Why are spinors not observables?

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  • Thread starter Silviu
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Hello! I am reading some QFT and it is a part about how causality implies spin-statistic theorem. In general, one needs 2 observables to commute outside the light-cone. For scalars, we have $$[\phi(x),\phi(y)]=0$$ outside the light-cone, and by using the operator form of the field you get that indeed scalars are bosons. However for spinors, we have that they don't commute, but the things that needs to commute in order to preserve causality is $$[\bar{\psi}(x)\psi(x),\bar{\psi}(y)\psi(y)]=0$$ and the author says that the spinors are not observables, hence why it is fine if they don't commute. I am not sure I understand why they are not observables. And why is the scalar field an observables? Is it because $$\phi(x)|0>=|x>$$? Also, why is $$[\bar{\psi}(x)\psi(x),\bar{\psi}(y)\psi(y)]=0$$ enough to impose causality? Thank you!
 

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  • #2
Demystifier
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All observables are hermitian operators. Scalar field for uncharged particles is a hermitian operator. Spinor field is not a hermitian operator. That's why!
 
  • #3
atyy
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Only self-adjoint operators are observables, So the commutation relation, which ensures that spacelike observables commute has to be between combinations of observables that are self-adjoint.
 
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Also, why is $$[\bar{\psi}(x)\psi(x),\bar{\psi}(y)\psi(y)]=0$$ enough to impose causality?
I wouldn't say that it's enough. One would also need to check similar commutators with other physically relevant bilinear combinations, such as ##\bar{\psi}(x)\gamma^{\mu}\psi(x)##. But once you checked it for ##\bar{\psi}(x)\psi(x)##, the explicit check for the other combinations should be easy.
 

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