Monday, 16 May 2016

symmetry breaking - Why is a nonzero VEV for a spinor field said to break Lorentz invariance?


Consider a spinor field $\psi(x)$. Its vacuum expectation value is given by $$v=\langle 0|\psi(x)|0\rangle.$$ Using the fact that the vaccum is invariant under Lorentz transformation, we get, $$v=\langle 0|\psi(0)|0\rangle.$$ Why is it that, if $v\neq 0$, the Lorentz invariance is broken?



Answer



The $v$ you write is itself a spinor, not a scalar. A non-zero spinor is obviously not invariant under Lorentz transformations, so a non-zero spinorial VEV breaks Lorentz invariance of the 1-point function.


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