Clebsch-Gordan Coefficients
A no-panic walkthrough of Clebsch-Gordan coefficient tables — because some absolute lunatic already did all the hard work, and all we have to do is learn to read it.
This is a continuation of posting #24, so if you haven’t read that, go back first.
OK, let’s just dive in.
For a single particle, if $s=k$, then the possible values of $m_s$ were $-k, -k+1, -k+2, \ldots, k-1, k$.
So here’s the thing:
$$S_{total} = k \;\Rightarrow\; m_{s_{total}} \text{ can be } -k,\; -k+1,\; -k+2,\; \ldots,\; k-1,\; k$$And here’s what you have to not get confused about: the possible values of $S_{total}$ itself are $S_1+S_2,\; S_1+S_2-1,\; S_1+S_2-2,\; \ldots,\; |S_1-S_2|$.
Feel your brain cramping up yet? Mine is.
So “some combined state”
$$|\; S_{total},\; m_{s_{total}} \rangle$$— this guy — gets built out of
$$|\; s_{1},\; m_{s_{1}} \rangle |\; s_{2},\; m_{s_{2}} \rangle$$as a linear combo of a bunch of these little guys. Right?!
Oh god… how the hell are you supposed to find those coefficients one by one?! WHAT!!!!
Relax, relax, relaaaax~~~~
Some absolute lunatic on this Earth has already done all of them and tabulated the coefficients like an integral table. For real.
It’s called the Clebsch-Gordan coefficient.
We just pull up the table when we need it and read off the coefficients. Done.
So really all we need to learn is how to read the table.
The professor said he’d definitely throw one problem from this on the exam, so I studied the hell out of it. Which means the explanation is also about to be amazing. Hype.
Here’s the basic layout:
Each position means something specific. We’ve been staring at this notation for aaages now, so at this point you should be able to tell what each number is, right??
OK, let’s read it step by step~~
First, you read here:
Basically I wrote what each number means inside the L-shaped cells.
Now let’s actually use this thing!!
Case: $S_1 = 2$, $S_2 = 1$~~
Then:
<finishing shot>
I went through this myself scrolling slowly, and… it’s fine???? It actually makes sense!!! hehhh hehhh hehhh
Am I the only one who gets it T_T T_T T_T lolololololol
OK so if we unpack what those coefficients actually mean a little more:
say we’ve got a system like this, and we know which particles it’s made of haha.
Anyway — we took $S_{total}^2$ and $S_{z,total}$, we measured, and if $|\; 3,\; 1\rangle$ came out, then the system is
$$|\; 3,\; 1\rangle \;=\; \sqrt{\frac{1}{15}}|\; 2,\; 2\rangle|\; 1,\; {-1}\rangle \;+\; \sqrt{\frac{8}{15}}|\; 2,\; 1\rangle|\; 1,\; 0\rangle \;+\; \sqrt{\frac{6}{15}}|\; 2,\; 0\rangle|\; 1,\; 1\rangle$$And what this says is:
$$\text{probability of being in } |\; 2,\; 2\rangle|\; 1,\; {-1}\rangle = \frac{1}{15} \\ \text{probability of being in } |\; 2,\; 1\rangle|\; 1,\; 0\rangle = \frac{8}{15} \\ \text{probability of being in } |\; 2,\; 0\rangle|\; 1,\; 1\rangle = \frac{6}{15}$$Right?!
But!!!!! That’s not all the Clebsch-Gordan coefficient does!!!!
It also works the other way!!
If so far we were going from combined state → uncombined state, now I’m going to show that uncombined state → combined state works too!!
So the real meaning of the Clebsch-Gordan coefficient is: it’s a link in both directions.
That is: combined state ↔ link ↔ uncombined state.
Let me try it with different numbers than the textbook. Going uncombined → combined this time.
The key move: where before we read the ‘coefficients’ block vertically, this time we read it horizontally!!!!
You got it, right?!!
Let me do the textbook’s example too, just verbally.
If we build $|\; \frac{3}{2},\; \frac{1}{2}\rangle|\; 1,\; 0\rangle$ using the table:
$$\left|\; \frac{3}{2},\; \frac{1}{2}\right\rangle\!\left|\; 1,\; 0\right\rangle \;=\; \sqrt{\frac{3}{5}}\left|\frac{5}{2},\; \frac{1}{2}\right\rangle \;+\; \sqrt{\frac{1}{15}}\left|\; \frac{3}{2},\; \frac{1}{2}\right\rangle \;+\; \sqrt{\frac{1}{3}}\left|\; \frac{1}{2},\; \frac{1}{2}\right\rangle$$Spelling out what this equation means one more time:
single particles —
$$S_{1} \;=\; \frac{3}{2} \qquad m_{s_{1}} \;=\; \frac{1}{2}$$one particle like this, plus
$$S_{2} \;=\; 1 \qquad m_{s_{2}} \;=\; 0$$another particle like this —
fluffy fluffy fluffy clouds
toss them in here, shake the cloud vigorously —
it becomes this. And if you now measure the Total against the combined system, the total state gives:
$$\text{prob. of measuring } \left|\; \frac{5}{2},\; \frac{1}{2}\right\rangle = \frac{3}{5} \\ \text{prob. of measuring } \left|\; \frac{3}{2},\; \frac{1}{2}\right\rangle = \frac{1}{15} \\ \text{prob. of measuring } \left|\; \frac{1}{2},\; \frac{1}{2}\right\rangle = \frac{1}{3}$$That’s what it’s saying. Right?!?!
And finally — the long, long Chapter 4 finally comes to an end……. hehhh hehhh hehhh
Let me crank through a few problems and put a period on Chapter 4!!
Prob 4.34
Apply $S_-$ to $|\; 1,\; 0\rangle$ (eq. 4.177) and verify that you get
$$\sqrt{2}\hbar \cdot |\; 1,\;{-1}\rangle$$$|\; 1,\; 0\rangle$ is $\frac{1}{\sqrt{2}}(\uparrow\downarrow + \downarrow\uparrow)$ — let’s slam $S_-$ against this.
$S_-$ acts on both the proton and the electron, so $S_- = S_-^{\text{proton}} + S_-^{\text{electron}}$. And we’ll use $S_\pm f_{s,m_s} = \hbar\sqrt{s(s+1) - m_s(m_s\pm 1)}\,f_{s,m_s\pm 1}$.
Let’s go:
$$\left(S_{-}^{\text{p}} + S_{-}^{\text{e}}\right)\!\left(\tfrac{1}{\sqrt{2}}(\uparrow\downarrow + \downarrow\uparrow)\right) \;=\; \tfrac{1}{\sqrt{2}}\!\left(S_{-}^{\text{p}}\uparrow\right)\!\downarrow + \tfrac{1}{\sqrt{2}}\!\left(S_{-}^{\text{p}}\downarrow\right)\!\uparrow + \tfrac{1}{\sqrt{2}}\uparrow\!\left(S_{-}^{\text{e}}\downarrow\right) + \tfrac{1}{\sqrt{2}}\downarrow\!\left(S_{-}^{\text{e}}\uparrow\right)$$The arrows are getting confusing… Dirac notation, let’s go:
$$=\; \tfrac{1}{\sqrt{2}}\!\left(S_{-}^{\text{p}}\left|\tfrac{1}{2},\tfrac{1}{2}\right\rangle\right)\!\downarrow + \tfrac{1}{\sqrt{2}}\!\left(S_{-}^{\text{p}}\left|\tfrac{1}{2},-\tfrac{1}{2}\right\rangle\right)\!\uparrow + \tfrac{1}{\sqrt{2}}\uparrow\!\left(S_{-}^{\text{e}}\left|\tfrac{1}{2},-\tfrac{1}{2}\right\rangle\right) + \tfrac{1}{\sqrt{2}}\downarrow\!\left(S_{-}^{\text{e}}\left|\tfrac{1}{2},\tfrac{1}{2}\right\rangle\right)$$$$=\; \tfrac{1}{\sqrt{2}}\!\left(\hbar\sqrt{\tfrac{1}{2}\cdot\tfrac{3}{2}-\tfrac{1}{2}(-\tfrac{1}{2})}\left|\tfrac{1}{2},-\tfrac{1}{2}\right\rangle\right)\!\downarrow + \tfrac{1}{\sqrt{2}}(0)\!\uparrow + \tfrac{1}{\sqrt{2}}\uparrow(0) + \tfrac{1}{\sqrt{2}}\downarrow\!\left(\hbar\sqrt{\tfrac{1}{2}\cdot\tfrac{3}{2}-\tfrac{1}{2}(-\tfrac{1}{2})}\left|\tfrac{1}{2},-\tfrac{1}{2}\right\rangle\right)$$Back to arrows:
$$=\; \tfrac{1}{\sqrt{2}}(\hbar\downarrow)\!\downarrow + \tfrac{1}{\sqrt{2}}\downarrow(\hbar\downarrow) \;=\; \tfrac{2}{\sqrt{2}}\hbar\downarrow\downarrow \;=\; \sqrt{2}\hbar\downarrow\downarrow$$Oh nice! Confirmed. How satisfying.
b) Apply to $|\; 0,\; 0\rangle$ and verify you get 0.
Hmm~
$|\; 0,\; 0\rangle$ is $\tfrac{1}{\sqrt{2}}(\uparrow\downarrow - \downarrow\uparrow)$, and we’re told to apply $S_\pm$. Doing both at once would be a mess, so let’s split it.
i) $S_+$:
$$\left(S_{+}^{\text{p}}+S_{+}^{\text{e}}\right)\cdot\tfrac{1}{\sqrt{2}}(\uparrow\downarrow - \downarrow\uparrow)$$$$=\; \tfrac{1}{\sqrt{2}}(S_{+}^{\text{p}}\uparrow)\!\downarrow - \tfrac{1}{\sqrt{2}}(S_{+}^{\text{p}}\downarrow)\!\uparrow + \tfrac{1}{\sqrt{2}}\uparrow(S_{+}^{\text{e}}\downarrow) - \tfrac{1}{\sqrt{2}}\downarrow(S_{+}^{\text{e}}\uparrow)$$$$=\; \tfrac{1}{\sqrt{2}}(S_{+}^{\text{p}}\left|\tfrac{1}{2},\tfrac{1}{2}\right\rangle)\!\downarrow - \tfrac{1}{\sqrt{2}}(S_{+}^{\text{p}}\left|\tfrac{1}{2},-\tfrac{1}{2}\right\rangle)\!\uparrow + \tfrac{1}{\sqrt{2}}\uparrow(S_{+}^{\text{e}}\left|\tfrac{1}{2},-\tfrac{1}{2}\right\rangle) - \tfrac{1}{\sqrt{2}}\downarrow(S_{+}^{\text{e}}\left|\tfrac{1}{2},\tfrac{1}{2}\right\rangle)$$$$=\; 0 - \tfrac{1}{\sqrt{2}}\!\left(\hbar\sqrt{\tfrac{1}{2}\cdot\tfrac{3}{2}+\tfrac{1}{2}\cdot\tfrac{1}{2}}\left|\tfrac{1}{2},\tfrac{1}{2}\right\rangle\right)\!\uparrow + \tfrac{1}{\sqrt{2}}\uparrow\!\left(\hbar\sqrt{\tfrac{1}{2}\cdot\tfrac{3}{2}-(-\tfrac{1}{2})(\tfrac{1}{2})}\left|\tfrac{1}{2},\tfrac{1}{2}\right\rangle\right) + 0$$$$=\; -\tfrac{1}{\sqrt{2}}(\hbar\uparrow)\!\uparrow + \tfrac{1}{\sqrt{2}}\uparrow(\hbar\uparrow) \;=\; 0$$ii) $S_-$:
$$\left(S_{-}^{\text{p}}+S_{-}^{\text{e}}\right)\cdot\tfrac{1}{\sqrt{2}}(\uparrow\downarrow - \downarrow\uparrow)$$$$=\; \tfrac{1}{\sqrt{2}}(S_{-}^{\text{p}}\uparrow)\!\downarrow - \tfrac{1}{\sqrt{2}}(S_{-}^{\text{p}}\downarrow)\!\uparrow + \tfrac{1}{\sqrt{2}}\uparrow(S_{-}^{\text{e}}\downarrow) - \tfrac{1}{\sqrt{2}}\downarrow(S_{-}^{\text{e}}\uparrow)$$$$=\; \tfrac{1}{\sqrt{2}}(S_{-}^{\text{p}}\left|\tfrac{1}{2},\tfrac{1}{2}\right\rangle)\!\downarrow - \tfrac{1}{\sqrt{2}}(S_{-}^{\text{p}}\left|\tfrac{1}{2},-\tfrac{1}{2}\right\rangle)\!\uparrow + \tfrac{1}{\sqrt{2}}\uparrow(S_{-}^{\text{e}}\left|\tfrac{1}{2},-\tfrac{1}{2}\right\rangle) - \tfrac{1}{\sqrt{2}}\downarrow(S_{-}^{\text{e}}\left|\tfrac{1}{2},\tfrac{1}{2}\right\rangle)$$$$=\; \tfrac{1}{\sqrt{2}}\!\left(\hbar\sqrt{\tfrac{1}{2}\cdot\tfrac{3}{2}-\tfrac{1}{2}(-\tfrac{1}{2})}\left|\tfrac{1}{2},-\tfrac{1}{2}\right\rangle\right)\!\downarrow - 0 + 0 - \tfrac{1}{\sqrt{2}}\downarrow\!\left(\hbar\sqrt{\tfrac{1}{2}\cdot\tfrac{3}{2}-(\tfrac{1}{2})(-\tfrac{1}{2})}\left|\tfrac{1}{2},-\tfrac{1}{2}\right\rangle\right)$$$$=\; \tfrac{1}{\sqrt{2}}(\hbar\downarrow)\!\downarrow - \tfrac{1}{\sqrt{2}}\downarrow(\hbar\downarrow) \;=\; 0$$Oh nice, confirmed.
Part c) was exactly what I already proved in the previous posting above, so I’ll skip it.
Prob 4.35
Quarks have spin $\tfrac{1}{2}$. Three quarks together make a Baryon (like a proton or neutron), and two quarks (more precisely, a quark + antiquark) together make a Meson (like $\pi$ or $\kappa$). Assume the quarks are in the ground state — so orbital angular momentum is 0, right?
b) What spin values can a meson have?
Two spin-$\tfrac{1}{2}$ things, so $S_{total} = 1$ or $0$, two options. That gives
$$|1,1\rangle,\;|1,0\rangle,\;|1,-1\rangle,\;|0,0\rangle$$a) What spin values can a baryon have?
$S_{total}=1$ combined with $S=\tfrac{1}{2}$ gives $\to \tfrac{3}{2},\tfrac{1}{2}$ $S_{total}=0$ combined with $S=\tfrac{1}{2}$ gives $\to \tfrac{1}{2}$
So:
$$\left|\tfrac{3}{2},\tfrac{3}{2}\right\rangle,\;\left|\tfrac{3}{2},\tfrac{1}{2}\right\rangle,\;\left|\tfrac{3}{2},-\tfrac{1}{2}\right\rangle,\;\left|\tfrac{3}{2},-\tfrac{3}{2}\right\rangle,\;\left|\tfrac{1}{2},\tfrac{1}{2}\right\rangle,\;\left|\tfrac{1}{2},-\tfrac{1}{2}\right\rangle$$The spin-$\tfrac{3}{2}$ ones are things like $\Delta$, $\Omega^-$. The spin-$\tfrac{1}{2}$ ones are… well, proton, electron, neutron…
Prob 4.36
A spin-1 particle and a spin-2 particle have total spin 3, and the $z$-component is $\hbar$. If you measure the $z$-component of the spin-2 particle’s angular momentum, what values can you get, and with what probabilities?
OK so. The system has an $s=1$ thing and an $s=2$ thing — we need to look at the $2\times 1$ table. And since $s_{total}=3$, $m_s=1$, it’s asking us to go combined → uncombined.
I don’t really think this one needs a diagram, so I’ll just write the equation straight off the Clebsch-Gordan table.
$$|\;S_{total},\;m_{s_{total}}\rangle \;=\; |\;3,\;1\rangle$$$$=\; \sqrt{\tfrac{1}{15}}|\;2,\;2\rangle|\;1,\;{-1}\rangle \;+\; \sqrt{\tfrac{8}{15}}|\;2,\;1\rangle|\;1,\;0\rangle \;+\; \sqrt{\tfrac{6}{15}}|\;2,\;0\rangle|\;1,\;1\rangle$$So:
$$\text{prob. } S_{2_z} = 2\hbar:\; \tfrac{1}{15} \\ \text{prob. } S_{2_z} = \hbar:\; \tfrac{8}{15} \\ \text{prob. } S_{2_z} = 0:\; \tfrac{6}{15}$$Originally written in Korean on my Naver blog (2015-12). Translated to English for gdpark.blog.