# Does Pauli Exclusion forbid two neutral fermions to occupy the same location in space?

I know that the Pauli Exclusion Principle does not allow two identical fermions to have the same set of quantum numbers. But can they share the same location in space if they are uncharged such as two neutrons? If not, why not because there is no quantum number associated with position. Thanks

The principle forbids two fermions of the same species from sharing the same quantum state. Let's first carefully look at where the idea comes from.

This principle arises simply because the spin statistics theorem implies that half integer spin particles - i.e. fermions - have antisymmetric multiparticle quantum states, i.e. that such states undergo a sign change under the swapping of any pair of members of the multipartite state. That needfully implies that a two-particle state, for example, must be of the form $\sum\limits_{i,j} \alpha_{i\,j}|i\rangle\otimes|j\rangle$, with $i,\,j$ ranging over all basis states for the individual particles, where $\alpha_{i\,j}=-\alpha_{j\,i}$ which immediately implies $\alpha_{i\,i}=0$: the bipartite state can never have the two members in the same state.

But state of course is the full quantum state. Therefore, identical fermions can perfectly well be at the same position if their other quantum numbers differ. On the other hand, the position is most definitely part of the quantum state, so that two fermions of the same species with all other quantum numbers being equal cannot be at the same position.

Now, of course, position is an observable with a continuous spectrum, so:

1. One doesn't wontedly think of it as a quantum number, but it is part of the full state nonetheless and so you will have an incorrect understanding of the principle if you think it says that positions "don't count because they have no quantum numner"and
2. Even though one might argue that the probability to measure two particles at exactly the same value of $x$ when $x$ is a continuous variable is nought, nonetheless the exclusion principle is still highly significant because it means that the amplitude must smoothly approach 0 as the position co-ordinates for the two particles approach equality. Thus, the requirement deeply influences the equation of state for a Fermi gas and thus the principle applied to position alone makes itself felt widely over the field of condensed matter physics.

The wavefunction for fermions has a spatial part too, and this is what describes the location of a particle at best in QM. Pauli Exclusion Principle forbids two identical fermions from having the very same wavefunction. For example, you can have spin-up fermions both in s- and p-waves, but of course no two spin-up fermions in an s-wave.

• But does it forbid two fermions to share the same position? For example, if the spin part of the wavefunction is antisymmetric singlet, the spatial part is symmetric. Does the principle forbid them to share the same coordinate? Thanks Commented Aug 12, 2017 at 13:14
• I think the confusing bit here is the meaning of "coordinate". How can you assign a coordinate to a quantum particle? Surely you can take the expectation value for $x$, but two different spatial wavefunctions might yield the same average, still allowing you to have two collinear fermions. So, instead of thinking in terms of "same coordinate" (as in the case of a classical particle), you really should think in terms of "same spatial wavefunction". Commented Aug 12, 2017 at 13:32
• You haven't really completed this answer until you explain that they can have the same position so long as the momentum are sufficiently different. Pauli-exclusion lives in phase-space. Commented Aug 12, 2017 at 17:38
• @Phoenix87 But what causes the Paul repulsion in degenerate fermi gas as in neutron star? What forbids the neutrons to come closer in a neutron star? What causes the degeneracy pressure? Commented Aug 12, 2017 at 19:01
• en.wikipedia.org/wiki/Exchange_interaction Commented Aug 12, 2017 at 19:03

Position is an undefined concept for a Particle in Quantum Mechanics. The thing which describes the Particle is the state-vector in Hilbert space and the position wavefunction can be considered as the position component of that vector.

So like the previous answer says, all the principle says is the impossibility of having the same wavefunction.

Two wavefunctions can overlap at a point is that what you mean by position? There is no meaning to position of a particle till you measure it. All you get is the probability distribution.

Recalling Szabo's Quantum Chemistry book, we can write a two-fermion wavefunction as a Slater determinant $$\Psi(\mathbf{x}_{1},\mathbf{x}_2)=|\chi_{1}(\mathbf{x}_{1})\chi_{2}(\mathbf{x}_2)\rangle$$, where $$\chi_{i}=\phi_{i}\sigma$$ is the $$i$$th spin orbital with $$\phi_{i}$$ the spatial orbital and $$\sigma=\{\alpha,\beta\}$$. Let $$\rho(\mathbf{r}_{1}, \mathbf{r}_{2})\mathrm{d}\mathbf{r}_{1}\mathrm{d}\mathbf{r}_{2}$$ be the probability of finding electron 1 between $$\mathbf{r}_{1}$$ and $$\mathbf{r}_{1}+\mathrm{d}\mathbf{r}_{1}$$ and electron 2 between $$\mathbf{r}_{2}$$ and $$\mathbf{r}_{2}+\mathrm{d}\mathbf{r}_{2}$$. This probability is obtained by "integrating" (or averaging) over the spin variables ($$\sigma$$) of the two fermions. If both fermions are in the same orbital ($$\phi_{1}=\phi_{2}$$) with opposite spins, then the probability is

\begin{aligned} \rho(\mathbf{r}_{1},\mathbf{r}_{2})\mathrm{d}\mathbf{r}_{1}\mathrm{d}\mathbf{r}_{2} &= \int \left(|\Psi|^{2}\mathrm{d}\mathbf{r}_{1}\mathrm{d}\mathbf{r}_{2}\right)\mathrm{d}\sigma_{1}\mathrm{d}\sigma_{2} \\ &=\frac{1}{2}\left(|\phi_{1}(\mathbf{r}_{1})|^{2}|\phi_{1}(\mathbf{r}_{2})|^{2} + |\phi_{1}(\mathbf{r}_{1})|^{2}|\phi_{1}(\mathbf{r}_{2})|^{2}\right)\mathrm{d}\mathbf{r}_{1}\mathrm{d}\mathbf{r}_{2} \\ &= |\phi_{1}(\mathbf{r}_{1})|^2|\phi_{1}(\mathbf{r}_{2})|^{2}\mathrm{d}\mathbf{r}_{1}\mathrm{d}\mathbf{r}_{2} \end{aligned}

However, if both fermions have the same spin (in any orbital), the probability density would be \begin{aligned} \rho(\mathbf{r}_{1},\mathbf{r}_{2}) &= \frac{1}{2}\left\{|\phi_{1}(\mathbf{r}_{1})|^{2}|\phi_{2}(\mathbf{r}_{2})|^{2} + |\phi_{2}(\mathbf{r}_{1})|^{2}|\phi_{1}(\mathbf{r}_{1})|^{2}\right. \\ & \qquad\left. - \phi_{1}^{*}(\mathbf{r}_{1})\phi_{2}^{*}(\mathbf{r}_{2})\phi_{2}(\mathbf{r}_{1})\phi_{1}(\mathbf{r}_{2}) - \phi_{2}^{*}(\mathbf{r}_{1})\phi_{1}^{*}(\mathbf{r}_{2})\phi_{1}(\mathbf{r}_{1})\phi_{2}(\mathbf{r}_{2})\right\} \end{aligned} In a single-Slater determinant approximation, where we only consider the correlation between fermions of the same spin (due to Pauli exclusion principle), there are extra terms whose consequence is that

$$\rho(\mathbf{r}_{1},\mathbf{r}_{1}) = 0$$ Therefore, the probability to find two fermions with parallel spins at the same point in space is zero. This lowered probability to find an electron near another with the same spin is usually called a Fermi hole in quantum chemistry.

And that's it, this argument is originally developed to explain electron wavefunctions, but since any many-fermion wavefunction can be approximated as a single Slater determinant (where only considering Pauli exclusion principle for correlated movement), then it also shows that there is zero probability of finding two fermions in the same place. Also, this argument extends to the exact many-fermion wavefunction, which can be expressed as a linear combination of Slater determinants.