# Vacuum expectation value of vector proper fields always zero

Let $$\phi$$ a non-scalar vector field. Why the Lorentz invariance of vacuum expectation value has as consequence that the vacuum expectation value $$v=\langle 0|\phi(x)|0\rangle$$ should be zero?

I conjecture that if $$v \neq 0$$ then something is going wrong with rotations, which form a subgroup of the Lorentz group, but what precisely it's going wrong?

Say $$U \neq I$$ is a non trivial rotation which acts by definition on vacuum expectation value as I know as $$\langle 0|U \phi(x) U^{\dagger} |0\rangle$$. How does the action of $$U$$ change $$v$$ if it's not zero?

Your idea is right. Consider a multiplet of fields $$\phi_i$$ that transforms under some representation of the Lorentz-Group, i.e., $$$$U \phi_i(x) U^{-1} = \mathcal{D}_{ij} \phi_j (\Lambda x).$$$$
Now we use that the vacuum is invariant under any Lorentz transformation, $$$$\langle 0 | \phi_i (x) | 0 \rangle = \langle 0 | U \phi_i (x) U^{-1}| 0 \rangle = \mathcal{D} _{ij} \langle 0 | \phi_j (x) | 0 \rangle.$$$$ Now, this is only possible if the multiplet transforms under the trivial representation or the vacuum expectation value is zero. Hence only scalar fields can have a non-zero VEV.
• how do you get the identity $U \phi_i(x) U^{-1} = \mathcal{D}_{ij} \phi_j (\Lambda x)$? How is $\Lambda$ related to $U$? Sep 27, 2021 at 18:37
• so $U= U(\Lambda)$ is the image of $\Lambda$ (as element of Lorentz group) under the representation map $U: L \to O(H)$ (with $H$ = Hilbert space of the vector fields $\phi$) in sense of Wigner's interpretation of spin representation? Sep 27, 2021 at 20:23
• Yes, that's correct. $U$ is a unitary representation of the Lorentz group. Since there is no finite-dimensional unitary representation it must be infinite-dimensional. $\mathcal{D}_{ij}$ is a finite-dimensional representation of the Lorentz group. For the vector representation (or fundamental representation) you have for instance $\mathcal{D}_{ij} = \Lambda_{\mu}^{\;\nu}$, while for the bispinor representation you have $D_{ij} = e^{i/2 \omega_{ij} \sigma_{ij}}$. Sep 28, 2021 at 7:43