Everything I've read online seems to say/imply that the Pauli exclusions principle only applies to fermions and not bosons.

As I understand it the Pauli exclusion principle arises when the spatial wavefunction is antisymmetric about particle interchange. I also thought that both fermions and bosons can have antisymmetric spatial wavefunctions as long as they have antisymmetric and symmetric total wavefunctions respectively. So if I am right in thinking that bosons can also have antisymmetric spatial wavefunctions then why does the Pauli exclusion principle not apply to them in this case?

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    $\begingroup$ You are correct that, if the spin part of a bosonic system happens to be antisymmetric, then the spatial part also has to be antisymmetric, and then some analogy of PEP would apply to them. But the salient feature of fermions that we point to PEP applying to them is that no two fermions may occupy the same quantum state, whereas bosons are actually attracted to being in the same quantum state, so that the spatially symmetric states ought to vastly outnumber the antisymmetric ones. Since the behaviour is so different, we do not want to claim that PEP applies to bosons. $\endgroup$ Commented May 5, 2023 at 19:06
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    $\begingroup$ The above comment is wrong. The spatial wavefunction on its own can be symmetric or anti-symmetric for fermions. It can also be symmetric or anti-symmetric for bosons. The difference is that fermions follow the Pauli exclusion principle (total wavefunction is anti-symmetric) while bosons do not (total wavefunction is symmetric). The closest thing to an answer for "why" that is would be looking at a proof of the spin-statistics theorem. $\endgroup$ Commented May 5, 2023 at 19:13
  • $\begingroup$ @ConnorBehan I specifically worded my comment above to cover the thing you were talking about. Instead, you are picking a technicality on the definition of PEP to dispute OP, which is not a useful thing to do. $\endgroup$ Commented May 6, 2023 at 2:59
  • $\begingroup$ The OP needs to be disputed because the PEP is about the full state. Saying "some analogy" of it applies to bosons is only correct if by that you mean "the exact opposite". Bose-Einstein condensation also assumes the atoms are weakly interacting and defining the PEP in this way is needlessly complicated. $\endgroup$ Commented May 6, 2023 at 12:35

4 Answers 4


The Pauli Exclusion principle talks about the entire state, that is, it's impossible for Fermions to occupy the same quantum state at the same time. It is indeed a direct consequence of the Spin-Statistics theorem that arises from Quantum Field Theory, but I don't think going that deep is necessary for what you want.

What is "spin statistics" at a surface level? Well, it says that a state of say two Fermions $|\psi_1\psi_2\rangle$ must be antisymmetric by exchange, that is, $|\psi_2\psi_1\rangle=-|\psi_1\psi_2\rangle$. How does this lead to the exclusion principle? Suppose the two fermions try to each occupy the same state, such that the total state is $|\psi_1\psi_1\rangle$, well they are still fermions so spin statistics tells us that under exchange we must have $|\psi_1\psi_1\rangle=-|\psi_1\psi_1\rangle$ the only solution to this is zero, such a state is not possible. This is the exclusion principle. (Notice that such a state is perfectly possible for Bosons, since spin-statistics says they are symmetric under exchange).

Going back to what you said. It is essentially correct, in principle, you can have Bosons described by an antisymmetric wave-function under exchange. But you would need another part of the state, say the spin component, to be antisymmetric as well in order for the entire state to be symmetric. This does not mean the Exclusion Principle is valid for Bosons, because it never said anything about specifically the spatial component of the state, it was always about the $\textbf{entire state}$.


If, for two identical bosons labelled 1 and 2, the total state is a product of spatial and spin parts, and if the spin part is antisymmetric, then, yes, the spatial part has to be antisymmetric and therefore it cannot take the form $| \phi \rangle_1 | \phi \rangle_2$. In other words you can't have two identical bosons in the same spatial state if their spin state is antisymmetric (and we are considering only the case where the total state is a product of spin and spatial parts as opposed to an entanglement).

This is not the Pauli Exclusion Principle because that principle is about the entire state. And anyway you could also point out that two identical fermions can be in the same spatial state if their spin state is antisymmetric. Indeed this situation is quite common; the ground state of helium (and many other atoms) is an example.


The question, answered by the Spin Statistics Theorem has three parts,


why there are only two types of tensor products of one variable as a basis of permutation symmetric functions in n variables, representing pure states for n identical particles.

This part is easy.

Physical expectations are quadratic in the wave functions and the observable physical numbers in experiments, obtained from an obervable linear operator $O$ in Hilbert space, are symmetric with respect to permutation of the variables in the permutation symmetric linear operator $O$ $$O(t,\ x_1\dots,\ x_n)=O(t,\Pi(\ x_1\dots,\ x_n),$$ where $\Pi$ means any renumbering of the finite set $\text{range}(1,n)$ $$\bigl< \ \psi(t,\ x_1\dots,\ x_n), \quad O( t,\ x_1\dots,\ x_n) \cdot \phi(t,\ x_1\dots,\ x_n) \ \bigr> = \bigl< \ \psi(t,\Pi( x_1\dots,\ x_n)), \quad O( t,\Pi( x_1\dots,\ x_n)) \cdot \phi(t,\ ( x_1\dots,\ x_n))\bigr>$$

Its evident, that this requirement is satified exactly if and only if for any two state functions $\phi, \psi$ with two neigbouring coordinates exchanged, in any fixed permutation, produce a factor in the state functions, that cancels out, while all observables have to be completely symmetric

$$\psi(t, \dots , x_j,x_k, \dots) = q_{jk} \ \psi(t, \dots , x_k,x_j, \dots) $$ with $$ \prod_{\text{any subsets of order 2}} |q_{kl}|^2 =1 $$

Its a subtle proof to show, that the value of q can be $\pm1 $ only for elementary systems.

The subsets of coordinates with $q=1$ are in the boson category, the antsymmetric ones are the fermion category.


Bosonic coordinates can be repeated, meaning that state functions can be build from symmetric sums of monoms of any power $$\psi = f_1(x)^nf_2(y)^m + f_1(y)^nf_2(x)^m $$
Fermionic coordinates can arise only once and products have to be antisymmetic in pair products $$ \psi = f_1(x) f_2(y)\ -\ f_1(y)f_2(x))$$

Keep in mind, that fixed bound pairs of fermions behave as bosons, because fixed pairs commute pairwise with any other single wave function with a factor of 1. So the pair in a given order can be shifted to be part of the bosonic factor of all products.


All free single particles states represent the Poincaré group of translations in space and time and spatial rotations.

The Lorenz boost representation is a very difficult task. So fix the system of reference as a universal space time at rest.

In quantum theory, this a the normal - in contradistinction to classical mechanics - because per definition the quantum systems occupy all of space (for a very short time). Boost representations are applied to the labaratory as a whole and not locally only to the set of variables.

The irreducible representions of the rotation group $SO(3)$, that cannot divided into invariant subspaces, are space of constant eigenvalues of the euclidean square of the three angular momentum operator$(j_1,j_2,j_3)$, that generate the rotations by $$e^{i \vec j \cdot \vec n}$$ with rotation plane perpendicular to $\vec n$ and rotation angle $\phi=|\vec n|$

And now, again, as for permutions, a full rotation around any axis by $e^{2 \pi i j_3}=\pm \mathbb 1 $ can be of both signs $\pm 1$ because of the bilinear expectations.

This again yields categories of eigenvalue of $(2j)^2$ as series of squares of even and odd integers.

Physical experience: Fermions carry charge and magnetic moment.

The magnetic moment of the coordinates is geometrically quantized in integer multiples of the Bohr magneton, coupled rigidly to the multiples of $\hbar$ of the angular momentum squared as the finite dimensions $2n+1$ with $L_z=0,m1, \dots \pm l $ of the Hilbert space with constant $j^2=L(L+1)$.

In a constant magnetic field, ferminoic waves, even with spatial angular mammentum zero interact with the magnetic field as a particle with angular momentum $\hbar/2$.

The splitting of the eigenvalues demands that there are two different wave functions, that allow for two identical space functions in product with different spin values.

In a sense the space can be thought to be doubled, by adding a fourth coordinate with values $\pm\frac{1}{2}$

The abstract proof of the Spin-Statistics Theorem cannot be done in nonrelativistic quantum theory. In a subtle way the first coordinate, the common general time $t$, is transferred to all space coordinates of all particle coordinates by a Lorrentz boost in any of the three space directions.

This fact finally, by continuity in complex space time at the origin, yields the famous theorem, that irreducible representations of the full Poincaré group of free fields are products of bosonic fields with integer spin and fermionic fields with half integer spin.

The half integer fields are built on 2-component wave functions carrying the representation $SL(2,C)$ of rotations and boosts.

The two equal (free) or different (magentic) wave functions in their components carry the representation of space-time translations.


The basis of the Pauli principle is the observation that in atoms, electrons that are identical in the three quantum numbers differ in a fourth quantum number. This was concluded from the splitting of the spectral lines of the excited electrons under the additional influence of a magnetic field.

As is known, electrons are equipped with a magnetic dipole. The direction of the dipoles of the electrons are changed under the influence of the external magnetic field in its direction. Since this results in different energy levels depending on the position of the electrons in the atom, the emission lines for otherwise identical electrons are split.

When we talk about the Pauli principle, we are always talking about bound electrons in the atom.

Photons do not have a bound state. They can be emitted in such a way that they are polarised and thus their electric and magnetic field components oscillate in the same direction (as with radio waves). However, a mutual influence cannot be observed at "normal" energy density.

If one knows the historical context, it seems absurd to want to apply to bosons something that applies to bound electrons in the atom.


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