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Is it possible to have a VEV (vacuum expectation value) for tensor field? I am mainly concerned about second rank tensors. It seems it can have a VEV which will be proportional to the metric tensor (in flat space-time). Does this make any sense at all? If it does, then what are it's implications?

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If we assume, as usual, that the vacuum state associated to state vector $|0\rangle$ is invariant under a unitary representation of (connected component subgroup of) Poincaré group $SO(3,1)_+\ni (\Lambda,a) \mapsto U(\lambda, a)$, we can write $U(\Lambda, a)|0\rangle = e^{i \alpha(\Lambda,a)}|0\rangle$ for some, irrelevant in the following, phases $\alpha(\Lambda,a)\in \mathbb R$.

A tensor operator $\hat{T}(x)$ transforms under the action of $U$ as follows: $$U(\lambda, a)^\dagger \hat{T}(x) U(\lambda, a) = \sum_{l} \sum_{B_l}L_l(\Lambda)^{A_l}_{B_l} \hat{T}^{B_l}_l(x-a)\:;$$ where the tensor field has been decomposed into its irreducible representation components with respect to the action of Lorentz group. The $L_l$ are the (finite dimensional) matrices describing the action of the group in the irreducible representation $V_l$ where the respective $\hat{T}_l(x)$ stay. We can therefore evaluate the VEV taking the invariance condition $U(\Lambda, a)|0\rangle = e^{i \alpha(\Lambda,a)}|0\rangle$ into account. $$\langle 0| \hat{T}(x) |0\rangle = \langle 0|U(\lambda, a)^\dagger \hat{T}(x) U(\lambda, a) |0\rangle\:.$$ That is, $$ \sum_{l} \sum_{B_l}L_l(\Lambda)^{A_l}_{B_l} \langle 0| \hat{T}^{B_l}_l(x-a)|0\rangle - \langle 0| \hat{T}^{A_l}_l(x)|0\rangle =0\:.$$ As the representations act separately, this is equivalent to say that, for every $l$, $$\sum_{B_l}L_l(\Lambda)^{A_l}_{B_l} \langle 0| \hat{T}^{B_l}_l(x-a)|0\rangle - \langle 0| \hat{T}^{A_l}_l(x)|0\rangle =0 \quad \forall a \in \mathbb R^4, \Lambda \in SO(3,1)_+\:.\tag{1}$$ Choosing $\Lambda=I$ we have that $$\langle 0| \hat{T}^{A_l}_l(x)|0\rangle = t_l^{A_l}\quad \mbox{constant in $x$}.$$ Consequently (1) implies: $$\sum_{B_l}L_l(\Lambda)^{A_l}_{B_l} t^{B_l}_l = t^{A_l}_l =0 \quad \forall \Lambda \in SO(3,1)_+\:.\tag{2}$$ In other words, each set of components $\{t^{A_l}_l\}_{A_l}$ must define an invariant tensor under the action of the corresponding irreducible rep. The invariant tensors are all obtained using the metric $g_{\mu\nu}$ and the Ricci (pseudo)tensor $\epsilon^{\alpha\beta \gamma \delta}$ and all possible tensorial combinations of them. It is impossible to produce a four vector this way, so no non-vanishing VEV exist for vectors. A second order tensor $T_{ab}$ can always decomposed as follows into its $SO(3,1)_+$-irreducible representation parts: $$T_{\mu\nu} =\left[ \frac{1}{2}(T_{\mu\nu}-T_{\nu\mu})\right] \:+ \:\left[ \frac{1}{2}(T_{\mu\nu}+T_{\nu\mu}) - \frac{1}{4}g_{\mu\nu} T^{\alpha}_\alpha \right]\:+ \: \frac{1}{4}g_{\mu\nu} T^{\alpha}_\alpha\:.$$ The first term in the RHS is the antisymmetric part. There is no way to construct an $SO(3,1)_+$ invariant second order antisymmtric tensor out of the metric and Ricci pseudotensor. Therefore this part of the VEV must vanish. The same result applies to the second term, too. Only the third one may survive.

So the answer is YES, as you guess, you may have a non-vanishing VEV for a second rank tensor operator. However that VEV must necessarily be of the form $$\langle 0| \hat{T}_{\mu\nu}(x)|0\rangle = c g_{\mu\nu}$$ for some constant $c$.

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Interpreting your question rigorously, it is possible. Every tensor field has a VEV, but in most cases it is identity. :-)

The Higgs-field is the only known field which has a non-zero VEV, but it is a scalar field.

A vector-field with non-zero VEV damaged the isotropy of the vacuum.

A non-zero tensor field: the vacuum value of most tensor fields are $I$ in their ground state and this is the case for vacuum as well.

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  • $\begingroup$ I know that a vector or spinor field cannot have VEV because of lorentz invariance. I am interseted in what cases tensor fields don't have VEV=0 and their implication. $\endgroup$
    – user44895
    May 9 '14 at 16:32
  • $\begingroup$ @user44895: The statement holds true for any representation of the Lorentz group other than the trivial (scalar) one. But I suppose that the moment you have an anisotropic metric, you have already broken the Lorentz invariance of the field equations. $\endgroup$
    – Siva
    May 9 '14 at 19:48
  • $\begingroup$ Existence of Minkowski metric in flat space-time (which is the case I am considering) assures that lorentz invariance is not broken. See the question directly for better information. $\endgroup$
    – user44895
    May 9 '14 at 19:57
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You can decompose your tensor according to irreducible representations of your symmetry group. The parts that transform under trivial representation can get non-zero VEV without damaging the symmetry of vacuum.

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