We know that fermions are identical particles and obey Pauli exclusion principle. But what is meant by distinguishable fermions? Does that mean, like proton and electron both are fermions but they are distinguishable because of charge? And if we put together both distinguishable fermions, will they obey Pauli exclusion principle ?
[Physics] Do distinguishable fermions obey the Pauli exclusion principle
fermionspauli-exclusion-principlequantum-statistics
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How does the Pauli Exclusion Principle actually create a force?
The Pauli exclusion principle doesn't really say that two fermions can't be in the same place. It's both stronger and weaker than that. It says that they can't be in the same state, i.e., if they're standing waves, two of them can't have the same standing wave pattern. But for bulk matter, for our purposes, it becomes a decent approximation to treat the exclusion principle as saying that if $n$ particles are confined to a volume $V$, they must each be confined to a space of about $V/n$. Since volume goes like length cubed, this means that their wavelengths must be $\lesssim (V/n)^{1/3}$. As $V$ shrinks, this maximum wavelength shrinks as well, and the de Broglie relation then tells us that the momentum goes up. The increased momentum shows up as a pressure, just as it would if you increased the momenta of all the molecules in a sample of air. A degenerate body like a neutron star or white dwarf is in a state where this pressure is in equilibrium with gravity.
Your intuition is correct.
Consider a system with a fixed number $N$ of nonrelativistic particles, all fermions. Ignoring spin for simplicity, the wavefunction of such a system is a function of $N$ points in space: $$ \newcommand{\bfx}{\mathbf{x}} \psi(\bfx_1,\bfx_2,...,\bfx_N). \tag{1} $$ If the $j$th and $k$th particles are the same species, then the wavefunction must satisfy $$ \newcommand{\bfx}{\mathbf{x}} \psi(\bfx_{\pi(1)},\bfx_{\pi(2)},...,\bfx_{\pi(N)}) = - \psi(\bfx_1,\bfx_2,...,\bfx_N) \tag{2} $$ for the permutation $\pi$ that exchanges $j\leftrightarrow k$ and leaves the other points unchanged. If the $j$th and $k$th particles are not the same species, then no such (anti)symmetry is required. In particular, if we have $5$ particles ($N=5$) all of different species, then the wavefunction doesn't need to have any (anti)symmetry at all. The fact that the particles are all fermions is irrelevant in that case. Nothing would change if they were bosons — in the strictly nonrelativistic and spinless model that we're using here for simplicity. (In a relativistic model, we can't ignore spin, and the number of particles is usually ill-defined, but I won't go into those complications here.)
We can consider all values of $N$ simultaneously using the formalism of creation/annihilation operators. Still ignoring spin for simplicity, we can describe a system of strictly nonrelativistic fermions using $K$ different creation operators $a_k^\dagger(\bfx)$ for each $\bfx$, with $k\in\{1,2,...,K\}$, where $K$ is the number of different species. If $|0\rangle$ is the state with no particles, then $$ \int_{\bfx,\bfx'} \psi(\bfx,\bfx') a_1^\dagger(\bfx)a_1^\dagger(\bfx')|0\rangle \tag{3} $$ is an example of a state with two particles of the same species, and $$ \int_{\bfx,\bfx'} \psi(\bfx,\bfx') a_1^\dagger(\bfx)a_2^\dagger(\bfx')|0\rangle \tag{4} $$ is an example of a state with two particles of different species. The assertion that the particles are all fermions can be expressed mathematically by the requirement that all of the creation operators anticommute with each other: $$ a_j^\dagger(\bfx)a_k^\dagger(\bfx') = -a_k^\dagger(\bfx')a_j^\dagger(\bfx). \tag{5} $$ For particles of the same species ($j=k$), this immediately implies the Pauli exclusion principle: only the antisymmetric part of $\psi$ matters in (3), because the minus sign in (5) eliminates any contribution from the symmetric part. This follows from the fact that in (3), exchanging the subscripts is the same as exchanging the points. But for particles of different species ($j\neq k$), that's no longer true, and the minus sign in (5) has no consequence in thiscase. In fact, the minus sign in (5) can be eliminated when $j\neq k$ using a Klein transform.
Best Answer
The simple way to think about it is to imagine that all fermions are excitations of a single field. These excitations can differ in their position, spin, charge, mass, and so on, and the Pauli exclusion principle applies to all of them. Mathematically, this is just the fact that all fermionic creation operators anticommute; the joint wavefunction of all fermions is antisymmetric.
So fermions that are far apart aren't affected, because they differ in position space. Fermions that have different spins aren't affected, because they differ in spin. And protons and neutrons don't affect each other because they differ in mass and charge. The Pauli exclusion principle always applies and there are no exceptions. Any two fermions must be different in some way to coexist.
Because spin and position are easily changed, and mass and charge aren't, sometimes people break the rule into two cases. They say that Pauli exclusion only looks at spin and position, and doesn't apply to things with different masses and charges (because they're automatically different). This works for simple situations, but it's dangerous, because if you take it too literally you'll get the wrong answer when constructing baryon wavefunctions. There you really do have to antisymmetrize over all degrees of freedom, including the type of quark. You cannot just apply it to each quark flavor individually.