In density functional theory (DFT) we know that the local density approximation (LDA) leads to an exchange-correlation energy kernel of the form:

$$E_\mathrm{xc}^\mathrm{LDA} =\int e_\mathrm{xc}\left( \rho \right) dr$$

If we want to be able to differentiate between $\alpha$ and $\beta$ spin we can introduce local spin density approximation (LSD), giving as kernel of the form:

$$E_\mathrm{xc}^\mathrm{LSD} =\int e_\mathrm{xc}\left( \rho_\alpha,\rho_\beta \right) dr$$

Further improvements can be made by introducing the gradient of the density, giving the rise to the generalized gradient approximation (GGA). Under the spin density anzats, if we want the size of density gradient, we can define it as (this is the square of the size):

$$\gamma_{\sigma,\sigma '} = \nabla\rho_\sigma \nabla\rho_{\sigma'}$$

Giving the kernel:

$$E_\mathrm{xc}^\mathrm{GGA} =\int e_\mathrm{xc}\left( \rho_\alpha,\rho_\beta, \gamma_{\alpha, \alpha},\gamma_{\beta, \beta},\gamma_{\alpha, \beta} \right) dr$$

Note here that $\gamma_{\alpha,\beta} = \gamma_{\beta,\alpha}$ due to the definition. Now going the next step, metaGGA can be introduced. Here we take the Laplacian of the density. Now this is where I get stuck.

How is the metaGGA extension defined, in a similar manner to how I defined the GGA? Since the Laplacian operator also only works on a single density, I guess the scalar quantity arising from the Laplacian can be given as:

$$\eta_{\sigma,\sigma'} = \nabla^2 \rho_\sigma \nabla^2 \rho_{\sigma'}$$

But I am not sure of this. Further, in the literature, the metaGGAs are almost always associated with the kinetic energy density. I know that the density is defined as:

$$\rho_\sigma = \sum_i\left| \psi_{i\sigma} \right|^2$$

I cannot figure out if the above definition directly gives rise to the kinetic energy density, when taking the Laplacian of $\rho_\sigma$. Or if the kinetic energy density is independent from the Laplacian of the density?

For the completeness of my question, I have some problems even when given the expression of the kinetic energy density, which is given as:

$$\tau_{\sigma} = \frac{1}{2}\sum_i^\mathrm{occ}\left| \nabla\psi_{i\sigma} \right|^2$$

From the literature, it would look like the kinetic energy density only depends on one spin at the time. I guess this is because $\left| \nabla \psi_\alpha \nabla \psi_\beta \right| = 0$, because the overlap of two spin functions is zero. But I am not fully sure about this argument.

Since I was unable to make the connection between the Laplacian of the density and the kinetic energy density I am not sure which of the following two kernel I expect to end up with:

$$E_\mathrm{xc}^\mathrm{metaGGA} =\int e_\mathrm{xc}\left( \rho_\alpha,\rho_\beta, \gamma_{\alpha, \alpha},\gamma_{\beta, \beta},\gamma_{\alpha, \beta}, \tau_{\alpha}, \tau_{\beta} \right) dr$$


$$E_\mathrm{xc}^\mathrm{metaGGA} =\int e_\mathrm{xc}\left( \rho_\alpha,\rho_\beta, \gamma_{\alpha, \alpha},\gamma_{\beta, \beta},\gamma_{\alpha, \beta}, \eta_{\alpha, \alpha}\eta_{\beta, \beta},\eta_{\alpha, \beta}, \tau_{\alpha}, \tau_{\beta} \right) dr$$

  • $\begingroup$ tandfonline.com/doi/abs/10.1080/00268976.2017.1333644 This article goes into some detail about how to express the exchange functional for spin polarized LDA, GGA, and mGGA. $\endgroup$
    – Tyberius
    Commented Jun 3, 2018 at 21:14
  • $\begingroup$ This might be my own ignorance on this subject, but what is the purpose of defining $\gamma_{\sigma\sigma'}$? It would seem you could write the GGA XC energy with just $\nabla \rho_{\alpha}$ and $\nabla\rho_{\beta}$. $\endgroup$
    – Tyberius
    Commented Jun 4, 2018 at 16:43
  • $\begingroup$ @Tyberius It appears in the XC contrib to the Fock matrix: chemistry.stackexchange.com/a/81832/194 $\endgroup$ Commented Jun 4, 2018 at 16:55
  • $\begingroup$ @Tyberius Yes. This is essentially the same as the depends of $\gamma_{\sigma \sigma '}$ that I wrote. Since this $\gamma$ also just depends on single index gradients. $\endgroup$ Commented Jun 4, 2018 at 17:55
  • $\begingroup$ Please note, that I made some major edits to the question. $\endgroup$ Commented Jun 4, 2018 at 18:06

1 Answer 1


The idea of a MGGA is not so different of a GGA. The Laplacian of the density or the kinetic energy density are employed to give a better description about the inhomogeneity of chemical systems. These parameters or "ingredients" can improve for instance the characterization of bonds. Take for instance the parameter $\alpha=(\tau - \tau^{W})/\tau^{unif}$ that is used in the SCAN density functional [1] where $\tau$, $\tau^{W}$ and $\tau^{unif}$ are the exact (in KS method) kinetic energy density, the Weizsaecker kinetic energy density and the kinetic energy density of the uniform electron gas, respectively. $\alpha$ can make a density functional recognize covalent, metallic and weak bonds.

Now, to the math...

First, the Laplacian operator is defined as,

$$ \nabla^{2} = \nabla \cdot \nabla = \left[ \frac{\partial}{\partial x} \hat i + \frac{\partial}{\partial y} \hat j + \frac{\partial}{\partial z}\hat k \right] \cdot \left[ \frac{\partial}{\partial x} \hat i + \frac{\partial}{\partial y} \hat j + \frac{\partial}{\partial z}\hat k \right] = \frac{\partial^{2}}{\partial x^{2}} + \frac{\partial^{2}}{\partial y^{2}} + \frac{\partial^{2}}{\partial z^{2}} . $$

We see that the laplacian operator is not a vector as the gradient operator. Second, we apply the laplacian to the density of a given spin,

$$ \nabla^{2} \rho_{\sigma}(\pmb{r}) = \nabla^{2} \sum_{i}^{occ} \psi_{i\sigma}^{*}(\pmb{r}) \psi_{i\sigma}(\pmb{r}) = 2\sum_{i}^{occ} [\psi_{i\sigma}^{*}(\pmb{r}) \nabla^{2} \psi_{i\sigma}(\pmb{r}) + \nabla \psi_{i\sigma}^{*}(\pmb{r}) \cdot \nabla \psi_{i\sigma}(\pmb{r})]. \label{laprho} $$

We found in the last term of $\nabla^{2} \rho_{\sigma}(\pmb{r})$ the kinetic energy density which we're going to discuss below. The Laplacian of the total density is defined by,

$$ \nabla^{2}\rho(\pmb{r}) = \nabla^{2} \left[ \rho_{\alpha}(\pmb{r}) + \rho_{\beta}(\pmb{r}) \right] = \nabla^{2} \rho_{\alpha}(\pmb{r}) + \nabla^{2} \rho_{\beta}(\pmb{r}). $$

Now, we turn to the kinetic energy density. Rearranging the terms in $\nabla^{2}\rho_{\sigma}(\pmb{r})$ and dividing by 4 yields,

$$ \frac{1}{2}\sum_{i=1}^{occ} \nabla \psi_{i\sigma}^{*}(\mathbf{r}) \cdotp \nabla \psi_{i\sigma}(\mathbf{r}) = \frac{1}{4} \nabla^{2} \rho_{\sigma}(\pmb{r}) - \frac{1}{2}\sum_{i}^{occ} \phi_{i\sigma}^{*}(\pmb{r}) \nabla^{2} \phi_{i\sigma}(\pmb{r}). $$

Finally, we obtain the kinetic energy density, $$ \tau_{\sigma} (\mathbf{r}) = \frac{1}{2}\sum_{i=1}^{occ} \nabla \psi_{i\sigma}^{*}(\mathbf{r}) \cdotp \nabla \psi_{i\sigma}(\mathbf{r}) = \frac{1}{2}\sum_{i=1}^{occ} |\nabla \psi_{i\sigma}(\mathbf{r})|^2. $$

Since we have alpha and beta spin contributions in the laplacian of the total density we get, $$ \tau(\pmb{r}) = \tau_{\alpha}(\pmb{r}) + \tau_{\beta}(\pmb{r}). $$

With the equations above you can define MGGA's that depend on $\nabla^{2} \rho$ and/or $\tau$.


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