Clebsch-Gordan Coefficients for two spin-1 particles - Why is there a ∣0⟩∣0⟩ ket? I have used the rules for addition of angular momenta to work out the Clebsch-Gordan coefficients, which all seem right except for state $\lvert0,0\rangle$:
For n = 1
\begin{align}
\lvert1,1\rangle & = \frac{1}{\sqrt 2} \left( \lvert0\rangle\lvert1\rangle - \lvert1\rangle\lvert0\rangle \right) \\
\lvert1,0\rangle & = \frac{1}{\sqrt 2} \left( \lvert-1\rangle\lvert1\rangle - \lvert1\rangle\lvert-1\rangle \right) \\
\lvert1,-1\rangle & = \frac{1}{\sqrt 2} \left( \lvert0\rangle\lvert-1\rangle - \lvert-1\rangle\lvert0\rangle\right)
\end{align}
Now the state $\lvert0,0\rangle$ must be perpendicular to $\lvert1,0\rangle$ and is a linear combination of the basis kets of $\lvert1,0\rangle$:
$$\lvert0,0\rangle = \frac{1}{\sqrt 2} \left(\lvert-1\rangle\lvert1\rangle + \lvert1\rangle\lvert-1\rangle\right)$$.
But in the table, there is an extra ket $\lvert0\rangle\lvert0\rangle$; Why is this so?
(From the table):
$$\lvert0,0\rangle = \frac{1}{\sqrt 3} \left(\lvert-1\rangle\lvert1\rangle + \lvert1\rangle\lvert-1\rangle - \lvert0\rangle\lvert0\rangle\right).$$
My intuition tells me that you need to include the $\lvert0\rangle\lvert0\rangle$ state in order for the entire set of basis to be complete. But how do I show this?
 A: If you have two spin-1 particles, there are $three$ states where the projection of the total angular momentum is zero:
$$
\begin{align}
\left|2,0\right> &= \frac1{\sqrt6} \left(
\big|1,1\big>\big|1,-1\big>
~~+~~
\big|1,-1\big>\big|1,1\big>
~~+~~
\sqrt4\cdot
\big|1,0\big>\big|1,0\big>
\right)
\\
\left|1,0\right> &= \frac1{\sqrt2} \left(
\big|1,1\big>\big|1,-1\big>
~~-~~
\big|1,-1\big>\big|1,1\big>
\right)
\\
\left|0,0\right> &= \frac1{\sqrt3} \left(
\big|1,1\big>\big|1,-1\big>
~~+~~
\big|1,-1\big>\big|1,1\big>
~~-~~
\big|1,0\big>\big|1,0\big>
\right)
\end{align}
$$
The state $\left|1,0\right>\left|1,0\right>$ enters $\left|2,0\right>$ state with positive sign from applying the lowering operator to $\left|2,1\right>$.  (If you haven't done this algebra yourself, do so — it's quite edifying.)  Therefore the symmetric zero-spin combination must contain some $\left|1,0\right>\left|1,0\right>$ with a negative sign, for orthogonality.
Alternatively, you can operate on your prospective total-spin-zero state with the angular momentum raising operator for your two spin-1 particles:
$$
\begin{array}{rclr}
L_+ & \left|1,1\right>\left|1,-1\right>
 &= & \sqrt2 \left|1,1\right>\left|1,0\right>\\
L_+ & \left|1,0\right>\left|1,0\right>
 &= \sqrt2 \left|1,1\right>\left|1,0\right> &{}+{} \sqrt2 \left|1,0\right>\left|1,1\right>\\
L_+ & \left|1,-1\right>\left|1,1\right>
 &= \sqrt2\left|1,0\right>\left|1,1\right>\\
\end{array}
$$
This should make it clear that a negative contribution from  $\left|0\right>\left|0\right>$, is required to make to construct an $m=0$ state that vanishes when it sees a raising or lowering operator.
