I've started reading many-particle systems in Quantum mechanics, and came across the concept of identical particles vs distinguishable particles.

However, I'm wondering, what happens in case of a set of distinguishable fermions. Are their any special rules, like the pauli exclusion principle for identical fermions, that we need to keep in mind while filling up these particles ?

Say, we have 5 distinguishable fermions of same mass in a $1D$ harmonic oscillator. Since the particles are distinguishable, we can use separation of variables to separate the wavefunctions for the $5$ fermions.

Suppose the system is in its ground state. Hence, the $5$ distinguishable fermions must also be in their respective ground states. However, since all of them have the same mass, they would have the same ground state energy level. Thus, in the system, we have an energy level that has $5$ distinguishable fermions in the ground state.

Aren't the fermions in here, just behaving as identical bosons would do in the same potential ? Is my intuition correct, or does the fermions not fill up their respective ground states in this manner. If they were identical, this would not have been the case, as it would have violated the exclusion principle. However, does the exclusion principle come to play even in case of distinguishable fermions.

From the energy level perspective, Distinguisable fermions seem to act exactly the same way as any distinguishable particles. The only difference comes in the wave functions, as we now have to consider the spin states too. However, from the energy level perspective, am I correct in saying that identical bosons, distinguishable particles and distinguishable fermions of the same mass, have the exact same energy values for different states and only their wave functions are different ?

  • $\begingroup$ I don't know any distinguishable fermions of the same mass. Is mass-equality a requirement? If not, you call look at muonic atoms or the deuteron. $\endgroup$
    – JEB
    Aug 2, 2021 at 14:54

1 Answer 1


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.

  • $\begingroup$ Thank you so much. However, I was wondering about the spin part. If we ignore that spin part, the wave function for distinguishable fermions should look the same as that for any distinguishable particles. The spin wavefunctions makes the difference here. However, for distinguishable fermions, is the spin wavefunction going to be antisymmetric? $\endgroup$ Aug 2, 2021 at 15:18
  • $\begingroup$ For example, say there are two fermions in a 1D well, one up spin and 1 down spin. Thus they are distinguishable, and their spacial wavefunction would just be a multiplication of both the individual spatial wavefunctions. However, I read that the spin is going to be a singlet state. I wonder why that is so. It has been done here : link $\endgroup$ Aug 2, 2021 at 15:20
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    $\begingroup$ @NakshatraGangopadhay Let $m\in\{$up,down$\}$. For two spin-1/2 fermions of the same species, the wavefunction must satisfy $\psi(\mathbf{x},m,\mathbf{x}',m')=-\psi(\mathbf{x}',m',\mathbf{x},m)$. If $m\neq m'$, then the function $\psi(\mathbf{x},m,\mathbf{x}',m')$ and $\psi(\mathbf{x}',m,\mathbf{x},m')$ don't need to be related to each other in any special way. But if the Hamiltonian is such that the lowest-energy state is symmetric under $\mathbf{x}\leftrightarrow\mathbf{x}'$, then it must be antisymmetric under $m\leftrightarrow m'$ (that is, must be a singlet state). $\endgroup$ Aug 2, 2021 at 15:35

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