Why spin-$1/2$ objects don't have quadrupolar magnetic moment? I'm asking myself more generally why a spin of size $S$ will feature multipolar states of degrees $k$ up to $2S$?
(This implies the question in the title: spin-$1/2$ can't have any quadrupolar contributions)
How can I (explicitly) derive this result? I'm looking for a "straightforward" connection with the field of the electron for example.
I'm studying this in the context of spin-nematic phases in quantum spin systems, any help in the subject would be very appreciated!
Edit: More precisely I am reading this text.
And my question is about what's on page 6. Also, I wonder if it is possible to demonstrate that expanding the projection of an arbitrary-spin wave function over a coherent state (we should find for spin-$1/2$ only a scalar term and a dipolar term, for spin-$1$ a scalar, a dipolar and a quadrupolar term, etc.)? If it is the case I don't know how to make it.
 A: Yes, the statement can be explicitly verified from the matrix representation of the spin operators acting on different spins. Acting on the spin-1/2 object, the spin operators read
$$S^x=\left(
\begin{array}{cc}
 0 & \frac{1}{2} \\
 \frac{1}{2} & 0 \\
\end{array}
\right), S^y=\left(
\begin{array}{cc}
 0 & \frac{i}{2} \\
 -\frac{i}{2} & 0 \\
\end{array}
\right), S^z=\left(
\begin{array}{cc}
 -\frac{1}{2} & 0 \\
 0 & \frac{1}{2} \\
\end{array}
\right).\qquad(1)$$
For spin-1 object, the spin operators read
$$S^x=\left(
\begin{array}{ccc}
 0 & \frac{1}{\sqrt{2}} & 0 \\
 \frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} \\
 0 & \frac{1}{\sqrt{2}} & 0 \\
\end{array}
\right), S^y=\left(
\begin{array}{ccc}
 0 & \frac{i}{\sqrt{2}} & 0 \\
 -\frac{i}{\sqrt{2}} & 0 & \frac{i}{\sqrt{2}} \\
 0 & -\frac{i}{\sqrt{2}} & 0 \\
\end{array}
\right), S^z=\left(
\begin{array}{ccc}
 -1 & 0 & 0 \\
 0 & 0 & 0 \\
 0 & 0 & 1 \\
\end{array}
\right).\qquad(2)$$
Using these operators, it is straight forward to verify that spin-1/2 object does not have quadrupole moment, but spin-1 object does have. For example, the quadrupole moment can be written in terms of the spin operators as $Q^{x^2-y^2}=(S^x)^2-(S^y)^2$ by definition. Plugging in Eq.(1) and complete the matrix multiplication, it can be verified that
$$Q^{x^2-y^2}=\left(
\begin{array}{cc}
 0 & 0 \\
 0 & 0 \\
\end{array}
\right),$$
meaning that the quadrupole moment vanishes for spin-1/2 object. However if we plug in Eq.(2), it can be found that
$$Q^{x^2-y^2}=\left(
\begin{array}{ccc}
 0 & 0 & 1 \\
 0 & 0 & 0 \\
 1 & 0 & 0 \\
\end{array}
\right),$$
meaning that the quadrupole moment is non-vanishing for spin-1 object. Similar calculations can be done for other components of the quadrupole moment straightforwardly. It can be verified that all the five quadrupole moment operators are zero matrices in the spin-1/2 representation, thus explicitly proven that spin-1/2 object has no quadrupole moment.

The answer is basically an expansion of @Meng Cheng's comment.
