What do CG coefficients signify?

Questions › Category: Quantum Mechancis › What do CG coefficients signify?

I want to know the significance or the importance of C.G.(Clebsch-Gordon) coefficients calculated for the addition of angular momenta. Like what physical quantity do they represent?

Question Tags: Angular Momentum, CG coefficients

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Big Bang Boffin Moderator answered 3 years ago

Let’s say you have two particles with angular momentum $j_1$ & $j_2$ and the z-components $m_1$ & $m_2$ respectively.

Then the combined state $\left| j m \right>$ with total angular momentum $j$ and z-component $m$ will be a linear combination of the composite states $\left|j_1 m_1\right>\left|j_2 m_2\right>$
$\left|j m \right> = C^{j_1j_2j}_{m_1m_2m}\left|j_1 m_1\right>\left|j_2 m_2\right>$

The constants $C^{j_1j_2j}_{m_1m_2m}$ are called Clebsch-Gordon coefficients.

Now let’s take an example to better understand what do these coefficients mean and represent.
Let’s say two particles of spin 2 and 3 are in a box. And the combined state has spin 3 and it’s z-component is 0, then we can write this state using the above equation as:
$\left|3 0\right> = \frac{1}{\sqrt{5}}\left|2 1\right>\left|1 -1\right> +\sqrt{\frac{3}{5}}\left|2 0\right>\left|1 0\right> +\frac{1}{\sqrt{5}}\left|2 -1\right>\left|1 1 \right>$

Therefore the square of the CG coefficients tell us the probability of the two particles being in a particular composite state. You can notice that the square of the coefficients adds upto 1. Since the state $\left|3 0\right>$ can be obtained in 3 ways that is the linear combination of three composite states, the CG coefficients would tell us that if a measurement were made on the system then which particular composite state would be observed with what probability.
So basically in other words lets say you know the total angular momentum and it’s z-component of a particular system of two particles, then the C.G. coefficients would give the probability of observing the system to be in a particular composite state upon a measurement.

Hope it clears it.

References: Griffiths: Introduction to Quantum Mechancis

Mohit Khurana Moderator replied 3 years ago

Niceeee…

Mohit Khurana Moderator replied 3 years ago

Niceeee…y

Mohit Khurana Moderator replied 3 years ago

Niceeee…y

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