# How can particles undergo EM interactions *and* have definite strong isospin?

1. Apr 21, 2011

### depeche1

I am deeply confused about the following and I'd really appreciate it if anyone could help! Consider a charged hadron such as a proton. Amongst the state-independent properties that define a proton are strong isospin Iz=1/2 and charge Q=e. Now, the total Hamiltonian for a proton is

Hs +Hem +Hw,

where these denote the strong, electromagnetic and weak interaction Hamiltonians respectively. And in the rest frame of the proton p, which has mass m, we have

Hs +Hem +Hw|p> = m|p>

where |p> is the wavefunction of the proton. Since Iz=1/2 and charge Q=e are two of the state-independent properties that define the proton, presumably this means that

Hs +Hem +Hw|Iz=1/2, Q=e> = m|Iz=1/2, Q=e>

- otherwise it wouldn't be the eigenvalue equation for a proton wavefunction. But the electromagnetic Hamiltonian Hem does not commute with Iz; so how can the proton be evolving in accordance with the above Hamiltonian *and* have definite isospin?!

Any help really appreciated!

2. Apr 21, 2011

### Bill_K

The charge operator Q = Iz + Y/2, where Y is the hypercharge. For protons and neutrons, Y = 1. The electromagnetic Hamiltonian does not commute with Iz by itself, or Y by itself, but it does commute with the combination Q.

It also commutes with I. Protons and neutrons form an isospin doublet with I = 1/2.

3. Apr 21, 2011

### metroplex021

That's certainly true. But as far as I can see, these observations don't resolve the original problem - namely that of why it is that, when listing the fundamental intrinsic properties of the proton, we include that it is an Iz=1/2 particle (in addition to being a *total* isospin I=1/2 particle) given that when it's evolving in accordance with its full (strong plus electroweak) Hamiltonian, the third component of isospin isn't even defined in it's own rest frame? Why *do* we regard it as an intrinsic property of the proton that it is an Iz=1/2 particle if that property isn't even well-defined along with its total energy? It seems so counter-intuitive (to me!)

4. Apr 21, 2011

### Bill_K

Sorry this is *not* true! Must eat my words. The strong Hamiltonian commutes with isospin, while the electromagnetic part does not.