Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Definition of the dielectric function in the linear response regime

  1. Oct 17, 2012 #1
    Sometimes the dielectric function is defined as the connection between the total electric field in a material and the external field,
    \mathbf{E}(\mathbf{r},\omega) = \int \epsilon^{-1}(\mathbf{r},\mathbf{r'},\omega) \mathbf{E}_{\text{ext}}(\mathbf{r'},\omega) d \mathbf{r'},
    and sometimes as the connection between the total effective potential and the externally applied potential,
    V_{\text{tot}}(\mathbf{r},\omega) = \int \epsilon^{-1}(\mathbf{r},\mathbf{r'},\omega) V_{\text{ext}}(\mathbf{r'},\omega) d \mathbf{r'}.
    I don't see how these two definitions are equivalent.
    See, e.g. "etsf.grenoble.cnrs.fr/dp/tutorial/dptutorial.pdf" [Broken] and "cms.mpi.univie.ac.at/mmars/ThesisJudithHarlChapter2.pdf" [Broken].
    Could somebody comment on that?
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Oct 17, 2012 #2


    User Avatar
    Science Advisor

    Good point, I think the relation holds only for the spatial Fourier components of the potential.
  4. Oct 17, 2012 #3
    Well, I guess the crucial piece of information is that the MACROSCOPIC dielectric function depends on [itex]\mathbf{r} - \mathbf{r}'[/itex], as it should be for transltionally invariant systems. Then:
    \mathbf{E}(\mathbf{r}, \omega) = -\nabla_{\mathbf{r}} V(\mathbf{r}, \omega) \\
    = -\int{d\mathbf{r}' \, \nabla_{\mathbf{r}} \varepsilon^{-1}(\mathbf{r} - \mathbf{r}', \omega) V_{\mathrm{ext}}(\mathbf{r}', \omega)} \\

    = \int{d\mathbf{r}' \, \nabla_{\mathbf{r}'} \varepsilon^{-1}(\mathbf{r} - \mathbf{r}', \omega) V_{\mathrm{ext}}(\mathbf{r}', \omega)} \\

    = \int{d\mathbf{r}' \, \varepsilon^{-1}(\mathbf{r} - \mathbf{r}', \omega) \left(-\nabla_{\mathbf{r}'} \, V_{\mathrm{ext}}(\mathbf{r}', \omega)\right)}
    where in the last step we applied Gauss's Divergence Theorem (integration by parts) and neglected the surface term.
  5. Oct 17, 2012 #4
    I see how Dickfore's observation about [itex]\mathbf{r} - \mathbf{r}'[/itex] somewhat helps, but from what I understand the relations are also supposed to work for the microscopic dielectric function. I think we should instead focus on the fact that we are in the linear regime.

    The external potential [itex]V_{\text{ext}}[/itex] perturbs the electron density, which in turn induces another coulomb field [itex]V_{\text{ind}}[/itex]. The total resulting potential seen by a test charge is
    V_{\text{tot}} (\mathbf{r}, \omega) = V_{\text{ext}}(\mathbf{r},
    \omega) + V_{\text{ind}}(\mathbf{r}, \omega).
    In the linear approximation, we assume the induced density to be proportional to the external potential,
    \delta n(\mathbf{r}, \omega) = \int d\mathbf{r'} \chi (\mathbf{r}, \mathbf{r'}, \omega)
    V_{\text{ext}}(\mathbf{r'}, \omega),
    where [itex]\chi[/itex] is the density-density response function. Using
    V_{\text{ind}}(\mathbf{r}, \omega) = \int d\mathbf{r'} \frac{\delta
    n(\mathbf{r'}, \omega)}{|\mathbf{r} - \mathbf{r'}|}
    we can express the total effective field as
    V_{\text{tot}}(\mathbf{r}, \omega) = \int d\mathbf{r'} \left[ \delta(\mathbf{r} -
    \mathbf{r'}) + \int d\mathbf{r''} \frac{\chi(\mathbf{r''}, \mathbf{r'},
    \omega)}{|\mathbf{r} - \mathbf{r''}|} \right] V_{\text{ext}} (\mathbf{r'},\omega).
    At this point everything in the square brackets is usually defined to be [itex]\epsilon^{-1}[/itex] in which case we arrive at the second equation of my first post. However, it is not clear to me that the dielectric function defined this way is the same as that in the first equation of my first post.
    Last edited: Oct 17, 2012
  6. Oct 17, 2012 #5


    User Avatar
    Science Advisor

    Also it is not possible to represent a transverse electric field as the gradient of a potential. So the expression in terms of potentials only yields the longitudinal dielectric constant.
  7. Oct 17, 2012 #6
    To be honest, I don't think the dielectric constant is defined on a microscopic scale, because the microscopic field in the material is what is felt by the atoms, so that field causes the polarization and its potential should be used in the linear response function.

    Now, if you perform a volume average of the microscopic field, you get the macroscopic field. The dielectric constant (a property of the dielectric, therefore a macroscopic quantity) is the proportionality between the existing field (potential) and the applied field (potential) that would have existed if the dielectric was absent (not necessarily equal to the field outside the dielectric due to boundary effects).

    I guess my point is that the dielectric constant is already a macroscopic quantity, and has the translational invariance.
  8. Oct 17, 2012 #7
    Sure it is. See, e.g., Chapter 2. While for the macroscopic response only the same Fourier component get affected as that of the perturbation,
    [tex]\mathbf{E}(\mathbf{q},\omega) = \epsilon_{\text{mac}}^{-1}(\mathbf{q},\omega) \mathbf{E}_{\text{ext}}(\mathbf{q},\omega), [/tex]
    at the microscopic level there are additional Fourier components,
    [tex]\mathbf{e}(\mathbf{q}+\mathbf{G},\omega) = \epsilon^{-1}(\mathbf{q}+\mathbf{G},\mathbf{q},\omega) \mathbf{E}_{\text{ext}}(\mathbf{q},\omega), [/tex]
    where [itex]\mathbf{G}[/itex] is a reciprocal lattice vector.
    [itex]\epsilon^{-1}(\mathbf{k}_1,\mathbf{k}_2,\omega)[/itex] describes the response between different Fourier components. These rapidly oscillating local fields cannot be directly measured. However, they do have an effect on the macroscopic dielectric function. The connection between the macroscopic and microscopic functions is [PRB 129, 62 (1963)]
    [tex]\epsilon_{\text{mac}}(\mathbf{q},\omega) = \frac{1}{\epsilon_{\mathbf{G} = 0,\mathbf{G'} = 0}^{-1}(\mathbf{q},\omega)}.[/tex]
    (In a crystal, the wavevectors can always be written as [itex]\mathbf{k}_1 = \mathbf{q} + \mathbf{G}[/itex] and [itex]\mathbf{k}_2 = \mathbf{q} + \mathbf{G'}[/itex]). Often, the approximation [itex]\epsilon_{\text{mac}}(\mathbf{q},\omega) = \epsilon_{\mathbf{G} = 0,\mathbf{G'} = 0}(\mathbf{q},\omega)[/itex] is made, which is referred to as "neglect of local fields".
  9. Oct 18, 2012 #8
    The two definitions of the dielectric function are equivalent if
    [tex] \alpha(\mathbf{r}, \mathbf{r'},\omega) = \int d\mathbf{r''} \frac{\chi(\mathbf{r''}, \mathbf{r'}, \omega)}{|\mathbf{r} - \mathbf{r''}|}, [/tex]
    where [itex] \alpha(\mathbf{r}, \mathbf{r'},\omega) [/itex] is the polarizability (microscopic) or electric susceptibility (macroscopic). In the linear response theory the following relation between polarizability and the density-density response function can be derived:
    [tex] \chi(\mathbf{k}, \mathbf{k'},\omega) = -\mathbf{k} \mathbf{k'}\alpha(\mathbf{k}, \mathbf{k'},\omega). [/tex]
    Assuming everything has been correct so far, all I need is to show that the double Fourier transform of [itex]\int \chi(\mathbf{r''}, \mathbf{r'}, \omega) |\mathbf{r} - \mathbf{r''}|^{-1} d\mathbf{r''} [/itex] is equal to [itex] -1/(\mathbf{k}\mathbf{k'})\chi(\mathbf{k}, \mathbf{k'}, \omega) [/itex]. However, I don't know how to calculate this.
  10. Oct 18, 2012 #9


    User Avatar
    Science Advisor

    This integral is a convolution of chi and 1/r. By the convolution theorem of fourier transform, the fourier transform of a convolution is equal to the product of the fourier transform of the individual terms.
    And the FT of 1/r is constant times 1/k^2.
  11. Oct 18, 2012 #10
    well, the Fourier transform of [itex]\frac{1}{r}[/itex] is [itex]-\frac{4\pi}{q^2}[/itex] and you can use the convolution theorem:
    \int{d\mathbf{r} \, d\mathbf{r}' \, e^{-i (\mathbf{k} \cdot \mathbf{r} - \mathbf{k}' \cdot \mathbf{r}')} \, \int{d\mathbf{r}'' \, \int{\frac{d\mathbf{q}}{(2\pi)^3} \left(-\frac{4\pi}{q^2} \right) e^{i \mathbf{q} \cdot (\mathbf{r} - \mathbf{r}'')}} \, \chi(\mathbf{r}'', \mathbf{r}')}}
    -\int \frac{dq}{(2\pi)^3} \frac{4\pi}{q^2} \int dr e^{-i (k - q) r} \int dr'' \, dr' \, e^{-i (q r'' - k' r')} \, \chi(r'', r')
    -\frac{4 \pi}{k^2} \, \tilde{\chi}(\mathbf{k}, \mathbf{k}')
  12. Oct 18, 2012 #11
    Thanks a lot. The convolution theorem gets me a lot closer to the result, but there are still some problems. First of all, I think both replies were a bit erroneous because the Fourier transform of 1/r should be 4pi/k^2. Using that I arrive at
    [tex] \frac{4\pi}{k^2} \chi(\mathbf{k},\mathbf{k}',\omega), [/tex]
    which is a bit different from
    [tex] -\frac{1}{\mathbf{k}\mathbf{k'}} \chi(\mathbf{k},\mathbf{k}',\omega). [/tex]
    What puzzles me the most is I don't see how [itex]\mathbf{k'}[/itex] in the denominator enters the game. Perhaps the relation [itex]\chi(\mathbf{k}, \mathbf{k'},\omega) = -\mathbf{k} \mathbf{k'}\alpha(\mathbf{k}, \mathbf{k'},\omega)[/itex] is incorrect after all. This is how I got it.
    Electric polarization can be expressed as
    [tex] i\mathbf{k}\mathbf{P}n(\mathbf{k},\omega) = -\delta n(\mathbf{k},\omega). [/tex]
    Using the relations
    [tex] \mathbf{P} = \sum_{\mathbf k'}\alpha(\mathbf k,\mathbf k',\omega)\mathbf E(\mathbf k',\omega), [/tex]
    where [itex] \alpha [/itex] is the polarizability, and
    [tex] \mathbf{E}(\mathbf k,\omega) = -\nabla \phi(\mathbf k,\omega) = -i\mathbf k \phi(\mathbf k,\omega) = -i\mathbf k V_{\text{ext}}(\mathbf k,\omega) [/tex]
    we can express
    [tex] \delta n(\mathbf k, \omega) = -\frac{1}{V} \sum_{\mathbf k'} \mathbf k \mathbf k' \alpha(\mathbf k,\mathbf k',\omega) V_{\text{ext}}(\mathbf k',\omega).[/tex]
    From the 2nd equation of my 2nd post (there should be some system of numbering equations so I could more easily refer to previous posts) we get
    [tex] \delta n(\mathbf k, \omega) = \frac{1}{V} \sum_{\mathbf k'} \chi(\mathbf k,\mathbf k',\omega) V_{\text{ext}}(\mathbf k',\omega). [/tex]
    Comparing the last two equations leads to
    [tex] \chi(\mathbf{k}, \mathbf{k'},\omega) = -\mathbf{k} \mathbf{k'}\alpha(\mathbf{k}, \mathbf{k'},\omega). [/tex]
    Perhaps I should check whether the 2nd relation for [itex] \mathbf{P} [/itex] really holds. Also, even though [itex] \epsilon^{-1}(\mathbf r, \mathbf r') \neq \epsilon^{-1}(\mathbf r - \mathbf r') [/itex] in general, it is cell periodic in both real and reciprocal space, e.g. [itex] \epsilon^{-1}(\mathbf r + \mathbf R, \mathbf r') = \epsilon^{-1}(\mathbf r, \mathbf r') [/itex]. I'll see if that helps.

    Finally, for the macroscopic dielectric function it is easy to show that
    [tex] \mathbf{E}(\mathbf{k},\omega) = \epsilon_{\text{mac}}^{-1}(\mathbf{k},\omega) \mathbf{E}_{\text{ext}}(\mathbf{k},\omega). [/tex]
    Using [itex] \mathbf E(\mathbf k,\omega) = -i\mathbf k V(\mathbf k,\omega) [/itex] we get
    [tex] V_{\text{tot}}(\mathbf{k},\omega) = \epsilon_{\text{mac}}^{-1}(\mathbf{k},\omega) V_{\text{ext}}(\mathbf{k},\omega) [/tex]
    and therefore
    [tex] V_{\text{tot}}(\mathbf{r},\omega) = \int d\mathbf r' \epsilon_{\text{mac}}^{-1}(\mathbf{r}-\mathbf{r}',\omega) V_{\text{ext}}(\mathbf{r}',\omega). [/tex]

    Edit: It appears I'm using Gauss units. My definition of polarization is therefore incorrect by a factor of 4pi. That explains partly the difference of the first two equations.
    Last edited: Oct 18, 2012
  13. Oct 19, 2012 #12
    Here's another attempt.
    Generally [itex] \mathbf E_{\text{tot}} = \mathbf E_{\text{ext}}\epsilon^{-1} = \mathbf E_{\text{ext}}(1+4\pi\alpha)^{-1} [/itex], but in the linear regime we take [itex] \mathbf E_{\text{tot}} = \mathbf E_{\text{ext}}(1-4\pi\alpha) [/itex]. From
    V_{\text{tot}}(\mathbf{r}, \omega) = \int d\mathbf{r'} \left[ \delta(\mathbf{r} -
    \mathbf{r'}) + \int d\mathbf{r''} \frac{\chi(\mathbf{r''}, \mathbf{r'},
    \omega)}{|\mathbf{r} - \mathbf{r''}|} \right] V_{\text{ext}} (\mathbf{r'},\omega)
    we see that (dropping the [itex] \omega [/itex]-dependence)
    -4\pi \int \alpha(\mathbf{r}, \mathbf{r'}) V_{\text{ext}}(\mathbf r') = \int d\mathbf{r''} \frac{\chi(\mathbf{r''}, \mathbf{r'})}{|\mathbf{r} - \mathbf{r''}|} V_{\text{ext}} (\mathbf{r'})
    has to hold if [itex] V_{\text{tot}}(\mathbf{r},\omega) = \int \epsilon^{-1}(\mathbf{r},\mathbf{r'},\omega) V_{\text{ext}}(\mathbf{r'},\omega) d \mathbf{r'} [/itex]. Double Fourier transforming both sides and using the convolution theorem (see previous posts) yields
    -\frac{4\pi}{V} \sum_{\mathbf k'} \alpha(\mathbf{k}, \mathbf{k'}) V_{\text{ext}}(\mathbf k') = \frac{4\pi}{V} \sum_{\mathbf k'} \frac{1}{k^2} \chi(\mathbf{k}, \mathbf{k'}) V_{\text{ext}} (\mathbf{k'}).
    Using [itex] \chi(\mathbf{k}, \mathbf{k'},\omega) = -\mathbf{k} \mathbf{k'}\alpha(\mathbf{k}, \mathbf{k'},\omega) [/itex] (see PRB 29, 4631 (1984) Eq. (A16)) we get
    \sum_{\mathbf k'} \alpha(\mathbf{k}, \mathbf{k'}) V_{\text{ext}}(\mathbf k') = \sum_{\mathbf k'} \frac{\mathbf k'}{\mathbf k} \alpha(\mathbf{k}, \mathbf{k'}) V_{\text{ext}} (\mathbf{k'}),
    which is almost true. I wonder if [itex] \mathbf k' \approx \mathbf k [/itex] in linear response.
  14. Oct 19, 2012 #13
    For time-dependent fields, the vector potential cannot be set to zero, and the electric field should be expressed as:
    \mathbf{e}(k) = -i \mathbf{k} \, V(k) + i \, \frac{\omega}{c} \, \mathbf{A}(k), \ k = (\mathbf{k}, \omega)
    If we work in the Coulomb gauge, it is clear that the scalar potential determines the longitudinal field, whereas the vector potential determines the transversal field.

    If you accept the "field definition" of the dielectric constant, then, it is a tensor. The "potential" definition, on the other hand, defines as a scalar, and exactly between the longitudinal components of the field.

    So, generally, the two definitions are not equal.
  15. Oct 19, 2012 #14
    Yes, I am only trying to show the equivalence for the longitudinal dielectric constant. But if we use the Coulomb gauge and consider only linear response, shouldn't then the two definitions be exactly equal?
  16. Oct 19, 2012 #15
    The longitudinal dielectric constant connects [itex]\hat{\mathbf{k}} \cdot \mathbf{E}(\mathbf{k}, \omega)[/itex] with [itex]\hat{\mathbf{k}} \cdot \mathbf{E}_{\mathrm{ext}}(\mathbf{k}, \omega)[/itex] in a form:
    \hat{\mathbf{k}} \cdot \mathbf{E}(\mathbf{k}, \omega) = \frac{1}{\mathcal{V}} \, \sum_{\mathbf{k}'} \alpha^{-1}_{l}(\mathbf{k}, \mathbf{k}'; \omega) \, \hat{\mathbf{k}'} \cdot \mathbf{E}_{\mathrm{ext}}(\mathbf{k}', \omega)
    Using the expression for the electric fields in terms of scalar and vector potentials, and the Coulomb gauge, we get:
    k \, V_{\mathrm{tot}}(\mathbf{k}, \omega) = \frac{1}{\mathcal{V}} \, \sum_{\mathbf{k}'} \alpha^{-1}_{l}(\mathbf{k}, \mathbf{k}'; \omega) \, k' \, V_{\mathrm{ext}}(\mathbf{k}', \omega)
    This would coincide with the "potential" definition of the dielectric constant if:
    (k - k') \, \alpha^{-1}_{l}(\mathbf{k}, \mathbf{k}'; \omega) = 0
    So, the dielectric constant is non-zero iff [itex]k = k'[/itex].
  17. Oct 19, 2012 #16
    So it seems the two definitions are not exactly equivalent. This issue came to my head when reading a text about how to calculate optical properties using ab initio methods. For DFT based methods often the Kohn-Sham orbitals are used to construct the polarizability, which then leads to the dielectric function, as defined in terms of potentials, and other optical quantities. However, expimentally determined [itex]\epsilon[/itex] is based on the other definition in terms of electric fields. How can we then compare the calculated and experimentally measured results if we are using two inequivalent definitions of the dielectric constant? Clearly, I must be missing something.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Definition of the dielectric function in the linear response regime
  1. Dielectric function (Replies: 6)

  2. Dielectric Function (Replies: 11)