I'm trying to understand the idea of $I_3$ component of the isospin.

If I get it right (check isospin or Thomson, Modern Particle Physics, chap. 9), the Isospin (or isobaric spin) is a kind of internal symmetry, i.e., one found in an auxiliary space. One example of isospin comes from Heisenberg's idea that if we turned off the electric charge, the proton and the neutron could be consired two states of the same particle, and though we've never seen any linear combination of the two it is worth imagining a space where $$ I=\alpha p+\beta n ,$$ where $$ p=\left(\begin{array}{c} 1\\0 \end{array}\right) ,\qquad n=\left(\begin{array}{c} 0\\1 \end{array}\right)$$ would exist. My questions are:

  1. What is the third component of the isospin, $I_3$ (maybe I should say where is it, since I see only two, p and n)?
  2. Why is it $I_3=\frac{1}{2}$ in this case?
  • 2
    $\begingroup$ $|p\rangle \equiv |I=1/2,I_{3}=1/2\rangle$ and $|n\rangle \equiv |I=1/2,I_{3}=-1/2\rangle$. They belongs to the fundamental representation of $SU(2)$ (strong) isopsin. $\endgroup$
    – Mass
    Sep 29, 2016 at 16:30
  • 1
    $\begingroup$ I think we have a confusion here about understanding a two-state quantum system : $\: |p\rangle, |n\rangle\: $ are not $\:I_{1},I_{2}\:$ but the two eigenstates of $\:I_{3}$. Following Gell-Man $\: |u\rangle, |d\rangle\: $ are the two eigenstates of $\:I_{3}$. $\endgroup$
    – Frobenius
    Sep 29, 2016 at 19:27

2 Answers 2


The proton and neutron are the two projections of the nucleon on the third axis of the isospin operator, like "spin up" and "spin down" are the two projections of angular momentum on your favorite spatial axis.

The other two axes for isospin have the eigenstates $1\choose{\pm1}$ and $1\choose{\pm i}$, which you can see by writing down the Pauli matrices. However the third axis corresponds to electric charge, and conservation of charge is an exact symmetry in nature. Because $I_{1,2}$ and $I_3$ don't commute, and any physical particle has a definite charge and is in an eigenstate of $I_3$, there aren't any observable states with definite $I_1$ or $I_2$.


Isospin operator, $\hat{\textbf{I}} \equiv \hat{I}_x \, \hat{x} + \hat{I}_y \, \hat{y} + \hat{I}_z \, \hat{z}$; where $\hat{I}_x$, $\hat{I}_y$ and $\hat{I}_z$ are operators and, they are the $x$, $y$ and $z$-components of $\hat{\textbf{I}}$, respectively.

Every nucleon has a specific value of isospin number $I$.

For a given value of $I$, a measurement of $\hat{I}_z$ yields a value of the form $$i_s$$ where $i_s$ is an integer or, a half-integer in the range $[-I, I]$: $$i_s = -I, \, -I+1, \cdots \cdots, I-1, \, I \, ;$$ $(2I+1)$ possibilities. That is, $i_s$ is a value of $I_z$ at a particular time.

$i_s$ is often called the $z$-component of $\textbf{I}$ and, therefore denoted by $I_z$ or $I_3$.

For this reason, only the eigenstates of $\hat{I}_z$ are denoted by $|I \quad i_s \rangle$ corresponding to the eigenvalues $i_s$: $$ \hat{I}_z |I \quad i_s \rangle = i_s |I \quad i_s \rangle .$$ Moreover, $\hat{I}^2 \, |I \quad i_s \rangle = I(I+1) |I \quad i_s \rangle$.

Now let us consider a nucleon with $I=\frac{1}{2}$. Then $i_s = I_z = I_3$ has values $\frac{1}{2}, -\frac{1}{2}$. Therefore $\hat{I}_z$ has two eigenstates $|\frac{1}{2} \quad \frac{1}{2} \rangle$ and $|\frac{1}{2} \quad -\frac{1}{2} \rangle$ with eigenvalues $\frac{1}{2}$ and, $-\frac{1}{2}$, respectively.

At time $t$ , upon measurement of $I_z$, the nucleon is either in $|\frac{1}{2} \quad \frac{1}{2} \rangle$ or, $|\frac{1}{2} \quad -\frac{1}{2} \rangle$.

If the nucleon is in the $|\frac{1}{2} \quad \frac{1}{2} \rangle$ state, then it is, by definition, called a proton. And $|\frac{1}{2} \quad \frac{1}{2} \rangle$ is called a proton state, which is $(1 \quad 0)^T$ in $\left\lbrace |\frac{1}{2} \quad \frac{1}{2} \rangle, |\frac{1}{2} \quad -\frac{1}{2} \rangle \right\rbrace$ basis.

If the nucleon is in the $|\frac{1}{2} \quad -\frac{1}{2} \rangle$ state, then it is, by definition, called a neutron. And $|\frac{1}{2} \quad -\frac{1}{2} \rangle$ is called a neutron state, which is $(0 \quad 1)^T$ in $\left\lbrace |\frac{1}{2} \quad \frac{1}{2} \rangle, |\frac{1}{2} \quad -\frac{1}{2} \rangle \right\rbrace$ basis.

Please note that the particles associated with the two different states of the nucleon has same spin $s$ and mass $m$ (approximately).


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