# Commutation relations for bosons and fermions

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## Main Question or Discussion Point

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.

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You can't create two fermions at the same place at the same time (but I bet you already knew that). $\psi (x) \psi (x) + \psi(x)\psi(x)=0$ so $\psi(x)\psi(x)=0$

Bosons, on the other hand, "like" to be in the same state.

You can't create two fermions at the same place at the same time (but I bet you already knew that). $\psi (x) \psi (x) + \psi(x)\psi(x)=0$ so $\psi(x)\psi(x)=0$

Bosons, on the other hand, "like" to be in the same state.
They are not created at the same place. Did you notice the prime on the x's`?

But $x$ and $x^\prime$ are arbitrary. They can be equal. When they are equal then they are created at the same place. So that is part of the physical meaning: they cannot be created at the same place. The commutation and anti-commutation relations reflect the Pauli principle, I believe.

Anyway, the full relation is $\psi(x)\psi(x^\prime)+\psi(x^\prime)\psi(x)=\delta(x-x^\prime)$ so that reflects equality of $x$ and $x^\prime$