The most general Hamiltonian of double group $T_d$ Consider the electron interaction Hamiltonian of the form
$$H= \sum A_{ijkl} c^\dagger_{k_R,i} c^\dagger_{k_L j} c_{k_L k} c_{k_R l}$$
on $j=3/2$ states. The index $i,j,k,l$ denotes the value of $m_j \in \{ 3/2, 1/2, -1/2, -3/2\}$, and $k_R, k_L$ denotes the left/right momentum in 1D.
I am trying to find the most general form of $A_{ijkl}$ that satisfy the double group $T_d$ symmetry. How can I find this?
 A: I do not know how to answer your question because it is unclear how the operators transform under the symmetry group, but I will give an example which is very similar and may help.
Consider four spins on a tetrahedron, interacting via a Hamiltonian
$$H = \frac{1}{2} \sum_{ij} \sum_{\alpha\beta} S_{i}^\alpha J_{ij}^{\alpha\beta} S_{j}^\beta$$
where $i,j$ label the corners of the tetrahedron and $\alpha,\beta$ label the spin components. There are twelve total degrees of freedom here (ignoring fixed spin length constraint), such that
$$S_{i}^\alpha = (S_1^x, S_1^y, S_1^z,S_2^x, S_2^y, S_2^z,S_3^x, S_3^y, S_3^z,S_4^x, S_4^y, S_4^z)$$
The components of $J_{ij}^{\alpha\beta}$ are initially undetermined. We wish to enforce that the Hamiltonian is invariant under the symmetries of the tetrahedron. To do so, we must first know how the $S_{i}^\alpha$ transform under the symmetries. The group $T_d$ (the symmetry group of the tetrahedron) has 24 elements, falling into five equivalence classes: 3 $C_2$ rotations, 8 $C_3$ rotations, 6 $\sigma_d$ reflections (reflection through a plane which bisects one of the 6 bonds), 6 roto-reflections ($\pm$90 degree rotation about a $C_2$ axis and a reflection through a plane perpendicular to the axis), plus the identity operation: 3 + 8 + 6 + 6 + 1 = 24.
The degrees of freedom $S_i^\alpha$ will in general transform in a 12-dimensional representation of the group $T_d$. Under a symmetry operation, the corner indices get permuted, and the spin vectors will be rotated/reflected. We already know how to represent a rotation of a 3-dimensional vector as a $3\times 3$ matrix, while a reflection through a plane with normal $\hat{\mathbf{n}}$ can be written as a $\pi$ rotation about $\hat{\mathbf{n}}$ and an inversion.
Since spin is a type of angular momentum, it transforms as a pseudo-vector. A normal vector simply changes sign under an inversion, while pseudo-vectors do not change sign.
So now we can construct our 24 $12\times 12$ matrices which represent each symmetry operation. Note this is a tensor product representation, i.e. each can be written as a tensor product of a $4\times 4$ permutation matrix (which permutes the corners) and a $3\times 3$ rotation matrix (we can ignore the sign change due to inversion in our case for the reflection operators).
So, under a symmetry operation, collapsing all of our indices into a single 12-dimensional index, we have that
$$H \rightarrow \frac{1}{2} \sum_{abcd}S_a U_{ab}^\mathrm{T} J_{bc} U_{cd} S_d$$
We demand that $H$ is invariant under this transformation for all 24 group elements. in practice, there are more ``elegant'' ways of doing it, and it is not necessary to check every single symmetry operation, for example we actually only need to consider the symmetry transformation for a single bond, and can obtain all the other bonds by a set of $C_3$ rotations, but in any case you will find in this process that $J$ has only five independent matrix elements after all symmetry constraints have been taken into account.
As an extra useful piece of information, using the character table of $T_d$ and the orthogonality relations for finite groups we can show that our twelve dimensional representation, which I will refer to as $\Gamma_{12}$, reduces into a set of irreps,
$$\Gamma_{12} = A_2 \oplus E \oplus T_1 \oplus T_1 \oplus T_2$$
and we can find a new basis in which the different irreps do not mix under a symmetry operation (note the dimensions of the above irreps in order give us $1+2+3+3+3 = 12$ total degrees of freedom). By a new basis, I mean we can construct linear combinations of the spin variables which only mix within an irrep. For example, consider the quantitiy
$$m_{A_2} = \sum_{i} \mathbf{S}_i \cdot \hat{\mathbf{z}}_i$$
where $\hat{\mathbf{z}}_i$ is the local $C_3$ axis of each corner. This quantity transforms in the $A_2$ irreducible representation of $T_d$, which is 1-dimensional, meaning it is invariant under the symmetries of the tetrahedron. We can similarly construct linear combinations for the other irreps, and these represent normal modes of the system.
All of the above discussion would also work if we were discussing a tetrahedral arrangement of masses connected by elastic bonds (e.g. springs), where the $S_i^\alpha$ would be replaced with the displacement from equilibrium of each mass, and would be treated as a vector instead of a pseudo-vector (in which case we would obtain $A_1\oplus E \oplus T_1 \oplus T_2 \oplus T_2$), then the irreps correspond to normal modes of vibration, e.g. for a molecule.
The Hamiltonian can be rewritten as
$$H = \frac{1}{2} \sum_{\mu} J_\mu \vert\mathbf{m}_\mu\vert^2$$
where $\mu$ labels the irreps and the components of $\mathbf{m}_\mu$ are the linear combinations of the original variables which transform in each irrep, and the $J_\mu$ are some linear combination of the five allowed free parameters of $J_{ij}^{\alpha\beta}$.
In your case, you need to know how the operators in your expression are to transform under the symmetries of $T_d$. In the case of spins, they transform as pseudo-vectors. If you can answer that question, then you can follow the above prescription to figure out which irreps are present in your case. Figuring out a change of basis to a basis for the irreps is always doable by brute force, but finding a "nice" basis will take some intuition based on symmetry considerations.
