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For the free boson, the field operators satisfies the commutation relation,

$${\varphi}_{x'}{\varphi}_{x} - {\varphi}_{x}{\varphi}_{x'} = 0$$ at equal times.

While the fermions satisfies,

$${\psi}_{x'}{\psi}_{x} + {\psi}_{x}{\psi}_{x'} = 0$$ at equal times.

I interpret ##{\varphi}_{x}## and ##{\psi}_{x'}## as creating a boson and a fermion at position x and x' respectively.

But what is the physical interpretation of the commutations relations? I'm trying to relate it to the fact that fermions changes sign when any two fermions are interchanged, while bosons do not.

$${\varphi}_{x'}{\varphi}_{x} - {\varphi}_{x}{\varphi}_{x'} = 0$$ at equal times.

While the fermions satisfies,

$${\psi}_{x'}{\psi}_{x} + {\psi}_{x}{\psi}_{x'} = 0$$ at equal times.

I interpret ##{\varphi}_{x}## and ##{\psi}_{x'}## as creating a boson and a fermion at position x and x' respectively.

But what is the physical interpretation of the commutations relations? I'm trying to relate it to the fact that fermions changes sign when any two fermions are interchanged, while bosons do not.

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