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In Halzen and Martin's Quarks and Leptons, on page 42, the $SU(2)$ isospin transformation represented by $e^{i\boldsymbol{\theta}\cdot\boldsymbol{\tau}/2}$ is said to act on the column represented by $$|\psi\rangle=\begin{pmatrix}\mathrm{p} \\\mathrm{n}\end{pmatrix}\tag{1}$$ with $$\mathrm{p}=\begin{pmatrix}1\\0\end{pmatrix}\hspace{0.3cm} \text{and}\hspace{0.3cm}\mathrm{n}=\begin{pmatrix}0 \\1\end{pmatrix}\tag{2}$$

I think this notation i.e., Eq, (1) used in conjunction with Eq.(2), is confusing. In the basis $|\mathrm{p}\rangle=\begin{pmatrix}1\\0\end{pmatrix}$, $|\mathrm{n}\rangle=\begin{pmatrix}0 \\1\end{pmatrix}$, the state $|\psi\rangle$ should be represented by a 2-component column (1) with its entries $\mathrm{p}$ and $\mathrm{n}$ being complex numbers. So are they using the notation that $$|\psi\rangle=\mathrm{p}|\mathrm{p}\rangle+\mathrm{n}|\mathrm{n}\rangle?\tag{3}$$

In this regard, Griffith's Introduction to Elementary particles, I think, uses a clearer notation. They represent general nucleon state $|N\rangle$ as $$|N\rangle=\alpha|p\rangle+\beta|n\rangle\tag{4}$$ with $$|p\rangle=\begin{pmatrix}1\\0\end{pmatrix}\hspace{0.3cm} \text{and}\hspace{0.3cm}|n\rangle=\begin{pmatrix}0 \\1\end{pmatrix}\tag{5}$$ so that the state $|N\rangle$ is indeed represented by a 2-component column vector with two complex numbers $\alpha$ and $\beta$.

My question Do Halzen and Martin use the notation I wrote in Eq. (3) with $\mathrm{p}$, $\mathrm{n}$ being complex numbers? I ask this because it is not clear from their notation.

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This is definitely bad (unclear) notation. There are a couple ways I can think of to interpret it: first, and most sensibly, they are doing exactly what you put forward in your question, namely reusing the same variables for both state labels and coefficients without distinguishing which are which.

The other possibility I can think of is that they are using the letters in $\begin{pmatrix}\mathrm{p}\\ \mathrm{n}\end{pmatrix}$ to represent particles. In this sense, you shouldn't think of $\begin{pmatrix}\mathrm{p}\\ \mathrm{n}\end{pmatrix}$ as a precisely defined mathematical expression; it's simply abusing the notation to convey how the isospin transformation mixes the particles (or really, their corresponding fields). Note that in this case, there are two almost entirely unrelated meanings for $\mathrm{p}$ and $\mathrm{n}$: $\mathrm{p}$ can mean either a proton, i.e. when it appears in the doublet, or the isospin state of a proton, i.e. when it appears in the definition $\mathrm{p} = \begin{pmatrix}1 \\ 0\end{pmatrix}$; and similarly for $\mathrm{n}$.

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  • $\begingroup$ Do you mean they are using $p,n$ as quantum fields (rather than numbers) arranged in a column $\psi$ which is multiplet like a quark doublet? But in that case, I think, they shouldn't have used $p=\begin{pmatrix}1\\0\end{pmatrix}$ and $n=\begin{pmatrix}0\\1\end{pmatrix}$. @David Z $\endgroup$
    – SRS
    Sep 15, 2017 at 7:41
  • $\begingroup$ Not quantum fields, but rather particles. I've edited to clarify this a bit. $\endgroup$
    – David Z
    Sep 15, 2017 at 7:44
  • $\begingroup$ "...particles (or really, their corresponding fields)" You are talking about spinor fields i.e., the proton field $p(x)$ and the neutron field $n(x)$. Right? $\endgroup$
    – SRS
    Sep 15, 2017 at 7:47
  • $\begingroup$ I wrote "(or really, their corresponding fields)" to acknowledge that particles cannot truly mix, only fields can mix. My point in the second paragraph is that the p and n in the doublet are not mathematical variables. I'm trying to avoid saying that they are fields because a field is a variable that can be assigned a number or spinor or some such thing; that's not what I'm getting at here. $\endgroup$
    – David Z
    Sep 15, 2017 at 7:52

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