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

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)?

• It means the neutron and proton states carry a fundamental representation of $SU(2)$. |proton> is like spin up and |neutron> is spin down. – Isidore Seville Feb 2 '14 at 3:06
• – user28355 Feb 2 '14 at 3:13
• An $SU(2)$ transformation is of the type $\psi\to e^{iT_a}\psi$ (a "gauge" transformation on the Hilbert space of states), and this signifies that the theory is invariant under unitary transformations (and as I'm sure you know, this is needed for a "consistent" quantum theory). – user28355 Feb 2 '14 at 3:28
• @SanathDevalapurkar The isospin transformation asked in this case is not a gauge transformation, i.e. it does not vary from point to point, but is rather just a global (internal) transformation. Weak isospin on the other hand is a gauge transformation – nervxxx Feb 2 '14 at 3:51
• Because the proton and neutron are not identical in the same sense in which the dog and cat and different. However, the proton and neutron are more similar and in many respects (when it comes to the strong nuclear force), they do behave according to identical rules, as opposed to dogs and cats. – Luboš Motl Feb 2 '14 at 6:42

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.

• this answer is fine except, at the level the question is asked, I would add that the usefulness for SU(2) symmetry started with nuclear physics where experimental observations showed that there was little difference between protons and neutrons, beginning with the mass. en.wikipedia.org/wiki/Isospin#Motivation_for_isospin . – anna v Feb 2 '14 at 4:29
• @annav, I expanded my answer. I hope it is more to your liking. – JeffDror Feb 2 '14 at 9:57
• Yes, it was fine and the extension is also good. Just, as an experimentalist, I have noticed that people whose interest starts with theory tend to put theory first, forgetting the long road of observations that pointed to a similarity to spin, and established isospin and finally to an elegant theory like QCD, which can then make powerful predictions. – anna v Feb 2 '14 at 10:16
• Could you expand your explanation on the example with reaction 2? I'm not clear on how the initial and final isospin states are calculated. – Whelp Apr 3 '14 at 13:09
• @JeffDror Hi Jeff. I think, in the second equation for approximate QCD Lagrangian where you wrote $\psi=(\psi_p,\psi_n)^T$ will be $\psi=(\psi_u,\psi_d)^T$? – mithusengupta123 Aug 14 '17 at 10:13

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.

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.

• You and I have a different opinion of "down-to-earth"...or maybe I'm not a physics guys. – Paul Draper Feb 2 '14 at 8:00
• Can another (perhaps better) geometric interpretation be given by using the fact that SU(2) represents $S^2$ as a manifold? How can you visualize the infinitesimal (one parameter?) transformations in this case. I don't like the idea of rotations in $\mathbb{C}^2$. – user7757 Feb 2 '14 at 13:31
• @ramanujan_dirac: You mean $S^3$ as the group manifold of $SU(2)$? That line of thinking should give some geometric picture for $SU(2)$, but if one wants to understand the doublet representation then $\mathbb{C}^2$ is the thing. – Siva Feb 2 '14 at 16:35
• @PaulDraper: Down-to-earth means different things to different people. I meant it as the answer I would give an undergrad who doesn't know representation theory; who, for the first time, has come across the statement that the $SU(2)$ "weak force" relates the proton and the neutron. – Siva Feb 2 '14 at 16:37