, two will form a singlet state with the exchange interaction and the other should strictly occupy another state orthogonal to this singlet state so that the overlap is zero avoiding this infinite coulomb energy.

$\endgroup$
2
  • $\begingroup$ Orthogonality just implies that in a given basis if a particle is in a state $\psi1$ then it cannot be simultaneously describes by wavefunction $\psi2$. Doesn't mean the overlap isn't there $\endgroup$ Jun 19, 2023 at 13:28
  • $\begingroup$ Orthogonality physically means that you cannot mesure a particule in this state $\psi_1$ you cannot have access to the state $\psi_2$. It means these states do not overlap in the sense that the particle will never be in a position decribes by these two states. Note that a WF has no physical meaning. $\endgroup$
    – M06-2x
    Jun 19, 2023 at 13:39
2
$\begingroup$

You're thinking this from a classical perspective where as the probability cloud is an entirely quantum mechanical (or a analogy would be classical wave mechanics) phenomenon. Here the point of 'position' breaks down.

Now you're considering the 1s and 2s orbital to overlap where as in any atom all the orbitals present do overlap and these orbitals are present in different energy levels which is anyways changing their quantum number. Pauli's exclusion principle talks about how two electrons being in a particular energy level should have different quantum numbers. It doesn't talk about position of an electron because that doesn't make any sense to talk about position physically.

$\endgroup$
0
$\begingroup$

Because of the Heisenberg uncertainty principle you cannot describe an electron as a point particle (with infinite precision on its position), but only as a probability cloud. Actually the atomic orbitals are just the regions where it is likely to find the electrons. So, you cannot have two electrons overlapping in the classical sense (like two tennis balls occupying the same position).

$\endgroup$
2
  • 1
    $\begingroup$ But now my doubt is that if it is not possible to describe an electron as a point particle then what does the probability cloud actually show? I mean if it is not a point particle then even though we know it will exist in some area the big question is "how" will it exist in that area/region? Although if we make a measurement on a particle, it exists as a point particle. Isn't all of this very contradicting? $\endgroup$ Jun 11, 2023 at 12:55
  • $\begingroup$ There is space for two particles to occupy the same position if their momenta are different, right? Measurements alter a particle's state, because of the interaction. $\endgroup$
    – Wookie
    Jun 11, 2023 at 21:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.