What does a $\rm SU(2)$ isospin doublet really mean?

Question

What do we really mean when we say that the neutron and proton wavefunctions together form an $\rm SU(2)$ isospin doublet? What is the significance of this? What does this transformation really doing to the wavefunctions (or fields)?

Answer 1

Two particles forming an $SU(2)$ doublet means that they transform into each other under an $SU(2) $ transformation. For example a proton and neutron (which form such a doublet) transform as, \begin{equation} \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) \end{equation} where $ \sigma _a $ are the Pauli matrices. It turns out the real world obeys certain symmetry properties. For example, the equations described the strong interactions of protons and neutrons are approximately invariant under unitary transformations with determinant 1 (the transformation shown above) between the proton and neutron. This didn't have to be case, but turns out that it is. Since the strong interaction is invariant under such transformations, each interaction term in the strong interaction Lagrangian is highly restricted. For one thing, this is useful since it allows one to make simple predictions about proton and neutron systems.

In order to get a better understanding of this transformation and why the symmetry holds. Consider the QCD Lagrangian for the up and down quarks (which, as for the proton and neutron, also make up an isospin doublet): \begin{equation} {\cal L} _{ QCD} = \bar{\psi} _{u,i} i \left( \left( \gamma ^\mu D _\mu \right) _{ ij} - m _u \delta _{ ij} \right) \psi _{u,j} + \bar{\psi} _{ d ,i} \left( \left( \gamma ^\mu D _\mu \right) _{ ij} - m _d \delta _{ ij} \right) \psi _{d ,j}% \\ % & \bar{\psi} _{i} i \left( \left( \gamma ^\mu D _\mu \right) _{ ij} - M \delta _{ ij} \right) \psi _{j} \end{equation} where $ D ^\mu $ is the covarient derivative and the sum over $ i,j $ is a sum over the color. Notice that if $ m _{ u} \approx m _d \equiv m $ we can write this Lagrangian in a more convenient form, \begin{equation} {\cal L} _{ QCD} = \bar{\psi} _{i} i \left( \left( \gamma ^\mu D _\mu \right) _{ ij} - m \delta _{ ij} \right) \psi _{j} \end{equation} where $ \psi \equiv \left( \psi _u \, \psi _d \right) ^T $. This Lagrangian is now invariant over transformations between up and down quarks ("isospin") since the color generators commute with the isospin generators. Since proton and neutron and only differ in their ratio of up to down quarks (the more precise statement is that their quantum numbers correspond to those of $uud$ and $udd$ respectively), we would expect these particles to behave very similarly when QED can be neglected (which is often the case because QED is much weaker then QCD at low energies).

As an explicit example of the use of the symmetry consider the reactions: \begin{align} & 1) \quad p p \rightarrow d \pi ^+ \\ & 2) \quad p n \rightarrow d \pi ^0 \end{align} where $ d $ is deuterium, an isospin singlet, and the pions form an isospin triplet. For the first interaction, the initial isospin state is $ \left| 1/2, 1/2 \right\rangle \otimes \left| 1/2, 1/2 \right\rangle = \left| 1, 1 \right\rangle $. The products have isospin $ \left| 0,0 \right\rangle \otimes \left| 1,1 \right\rangle = \left| 1,1 \right\rangle $. The second interaction has an initial isospin state, $ \frac{1}{\sqrt{2}} \left( \left| 0,0 \right\rangle + \left| 1,0 \right\rangle \right) $, and final isospin, $ \left| 0,0 \right\rangle $.

Since both cases have some overlap between the isospin wavefunctions, both can proceed. However, the second process has a suppression factor of $ 1/ \sqrt{2} $ when contracting the isospin wavefunctions. To get the probabilities this will need to be squared. Thus one can conclude, \begin{equation} \frac{ \mbox{Rate of 1} }{ \mbox{Rate of 2}} \approx 2 \end{equation}

Notice that even without knowing anything about specifics of the system we were able to make a very powerful prediction. All we needed to know is that the process occurs through QCD.

Answer 2

I don't know what background you bring to the question. So at the risk of sounding patronizing, let me give a down-to-earth answer. I wonder if this helps.

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Think of rotations on the (Real) 2-dimensional plane $\mathbb{R}^2$. You can rotate the X-axis into the Y-axis, and the Y-axis into the negative X-axis. This group of 2d rotations is called $SO(2)$. Note that here, each axis consists of the set of real numbers. If, instead, each axis corresponded to the set of complex numbers, then we would have the 2d complex plane $\mathbb{C}^2$. Rotations in this plane would correspond to the group $SU(2)$ and you can think of protons and neutrons (rather, their wavefunctions) as the basis elements forming the two axes in this $\mathbb{C}^2$ space. The "doublet" refers to having two axes.

This $\mathbb{C}^2$ does not refer to actual physical dimensions, but just some property of protons and neutrons.

I'm at a loss for explaining the "significance" of this, apart from the fact that this is how nature behaves. One consequence is the fact that the proton and neutron have approximately equal masses, because apart from being different axes corresponding to this property, they are supposed to be pretty similar otherwise.

Usually when you write the wavefunction (of a particle), you concentrate on its spatial profile (in introductory quantum mechanics) and neglect other properties which characterize it. Similarly, every particle also carries a wavefunction corresponding to every other characteristic and the complete description involves writing down all the wavefunctions (you could "multiply" the wavefunctions corresponding to all the different properties, for what it's worth). Just like you might have seen operators acting on the spatial part of a wavefunction, you will also have operators acting on the wavefunction corresponding to every property.