# Transition matrix element and Isospin

• I
Hi, guys.

A type of problem that often appears is to find the relation between cross sections of some processes. An example would be:

$$\pi _{- }+ p \rightarrow K_0 + \Sigma_0$$
$$\pi _{- }+ p \rightarrow K_+ + \Sigma_-$$
$$\pi _{+}+ p \rightarrow K_+ + \Sigma_+$$

To do this, I argue that

$$\sigma \, \alpha \, \Gamma \, \alpha \, M$$

with ##M## the transition matrix element.

In this case, the interactions are strong. I write the initial and final states for each process in the ##{I,i}## basis and I then write ##M=\langle i | H_s | f \rangle##.

I end up getting, for example for the third process: ##M=\langle 3/2,3/2 | H_s | 3/2, 3/2 \rangle##.
For the first process, I get one term of the form ##\langle 3/2,-1/2 | H_s | 3/2,-1/2 \rangle##.

My question: are there two expressions I just wrote equal? i.e. does ##M=\langle I,i | H_s | I,i' \rangle## depend only on the value of ##I##? Why?

samalkhaiat
Yes. According to the Wigner-Eckart theorem, the matrix elements of a tensor operator $\mathcal{H}^{M}_{I}$, having isospin $I$ and third component value $M$, factorize as
$$\langle I_{2}, m_{2}| \mathcal{H}^{M}_{I}| I_{1},m_{1} \rangle = \langle I_{2},m_{2}| I,M ; I_{1},m_{1}\rangle \langle I_{2}|| \mathcal{H}_{I}|| I_{1}\rangle ,$$ where the first factor on the RHS is the Clebsch-Gordon coefficient and the second factor is the reduced matrix element, which is independent of $m_{1},m_{2}$ and $M$. The processes, you listed, proceed through either $I = 3/2$ or $I = 1/2$ channels. So, your strong Hamiltonian operator can be decomposed into two pieces with definite isospins: $\mathcal{H}_{s} = \mathcal{H}_{3/2} + \mathcal{H}_{1/2}$, and the problem is solved by reading the corresponding C-G cofficients.

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• Xico Sim
Yes. According to the Wigner-Eckart theorem, the matrix elements of a tensor operator $\mathcal{H}^{M}_{I}$, having isospin $I$ and third component value $M$, factorize as
$$\langle I_{2}, m_{2}| \mathcal{H}^{M}_{I}| I_{1},m_{1} \rangle = \langle I_{2},m_{2}| I,M ; I_{1},m_{1}\rangle \langle I_{2}|| \mathcal{H}_{I}|| I_{1}\rangle ,$$ where the first factor on the RHS is the Clebsch-Gordon coefficient and the second factor is the reduced matrix element, which is independent of $m_{1},m_{2}$ and $M$. The processes, you listed, proceed through either $I = 3/2$ or $I = 1/2$ channels. So, your strong Hamiltonian operator can be decomposed into two pieces with definite isospins: $\mathcal{H}_{s} = \mathcal{H}_{3/2} + \mathcal{H}_{1/2}$, and the problem is solved by reading the corresponding C-G cofficients.

What if it's the weak interaction? Should't the matrix elements behave the same way?

samalkhaiat
Okay, here is an exercise for you. Use the Wigner-Eckart theorem to calculate the (non-leptonic) weak decay amplitudes for $$K^{0} \to \pi^{+}\pi^{-} , \ \ \ \ K^{0} \to \pi^{0}\pi^{0} .$$
Experimentally, it has been observed that $$\frac{T \left( K^{0} \to \pi^{+}\pi^{-}\right)}{T \left( K^{0} \to \pi^{0}\pi^{0} \right)} \approx \sqrt{2} .$$ Translate this into relation between amplitudes.
Hint: Ignoring Dirac’s gamma matrices, the low-energy (non-leptonic) weak Hamiltonian can be written as $$\mathcal{H}^{\Delta S = 1}_{(w)} = \frac{G}{\sqrt{2}}\left( [\bar{u}d][\bar{s}u] + h.c. \right) .$$ The first factor has $I = 1$, and the second factor has $I = 1/2$. So, you have the following isospin decomposition $$\mathcal{H}^{\Delta S = 1}_{(w)} = \mathcal{H}_{1 \otimes \frac{1}{2}} = \mathcal{H}_{\frac{3}{2} \oplus \frac{1}{2}} = \mathcal{H}_{3/2} + \mathcal{H}_{1/2} .$$
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