Can two electrons (with different quantum numbers) exist at the same place in space?

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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.

Answer 5

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.

Answer 6

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).