# 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?

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.
• @ACM But are 1-point functions related to measurables. Right? The LSZ reduction formula does not contain 1-point function (but $n$-point functions in general with $n>1$) and therefore, is not related to scattering amplitude. Am I wrong? So if 1-point functions are not measurables, should one care? – SRS Jan 18 '17 at 7:54
• @SRS the derivation of the LSZ formula explicitly assumes that $v\equiv 0$. – AccidentalFourierTransform Jan 18 '17 at 14:09
• @SRS because if $v$ is not invariant then $U(\Lambda)$ cannot exist (as in my answer below), which means that Lorentz transformations are not a symmetry of the theory (and, in particular, of the $S$ matrix). – AccidentalFourierTransform Jan 18 '17 at 14:16
To make ACM's argument more explicit, consider \begin{align} v&=\langle 0|\psi|0\rangle\\ &=\langle 0|\overbrace{UU^\dagger}^1\psi\overbrace{UU^\dagger}^1|0\rangle\\ &=\overbrace{\langle 0|U}^{\langle 0|}\overbrace{U^\dagger\psi U}^{D_\Lambda \psi}\overbrace{U^\dagger|0\rangle}^{|0\rangle}\\ &=D_\Lambda v \end{align} where $U=U(\Lambda)$ is the unitary operator that corresponds to Lorentz transformations in the Hilbert space, and $D_\Lambda$ its representation in the space of spinors.
Considering $\Lambda$ to be, say, a rotation around the $z$ axis with angle $\theta$, and expanding to first order in $\theta$, we get $$S^zv=0$$ which is impossible for representations of the Lorentz Group with half-integer spin, as $S^z$ has eigenvalues $$-j,-j+1,\cdots,+j$$ none of which is zero.
Therefore, we must conclude that $U(\Lambda)$ doesn't exist, that is, the Lorentz symmetry is broken.