2305NSC Quantum Mechanics, Relativity Assignment 4 Solution

This document presents the complete solution to Assignment 4 for the 2305NSC course at Griffith University, focusing on Quantum Mechanics, Relativity and Elementary Particles. The solution addresses multiple questions, including calculations of total energy and momentum of photons, determination of particle mass and speed, and analysis of particle interactions based on conservation laws. The assignment also includes an investigation of quark composition and structure, especially in the context of the Pauli Exclusion Principle and the properties of various baryons, such as the Xi and Sigma baryons. The document provides detailed explanations, calculations, and references to support the answers, making it a valuable resource for students studying these complex topics. The solution covers topics like the Lorentz transformations, d’Alembertian invariance, photon energy, momentum calculations, mass determination, relativistic speed calculation, conservation laws, and quark composition.

Assignment 4
(2305NSC Trimester 2 2019)
Q. 3. Suppose one photon has an energy of 200 MeV and is traveling along the x axis. Suppose
another has an energy of 100 MeV and is traveling along the y axis.
(a) What is the total energy of this system?
(b) What is the total momentum?
(c) If a single particle had this same total energy and momentum:
i. What would be its mass?
ii. In what direction would it be traveling?
iii. With what speed?
Solution:
(a) Given, E1 = 200 MeV, E2 = 100 MeV
Therefore,
Total energy, ETotal = E1+ E2 = (200+100) MeV = 300 MeV.
(b) For momentum calculation,
Energy of a photon, E=hc/λ = hc/(h/p), (λ=h/p)
Therefore, E = pc
For Particle having energy 200 Mev,
⇒ p1 = E/c
⇒ p1 = 200 x 106 x (1.6×10−19)/ (3× 108) kgm/s
⇒ p1 = 106.66 x 10-21 kgm/s
And, ⇒ p2 = 100 x 106 x (1.6×10−19)/ (3× 108) kgm/s
⇒ p2 = 53.33 x 10-21 kgm/s
Therefore, Total momentum,
P = p1 + p2
= 106.66 – 53.33
= 53.33 kgm/s.

(Note: Momentum is a vector quantity, so total momentum of the two photons is calculated by
vectors addition rule of two vectors which are in opposite directions.)
(c) i. the photon which has mass when moving but a zero mass at rest
Therefore, the relativistic mass can be calculated by below formula,
E = m.c2
⇒ m =E/c2
= 200 MeV.
ii. 27o
iii. Speed(relativistic),
If the rest mass of a particle is m and the total energy is E , then
and,
Therefore, 1− v
c = 1
2 (200
300 ) power ( 2 )
Finally, speed, v = 0.74 c.
Q. 4. State which of the following processes can be observed and why. Are the corresponding
charge conjugate processes observed? why?:
Solution:
(a) p + n p + μ+ + μ-
Nature has specific rules for particle interactions and decays, and these rules have been
summarized in terms of conservation laws. One of the most important of these is the
conservation of baryon number.
So, we now check Baryon number conservation,
Therefore, p + n p + μ+ + μ-
Charge, Q: 1 + 0 = 1 + 1 -1 (conserved)
Baryon no., B: 1 + 1 ≠ 1 + 0 + 0 (conserved)

From the above inequality we can conclude that this reaction is not allowed because non-
conservation of baryon number.
(b)
Charge, Q: -1 = -1 (conserved)
Baryon number, B: 0 = 0 + 0 (conserved)
(c) n + P e-
Charge(Q): 0 + 1 ≠ -1 (Not conserved, therefore this reaction not allowed)
(d)
Q: -1 = -1 + 0 + 0 (conserved)
B: 0 = 0 + 0 + 0 (conserved)
L:1 = +1 -1 +1 (Conserved)
(e)
Q: 0 = 0 +0 (Conserved)
B: 1= 1 + 0 (Conserved)
Q. 5. The four particles are charmless, topless and bottomless baryons.
They have a charge of Q = -1. The strangeness S, ordinary spin J and measured mass in MeV
are given by the following numbers in brackets respectively:
(a) Find the quark composition of the four particles.
(b) How can the quark structure of Ώ- be reconciled with Pauli's exclusion principle?
(c) How does the particle differ from the particle in a way which might explain
qualitatively why their masses are different?
Solution:
(a) Δ- = ddd
= sdd
= dss

Ώ- = sss
(b)
The quark model and the exclusion principle were reconciled as a result of ideas
developed by Oscar W. Greenberg of the University of Maryland at College Park and,
independently, by Moo-Young Han of Duke University and Yoichiro Nambu of the
University of Chicago. What is needed is to assume that each kind of quark can exist in
any of three states. For example, if an s quark in state A is combined with an s quark in
state B and an s quark in state C to form the omega-minus particle, the exclusion
principle is saved (Rebbi, 1983).
(c)
The Xi baryons or cascade particles are a family of subatomic hadron
particles which have the symbol Ξ and may have an electric charge (Q) of +2 e, +1 e, 0,
or −1 e, where e is the elementary charge. Like all conventional baryons, they contain
three quarks. Xi baryons, in particular, contain one up or down quark plus two more
massive quarks: either strange, charm or bottom. They are historically called the
cascade particles because of their unstable state; they decay rapidly into lighter particles
through a chain of decays.
The Sigma baryons are a family of subatomic hadron particles which have
two quarks from the first flavour generation (up and/or down quarks), and a third quark
from higher flavour generations, in a combination where the wavefunction does not
swap sign when any two quark flavours are swapped. They are thus baryons, with total
Isospin of 1, and can either be neutral or have an elementary charge of +2, +1, 0, or −1.
They are closely related to the Lambda baryons, which differ only in the wavefunction's
behavior upon flavour exchange
Reference:
[1] Rebbi, C. (1983). The Lattice Theory of Quark Confinement. Scientific American, 248(2), 54–
65. https://doi.org/10.1038/scientificamerican0283-54
[2] Warfreak. (2010, September 29). Physics Forums: Science Articles, Homework Help, Discussion.
Retrieved from https://www.physicsforums.com/threads/photon-energy-momentum-and-
velocity.433429/.
[3]Physics Stack Exchange, Relativistic speed/energy relation,
https://physics.stackexchange.com/questions/716/relativistic-speed-energy-relation-is-this-
correct.