- Addition of angular momenta

Heisenberg observed that the proton and neutron had such similar mass, and similar properties as regards the strong force, that they might be considered as different states of a single particle dubbed the nucleon.
A new quantum number was needed to denote the 2 possible states of a nucleon as either a proton or a neutron.
In analogy with the two possible z-spin states of a spin ½ particle, this new quantum number was called isospin.
For the proton and neutron, I=½ and the z component of isospin called I_{3} is I_{3}=+½ for the proton and I_{3}=-½ for the neutron.
This is **not** a "spin" in physical space with a corresponding angular momentum.
This is a spin in an abstract space with no associated angular momentum.

Isospin symmetry helps to explain the existence of mirror nuclei, nuclei that differ in the exchange of a neutron and proton.
For instance, I will shortly do an example involving tritons (^{3}H) and helium-3 (^{3}He) nuclei.

In the context of the quark model, isospin symmetry arises from the nearly identical masses of the up and down quarks.
A proton becomes a neutron when one of the up quarks is replace by a down quark.
Therefore, we assign the up quark I_{3}=+½, and the down quark I_{3}=-½.

We can extend isospin symmetry to other particles, for instance, the pions.
Being mesons, pions are a quark and an anti-quark pair constructed using only up and down quarks.
By analogy with the combination of two spin-½ particles, we can construct 4 states from the combination of two isospin-½ particles.
(Careful, there is a phase convention that enters s.t. the I_{3}=0 states come out with the opposite signs, see Sec. 4.6.)

|½,½> = |I=1, I_{3}=1> : u dbar = π^{+}

(|½,-½>-|-½,½>)/sqrt{2} = |I=1, I_{3}=0> : (u ubar - d dbar)/sqrt{2} = π^{0}

|-½,-½> = |I=1, I_{3}=-1> : d ubar = π^{-}

(|½,-½>+|-½,½>)/sqrt{2} = |I=0, I_{3}=0> : (u ubar + d dbar)/sqrt{2} = η

The mass of π(|½,-½>-|-½,½>)/sqrt{2} = |I=1, I

|-½,-½> = |I=1, I

(|½,-½>+|-½,½>)/sqrt{2} = |I=0, I

Now we apply isospin to some situations. The first is the deuteron, a nucleus containing a proton and a neutron.

The deuteron consists of two nucleons, so let's consider the possible isospin combinations of two I=½ nucleons.
Again we have four possible states, denoting them as being either p (I_{3}=+½) or n (I_{3}=-½) and labeling the particles as (1) or (2):

χ(1,1) = p(1)p(2)

χ(1,0) = [p(1)n(2) + n(1)p(2)]/sqrt{2}

χ(1,-1) = n(1)n(2)

χ(0,0) = [p(1)n(2) - n(1)p(2)]/sqrt{2}

The first state are members of an I=1 triplet, and the last state is an I=0 singlet.
Bound nuclei consisting of two protons or two neutrons don't exist, so it is natural to associate the singlet state with the deuteron.
χ(1,0) = [p(1)n(2) + n(1)p(2)]/sqrt{2}

χ(1,-1) = n(1)n(2)

χ(0,0) = [p(1)n(2) - n(1)p(2)]/sqrt{2}

We can further justify this with an analysis of the symmetry of the wave function.
The wave function for a non-relativistic two nucleon state (a state of two *identical* fermions) can be decomposed into a spatial part, a spin part, and an isospin part:

ψ = φ(**r**) α(**S**) χ(**I**)

In a relativistic state, the orbital and spin angular momenta cannot be separated -- one must use the full Dirac equation.
Recall that the deuteron has spin (total angular momentum) J=1, and it is known that L=0, therefore S=1.
The parity of the spatial part is (-1)Now let's apply isospin to some strong interactions.
It turns out that we can break a reaction down into separate isospin components (channels), and we find that, generally speaking, *reactions with the same I but different I _{3} have the same amplitude (matrix element)* at a given energy.
Consider the following reactions

(1) pp --> dπ^{+}

(2) pn --> dπ^{0}

The reactions differ in that the first has I(2) pn --> dπ

σ(1)/σ(2) = 1/2

which is observed.
(This is not a high precision prediction since electromagnetic effects which break isospin symmetry are ignored.
It should be correct at the level of a few percent.)
Next consider the strong scattering of a pion and a nucleon to a pion and a nucleon. The initial and final states are by the rules for combining angular momenta. (The above notation means if I combine states of isospin 1 and 1/2, then I'll get states of total isospin 3/2 or 1/2. The exact amounts of each are determined by the appropriate Clebsch-Gordon coefficients.)

There are 3×2 different scattering processes (listed below). The hypothesis of isospin symmetry leads to the conclusion that these 6 processes can be described using only two amplitudes, one for the I=3/2 channel and one for the I=1/2 channel.

Two of the processes are I_{3} = ±3/2 and therefore only occur through the I=3/2 channel:

(a) π^{+}p --> π^{+}p , and

(b) π^{-}n --> π^{-}n

For identical energies, these process will have identical cross sections (to the precision allowed by electromagnetic breaking of isospin).
(b) π

The remaining processes can proceed through both channels:

(c) π^{-}p --> π^{-}p

(d) π^{-}p --> π^{0}n

(e) π^{+}n --> π^{+}n

(f) π^{+}n --> π^{0}p

We use a table of Clebsch-Gordon coefficients to find the relative amounts of I=1/2 and I=3/2 in the initial and final states of these reactions.
(This will be done interactively in lecture.)
(d) π

(e) π

(f) π

Recall that the cross section depends on what I called phase space factors and the square of the matrix element:

σ(i-->f) = |M_{if}|^{2} × (phase space)

where
M_{if} = <ψ_{f} | H | ψ_{i}>

Our hypothesis is that these reactions are dominated by strong interactions, and that the strong interaction conserves isospin.
Therefore, we can separate the total Hamiltonian into parts that only operates between states of equal isospin, H = HWriting the wave functions as separate I=1/2 and I=3/2 parts,

ψ_{i} = ψ_{i}(1/2) + ψ_{i}(3/2), and

ψ_{f} = ψ_{f}(1/2) + ψ_{f}(3/2)

and inserting these pieces into the formula for Mψ

M_{if} = <ψ_{f}(1/2) + ψ_{f}(3/2) | H_{1} + H_{3} | ψ_{i}(1/2) + ψ_{i}(3/2)>

= <ψ_{f}(1/2) | H_{1} | ψ_{i}(1/2)> + <ψ_{f}(3/2) | H_{3} | ψ_{i}(3/2)>

= k_{1} M_{1} + k_{3} M_{3}

Where k= <ψ

= k

Now we can work out the constants, k_{1} and k_{2}, for reactions (a) through (f).
Again, this will be done interactively in class.

Then consider the limiting cases where one or the other amplitude dominates, M_{1}>>M_{3}, or M_{3}>>M_{1}.
The results for reactions (a) through (c) are:

M_{3}>>M_{1} σ_{a}:σ_{b}:σ_{c} = 9 : 1 : 2

M_{1}>>M_{3} σ_{a}:σ_{b}:σ_{c} = 0 : 2 : 1

The ratio of the cross sections for πM

By the way, the resonance is called the Δ(1232) and is assigned the quantum numbers I(J^{P}) = 3/2(3/2^{+}).

Copyright © Robert Harr 2003