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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?

Joman
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1 Answers1

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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.

Edward Hughes
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Arnold Neumaier
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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 08 '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