Characterization of states for "-wave" in discrete lattice system Assuming there exists four states, i.e. $|1\rangle, |2\rangle,|3\rangle,|4\rangle$, localized in four sites, which satisfy $C_4$ symmetry:

And we can characterize the combination of them via the eigenvalue of $J_z$, i.e. $e^{-iJ_z\theta/\hbar}|m\rangle=e^{-im\theta}|m\rangle$:
$$|m=0\rangle=\frac{1}{\sqrt{2}}(|1\rangle +|2\rangle +|3\rangle+|4\rangle)$$
$$|m=1\rangle=\frac{1}{\sqrt{2}}(|1\rangle +i|2\rangle -|3\rangle-i|4\rangle)$$
$$|m=-1\rangle=\frac{1}{\sqrt{2}}(|1\rangle -i|2\rangle -|3\rangle+i|4\rangle)$$
$$|m=2\rangle=\frac{1}{\sqrt{2}}(|1\rangle -|2\rangle +|3\rangle-|4\rangle)$$
Finally, we often label the above states as the label of partial wave of spherical symmetry:
$$m=0 \rightarrow s\text{-wave}$$$$m=\pm1 \rightarrow p\text{-wave}$$$$m=2 \rightarrow d\text{-wave}$$
Question:


*

*The labels of states here are the eigenvalues of $J_z$, i.e. magnetic quantum number $m$, but "-wave" character is labeled  by the angular    quantum number $l$ as in the spherial symmetry problem. Thus, I am confused of the relation between them.

*For $C_4$ symmetry, the eigenvalue $m$ need to mod $4$, which means $m=-1$ can also be considered as $m=3$, so why we don't say the $|m=-1\rangle$ as $f$-wave?


Addition
Here is the classification for states, but there also exists the classification for order parameter by symmetry via group representation theory, are there any reference/books recommendation in term of it?
 A: You are doing quantum mechanics around the clock in a clock of only 4 hours (period 4). Apart from an excessive square root in your normalizations (which I suspect should be just 1/2), your are Discrete Fourier Transforming your "hour" states $| 1\rangle, | 2\rangle,| 3\rangle,| 4\rangle,$ (the discrete analog of position angle eigenstates) to "shift" eigenstates $|m=i\rangle$, the discrete analog of angular momentum eigenstates. 
You are then inspecting these shift eigenstates as a 4-d representation of the (Abelian) cyclic group $C_4$,
$$|m=0\rangle=1\!\! 1 ~ \begin{pmatrix} 1\\1\\1\\1 \end{pmatrix} \frac{1}{ 2} ;\\
|m=1\rangle= \begin{pmatrix} 1\\i\\1\\-i \end{pmatrix} \frac{1}{ 2} = A \begin{pmatrix} 1\\1\\1\\1 \end{pmatrix} \frac{1}{ 2};\\  
 |m=-1\rangle= \begin{pmatrix} 1\\-i\\1\\i \end{pmatrix} \frac{1}{ 2} = C \begin{pmatrix} 1\\1\\1\\1 \end{pmatrix} \frac{1}{ 2}; \\  
 |m=2\rangle= \begin{pmatrix} 1\\-1\\1\\-1 \end{pmatrix} \frac{1}{ 2} = B \begin{pmatrix} 1\\1\\1\\1 \end{pmatrix} \frac{1}{ 2}~.$$
Check these four 4-vectors are (complex) orthonormal.  A represents a rotation (cyclic shift) by $\pi/2$, B by  $\pi$, and C by $3\pi/2$, and confirm their multiplication table in the link provided. 
Normally, $\ell$-wave denotes the spin of SO(3), into which your U(1) "rotation" generated by $J_z$ is embedded, but, here, I don't see such embeddings (but I am not experienced in the application you have in mind; you give little to no context).
Using the mainstream notation, it is, indeed, evident by inspection that $C=A^3$, so your $m=-1$ state is, indeed, equivalent to "m=3". Now, both
m=1 and  m=-1 have odd parity (reflection symmetry) like $\ell=1$ and 
m=2 even parity, (together with the identity m=0 a "spinless" rep!) , so it would make sense to identify it with d-wave, orthogonal to the rest. That is, when embedded into an SO(3), they transform just like the azimuthal phase factor $e^{-im\theta}$ of spherical harmonics.  I see no compelling reason to introduce f-waves, anywhere, but, then again, using SO(3) representation labels here appears like overkill.
