Does free phonon propagator always vanish at zero momentum? When discussing electron-phonon interaction, the simplest leading electron self-energy usually has only the simplest phonon bubble diagram and excludes the 'tadpole' diagram below with a phonon propagator $D(q=0)$ connected to an electron loop. The argument is $D(q=0)$ should vanish somehow because

*

*the phonon field $\phi(x)$ is a spatial divergence, or


*$q=0$ phonon is just an overall translation.

This is what many textbooks say (e.g., page 82 top of Mahan's book (3rd ed.) and page 401 top of Fetter & Walecka's book). But it is not clear to me whether this is only true for acoustic phonons. How about optical phonons? These books seem to only vaguely refer to 'electron-phonon interaction' in terms of this issue.
 A: TL;DR: The answer to OP's title question is Tadpoles does not necessarily vanish.

*

*Tadpoles $\langle\tilde{\phi}(k)\rangle_{J=0}$ can only be non-zero for crystal momenta $k\in\tilde{\Lambda}$ in the reciprocal lattice due to momentum conservation, cf. Ref. 1.


*Phonon tadpoles describe global translation/deformation/permanent strain of the lattice structure, e.g. thermal expansion. There can be both acoustic and optical phonon tadpoles, cf. e.g. Refs. 3-4. Optical phonon tadpoles are associated with that different parts of a unit-cell may deform differently.


*Ref. 2 argues that the phonon field $\phi =  \vec{\nabla}\cdot \vec{d}$ is a spatial divergence, so that the tadpole has to vanish. It seems the conclusion is rather the opposite: A possible tadpole may break down the spatial divergence picture.


*Tadpoles can vanish automatically if the system has symmetry, e.g. $\mathbb{Z}_2$-symmetry $\phi\to-\phi$, or charge conjugation symmetry, cf. Furry's theorem.


*The vanishing of tadpoles is often imposed as a renormalization condition, cf. e.g. my Phys.SE answer here.
References:

*

*G.D. Mahan, Many-Particle Physics, 2000; p. 82.


*A.L. Fetter & J.D. Walecka, Quantum Theory of Many-Particle Systems, 2003; p. 401-402.


*David Snoke, Solid State Physics: Essential Concepts, 2020; p. 471 exercise 8.10.1.


*L. Paulatto, I. Errea, M. Calandra & F. Mauri, arXiv:1411.5628; p. 1.
A: The argument here is physical rather than mathematical: phonons, by their physical nature, are deformation waves, and vector $q$ describes how deformation is distributed within the material. $q=0$ means that the deformation is spatially independent, i.e., that the deformation either does not exist or the material is shifted as a whole.
However, let me point out the following:

*

*Given different possible nature of the deformation (acoustic vs. optical phonons, transverse versus longitudinal polarization) one should be cautious about interpreting $q=0$ as translation. It is better to assume that we are working from the beginning in the center-of-mass reference frame and so deformation is absent.

*It is implied that $q$ lies in the first Brilouin zone. If not - as @Qmechanic has correcly pointed out, propagator is periodic in respect to the reciprocal lattice translations.

*Same argument with the appropriate adjustments may work for many other types of collective excitations.

