I just want to get some clarity a simple point regarding simultaneous eigenkets of angular momentum operators. With regards to simultaneous eigenstates of angular momentum operators, we have that $$J^2_1,~J^2_2,J_{1z}~~~\text{and }~~J_{2z}$$ have simultaneous eigenket $|j_1 j_2; m_1 m_2 \rangle$ and $J^2, J_{1}^2, J_{2}^2$ and $J_{z}$ has simultaneous eigenket $|j_1 j_2; j m \rangle$.

Am I correct in stating that the one eigenket is really a tensor product i.e. $$| j_1 j_2; m_1 m_2 \rangle = | j_1 m_1 \rangle \otimes |j_2, m_2 \rangle.$$

Hence would it be valid to write the Clebsch-Gordan coefficient as $\langle j_1 j_2; j m| (| j_1 m_1 \rangle \otimes | j_2, m_2 \rangle )$?


1 Answer 1


Your notation is a little redundant as you list $j_1$ and $j_2$ twice. Technically it is correct and it is also correct to write $$ \vert j_1j_2;m_1m_2\rangle=\vert j_1m_1\rangle\otimes \vert 2_1m_2\rangle \, . $$

If I may suggest:

Take the state $\vert j_1j_2;m_1m_2\rangle$ and expand it in terms of the coupled states $\vert JM\rangle$: \begin{align} \vert j_1j_2;m_1m_2\rangle=\sum_{J(M)} \vert JM\rangle \langle JM\vert j_1j_2;m_1m_2\rangle\, . \end{align} The coefficient $\langle JM\vert j_1j_2;m_1m_2\rangle$ is the Clebsch. Note that the sum over $M$ is actually just formal since the CG is $0$ unless $M=m_1+m_2$.

Because matrix elements of $\hat J_\pm$ and $\hat J_z$ can always be chosen to be real, the CG can also always chosen to be real, so that $$ \langle JM\vert j_1j_2;m_1m_2\rangle= \langle j_1j_2;m_1m_2\vert JM\rangle $$ and therefore $$ \vert JM\rangle = \sum_{j_1m_1j_2m_2} \vert j_1j_2;m_1m_2\rangle \langle j_1j_2;m_1m_2\vert JM\rangle\, . $$ Writing the CG in this way emphasizes it's an overlap from which several properties immediately follow, like \begin{align} \sum_{JM} \vert \langle j_1j_2;m_1m_2\vert JM\rangle\vert^2&=\delta_{m_1+m_2,M}\, ,\\ \sum_{j_1m_1j_2m_2} \vert \langle j_1j_2;m_1m_2\vert JM\rangle\vert^2&=\delta_{m_1+m_2,M}\, , \end{align} as well as orthogonality properties such as \begin{align} \sum_{J'M'JM} \langle j_1j_2;m_1m_2\vert JM\rangle \langle j_1j_2;m_1m_2\vert J'M'\rangle =\delta_{JJ'}\delta_{MM'} \delta_{m_1+m_2,M} \end{align}

Edit: The CGs are real because:

  1. The state $\vert J,M=J\rangle $ is killed by $J_+$. Write this state as $$ \vert J,M=J\rangle=\sum_{m_1m_2}\vert j_1j_2;m_1m_2\rangle \langle j_1j_2;m_1m_2 \vert JM\rangle $$ and act on this using $\hat J_+$. The result is a recursion relation for the CG which contains real coefficients since the matrix elements of $J_+$ are real. The CGs must satisfy this and so can be chosen as real.
  2. Once you have the highest weight $\vert J,M=J\rangle$ state as a real linear combination of $\vert j_1j_2;m_1m_2\rangle$, one gets the remaining $\vert JM\rangle$ states by lowering with $\hat J_-$. Since the matrix elements of $\hat J_-$ are real, it follows that the CGs will also be real.

As an example, consider the state $$ \vert 1/2,1/2\rangle = \vert 1,1/2;1,-1/2\rangle \langle 1,1/2;1,-1/2\vert 1/2,1/2\rangle + \vert 1,1/2;0,1/2\rangle \langle 1,1/2;0,1/2\vert 1/2,1/2\rangle\, . $$ Act on it with $\hat J_+$ to get $$ 0=\vert 1,1/2;1,1/2\rangle \langle 1,1/2;1,-1/2\vert 1/2,1/2\rangle +\sqrt{2}\vert 1,1/2;1,1/2\rangle\langle 1,1/2;0,1/2\vert 1/2,1/2\rangle $$ from which the ratio of the CG's can be obtained from \begin{align} \vert 1,1/2;1,1/2\rangle\langle 1,1/2;0,1/2\vert 1/2,1/2\rangle&=-\frac{1}{\sqrt{2}} \vert 1,1/2;1,1/2\rangle \langle 1,1/2;1,-1/2\vert 1/2,1/2\rangle\, ,\\ \langle 1,1/2;0,1/2\vert 1/2,1/2\rangle&=-\frac{1}{\sqrt{2}} \langle 1,1/2;1,-1/2\vert 1/2,1/2\rangle\ \tag{1} \end{align} where $\sqrt{2}=\langle 1,1\vert \hat J_+\vert 1,0\rangle$.

Now the CGs must satisfy $$ \vert\langle 1,1/2;1,-1/2\vert 1/2,1/2\rangle\vert^2+ \vert\langle 1,1/2;0,1/2\vert 1/2,1/2\rangle\vert^2=1\, . $$ Using (1) we can rewrite the above as \begin{align} &\vert\langle 1,1/2;1,-1/2\vert 1/2,1/2\rangle\vert^2+ \frac{1}{2}\vert\langle 1,1/2;1,-1/2\vert 1/2,1/2\rangle\vert^2\\ &\quad =\frac{3}{2}\vert\langle 1,1/2;1,-1/2\vert 1/2,1/2\rangle\vert^2=1\, . \end{align} with solution $$ \langle 1,1/2;1,-1/2\vert 1/2,1/2\rangle=\sqrt{\frac{2}{3}} $$ where the phase of $\langle 1,1/2;1,-1/2\vert 1/2,1/2\rangle$ has been chosen to be $+1$. Clearly (1) then implies $\langle 1,1/2;1,-1/2\vert 1/2,1/2\rangle=-1/\sqrt{3}$, which is also real.

Once you have $\vert J,M=J\rangle$, acting with $\hat J_-$ produces a recursion between the different CGs with coefficients depending on the matrix elements of $\hat J_-$, which are real, so the solutions of the recursions that give the CGs will also be real.

  • $\begingroup$ Thanks. The notation is used in Sakurai, although he does supress the $j_1 j_2$ further on in the text. Anyway, I understand your reasoning except for how $\hat{J}_{pm}$ and $\hat{J}_z$ implies that CG can be chosen to be real? $\endgroup$
    – user101311
    May 22, 2017 at 14:11
  • $\begingroup$ Shouldn't the second term in your example after you act on $| 1/2, 1/2 \rangle$ with $J_{+}$ be $$0 = ...+ \sqrt{2} |1, 1/2; 1, 1/2 \rangle \langle 1,1/2;0, 1/2 | 1/2, 1/2 \rangle?$$ Also, how are what we did with $J_{+}$ imply that CG $\langle 1 1/2; 1 , 1/2 | 1/2 , 1/2 \rangle$ is real? $\endgroup$
    – user101311
    May 22, 2017 at 15:09
  • $\begingroup$ You use $\langle 1,1/2;1,1/2\vert 1/2,1/2\rangle$ when you state "As the CG..." I don't see where you have this CG before that? $\endgroup$
    – user101311
    May 22, 2017 at 19:33
  • $\begingroup$ Thanks. But how do you use normalization to imply that it is real with phase $+1$? $\endgroup$
    – user101311
    May 22, 2017 at 21:20
  • 1
    $\begingroup$ @Alex You could choose this CG to be $-\sqrt{2/3}$ is you wanted, and all the other CGs connected to this one by recursion would have a $-$ sign. You could even choose the phase to be $+i$ so this CG is $i\sqrt{2/3}$, and all other CGs connected to this one by recursion would have an extra $i$... but why make life complicated? It's like normalizing a wavefunction... you can choose the normalization constant to have a factor of $i$ in it, but why? The choice $+1$ is the most convenient. $\endgroup$ May 23, 2017 at 19:21

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