# What is the effect of isospin on the proton or neutron alone, i.e. not in a doublet?

So I start with a proton $$p$$.

I extend my "physical" space by means of the internal degree of freedom of isospin, so that I know write $$p$$ in a higher dimensional space:

$$p = \left( \begin{array}{c} 1 \\ 0 \end{array} \right).$$

I notice that in this representation I can also have the neutron $$n$$:

$$n = \left( \begin{array}{c} 0 \\ 1 \end{array} \right),$$

which is great because now in this isospin extended space $$p$$ and $$n$$ form an (approximate) $$SU(2)$$ doublet:

$$$$\left( \begin{array}{c} p \\ n \end{array} \right) \xrightarrow{SU(2)} \exp \left( - \frac{ i }{ 2} \theta_a \sigma_a \right) \left( \begin{array}{c} p \\ n \end{array} \right).$$$$

Question

Back in my "physical space" where $$p$$ is just one-dimensional, i.e. it is not embedded in the higher dimensional space derived from its internal state, what is the effect/consequence of the isospin symmetry?
Does it appear as a phase factor?

• The analogy between isospin and ordinary angular momentum is incomplete, because rotations which change the third component of isospin are forbidden by conservation of electric charge, while operations which change the total isospin are permitted. I'm not sure whether that observation brings you any closer to an answer, though. – rob Feb 2 '19 at 15:58
• I'm not really clear on what the question is. If the question is "what is the effect," then I think the answer is probably "there is none." – user4552 Feb 2 '19 at 16:31
• Can you use the isospin quantum number without using the doublet representation? – SuperCiocia Feb 2 '19 at 17:02
• A "nisospin" singlet has "nisospin" 0, so it is pointless involving it in "nisospin" considerations. Is that what you are asking? – Cosmas Zachos Feb 2 '19 at 21:12
• What I mean is: does the 1D representation of the proton carry any information of the fact that, in its 2D extended space (1,0) there is a SU(2) symmetry? – SuperCiocia Feb 2 '19 at 22:41