0
$\begingroup$

So let's consider the case of $j_1=3/2,j_2=1/2$. We know from a table that the Clebsch Gordan Coefficient should be one. However, we also know that the recursion relation, referencing Zettili Quantum Mechanics, is given by $$\sqrt{(j\mp m)(j\pm m+1)}\langle j_1 j_2;m_1 m_2|j,m\pm 1\rangle=\sqrt{(j_1\pm m_1)(j_1\mp m_1+1)}\langle j_1 j_2;m_1\mp 1,m_2|jm\rangle+\sqrt{(j_2\pm m_2)(j_2\mp m_2+1)}\langle j_1 j_2;m_1,m_2\mp 1|jm\rangle$$ However, if we plug in, for $m=2$ with $m_1=3/2,m_2=1/2$, and the aforementioned $j_1,j_2$ values, we get $$\sqrt{(3/2+3/2)(3/2-3/2+1)}\langle3/2,1/2;1/2,1/2\rangle+\sqrt{(1/2+1/2)(1/2-1/2+1)}\langle{3/2,1/2; 3/2,-1/2\rangle} $$ Which is equal to $$\sqrt{3}\langle3/2,1/2;1/2,1/2\rangle+\langle{3/2,1/2; 3/2,-1/2\rangle} $$ Invoking the normalization condition, $$\left(\sqrt{(3/2+3/2)(3/2-3/2+1)}\langle3/2,1/2;1/2,1/2\rangle\right)^2+\left(\sqrt{(1/2+1/2)(1/2-1/2+1)}\langle{3/2,1/2; 3/2,-1/2\rangle}\right)^2=1 $$ We get that the CGC is not 1, but $\frac{1}{4}$.

There is obviously some mistake I am making here, but I am following the recursion relation formula. So is there some kind of restriction on using it as opposed to simply using the annihilation operator $\hat{J}_{-}$?

$\endgroup$
5
  • $\begingroup$ Where did you find that normalization condition? $\endgroup$ Feb 3, 2023 at 1:30
  • $\begingroup$ It is in Zettili Quantum Mechanics, equation 7.125 $\endgroup$ Feb 3, 2023 at 1:32
  • 1
    $\begingroup$ Aren't your expectation values missing the $|jm\rangle$ or $|j, m\pm 1\rangle$ parts? Also how are you computing $j$ from $j_1$ and $j_2$? $\endgroup$
    – Andrew
    Feb 3, 2023 at 2:40
  • $\begingroup$ As $|jm\rangle$ are constant and I am focused on the coefficient rather than the ket, I have excluded them. And $j=2$, as $j_1=s_1=3/2$ and $j_2=s_2=1/2$, this means $j=j_1+j_2=2$(this is a case for orbital AM =0) $\endgroup$ Feb 3, 2023 at 2:52
  • $\begingroup$ The kets you skipped would have alerted you to the entire brackets vanishing, by virtue of mismatched m s. $\endgroup$ Feb 3, 2023 at 23:06

1 Answer 1

2
$\begingroup$

You are misusing the recursion. All brackets where the (summed) ms don't match up between bras and kets vanish.

It looks like you are choosing the upper sign, so, indeed, you are verifying that 0=0+0. You did not witness a failure of the identity.

To find a nontrivial example, instead, consider $m_1=3/2; ~~m_2=1/2; ~~m=1;~~j=2$, and take the upper signs again, for the sake of argument: $$ 2\langle 3/2~1/2; 3/2~1/2| 2~2\rangle = \sqrt{3} \langle 3/2~1/2 ; 1/2~1/2 | 2~1\rangle +\langle 3/2~1/2 ; 3/2~-\!1/2 | 2~1\rangle \\ 2= 3/2 +1/2. $$ Can you replicate the CG table, now?

$\endgroup$
1
  • $\begingroup$ Yes, I realized a bit later. Thanks for the answer anyways $\endgroup$ Feb 4, 2023 at 0:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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