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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?

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2 Answers 2

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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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  • $\begingroup$ @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? $\endgroup$
    – SRS
    Jan 18, 2017 at 7:54
  • $\begingroup$ @SRS the derivation of the LSZ formula explicitly assumes that $v\equiv 0$. $\endgroup$ Jan 18, 2017 at 14:09
  • $\begingroup$ @AccidentalFourierTransform Hmm. But if 1-point functions are Lorentz non-invariant, how does that make (the predictions of) the theory problematic? $\endgroup$
    – SRS
    Jan 18, 2017 at 14:13
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    $\begingroup$ @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). $\endgroup$ Jan 18, 2017 at 14:16
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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.

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  • $\begingroup$ But how do you that $S^z v = 0$? I don't see the expansion to 1st order to prove this $\endgroup$
    – Vicky
    Jan 14, 2021 at 20:45
  • $\begingroup$ @Vicky $D_\Lambda=e^{i\theta S^z}$, so expanding $D_\Lambda v=v$ to first order in $\theta$ we get $S^zv=0$. $\endgroup$ Jan 16, 2021 at 19:09

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