Transition matrix element and Isospin

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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?
 

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  • #2
samalkhaiat
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Yes. According to the Wigner-Eckart theorem, the matrix elements of a tensor operator [itex]\mathcal{H}^{M}_{I}[/itex], having isospin [itex]I[/itex] and third component value [itex]M[/itex], factorize as
[tex]\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 ,[/tex] 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 [itex]m_{1},m_{2}[/itex] and [itex]M[/itex]. The processes, you listed, proceed through either [itex]I = 3/2[/itex] or [itex]I = 1/2[/itex] channels. So, your strong Hamiltonian operator can be decomposed into two pieces with definite isospins: [itex]\mathcal{H}_{s} = \mathcal{H}_{3/2} + \mathcal{H}_{1/2}[/itex], and the problem is solved by reading the corresponding C-G cofficients.
 
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Yes. According to the Wigner-Eckart theorem, the matrix elements of a tensor operator [itex]\mathcal{H}^{M}_{I}[/itex], having isospin [itex]I[/itex] and third component value [itex]M[/itex], factorize as
[tex]\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 ,[/tex] 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 [itex]m_{1},m_{2}[/itex] and [itex]M[/itex]. The processes, you listed, proceed through either [itex]I = 3/2[/itex] or [itex]I = 1/2[/itex] channels. So, your strong Hamiltonian operator can be decomposed into two pieces with definite isospins: [itex]\mathcal{H}_{s} = \mathcal{H}_{3/2} + \mathcal{H}_{1/2}[/itex], 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?
 
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samalkhaiat
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What if it's the weak interaction? Should't the matrix elements behave the same way?
The Wigner-Eckart theorem is a general statement about tensor operators. These tensor operators can (but they don’t have to) describe interactions. And, when the tensor operators do describe interactions, the type of interaction is not relevant. So, when the assumptions of the theorem are satisfied, you can apply it to the strong, weak or electromagnetic interactions.
Okay, here is an exercise for you. Use the Wigner-Eckart theorem to calculate the (non-leptonic) weak decay amplitudes for [tex]K^{0} \to \pi^{+}\pi^{-} , \ \ \ \ K^{0} \to \pi^{0}\pi^{0} .[/tex]
Experimentally, it has been observed that [tex]\frac{T \left( K^{0} \to \pi^{+}\pi^{-}\right)}{T \left( K^{0} \to \pi^{0}\pi^{0} \right)} \approx \sqrt{2} .[/tex] Translate this into relation between amplitudes.
Hint: Ignoring Dirac’s gamma matrices, the low-energy (non-leptonic) weak Hamiltonian can be written as [tex]\mathcal{H}^{\Delta S = 1}_{(w)} = \frac{G}{\sqrt{2}}\left( [\bar{u}d][\bar{s}u] + h.c. \right) .[/tex] The first factor has [itex]I = 1[/itex], and the second factor has [itex]I = 1/2[/itex]. So, you have the following isospin decomposition [tex]\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} .[/tex]
 
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