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For fun I am trying to find the propagator of a free acoustic longitudinal phonon, using the vectorial displacement field. The Hamiltonian for our system is $$ H_0 = \sum_{\vec{q}} \omega_{\vec{q}} ( b_{\vec{q}}^\dagger b_{\vec{q}}^{\vphantom{\dagger}} + \frac{1}{2}), $$ with $b_q^{\vphantom{\dagger}}$ and $b_q^\dagger$ being the bosonic phonon creation and annihilation operators. Let's define the real phonon displacement field

$$ \vec{\varphi} (x) = \int \frac{d^3{\vec{q}}}{(2\pi)^3}\sqrt{\frac{1}{2\omega_{\vec{q}}}} \frac{\vec{q}}{|{\vec{q}}|} \left[b_{\vec{q}} e^{-iq x} + b_{\vec{q}}^\dagger e^{i q x} \right], $$ with $\omega_{\vec{q}} = |{\vec{q}}|$.

We can rewrite $H_0$ in term of $\vec{\varphi}$ ; we get

$$ H_0 = \int d^3x \frac{\dot{\vec{\varphi}}^2}{2} +\frac{(\partial_i \varphi^i)^2}{2}. $$

The corresponding Lagrangian density is given by

$$ \mathcal{L} = \frac{\dot{\vec{\varphi}}^2}{2} -\frac{(\partial_i \varphi^i)^2}{2} = -\frac{1}{2} \vec{\varphi}^T D_0^{-1} \vec{\varphi}$$ where we defined the operator $$ D_0^{-1} = \begin{pmatrix} \Box _x & -\partial _x \partial_y &-\partial _x \partial_z \\ -\partial _x \partial_y & \Box_y & -\partial _y \partial_z \\ -\partial _x \partial_z & -\partial _y \partial_z & \Box_z \end{pmatrix} $$ with $\Box_i = \partial_t^2 - \partial_i^2$. Then the propagator for $\vec{\varphi}$ in momentum space in given by

$$ D_0 (\vec{q}, \omega)_{ij} = \frac{1}{\omega ^2}\left[\delta_{ij} + \frac{q_i q_j}{\omega^2 - \omega_{\vec{q}}^2} \right].$$

I don't really understand this propagator. Why is there a pole at $\omega = 0$ ? I would instead expect only a $\propto 1/q^2$ dependance with $q^2 = \omega^2 - \omega_{\vec{q}}^2$ (like in the "traditional" phonon propagator $D_0 = \omega_q^2 / (\omega ^2 - \omega_{\vec{q}}^2 + i \varepsilon)$ or in the photon propagator).

EDIT : So I try another approach, calculating directly

$$ D_0(x_1, x_2)_{ij} = \langle T \left\{\varphi_i (x_1) \varphi_j(x_2) \right\} \rangle = \langle \varphi_i (x_1) \varphi_j(x_2) \rangle \Theta(t_1 - t_2) + \langle \varphi_j (x_2) \varphi_i(x_1) \rangle \Theta(t_2 - t_1) = \int \frac{d^3{\vec{q}}}{(2\pi)^3}\frac{1}{2\omega_{\vec{q}}^3} q_i q_j\left[e^{-iq (x_1 - x_2)} \Theta(t_1 - t_2) + e^{-iq (x_2 - x_1)} \Theta(t_2 - t_1)\right] $$ The next step is to take the FT w.r.t. $x_1-x_2 = (\tau, \vec{x_1} - \vec{x_2})$. We get $$ D_0(\vec{k}, \omega)_{ij} = \int d\tau \frac{1}{2\omega_{\vec{k}}^3} k_i k_j\left[e^{-i(\omega_{\vec{k}} - \omega) \tau} \Theta(\tau) + e^{i(\omega_{\vec{k}} + \omega) \tau} \Theta(-\tau)\right] = \frac{k_i k_j}{\omega_{\vec{k}}^2} \left[ \frac{i}{\omega^2 - \omega_{\vec{k}}^2 + i \varepsilon} \right].$$ This looks more reasonable than the previous result... except that it is not invertible ! I surely did mistakes but I don't see where.

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    $\begingroup$ I suggest checking the answer in AGD or Mahan or other QFT for condensed matter tetxbook. As now it looks poorly researched (I am not downvoting, since there is a real question in there.) $\endgroup$
    – Roger V.
    Jun 22, 2022 at 8:00
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    $\begingroup$ @RogerVadim In fact I tried to do this little exercise because of the passage in AGD where he introduces his phonon field (p.53 in the second edition). They use a different definition for the phonon, but they point out that another definition (the one I took) is also valid to describe the phonon field. I then wanted to re-express the phonon-electron interactions with this new field. I just looked in the Mahan, its definition is similar to AGD (actually I didn't know Mahan's book, thanks for the recommendation !). $\endgroup$
    – Jean
    Jun 22, 2022 at 9:17
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    $\begingroup$ Not sure, if Kittel has it, but it is a definitive source regarding the solid state. ANother approach is to look up the equation for elastic waves in a solid and quantize it/find the propagator - after all, these are acoustic phonons in continuum approxiation. $\endgroup$
    – Roger V.
    Jun 22, 2022 at 9:22
  • $\begingroup$ @RogerVadim I read the part on Mahan's book about the phoTon propagator. It's funny because his equation 2.175 is exactly what I found for the phoNon propagator in the last edit. Except that Mahan sums over the polarization. If he doesn't do that he gets a non-invertible matrix, like me. Now if I sum over all possible polarization I found that $D_0(\vec{k}, \omega)_{ij} = i \delta_{ij} / (\omega^2 - \omega_k^2+i\varepsilon)$, which is of course invertible. Maybe defining the phonon field without summing over all polarizations is non sensical ? Kittel's book didn't help unfortunately. $\endgroup$
    – Jean
    Jun 22, 2022 at 20:58

1 Answer 1

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The reason why you get a weird pole at $\omega = 0$ in your first attempt to find the propagator of a free acoustic phonon is that the transversal oscillations which are always part of the displacements vectorial field are not considered in your approach.It appears that in the formula you obtained the propagation velocity of transversal oscillations is taken equal to $0$. Instead of the Hamiltonian $$H_0 = \int d^3 x \left( \frac {\dot{\vec{\varphi}^2} } 2 + c_p^2 \frac{(\partial_i\varphi^i)^2}2\right) = \sum_{\vec q} \omega(\vec q) \left( b^{\dagger}_{\vec q} b_{\vec q} + \frac {1} 2 \right)$$ where $\omega(\vec q)=c_p\Vert{\vec q}\Vert$ is considered, it is more accurate to take the Hamiltonian $$ H = \sum_{\vec q , s}\omega_s(\vec q)\left(b^{\dagger}_{\vec q , s}b_{\vec q , s}+ \frac{1} 2 \right) $$ where $\omega_1(\vec q)=c_p\Vert{\vec q}\Vert$ and $\omega_2(\vec q)=\omega_3(\vec q)= c_T\Vert{\vec q}\Vert$ with $s = 1,2,3$.

In the case of a covalent crystal assimilable to an elastic isotropic homogeneous medium, the small oscillations are represented by acoustic phonons having three orthogonal each to other polarization directions: one parallel to the propagation direction $(s=1)$ , with propagation velocity of longitudinal elastic waves $c_p = \sqrt{\frac{\lambda+ 2\mu} {\rho}}$ and two perpendicular to the propagation direction $(s = 2,3)$ with propagation velocity of transversal elastic waves $c_T = \sqrt{\frac{\mu}{\rho}}$. The $\lambda$ , $\mu$ are the Lame coefficients of the elastic isotropic medium which has density $\rho$.
The $H$ Hamiltonian can be written as $$H=\int d^3 x \left( \frac{1} 2 (\partial_t {\vec{\varphi}})^2 + \frac{1} 2 (c_p^2 (\nabla . {\vec \varphi})^2 + c_T^2 (\nabla \times {\vec \varphi})^2) \right)$$ with a corresponding Lagrangian density $$\mathscr L = \frac{1} 2 (\partial_t {\vec \varphi})^2 - \frac{1} 2 (c_p^2 (\nabla . \vec{\varphi})^2 + c_T^2 (\nabla \times \vec{\varphi})^2) $$ the phonon field being $$ \vec {\varphi}(t , x) = \int \frac{d^3\vec q}{(2\pi)^3} \sum_s \left( \frac{V}{2\omega_2(\vec q)} \right)^{\frac{1} 2} \mathbf e_s(\vec q)(b_{\vec q} e^{-i \omega_s(q) t + i \vec q \vec x} + b^{\dagger}_{\vec q} e^{i \omega_s(q) t - i \vec q \vec x} ) $$ where $\mathbf e_s$ , $s = 1,2,3$ are the polarization versors and $V$ is the volume.

We find out that the propagator of the phonon field is given in the momentum space by $$ D_{i j} (\omega , \vec q) = -{\frac{1}{\omega^2 - c_T^2 {\vec q}^2}} \left( \delta_{i j} + \frac{c_p^2 - c_T^2}{\omega^2 - c_p^2 {\vec q}^2} q_i q_j \right) $$ You can see that your propagator formula takes $ c_T = 0 $ .

I will mention that the electron-phonon interaction term of the system Lagrangian density,in the case of an covalent crystal assimilable to an isotropic homogeneous elastic medium can be taken as $ \sigma \psi^{\dagger} \psi (\nabla .\vec \varphi) $ where $\psi$ and $\vec \varphi$ are respective the electron and the phonon field and $\sigma$ is a coupling constant , leading to a coupling $-\sigma q_i$ with $q_i$ the incoming phonon momentum component in the computation of the Feynman amplitude for two electrons interchanging a phonon process. With this coupling the phononic factor of the Feynman amplitude is $ \frac{{\vec q}^2}{\omega^2 - c_p^2 {\vec q}^2 + i \varepsilon} $ and as expected has no participation of the transversal component of the phonon field (the $c_T$ containing elements disappear from the amplitude by cancelling out during the calculation).

As a source I can give you "Theory of Solids", Nauka,Moscow by A.S.Davydov which I consulted in a translation and "Quantum Field Theory in a Nutshell", Priceton University Press, by A.Zee.

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  • $\begingroup$ Thank you very much, nice answer. $\endgroup$
    – Jean
    Mar 1 at 15:03

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