Question

Explain why a particularly good way of identifying mesons coupled to the $\pi \pi$-channel is to study the reaction $\pi \mathrm{N} \rightarrow(\pi \pi) \mathrm{N}$ at high energies. Show that $I+J$ must be an even integer for these mesons. In passing, we should note that particles that decay by strong interactions do not live long enough to leave tracks in an experimentalist's detector. Rather, they are identified by tracking their decay products. The mass of the decaying particle is the total energy of these products as measured in its rest frame. Due to its short lifetime, the uncertainty in its mass $(\sim \hbar / \Delta t)$ is sufficiently large to be directly observable. For example, the $\Delta$ is formed and rapidly decays in $\pi \mathrm{N}$ scattering. $\pi \mathrm{N} \rightarrow \Delta \rightarrow \pi \mathrm{N}$. Such an unstable particle decays according to the exponential law $|\psi(t)|^{2}=|\psi(0)|^{2} e^{-\Gamma t}$, where $\tau \equiv 1 / \Gamma$ is called the lifetime of the state. Thus, the time depend $\psi(t)$ for an unstable state must include the decay factor $\Gamma / 2 ;$ that is, $\psi(t)$ for an unstable state must include the decay where $M$ is the rest mass energy of the state. As a function of the center-of-mass energy $E$ of the $\pi \mathrm{N}$ system, the state is described by the Fourier transform $$ \begin{aligned} \chi(E) &=\int \psi(t) e^{i E t} d t \\ & \sim \frac{1}{E-M+(i \Gamma / 2)} \end{aligned} $$ The experimenter thus sees a $\pi \mathrm{N}$ reaction rate of the form $$ |\chi(E)|^{2}=\frac{A}{(E-M)^{2}+(\Gamma / 2)^{2}} $$ This function has a sharp peak centered at $M$ with a width determined by $\Gamma$. Equation (2.57) is called a Breit-Wigner resonance form, and $M$ and $\Gamma$ are known as the mass and width of the resonance, respectively. In a detailed resonance analysis, the form (2.55) will include kinematic factors. For instance, resonance production and decay near threshold are inhibited by phase space; its observed width is suppressed by kinematic factors.

Name: Problem 10, Chapter 2: Quarks And Leptons. An Introductory Course In Modern Particle Physics
Uploaded: 2021-12-27T03:57:50-08:00
Duration: 20 min 37 s
Channel: Amit Srivastava

   Explain why a particularly good way of identifying mesons coupled to the $\pi \pi$-channel is to study the reaction $\pi \mathrm{N} \rightarrow(\pi \pi) \mathrm{N}$ at high energies. Show that $I+J$ must be an even integer for these mesons.
In passing, we should note that particles that decay by strong interactions do not live long enough to leave tracks in an experimentalist's detector. Rather, they are identified by tracking their decay products. The mass of the decaying particle is the total energy of these products as measured in its rest frame. Due to its short lifetime, the uncertainty in its mass $(\sim \hbar / \Delta t)$ is sufficiently large to be directly observable. For example, the $\Delta$ is formed and rapidly decays in $\pi \mathrm{N}$ scattering. $\pi \mathrm{N} \rightarrow \Delta \rightarrow \pi \mathrm{N}$. Such an unstable particle decays according to the exponential law
$|\psi(t)|^{2}=|\psi(0)|^{2} e^{-\Gamma t}$, where $\tau \equiv 1 / \Gamma$ is called the lifetime of the state. Thus, the time depend $\psi(t)$ for an unstable state must include the decay factor $\Gamma / 2 ;$ that is,
$\psi(t)$ for an unstable state must include the decay
where $M$ is the rest mass energy of the state. As a function of the center-of-mass energy $E$ of the $\pi \mathrm{N}$ system, the state is described by the Fourier transform
$$
\begin{aligned}
\chi(E) &=\int \psi(t) e^{i E t} d t \\
& \sim \frac{1}{E-M+(i \Gamma / 2)}
\end{aligned}
$$
The experimenter thus sees a $\pi \mathrm{N}$ reaction rate of the form
$$
|\chi(E)|^{2}=\frac{A}{(E-M)^{2}+(\Gamma / 2)^{2}}
$$
This function has a sharp peak centered at $M$ with a width determined by $\Gamma$. Equation (2.57) is called a Breit-Wigner resonance form, and $M$ and $\Gamma$ are known as the mass and width of the resonance, respectively. In a detailed resonance analysis, the form (2.55) will include kinematic factors. For instance, resonance production and decay near threshold are inhibited by phase space; its observed width is suppressed by kinematic factors.

Quarks And Leptons. An Introductory Course In Modern Particle Physics

Francis Halzen, Alan… 1984 Edition

Chapter 2, Problem 10

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Solved by Expert Amit Srivastava

12/27/2021

Step 1

This reaction involves a pion ($\pi$) interacting with a nucleon ($N$) to produce a pair of pions ($\pi \pi$) and a nucleon. At high energies, this process can reveal information about the mesons that couple to the $\pi \pi$ channel. Show more…

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Explain why a particularly good way of identifying mesons coupled to the $\pi \pi$-channel is to study the reaction $\pi \mathrm{N} \rightarrow(\pi \pi) \mathrm{N}$ at high energies. Show that $I+J$ must be an even integer for these mesons. In passing, we should note that particles that decay by strong interactions do not live long enough to leave tracks in an experimentalist's detector. Rather, they are identified by tracking their decay products. The mass of the decaying particle is the total energy of these products as measured in its rest frame. Due to its short lifetime, the uncertainty in its mass $(\sim \hbar / \Delta t)$ is sufficiently large to be directly observable. For example, the $\Delta$ is formed and rapidly decays in $\pi \mathrm{N}$ scattering. $\pi \mathrm{N} \rightarrow \Delta \rightarrow \pi \mathrm{N}$. Such an unstable particle decays according to the exponential law $|\psi(t)|^{2}=|\psi(0)|^{2} e^{-\Gamma t}$, where $\tau \equiv 1 / \Gamma$ is called the lifetime of the state. Thus, the time depend $\psi(t)$ for an unstable state must include the decay factor $\Gamma / 2 ;$ that is, $\psi(t)$ for an unstable state must include the decay where $M$ is the rest mass energy of the state. As a function of the center-of-mass energy $E$ of the $\pi \mathrm{N}$ system, the state is described by the Fourier transform $$ \begin{aligned} \chi(E) &=\int \psi(t) e^{i E t} d t \\ & \sim \frac{1}{E-M+(i \Gamma / 2)} \end{aligned} $$ The experimenter thus sees a $\pi \mathrm{N}$ reaction rate of the form $$ |\chi(E)|^{2}=\frac{A}{(E-M)^{2}+(\Gamma / 2)^{2}} $$ This function has a sharp peak centered at $M$ with a width determined by $\Gamma$. Equation (2.57) is called a Breit-Wigner resonance form, and $M$ and $\Gamma$ are known as the mass and width of the resonance, respectively. In a detailed resonance analysis, the form (2.55) will include kinematic factors. For instance, resonance production and decay near threshold are inhibited by phase space; its observed width is suppressed by kinematic factors.

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Key Concepts

Scattering Resonance Identification

High-energy scattering experiments, such as ?N ? (??)N reaction studies, are a cornerstone in identifying meson resonances. By using reactions where intermediate unstable particles are formed, researchers can analyze the energy distribution of the decay products, revealing peaks corresponding to resonances. The high energy ensures that many resonant states, even those with short lifetimes, can be produced and studied effectively through their decay channels.

Meson Decay Channels and Reconstruction

Unstable mesons cannot be directly observed because they decay almost immediately via strong interactions. Instead, they are identified by reconstructing the energy and momenta of their decay products, such as ?? pairs. The mass of the decaying meson is determined from the total energy of its decay products measured in its rest frame, allowing experimentalists to infer its properties despite its short lifetime.

Angular Momentum and Isospin Coupling

Symmetry properties of mesons, including total angular momentum (J) and isospin (I), impose restrictions on the allowed quantum states. For mesons decaying into a two-pion system, the combination I+J must be an even integer. This arises due to the symmetry requirements of the pion pair, which are identical bosons, and ensures that the overall wavefunction has the correct symmetry under particle exchange.

Unstable State Lifetime and Decay Width

The short lifetime of unstable particles is intimately related to their decay width (?) via the uncertainty principle. The energy uncertainty, which manifests as the observed width of the resonance peak, is inversely proportional to the lifetime (? = 1/?). This relationship explains why rapidly decaying resonances exhibit broad energy distributions, a fundamental concept in understanding particle decay processes in quantum mechanics.

Breit-Wigner Resonance

The Breit-Wigner resonance formula describes the distribution of the reaction rate as a function of energy in the vicinity of a resonance. It characterizes how the probability amplitude for forming an unstable state peaks at its mass value (M) and falls off with a width determined by the decay rate (?). This resonance shape is a critical tool for extracting the mass and width of short-lived particles from experimental data.

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Transcript

00:01 Depending the temperature through a derivative of a log of number of ways may be hard to embrace, but it's built into the boltzmann distribution and all others.

00:13 We can at least verify it in the oscillator case.

00:19 The average energy mh omega -0 by n of a collection of oscillators through the average predicted by the bolzman precision demands a fairly much.

00:33 Our system assuming the number of particles equals to 10 ,000 and the total energy is 49h omega not it's easiest to choose the energy units in which h omega is equal to 1 so that e and m e and m are both 49 ,000 first calculate the number of particles n at each level by multi -plan capital n by the exact probabilities using instead n e is the total number at a given level not a specific particle.

01:46 Here we know that h -omeagnot is equal to 1.

01:48 Small n is the level's energy round each capital n n to the nearest integer then determine the total energy by simply multiplying capital n n by small n the number times the energy at each level and summing over all the levels it might seem that the total should be 49 ,000.

02:19 Due to running it won't be but it should differ by less than 1 % entropy is fairly easy to calculate here the number of ways of rearranging particle labels when n particles are distributed in different boxes so it will be n factorial by and now calculate the entropy for b part repeat the a part changing of only capital m, marking it to 51 ,000 instead of 49 ,000.

03:07 For c, calculate the quotient, that is change in entropy over changing energy, bearing in mind that we are considering the system of n equals to 10 ,000, and m equals to 50 ,000.

03:34 How does it compare with the given equation? in order to determine the probabilities, energies, occupants, levels and entropy the equation for the probability to find a particle in the particular state, the number of particles expected per level, the energy per level, the number of ways of counting numbers, the probability, p, n is given by m minus n plus n minus 1 minus 1 m minus 1 m minus n divided by m plus n minus 1 m the number of particles expected number of particles expected capital n n per state n is given by n n is equal to n into p n is the probability of finding the particle in that state the energy level, energy per level, en for some state n is given by small nn to capital m.

05:18 Your capital n is the expected number of particles per level.

05:22 The number of ways of counting n number of particles by the binomial coefficient w is given by n factorial by pi n n factorial.

05:44 The entropy s is defined by s is equal to kb into ln of the binomial coefficient.

05:56 Your kb is the bolzman constant and w is the number of ways that system can be arranged.

06:04 Sterling approximation involving the n factorial with capital n being very large number will also be useful.

06:17 That is given by ln of n factorial is approximately capital n ln n minus n as well as the approximation using a factorial given by k factorial by k minus a small k factorial is equal to k to the power.

06:42 This is well as well.

06:44 When small k is very very less than capital k.

06:49 The equation relating the temperature, number of particles and sum of numbers of the particle capital m will be used as a comparison given by kv t is equal to h omega not by l n 1 plus n by m.

07:10 With k b being the bolzman constant, h is the plume, reduced plank constant, and omega -0 is fundamental angular frequency of the particle.

07:19 The relationship between entropy and the temperature is t is equals to differential of s by differential of e raised to to the power minus 1.

07:37 This shows that it's the inverse of change in entropy over the change in energy.

07:45 For first in order to find the probability of finding a particle in any particular another state we will start over by the probability formula.

07:56 This can be rewritten as m factorial by m minus n factorial multiplied by n minus 1 factorial by n minus 1 minus 1 multiplied by divided by m plus n minus 1 factorial factorial and it is divided by m plus n minus 1 minus of n plus 1 factorial each of those terms can be rewritten and the probability formula will be respectively m raised to the power n n minus 1 divided by m plus n minus 1 whole raise to the power n plus 1 since m and n will be very large subtracting 1 will not affect it to ignoring it we have the expression n raised to the power n capital n divided by m plus n raised to the power n plus 1.

09:46 This can again be rearranged n by m plus n into m by m plus n raised to the power small n.

10:09 The parenthesis can be written with a negative exponent so as to simplify this expression a little so the probability formula will be equals to n by m plus n, 1 plus n by m raised in power minus n.

10:38 So that is the simpler form that will be used in the calculations.

10:42 Instead, substituting the value n equals to 10 ,000 and m equals to 49 ,000 will have the equation in the reduced form in terms of small n will be 10 by 59 into 59 by 49 raised to the power minus n...

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