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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

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

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

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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.

xiaohuamao
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2 Answers2

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TL;DR: The answer to OP's title question is Tadpoles does not necessarily vanish.

  1. 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.

  2. 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.

  3. 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.

  4. 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.

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

References:

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

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

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

  4. L. Paulatto, I. Errea, M. Calandra & F. Mauri, arXiv:1411.5628; p. 1.

Qmechanic
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  • I presume that only acoustic phonon tadpole describes global translation/deformation etc., but not for optical phonon tadpole. And hence neglecting tadpole is only justified for acoustic phonon. Is this understanding correct? – xiaohuamao Dec 18 '21 at 13:05
  • No, cf. e.g. Ref. 4. – Qmechanic Dec 18 '21 at 13:09
  • Ref 4 talks about tadpole for phonon self energy, but I’m asking about the tadpole for electron self energy. Not so relevant? Or I misunderstood? – xiaohuamao Dec 18 '21 at 13:13
  • Well, pt. 2. was meant to be more general. – Qmechanic Dec 18 '21 at 13:23
  • @Qmechanic. I am not sure I have understood. You are saying that Fetter & Walecka is wrong? At pag. 401 it says that the general tadpole diagram vanishes. – Gippo Jun 03 '22 at 18:23
  • Yes, Fetter & Walecka's argument is backwards, cf. pt. 3. – Qmechanic Jun 03 '22 at 21:05
  • Dunno, in literature you never find the 1-phonon tadpole diagram (which is real, i.e. no imaginary part) evaluated to make a correction to the electron energy, whereas other 2nd order diagrams (like the so called Debye-Waller and Midgal-Fan diagrams) are included. Moreover, I think phonons and photons are pretty similar. In the Peskin Schroeder book it is stated that the 1-photon tadpole is zero (see problem 10.1), and here a demonstration https://higgs.physics.ucdavis.edu/peskin10_1.pdf – Gippo Jun 04 '22 at 13:12
  • Photons and phonons are different: 1. The 1-photon tadpole is manifestly zero because of $C$-symmetry, cf. pt. 4. 2. The 1-phonon tadpole in a theory is often zero. It could be for various reasons, e.g. a symmetry or a renormalization condition. – Qmechanic Jun 04 '22 at 13:23
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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.
Roger V.
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  • But I think q=0 optical phonon is always valid and physical, not just global deformation/translation. So I feel the argument is only true for acoustic phonon, isn’t it? – xiaohuamao Dec 18 '21 at 13:49
  • @xiaohuamao what does this deformation correspond to? A sublattice shift? A salient point here is also the choice of the point$q=0$ in the Brillouin zone. – Roger V. Dec 18 '21 at 14:11
  • Say, diatomic molecules forming a lattice. Just two atoms oscillating out of pi phase. And yes, let’s restrict to the 1st Brillouin zone. – xiaohuamao Dec 18 '21 at 14:14
  • Yes, to have acoustic phonons one needs at least diatomic lattice, and optical phonons correspond to the sublattice atoms oscillating out of phase. But still, what kind of deformation do we have at $q=0$? – Roger V. Dec 18 '21 at 14:39
  • Oh, I did mean that optical phonon at q=0 doesn't look like any deformation/translation, although the atoms are locally oscillating inside unit cells. And that's exactly why I presume vanishing free phonon propagator at q=0 (and hence vanishing tadpole diagram in my question) is only true for acoustic phonon. Do you agree? BTW, pls @me when replying. – xiaohuamao Dec 19 '21 at 03:06