# 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?

• Are you using Griffiths by chance? – BMS May 13 '14 at 4:33
• What are the angular momenta that you are adding? – suresh May 13 '14 at 4:37

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.
• Is the message here that the $\vert j=0, m=0 \rangle$ state needs to be orthogonal to $\vert 1, 0 \rangle$ and $\vert 2, 0 \rangle$? – BMS May 13 '14 at 5:20