# Lorentz invariance and the vacuum expectation value of fields with spin > 0

I had a question about Moduli space, which I was reading about here, but then I read this sentence:

"Lorentz invariance forces the vacuum expectation values of any higher spin fields to vanish."

Can someone explain how exactly this happens? Or at least suggest an exercise to carry out?

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## 1 Answer

There is a Lorentz transformation that maps a spacelike vector $u$ to $-u$. If $A(x)$ is a field of spin 1 with $\langle u \cdot A(0)\rangle = c$ then applying the Lorentz transform we find $-c=c$ and hence $c=0$. Doing this for all spacelike vectors implies $\langle A(0)\rangle = 0$, and translation invariance then gives $\langle A(x)\rangle =0$ for all $x$.

For other spins the argument is similar. You are welcome to try the spinor case as an exercise.

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Thanks Arnold, but can I ask two silly questions before I attempt the spinor case. Why did you choose a space-like vector to dot it with A(0)? Is this how a spin 1 field is minimised in general? –  Joman Jun 8 '12 at 17:42
Because timelike vectors form an orbit, hence cannot be changed in their sign only using a Lorentz transformation. So one has to work with space-like vectors, where (exercise) this is possible. –  Arnold Neumaier Jun 10 '12 at 10:01
Okay, I have too many questions now! So you used vector.vector to produce a scalar and showed this will be zero due to Lorentz invariance. What about chiral condensates where you have $<\bar{\psi} \psi> \neq 0$? –  Joman Jun 19 '12 at 14:13
@Joman: $\overline\psi\psi$ is a scalar, hence you cannot apply the same argument. –  Arnold Neumaier Jun 19 '12 at 15:23