Pauli’s exclusion principle

So here’s what the exclusion principle states: “No 2 fermions( let’s say electrons) can have the same quantum states”

Consider the following hypothetical situation:

Let’s have 2 free electrons in a vacuum. They are also free from gravity. Is it possible to modify the exclusion principle in to the following statement:

“ No 2 fermions (electrons) can occupy the the same position in space and time simultaneously”

I state the above because the electrons are not bound to any atoms. The 2 electrons also have different spins.

• You have spin to worry about too. – NickD Mar 18 '18 at 17:28
• @Nick let’s say they have different spins. Then can the above be applied? – physics2000 Mar 18 '18 at 17:28

You need also to account for spin.

One way to answer your question is rephrase it as the statement that the total wavefunction of a 2-fermion system must be fully antisymmetric. If the spin part of this system is antisymmetric, i.e. $$\vert\chi\rangle = \frac{1}{\sqrt{2}}\left( \vert +\rangle \vert -\rangle - \vert -\rangle \vert +\rangle\right) \tag{1}$$ then the spatial part can be symmetric and then two fermions could occupy the same position (if they didn't interact) in the sense that the probability density for $\vec r_1=\vec r_2$ would not be $0$. The probability of finding them both at exactly the same point in space is $0$ since the integral of this distribution over an area of size $0$ (i.e. the line $\vec r_1=\vec r_2$) will be $0$.

Of course the electrons would interact through electromagnetic repulsion, which would push them one away from the other.

• The spins of both electrons are different. – physics2000 Mar 18 '18 at 17:32
• Yes indeed. They are different in (1): the spin of the first is never the same as the spin of the second. In fact if they were the same the spatial part of the wavefunction would have to be antisymmetric and would thus be automatically $0$ when the positions would be the same. – ZeroTheHero Mar 18 '18 at 17:33
• so what you are trying to say is that if the coulomb force would not exist, then electrons could occupy the same position in space? – physics2000 Mar 18 '18 at 17:36
• @physics2000 what does "occupy the same position in space" means when one is dealing with wavefunctions? The probability density would not be $0$ there but of course the probability of finding any particle at any specific point in space is $0$, and the probability of finding them both at exactly the same point in space would also be $0$. – ZeroTheHero Mar 18 '18 at 17:40
• Note that even though electrons do interact via Coulomb repulsion, probability density of $\vec r_1=\vec r_2$ is still nonzero for the case of different spins. See e.g. this discussion. – Ruslan Mar 18 '18 at 21:24

This is exactly what happens in the orbitals of an atom or molecule. The orbital describes the probability (amplitude) distribution of the electron, so when two electrons occupy the same one, they do have the same position state at the same time. There is nothing fundamental that prevents them from being found in the same region of space at the same time, so your modified statement is false. Pauli exclusion doesn't forbid this because the full state of an electron also includes its spin. Since electrons have only two orthogonal spin states, no more than two electrons can occupy an orbital (i.e., exist simultaneously in the same position state).

Another point that may be of interest to you: when two electrons occupy the same orbital, Pauli exclusion does require that they have opposite spin. Spin is a vector that can be measured along any direction, and Pauli exclusion must be satisfied regardless of which direction is chosen by the observer. The unique spin state of two electrons that meets this condition is an entangled state called the singlet: $\frac{1}{\sqrt{2}}(\lvert\uparrow,\downarrow\rangle-\lvert\downarrow,\uparrow\rangle)$.

The question is really hypothetical because any two electrons will repulse each other due to their identical charges. But let’s consider they will not feel the Coulomb force.