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In this post Arnold Neumaier wrote on necessity of vanishing vacuum expectation value if we want to have a particle interpretation of our field $\phi$:

As is easily checked, fields linear in creation and annihilation operators (and hence amenable to a particle interpretation) have zero vacuum expectation value. Thus the $\phi$ field with its non-vanishing vacuum expectation value cannot be given a particle interpretation.

I not see why it's necessary if we want have a meaningful interpretation as particle to demand that the vacuum expectation value should be zero. What is precisely running wrong if we deal with a field with non-vanishing expectation value physically and mathematically?

What are moreover in general the main drawbacks when we dealing with field theories with non-vanishing expectation values? I saw several times stated that this causes also troubles if we apply perturbative methods. Could somebody give a concrete example where perturbative calculations/reasonings going wrong if we apply them to such field theories?

Qmechanic
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user267839
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1 Answers1

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First, the only kinds of fields that can possibly have non-zero vacuum expectation values are scalar fields. This follows by Lorentz invariance.

Now let $\phi(x)$ be such a scalar field. Let us consider $\langle \phi(x)\rangle$. Let us first use translation invariance to argue that this must be constant. Indeed

$$\langle \phi(x)\rangle = \langle 0|\phi(x)|0\rangle = \langle 0|e^{-iP\cdot x}\phi(0)e^{iP\cdot x}|0\rangle=\langle 0|\phi(0)|0\rangle = \langle \phi(0)\rangle\tag{1}.$$

The main point here is that the vacuum is Poincaré invariant, and therefore $e^{iP\cdot x}|0\rangle = |0\rangle$.

Now we want to talk about particle interpretation. To do so we must look to the asymptotic region of spacetime and connect $\phi(x)$ properly with in/out fields $$\phi(x)\to \sqrt{Z}\phi_{\rm in/out}(x)\tag{2}$$

where we recall that $$\phi_{\rm in/out}(x)=\int \dfrac{d^3p}{(2\pi)^32\omega_p}(a_{\rm in/out}(p)e^{ipx}+a_{\rm in/out}^\dagger(p)e^{-ipx})\tag{3}.$$

Now because of (1) we can evaluate $\langle \phi(x)\rangle$ wherever we want. So push the point to the asymptotic region. There you can use (3) to write $$\langle \phi(x)\rangle = \sqrt{Z}\langle \phi_{\rm in/out}(x)\rangle\tag{4}$$

and now you can use (3). But now there is a tension. In fact, $a_{\rm in/out}(p)$ and $a^\dagger_{\rm in/out}(p)$ should annihilate the vacuum, when acting respectively to the right and to the left, and so the result should be zero. So what is wrong with the reasoning? Well, it is just that we are looking at the wrong variable. Define $$\varphi(x)=\phi(x)-\langle \phi(x)\rangle\tag{5}.$$

This field has zero vacuum expectation value and it does not run in that problem. So my take on this is that:

  1. Particle interpretation comes from (2) and (3) and you run into one problem if you try applying this to a field with non-zero VEV;

  2. Nevertheless there is nothing wrong. There is a particle interpretation associated to this field, provided you look to the right variable. Using (5) the appropriate variable is the deviation from the VEV. And in a sense this makes sense: particles are meant to be excitations about a ground state and that is exactly what $\varphi$ should capture.

Nihar Karve
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Gold
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  • Thank you for the answer. I have two questions.
    1. Why proper vector/tensor fields must have zero

    vacuum expectation values as a consequence of Lorentz invariance? Assume that $0 \neq v= \langle 0|\phi|0\rangle $ for vector field $\phi$. Then $v$ is vector. Lorentz group contains shift and proper rotation operations in vector case. I guess that in non scalar case something is going wrong with rotations. Assume $U$ is such (non trivial) rotation. Then $U$ acts on $\langle 0|\phi|0\rangle $ by $\langle 0|U \phi U^{\dagger} |0\rangle $. What does $U$ with $v$ if it's non zero?

     – user267839 Sep 27 '21 at 17:03
  • You wrote in 1. that one runs into problems if one tries to

    • apply it to fields with VEV non-zero. Could you sketch what fails there concretely? Would $a_{\rm in/out}(p)$ not be able to annihilate the vacuum in that case?

    – user267839 Sep 27 '21 at 17:03
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    Regarding your first question, consider a field $\Phi$ which transforms in some representation of the Lorentz group with non-zero spin. The vacuum $|0\rangle$ is, by supposition, Poincaré invariant, so it is a state with zero spin. When you act with the field $\Phi(x)|0\rangle$ you produce a state which has the same spin as the field. Such a state is orthogonal to the vacuum, which transform in the spin zero representation. That would be the argument. – Gold Sep 27 '21 at 17:23
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    Regarding your second question, in/out fields must have zero VEV because they are linear in creation and annihilation operators. But since the VEV is actually constant you can compute it on any point. If (4) holds the LHS will be non-zero and the right-hand side will be zero. The conclusion is that (4) can't hold for a field with non-zero VEV and in my opinion this justifies the claim you quote that such a field does not have a particle interpretation, since (4) is the key to particle interpretation. To get a particle interpretation you must study deviations from the VEV. – Gold Sep 27 '21 at 17:27
  • another question which is not directly related to this problem, but which deals with the expansion of in/out field in terms of creation and annilation

    $$\phi_{\rm in/out}(x)=\int \dfrac{d^3p}{(2\pi)^32\omega_p}(a_{\rm in/out}(p)e^{ipx}+a_{\rm in/out}^\dagger(p)e^{-ipx})\tag{3}$$

    Is there a deeper reason involved why to $a_{\rm in/out}(p)$ is associated the exponential $e^{ipx}$ (and to $a^{\dagger}_{\rm in/out}(p)$ the $e^{-ipx}$ with minus sign) and not vice verse? Can it be formally "proved" or is it just a convention? (every book/script I read used it without explanation)

     – user267839 Sep 27 '21 at 18:49
  • This follows from translation symmetry. I strongly suggest you give a look into chapter 5 of Weinberg's "The Quantum Theory of Fields". In summary, if you look for fields linear in creation and annihilation operators which under the Poincaré action defined on the Hilbert space transform in some definite representation of the Lorentz group you'll find that the transformation law of the creation and annihilation operator lead to that. – Gold Sep 27 '21 at 20:42