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Consider the sum of angular momenta $J=J_1+J_2$.

When one has a state $|J,M\rangle$ of an eigenbasis of a common eigenspace of $J^2$ and $J_z$, one can write it in terms of the elements $|j_1,m_1;j_2,m_2\rangle$ of an eigenbasis of a common eigenspace of $J_1^2,\,J_2^2,\,J_{1z},\,J_{2z}$: $$|J,M\rangle=\underset{m_1,m_2}{\sum} C^{j_1,\,j_2,\,J}_{m_1,\,m_2,\,M}\ \ |j_1,m_1;j_2,m_2\rangle$$ Where $C^{j_1,\,j_2,\,J}_{m_1,\,m_2,\,M}$ are the Clebsch Gordan coefficients.

I can do this easily using a Clebsch Gordan Coefficient's table. What if I want to express the kets $|j_1,m_1;j_2,m_2\rangle$ in terms of the kets $|J,M\rangle$? In this case, I can write all the $|J,M\rangle$ in terms of the states $|j_1,m_1;j_2,m_2\rangle$ and combine them in such a way that I get one of the $|j_1,m_1;j_2,m_2\rangle$ states. This can lead to long calculations.

I wonder if there is a table like the one for Clebsch Gordan coefficients but for a kind of "reversed" Clebsch Gordan coefficients $B^{j_1,\,j_2,\,J}_{m_1,\,m_2,\,M}$ such that $$|j_1,m_1;j_2,m_2\rangle=\underset{J,M}{\sum} B^{j_1,\,j_2,\,J}_{m_1,\,m_2,\,M}\ \ |J,M\rangle$$

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3 Answers 3

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If I understand well, the coefficients $B$ are just in fact CG's: to be explicit $$ \vert j_1m_1; j_2m_2\rangle = \sum_{J(M)} C^{j_1j_2J}_{m_1m_2M} \vert JM\rangle $$ Note the sum of $M$ is not really a sum as $M$ must satisfy $M=m_1+m_2$.

The best way to see this is to start with $\vert j_1m_1; j_2m_2\rangle$ and just insert the unit $I=\sum_{JM}\vert JM\rangle\langle JM\vert$. One then has $$ \vert j_1m_1; j_2m_2\rangle=\sum_{JM}\vert JM\rangle\langle JM\vert j_1m_1;j_2m_2\rangle $$ with $$ \langle JM\vert j_1m_1;j_2m_2\rangle = \langle j_1m_1;j_2m_2\vert JM\rangle = C^{j_1j_2J}_{m_1m_2M} $$ since the CGs are real.

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  • $\begingroup$ Ah, we posted at the same time. I agree with you. $\endgroup$
    – Lagerbaer
    Jan 23, 2017 at 17:25
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The way you ask the question now, it can't quite be done. That's because if you tell me $J$ and $M$, there's lots of ways for $j_1$ and $j_2$ to get to that. If the total angular momentum is $0$, for example, then there's an infinite number of possibilities for $j_1$ and $j_2$ to get me to $0$. They could both be spin 1/2, or they could both be spin 1. Or they could both be spin 42.

However, if you specify what $j_1$ and $j_2$ are, then it is easy to reverse the Clebsch-Gordan coefficients via the insight that they form an orthogonal relationship.

Basically:

$$|j_1, m_1, j_2, m_2\rangle = \sum_{J,M'} |J,M\rangle \langle J,M | j_1, m_1, j_2, m_2\rangle$$

where the states $|J, M\rangle$ are restricted to those that you can actually form out of $j_1$ and $j_2$.

And, well that braket is just the Clebsch-Gordan coefficient $C_{m_1, m_2, M}^{j_1, j_2, J}$. Well, technically it's its complex conjugate, but since the coefficients are real-valued in the typical phase convention, that distinction doesn't matter.

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The Clebsch-Gordan coefficients contain enough information to calculate the inverse, since $$\langle j_1 j_2 m_1 m_2 | jm \rangle = C^{j_1, j_2, j}_{m_1, m_2, m}.$$ Conjugating, the matrix elements you're looking for are $$\langle jm | j_1 j_2 m_1 m_2 \rangle = (C^{j_1, j_2, j}_{m_1, m_2, m})^*.$$

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  • $\begingroup$ Note some inconsistent capitalization. $\endgroup$ Jan 23, 2017 at 19:25

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