Why are fermion states anti-symmetric under exchange operator?

[LCSphysicist]

Homework Statement

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Relevant Equations

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Let $L$ be the state space of two identical/indistinguishable particles. Let $L \otimes L$ be the state space of the combined system formed by both particle.

If the particles were distinguishable, $LxL$ would have four mutually orthogonal states: $|\phi\rangle|\phi\rangle, |\omega\rangle|\omega\rangle, |\phi\rangle|\omega\rangle, |\omega\rangle|\phi\rangle$.

But, if the particles are bosons, we have actually three states: $|\phi\rangle|\phi\rangle, |\omega\rangle|\omega\rangle, |\omega\rangle|\phi\rangle + |\phi\rangle|\omega\rangle$

My question is for the fermions case, the only state would be $|\omega\rangle|\phi\rangle - |\phi\rangle|\omega\rangle$

But i can not understand why. (I think i understand the signal, it is because system of fermions are anti symmetric under exchange operator, right?)

OBS (This is being used in the book to introduce/derive the Pauli exclusion, so it is not a good idea to answer "because Pauli exclusion (...)")

 

[PeroK]

But i can not understand why. (I think i understand the signal, it is because system of fermions are anti symmetric under exchange operator, right?)

Yes. Witin QM the anti-symmetry of mutiple fermion states is taken as an axiom/law of nature. It can be derived within QFT.

Symmetry and anti-symmetry laws of nature tend to dominate particle physics. For example, it's local-gauge invariance (a form of symmetry requirement) that brings together charged particles and the electromagnetic field.

You should start to see these symmetry axioms in the same light at $F = ma$ and Coulomb's law etc. Fundamental laws that we take as a starting point.