# Order parameter for liquid to crystalline solid transition under the broken continuous group of translation

The order parameter for liquid to crystalline solid transition is given by the Fourier transform of the density $$\tilde{\rho}(\textbf{k})=\int\rho(\textbf{r})e^{i\textbf{k}\cdot\textbf{r}}d^3\textbf{r}.\tag{1}$$ One of the symmetries that get broken in this process is the continuous transnational symmetry. But an order parameter must transform as an irreducible representation of the unbroken symmetry group i.e., the continuous group of translation. Let us ask does the order parameter behave under arbitrary translation $\hat{\mathbb{T}}({\textbf{a}})$ where $\textbf{a}\in\mathbb{R}^3$. Applying (also as @noah suggested) $\hat{\mathbb{T}}({\textbf{a}})$, $$\hat{\mathbb{T}}({\textbf{a}})\Big[\tilde{\rho}(\textbf{k})\Big]=\tilde{\rho}(\textbf{k})\tag{2}$$

Question: Can we interpret Eq. (2) as the order parameter $\tilde{\rho}(\textbf{k})$ transforming as a scalar (one dimensional irreducble) representation of the unbroken continuous translation group?

I am not sure your expression in (2) even makes sense, since you are applying a spatial shift to something that is not position-dependent. The way you wrote it down it is simply $$\hat{\mathbb{T}}({\mathbf{a}}) \left[\tilde{\rho}(\mathbf{k})\right] \equiv \tilde{\rho}(\mathbf{k})$$ since the integration goes over all $\mathbf{r}$, so a finite shift does not change the integral. You can also see this by making a simple substitution like $\mathbf{y} = \mathbf{r} + \mathbf{a}$.
However, I assume what you meant to write down is $$\tilde{\rho}'(\mathbf{k}) = \mathcal{F}\left\{\hat{\mathbb{T}}({\mathbf{a}}) \left[\rho(\mathbf{r})\right]\right\} = \int \rho(\mathbf{a} + \mathbf{r}) e^{i\mathbf{k}\mathbf{r}} d^3\mathbf{r} = e^{i\mathbf{k}\mathbf{a}}\int \rho(\mathbf{r}) e^{i\mathbf{k}\mathbf{r}} d^3\mathbf{r} = e^{i\mathbf{k}\mathbf{a}} \tilde{\rho}(\mathbf{k}).$$ This is a basic property of the Fourier transform. It picks up a phase factor when its argument it is translated some distance.
$\tilde{\rho}(\mathbf{k})$ qualifies as an order parameter, because whenever there is anything periodic in the system, there will be a $\mathbf{k}$ for which $\tilde{\rho}(\mathbf{k}) \neq 0$. If there is no periodicity at all, $\tilde{\rho}$ will vanish for all $\mathbf{k}$ except for $\mathbf{k} = \mathbf{0}$, which will be the total amount of whatever $\rho(\mathbf{r})$ is the density of.
• "No, the absolute square of the magnitude of the order parameter must be invariant by all symmetry operations of the unbroken symmetry group". How does it contradict what I say? Magnetization $\textbf{M}$ is transforms as $\textbf{M}\to R\textbf{M}$ under SO(3) rotation so that $|\textbf{M}|^2=\textbf{M}\cdot\textbf{M}$ is invariant because $R^TR=\rm identity$. But I agree with you, I have been careless in writing Eq. (2). It is meaningless. I'll correct that. @noah – SRS Aug 28 '17 at 14:31