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?


1 Answer 1


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.

  • $\begingroup$ What about the fact that the order parameter transforms according to an n-dimensional representation of the unbroken symmetry group?" @noah $\endgroup$
    – SRS
    Aug 26, 2017 at 21:20
  • $\begingroup$ The order parameter does not transform according to the unbroken symmetry group because of the phase factor for an arbitrary translation. I disregarded this fact as irrelevant, but you are right, the shift property is relevant for it being an order parameter. $\endgroup$
    – noah
    Aug 26, 2017 at 21:57
  • $\begingroup$ But isn't it true that the order parameter must transform according to an irreducible representation of the unbroken symmetry group? The magnetization, the order parameter for paramagnetic to ferromagnetic transition is a vector transforms like an irreducible representation of SO(3). @noah $\endgroup$
    – SRS
    Aug 28, 2017 at 13:39
  • $\begingroup$ No, the absolute square of the magnitude of the order parameter must be invariant by all symmetry operations of the unbroken symmetry group. Stated also in these lecutre notes on page 7, and references therein. Since the absolute magnitude of the phase factor from the fourier transform is 1, this condition is fulfilled. $\endgroup$
    – noah
    Aug 28, 2017 at 14:22
  • $\begingroup$ "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 $\endgroup$
    – SRS
    Aug 28, 2017 at 14:31

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