Edit and TLDR: For classical lattice vibrations, Ashcroft and mermin (pg 424) determine that the harmonic term in the potential energy expansion is

$$ U^{harm}=\frac{1}{4}\sum_{R\, R', \mu \,v}[u_{\mu}(R)-u_{\mu}(R')]\phi_{\mu v}(R-R')[u_{v}(R)-u_{v}(R')] \tag{1} $$

where they define $\phi_{\mu v} (\vec{r})\equiv \frac{\partial^2 \phi(\vec{r})}{\partial r_{\mu} \partial r_{v}}$. They then reduce this to the supposedly equivalent form

$$ U_{harm}=\frac{1}{2}\sum_{R, R'}u_{\mu}(R)D_{\mu v}(R-R')u_{v}(R')\tag{2} $$

Then in eq (22.48) (on pg 438) they state $D_{\mu, v}(R-R')$ should take the form

$$ D_{\mu,v}(R-R')= \frac{\partial^2 U}{\partial u_{\mu}(R)\partial u_{v}(R')}\Bigg|_{u=0}\tag{3} $$

Firstly how do we get from Eq 1 to Eq 2 but primarily and most importantly, how do we get from Eq 2 to Eq 3?


In Ashcroft and Mermin (Pg 424), a multidimensional taylor series approximation of the potential energy function is used. The textbook formally uses the equation

$$ f(\vec{r}+\vec{a})=f(\vec{r})+\vec{a}\cdot \nabla f(\vec{r})+\frac{1}{2}(\vec{a}\cdot \nabla)^2f(\vec{r})+ \cdot \cdot \cdot $$

Then $\vec{r}=R-R'$ and $\vec{a}=u(R)-u(R')$ (I drop the vector notation from here on for ease of typing) are substituted into the above to eventually yield

$$ U^{harm}=\frac{1}{4}\sum_{R\, R', \mu \,v}[u_{\mu}(R)-u_{\mu}(R')]\phi_{\mu v}(R-R')[u_{v}(R)-u_{v}(R')] \tag{1} $$

where he defines $\phi_{\mu v} (\vec{r})\equiv \frac{\partial^2 \phi(\vec{r})}{\partial r_{\mu} \partial r_{v}}$.

My first question is regarding the variables that these partial derivatives are respect to. If $\phi_{\mu v}(\vec{r})$ is defined in this way, then the $\phi_{\mu v}(R-R')$ used in eq (1) should have the form

$$ \phi_{\mu v} (R-R')= \frac{\partial^2 \phi(R-R')}{\partial (R-R')_{\mu} \partial (R-R')_{v}} $$

However, on page 438, equation 22.48, the textbook indicates that in fact we have that

$$ D_{\mu,v}(R-R')= \frac{\partial^2 U}{\partial u_{\mu}(R)\partial u_{v}(R')}\Bigg|_{u=0} $$

But because $D_{\mu v}(R-R')$ is defined in terms of the $\phi _{\mu v}$ present in eq (1)(Ashcroft defines it in eq 22.11 on pg 425), this strongly indicates that the partial derivatives in eq (1) above should actually be something of the form

$$ \phi_{\mu v} (R-R')= \frac{\partial^2 \phi(R-R')}{\partial u_{\mu}(R)\partial u_{v}(R')}\Bigg|_{u=0} $$

This is not the same as the form present in eq (1) . So either something has to be wrong or these two forms have to be equivalent in some way. If they are equivalent, how so? and if they are not equivalent then where am I going wrong?

Sorry if this query seems overly pedantic or even unclear, its just that this issue has been driving me mad recently. Any help on this problem would be greatly appreciated as this issue is giving me nightmares!


From equation (22.1) $\vec{r}(\vec{R}) = \vec{R} + \vec{u}(\vec{R})$ so that $r_\mu = R_\mu + u_\mu$ and hence, $$\tag{1} \frac{\partial\phi}{\partial u_\mu} = \frac{\partial\phi}{\partial r_\mu} \frac{\partial r_\mu}{\partial u_\mu} = \frac{\partial\phi}{\partial r_\mu} $$ so that the two definitions of $\phi_{\mu\nu}$ are identical. $\vec{R}$ being a fixed point in the Bravais lattice, is a constant in this analysis.

Secondly, I think, equation (2) does not follow from equation (1). Rather, equation (1) is a special case of equation (2) when $D_{\mu\nu}$ takes the form $$\tag{2} D_{\mu\nu}(\vec{R} - \vec{R}^\prime) = \delta_{\vec{R}\vec{R}^{\prime}} \sum_{\vec{R}^{\prime\prime}}\phi_{\mu\nu}(\vec{R} - \vec{R}^{\prime\prime}) - \phi_{\mu\nu}(\vec{R} - \vec{R}^{\prime}) $$

When you substitute this in your equation (2), you will have five summations, one each over $\vec{R}, \vec{R}^\prime, \vec{R}^{\prime\prime}, \mu$ and $\nu$. The term $\delta_{\vec{R}\vec{R}^{\prime}}$ allows you to simplify the sum by either summing first over $\vec{R}$ or over $\vec{R}^\prime$.

If you sum over $\vec{R}^\prime$, you will get $$\tag{3} U = \frac{1}{2}\sum_{\vec{R}\vec{R}^\prime}\sum_{\mu\nu}u_\mu(\vec{R})\phi_{\mu\nu}(\vec{R} - \vec{R}^\prime)[u_\nu(\vec{R})-u_\nu(\vec{R}^\prime)]. $$ In deriving this sum, you will first get two terms, one of which will have sum over $\vec{R}$ and $\vec{R}^{\prime\prime}$. Rename $\vec{R}^{\prime\prime}$ as $\vec{R}^{\prime}$ to get this form.

If, instead, you sum over $\vec{R}$, you will get $$\tag{4} U = \frac{1}{2}\sum_{\vec{R}\vec{R}^\prime}\sum_{\mu\nu}[u_\mu(\vec{R}^\prime)-u_\mu(\vec{R})]\phi_{\mu\nu}(\vec{R}^\prime - \vec{R})u_\nu(\vec{R}^\prime). $$ or $$\tag{4} U = -\frac{1}{2}\sum_{\vec{R}\vec{R}^\prime}\sum_{\mu\nu}[u_\mu(\vec{R})-u_\mu(\vec{R}^\prime)]\phi_{\mu\nu}(\vec{R}^\prime - \vec{R})u_\nu(\vec{R}^\prime). $$ Interchange $\mu$ and $\nu$, since $\phi_{\mu\nu} = \phi_{\nu\mu}$, $$\tag{5} U = -\frac{1}{2}\sum_{\vec{R}\vec{R}^\prime}\sum_{\mu\nu}[u_\nu(\vec{R})-u_\nu(\vec{R}^\prime)]\phi_{\mu\nu}(\vec{R}^\prime - \vec{R})u_\mu(\vec{R}^\prime). $$ Rearranging a bit, taking into account that $\phi(\vec{R}-\vec{R}^\prime) = \phi(\vec{R}^\prime-\vec{R})$, $$\tag{6} U = -\frac{1}{2}\sum_{\vec{R}\vec{R}^\prime}\sum_{\mu\nu}u_\mu(\vec{R}^\prime)\phi_{\mu\nu}(\vec{R} - \vec{R}^\prime)[u_\nu(\vec{R})-u_\nu(\vec{R}^\prime)]. $$ Add equations (3) and (6) so that $$\tag{7} 2U = \frac{1}{2} \sum_{\vec{R}\vec{R}^\prime}\sum_{\mu\nu} [u_\mu(\vec{R})-u_\mu(\vec{R}^\prime)]\phi_{\mu\nu}(\vec{R} - \vec{R}^\prime)[u_\nu(\vec{R})-u_\nu(\vec{R}^\prime)] $$ or $$\tag{8} U = \frac{1}{4} \sum_{\vec{R}\vec{R}^\prime}\sum_{\mu\nu} [u_\mu(\vec{R})-u_\mu(\vec{R}^\prime)]\phi_{\mu\nu}(\vec{R} - \vec{R}^\prime)[u_\nu(\vec{R})-u_\nu(\vec{R}^\prime)]. $$

There's a lot of LaTeX typing. Should there be a typo, please let me know.

In order to get your equation (3) from your equation(2), note that $u_\mu(\vec{R})$ is different for every $\vec{R}$. You can treat it as if it were a different, independent variable. Therefore, when you take the partial derivative of $U$ with respect to one particular $u_\mu(\vec{R})$, only the term with this $u_\mu(\vec{R})$ contributes.

  • $\begingroup$ Thanks for the response! I'll defs give you the bounty. Just a small issue on my side though. We begin by noting that the taylor series of a function $f(x,y,z)$ of the spatial variables x,y and z is $f(\vec{r}+\vec{a})=f(\vec{r})+\vec{a}\cdot \nabla f(\vec{r})+..$ where the gradient is with respect to the spatial variables x,y and z. We now apply this to eq 22.3 and eventually get your eq 8 which contains the $\phi_{\mu,v}(R-R')$ term. Is this term of the form $\frac{\partial^2 \phi (R-R')}{\partial \mu \partial v}$ where u and v can take on any of the three spatial variables x,y or z? $\endgroup$ Sep 19 at 8:51
  • $\begingroup$ I think a good way to answer this question is by finding variables in the Taylor series you have mentioned in your comment and the one A&M have used. The constant $\vec{r}$ corresponds to $\vec{R}$, $\vec{a}$ to $\vec{u}$ and $\vec{r} + \vec{a}$ to $\vec{R} + \vec{u} = \vec{r}$ (this is (22.1)). The derivative $\phi_{\mu\nu}$ is with respect to $u_\mu$ or equivalently $r_\mu$. $\endgroup$
    – Amey Joshi
    Sep 19 at 13:02

The only change you need to make is to write your last equation as \begin{equation} \phi_{\mu \nu}(R-R') = \left . \frac{\partial^2 \phi(R-R'+u(R)-u(R'))} {\partial u_\mu(R) \partial u_\nu(R')}\right |_{u=0} \end{equation} which you can see is the contributing term from $U$. Note that $U$ is the full potential, not the potential at the equilibrium points in your equation defining $D$.

You then have \begin{equation} \phi_{\mu\nu}(R-R') = D_{\mu\nu}(R-R') \end{equation} for this special case of the pairwise potential between identical atoms.


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