Why do bosonic particles behave the way they do? Why doesn't the Pauli exclusion principle affect the electrons, protons, and neutrons of the atoms at temperatures close to absolute zero? How does spin factor in?


How can an atom composed of fermions (electrons, protons, and neutrons) behave as a boson, despite the Pauli exclusion principle?

The answer is related to entanglement. The Pauli exclusion principle still applies to the atom's constituents, but this does not prevent many of these atoms from occupying the same state, because the fermionic constituents are entangled with each other.

Here's a toy model that demonstrates the basic idea. Suppose that $e_n$ is the operator that creates an electron with a wavefunction specified by $n$, and and suppose that $p_k$ is the operator that creates an electron with a wavefunction specified by $k$. Saying that the electron and proton are fermions means that all of these operators anticommute with each other: \begin{align*} \big\{e_n,\,e_k\big\} &= 0 \\ \big\{p_n,\,p_k\big\} &= 0 \\ \big\{e_n,\,p_k\big\} &= 0 \tag{1a} \end{align*} with $$ \{A,B\}\equiv AB+BA. \tag{1b} $$ These anticommutation relations imply $$ e_n e_n = 0 \hskip2cm p_k p_k = 0, \tag{2} $$ which expresses the Pauli exclusion principle: we cannot put two electrons in the same state, or two protons.

As a toy model of a bosonic atom composed of two fermions (such as a hydrogen atom composed of an electron and a proton), suppose that the operator $$ a(f) = \sum_{n,k} f_{nk} e_n p_k. \tag{3} $$ creates a single-atom state. The coefficients $f$ characterize the atom's internal structure as well as its external characteristics (such as its overall momentum). The fermion anticommutation relations (1) imply that the operators (3) satisfy the boson commutation relations $$ \big[a(f),\,a(g)\big] = 0 \tag{4a} $$ with $$ [A,B]\equiv AB-BA. \tag{4b} $$ This doesn't answer the question yet, though. To answer the question, we have to prove that we can put two of these atoms in the same state without annihilating the state. In other words, we have to prove $$ a(f)\,a(f) \neq 0. \tag{5} $$ To prove this, use the preceding equations to get \begin{align*} a(f)\,a(f) &= \sum_{n,k} f_{nk} \sum_{m,j} f_{mj} e_n p_k e_m p_j \\ &= -\sum_{n,m} \sum_{k,j} f_{nk} f_{mj} e_n e_m p_k p_j. \tag{6} \end{align*} If the coefficients $f_{nk}$ had the form $$ f_{nk} = w_n z_k, \tag{7} $$ then the anticommutation relations (1) would imply that (6) is zero. But when we consider a system of atoms at low temperature (for example), then the coefficients $f$ do not have the form (7), because the atom has an overall momentum concentrated near zero. This implies that the atom cannot be sharply localized, which in turn implies the electron and proton are entangled with each other. In the generic case when the coefficients $f$ do not have the form (7), the quantity (6) will not be zero.

To see this more simply, suppose that each index takes only two values, say 1 and 2, and that $$ a(f) = \sum_{n,k} f_{nk} e_n p_k = e_1 p_2 + e_2 p_1. \tag{8} $$ Then the anticommutation relations (1) imply \begin{align*} a(f)\,a(f) &= (e_1 p_2 + e_2 p_1)^2 \\ &= 2 e_1 p_2 e_2 p_1 \\ &= 2 e_1 e_2 p_1 p_2. \tag{9} \end{align*} Terms with two factors of $e_1$ are gone because of the Pauli exclusion principle (2), and likewise for all other terms with two identical fermion factors. However, because of the entanglement between $e$ and $p$ in equation (8), a cross-term survives.

In the simplified case (8), putting three atoms in the same state would give zero, because then we cannot avoid having at least two identical fermions in each product. However, if we use a more realistic model in which the sum in (3) has infinitely many terms (because the "sum" includes an integral over the electron/proton momenta, for example), then we can put an unlimited number of atoms in the same state, because there will always be plenty of cross-terms that survive.

In summary, an atom made of fermions can behave as a boson because the constituent fermions are entangled with each other. This entanglement becomes extreme at low temperatures, where the overall wavefunction of the atom becomes highly delocalized.

How does spin factor in?

Spin is not needed for understanding this phenomenon. The spin-statistics theorem in relativistic QFT says that fermions have half-integer spin and bosons have integer spin; but even if we use a nonrelativistic model that ignores this connection, an atom made of fermions can still behave as a boson.


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