Strong nuclear force

Isospin

Isospin symmetry: up and down quarks are indistinguishable to the strong nuclear force

Isospin is a non-physical property of hadrons that is mathematically analogues to spin, ie

✏️ Notation

Spin	Isospin
$\bold {\hat L}^2$	$\bold {\hat I}^2$
$L_x$	$\hat I_1$
$L_y$	$\hat I_2$
$L_z$	$\hat I_3$
$m$	$I_3$
$l$	$I$

📜 Properties

$$ \begin{aligned} \bold {\hat I}^2&=\hat I^2_1+\hat I^2_2+\hat I^2_3 \\ [\hat I_i,\hat I_j]&=i\hbar \epsilon_{ijk} \hat I_k \quad \text{distinct} \; i,j,k\in \{1,2,3 \} \\ [\bold {\hat I}^2, \hat I_i]&=0 \quad \forall i \in\{1,2,3\}\\ \bold {\hat I}^2\ket{I, I_3}&=I(I+1)\hbar^2 \ket{I, I_3} \\ I_3&\in[-I,-I+1,\ldots,I] \end{aligned} $$

$I_3$ has $(2I+1)$ states for each multiplet (same $I$)

🗒️ Note: similar to $L_z$ by convention we use $I_3$

🗒️ Note: notation is a bit ambiguous because the usual lower case convention doesn’t work for $i$

Quantum numbers

📖 Definition:

Up quark: $I_3=+\frac 12$

Down quark $I_3=-\frac 12$

all other quarks and leptons: $I=0$

⚡ Isospin is conserved in strong and electromagnetic interacts

From the definitions we write

$$ \begin{aligned} \text{Proton}: \;I_3&=+\frac{1}{2} \\ \text{Neutron}: \;I_3&=-\frac{1}{2} \end{aligned} $$

They form a doublet

💃 Example: pions $\pi^+$, $\pi^-$ and $\pi^0$

They have

$$ \left . \begin{aligned} \pi^+&:u\overline d\\ \pi^-&: \overline u d \\ \pi^0&:u \overline u +d\overline d

\end{aligned} \; \right \} \quad \Rightarrow \quad \left \{ \begin{aligned} \pi^+&:I_3=+1\\ \pi^-&:I_3=-1 \\ \pi^0&:I_3=0

\end{aligned} \right . $$

💎 Conclusion: they form an isospin triplet with $I=1$ with multiplicity of $(2I+1)=3$

Hadron Supermultiplets

📖 Definition: Hadrons are divided into baryons and mesons

Baryons: quark content $qqq$

Anti Baryons: quark content $\overline q \overline q \overline q$

mesons: $q\overline q$

🗒️ Note: $q$ represent arbitrary quark flavour that are not necessarily the same

💼 Case: Consider $u,d,s$ the lightest quarks in there ground stay $\vec L=0$ thus $\vec J=\vec L +\vec s=\vec s$

Looking at a baryon with arbitrary quark flavour $q_iq_jq_k$ where $i,j,k$ are indices indicating flavour

Each quark has spin $s=\frac 12$ and m$_s=\pm \frac 12$ or $\uparrow \downarrow$ respectively
These can be in a mixed configuration like $\uparrow \uparrow \downarrow$ or any other permutations giving $|\vec s|=\frac 12 = |\vec J|$
Or in an aligned configuration like $\uparrow\uparrow\uparrow$ which gives $|\vec s|=\frac 32=|\vec J |$

Without consider Pauli's principle we could think there are 27 combinations of $u,d,s$ quarks