4
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Are you using Griffiths by chance? $\endgroup$ – BMS May 13 '14 at 4:33
  • $\begingroup$ What are the angular momenta that you are adding? $\endgroup$ – suresh May 13 '14 at 4:37
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Is the message here that the $\vert j=0, m=0 \rangle$ state needs to be orthogonal to $\vert 1, 0 \rangle$ and $\vert 2, 0 \rangle$? $\endgroup$ – BMS May 13 '14 at 5:20
  • $\begingroup$ That's it exactly. $\endgroup$ – rob May 13 '14 at 5:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.