Return to Answer

Isospin. It looks like a homework problem, with a summary answer in Okun's book, p 63. You need quark diagrams like a hole in the head.

This strangeness-changing decay violates isospin by 1/2, so assuming the $\Delta T=1/2$ piece of the hamiltonian dominates, and, since Λ is an isosinglet (so irrelevant to the isospin amps), you just consider the addition of a spurion s of isospin $|1/2,-1/2\rangle$ added to the cascade isodoublet $|1/2,\pm 1/2\rangle$, to yield pion states $|1,0\rangle$ and $|1,-1\rangle$ respectively. Isospin is preserved in the subsequent hadronization sequence.

The ratio of the respective decay amplitudes, then, is the simplest Clebsch ever, wich is exactly why your PDG booklet you no doubt are staring at has a Clebsch table: the most useful page of it! $$ \frac{\langle \pi^- \Lambda | \Xi^- \rangle} {\langle \pi^0 \Lambda | \Xi^0 \rangle} = \frac { \langle J_\pi=1 , M_\pi=-1 | j_s=1/2 , m_s=-1/2; j_\Xi=1/2 , m_\Xi=-1/2 \rangle} { \langle J_\pi=1 , M_\pi=0 | j_s=1/2 , m_s=-1/2; j_\Xi=1/2 , m_\Xi=1/2 \rangle} =\sqrt{2}. $$

Thus, squaring the amplitude...

Response to comment:

Here is an "explanation" of the basic ΔI=1/2 rule of weak interactions: Assuming the short distance weak hamiltonian is dominated by its ΔI=1/2 piece, the QCD-driven hadronization preserves isospin, so the weak blob is regarded as an "act of god", simulated/summarized by a I=1/2 spurion. Okun gives you the practical application rules for these spurions, but, frankly, you should have learned all about them in your HEP course, not here...

When it comes to "math is great", of course the $\sqrt 2$ is the normalization of the neutral pion wave function, because it is basically the Clebsch you get by adding two spin 1/2s, and descending by one step. I assume you appreciate that, isoladder-wise, the pion is a composite of a spurion and a cascade! I would call this mathematical analogy, rather than the "underlying physics". The "physics" of this is just the cornerstone ΔI=1/2 rule.

Isospin. It looks like a homework problem, with a summary answer in Okun's book, p 63. You need quark diagrams like a hole in the head.

This strangeness-changing decay violates isospin by 1/2, so assuming the $\Delta I=1/2$ piece of the hamiltonian dominates, and, since Λ is an isosinglet (so irrelevant to the isospin amps), you just consider the addition of a spurion s of isospin $|1/2,-1/2\rangle$ added to the cascade isodoublet $|1/2,\pm 1/2\rangle$, to yield pion states $|1,0\rangle$ and $|1,-1\rangle$ respectively. Isospin is preserved in the subsequent hadronization sequence.

The ratio of the respective decay amplitudes, then, is the simplest Clebsch ever, wich is exactly why your PDG booklet you no doubt are staring at has a Clebsch table: the most useful page of it! $$ \frac{\langle \pi^- \Lambda | \Xi^- \rangle} {\langle \pi^0 \Lambda | \Xi^0 \rangle} = \frac { \langle J_\pi=1 , M_\pi=-1 | j_s=1/2 , m_s=-1/2; j_\Xi=1/2 , m_\Xi=-1/2 \rangle} { \langle J_\pi=1 , M_\pi=0 | j_s=1/2 , m_s=-1/2; j_\Xi=1/2 , m_\Xi=1/2 \rangle} =\sqrt{2}. $$

Thus, squaring the amplitude...

Response to comment:

Here is an "explanation" of the basic ΔI=1/2 rule of weak interactions: Assuming the short distance weak hamiltonian is dominated by its ΔI=1/2 piece, the QCD-driven hadronization preserves isospin, so the weak blob is regarded as an "act of god", simulated/summarized by a I=1/2 spurion. Okun gives you the practical application rules for these spurions, but, frankly, you should have learned all about them in your HEP course, not here...

When it comes to "math is great", of course the $\sqrt 2$ is the normalization of the neutral pion wave function, because it is basically the Clebsch you get by adding two spin 1/2s, and descending by one step. I assume you appreciate that, isoladder-wise, the pion is a composite of a spurion and a cascade! I would call this mathematical analogy, rather than the "underlying physics". The "physics" of this is just the cornerstone ΔI=1/2 rule.

Isospin. It looks like a homework problem, with a summary answer in Okun's book, p 63. You need quark diagrams like a hole in the head.

This strangeness-changing decay violates isospin by 1/2, so assuming the $\Delta T=1/2$ piece of the hamiltonian dominates, and, since Λ is an isosinglet (so irrelevant to the isospin amps), you just consider the addition of a spurion s of isospin $|1/2,-1/2\rangle$ added to the cascade isodoublet $|1/2,\pm 1/2\rangle$, to yield pion states $|1,0\rangle$ and $|1,-1\rangle$ respectively.

The ratio of the respective decay amplitudes, then, is the simplest Clebsch ever, wich is exactly why your PDG booklet you no doubt are staring at has a Clebsch table: the most useful page of it! $$ \frac{\langle \pi^- \Lambda | \Xi^- \rangle} {\langle \pi^0 \Lambda | \Xi^0 \rangle} = \frac { \langle J_\pi=1 , M_\pi=-1 | j_s=1/2 , m_s=-1/2; j_\Xi=1/2 , m_\Xi=-1/2 \rangle} { \langle J_\pi=1 , M_\pi=0 | j_s=1/2 , m_s=-1/2; j_\Xi=1/2 , m_\Xi=1/2 \rangle} =\sqrt{2}. $$

Thus, squaring the amplitude...

Isospin. It looks like a homework problem, with a summary answer in Okun's book, p 63. You need quark diagrams like a hole in the head.

This strangeness-changing decay violates isospin by 1/2, so assuming the $\Delta T=1/2$ piece of the hamiltonian dominates, and, since Λ is an isosinglet (so irrelevant to the isospin amps), you just consider the addition of a spurion s of isospin $|1/2,-1/2\rangle$ added to the cascade isodoublet $|1/2,\pm 1/2\rangle$, to yield pion states $|1,0\rangle$ and $|1,-1\rangle$ respectively. Isospin is preserved in the subsequent hadronization sequence.

The ratio of the respective decay amplitudes, then, is the simplest Clebsch ever, wich is exactly why your PDG booklet you no doubt are staring at has a Clebsch table: the most useful page of it! $$ \frac{\langle \pi^- \Lambda | \Xi^- \rangle} {\langle \pi^0 \Lambda | \Xi^0 \rangle} = \frac { \langle J_\pi=1 , M_\pi=-1 | j_s=1/2 , m_s=-1/2; j_\Xi=1/2 , m_\Xi=-1/2 \rangle} { \langle J_\pi=1 , M_\pi=0 | j_s=1/2 , m_s=-1/2; j_\Xi=1/2 , m_\Xi=1/2 \rangle} =\sqrt{2}. $$

Thus, squaring the amplitude...

Response to comment:

Here is an "explanation" of the basic ΔI=1/2 rule of weak interactions: Assuming the short distance weak hamiltonian is dominated by its ΔI=1/2 piece, the QCD-driven hadronization preserves isospin, so the weak blob is regarded as an "act of god", simulated/summarized by a I=1/2 spurion. Okun gives you the practical application rules for these spurions, but, frankly, you should have learned all about them in your HEP course, not here...

When it comes to "math is great", of course the $\sqrt 2$ is the normalization of the neutral pion wave function, because it is basically the Clebsch you get by adding two spin 1/2s, and descending by one step. I assume you appreciate that, isoladder-wise, the pion is a composite of a spurion and a cascade! I would call this mathematical analogy, rather than the "underlying physics". The "physics" of this is just the cornerstone ΔI=1/2 rule.

Isospin. It looks like a homework problem, with a summary answer in Okun's book, p 63. You need quark diagrams like a hole in the head.

This strangeness-changing decay violates isospin by 1/2, so assuming the $\delta T=1/2$ piece of the hamiltonian dominates, and, since Λ is an isosinglet (so irrelevant to the isospin amps), you just consider the addition of a spurion s of isospin $|1/2,-1/2\rangle$ added to the cascade isodoublet $|1/2,\pm 1/2\rangle$, to yield a pion states $|1,0\rangle$ and $|1,-1\rangle$ respectively.

The ratio of the respective decay amplitudes, then, is the simplest Clebsch ever, wich is exactly why your PDG booklet you no doubt are staring at has a Clebsch table: the most useful page of it! $$ \frac{\langle \pi^- \Lambda | \Xi^- \rangle} {\langle \pi^0 \Lambda | \Xi^0 \rangle} = \frac { \langle J_\pi=1 , M_\pi=-1 | j_s=1/2 , m_s=-1/2; j_\Xi=1/2 , m_\Xi=-1/2 \rangle} { \langle J_\pi=1 , M_\pi=0 | j_s=1/2 , m_s=-1/2; j_\Xi=1/2 , m_\Xi=1/2 \rangle} =\sqrt{2}. $$

Thus, squaring the amplitude...

Isospin. It looks like a homework problem, with a summary answer in Okun's book, p 63. You need quark diagrams like a hole in the head.

This strangeness-changing decay violates isospin by 1/2, so assuming the $\Delta T=1/2$ piece of the hamiltonian dominates, and, since Λ is an isosinglet (so irrelevant to the isospin amps), you just consider the addition of a spurion s of isospin $|1/2,-1/2\rangle$ added to the cascade isodoublet $|1/2,\pm 1/2\rangle$, to yield pion states $|1,0\rangle$ and $|1,-1\rangle$ respectively.

The ratio of the respective decay amplitudes, then, is the simplest Clebsch ever, wich is exactly why your PDG booklet you no doubt are staring at has a Clebsch table: the most useful page of it! $$ \frac{\langle \pi^- \Lambda | \Xi^- \rangle} {\langle \pi^0 \Lambda | \Xi^0 \rangle} = \frac { \langle J_\pi=1 , M_\pi=-1 | j_s=1/2 , m_s=-1/2; j_\Xi=1/2 , m_\Xi=-1/2 \rangle} { \langle J_\pi=1 , M_\pi=0 | j_s=1/2 , m_s=-1/2; j_\Xi=1/2 , m_\Xi=1/2 \rangle} =\sqrt{2}. $$

Thus, squaring the amplitude...