2D Ising Landau theory and the term $A_i( m, T)\nabla_i m$? In (Sethan, 2007; pg$\sim$206) it is said that in the 2D Ising terms of the form e.g.
$$A_i( m, T)\nabla_i  m$$
are not allowed in the free energy since there is no vector $A_i(m,T)$ that is invariant under $\pi/2$ rotations. I am confused why we need this since:
$$A_i(m, T)\nabla_i$$
is invariant under $\pi/2$ rotations why do we need  $A_i( m, T)$ alone to be? I have a feeling the answer is to do with active and passive transforms but can't quite get my head around it.
 A: In the 2d Ising there are two symmetries we need to concern ourselves with:


*

*Symmetry of the Lattice: The lattice is symmetric under $\pi/2$ rotations as stated in the question.

*Symmetry of the Order Parameter: The order parameter has summetry under $m\rightarrow -m$
It is the symmetry of the lattice that concerns us here. From this consider $A_i \partial_i$:
$$ A_x \frac{\partial}{\partial x}+A_y \frac{\partial }{\partial y}\tag{1}$$
I will here look at a passive rotation rather then an active since I think it is easier to understand. So consdier the change of basis:
\[ x \mapsto y, \quad y \mapsto -x\tag{2}\]
under this (1) becomes:
$$ A_x \frac{\partial}{\partial y}-A_y \frac{\partial }{\partial x}\tag{3}$$
But (2) simply corresponds to a rotation of our coordinate system by $\pi/2$ and as such we require (1) to be equivalent to (3). the only way this can work is if $A_i=0$. Thus proving that such a term is not allowed.
Summary
The statement that $A_i \partial_i$ is not allowed is a statement from the required symmetry of the lattice and not the order parameter. If the lattice was say a triangular lattice this rotational invariance condition would not hold.
