Relativistic Kinematics, Classical Electromagnetism and Quantum Mechanics

Question 1

Invariant Mass Calculations

Overview

Answer 1

-looking for evidence of a new particle X
-suspect that X decays to A+B
-X is short-lived so can’t observe it directly
-in experiment, collide e- and e+ and regularly observe:
e- + e+ -> A + B (+ other stuff)
-want to know if the A and B come from X
-plot the relative frequency of invariant mass of A and B and see if there is a peak at mx, the mass of X

Question 2

Invariant Mass

Definition

Answer 2

W² = (ΣEi)² - (Σpi_)²

Question 3

Wavefunction of X Particle

Answer 3

ψ(t) = ψ(0) e^(-iEt) e^(-Γt/2)
-for a particle at rest, E=mx:
ψ(t) = ψ(0) e^(-iEmx) e^(-Γt/2)
-the second exponential is the real part required to ensure an overall decrease in probability over time since the particle is likely to decay
-Γ is the decay constant, factor of 1/2 since the probability equals |ψ|² so factor of two cancels

Question 4

Breit-Wigner Formula

Answer 4

ϱ(E) = R / [(E-mx)² + Γ²/4]

-Γ is the full width half maximum

Question 5

If A and B haven’t come from X decay, what does W represent?

Answer 5

-the effective mass at the centre of mass of the composite system of A and B

Question 6

Derive the Maxwell Equation

Answer 6

-combine the field strength tensor:
Fμν = ∂μAν - ∂νAμ
-and 
∂μFμν = Jν
=>
∂μ∂μAν - ∂μ∂νAμ = Jν

Question 7

Derive the Charge Continuity Equation

Answer 7

-start with the Maxwell Equation
∂μ∂μAν - ∂μ∂νAμ = Jν
-differentiate with respect to xν:
∂ν∂μ∂μAν - ∂ν∂μ∂νAμ = ∂νJν
-differentials commute so LHS is 0:
∂νJν = 0

Question 8

Derive Charge Conservation

Answer 8

-start with the continuity equation:
∂νJν = 0
-integrate over some volume, V
-split the differential into the 0th element and 1,2,3 elements
-apply divergence theorem to RHS and definition of charge on LHS:
∂Q/∂t = - ∫ Ji dsi
-i.e. charge in a region only changes if current density flows across the boundary, charge is conserved
-in particular if V is the universe then total charge Q is constant

Question 9

Gauge Invariance

Definition

Answer 9

-the Maxwell Equation is invariant under transformations of the form:
Aμ -> Aμ’ = Aμ + ∂μχ
-for any scalar function χ
-check by subbing into the Maxwell equation, terms cancel and the original equation is recovered
-can also show that Fμν=Fμν’

Question 10

Gauge Invariance

Fixing the Gauge

Answer 10

• in practice, we must choose some condition that A satisfies in order to ‘fix the gauge’
• i.e. a condition that uniquely defines A

Question 11

Coulomb Gauge

Answer 11

∂iAi = 0

Question 12

Lorenz Condition

Answer 12

-not technically a gauge as it only restricts A and doesn’t uniquely define it
∂μAμ = 0
-equivalent to:
∂ . A = 0

Question 13

Maxwell Equation Under the Lorenz Transformation

Answer 13

∂²Aμ = Jμ

Question 14

Quantum Mechanics

Wave Mechanics Formulation

Answer 14

-the wave mechanics formulation of quantum mechanics postulates a wave function to describe the system
-each observable quantity from classical mechanics is promoted to an operator
-making a measurement of an observable results in an eigenvalue and immediately after measurement the system takes on the corresponding eigenstates
-between measurements the wavefunction evolves
-in non-relativistic quantum mechanics this evolution is described by the time dependent Schrodinger equation:
i ∂/∂t Ψ = -1/2m ∂i∂iΨ + VΨ = H^Ψ

Question 15

Where does the time dependent Schrodinger equation come from?

Answer 15

-in natural units, quantum theory equates energy of a particle with frequency and momentum with wave vector:
E=ω
p_=k_
-assuming the time-evolution equation to be linear , any wavefunction can be constructed out of a complete set of plane-wave solutions as they form a basis
-so can deduce an appropriate equation by looking at plane-wave solutions:
Ψ = Ψoe^[ip_.x_] = Ψoe^[i(Et-pxx-pyy-pzz)]
-differentiate with respect to time to find the energy operator
-take the gradient to find the momentum operator
-sub into energy = KE + PE equation for Schrodinger equation

Question 16

Probability Amplitude

Answer 16

-the interpretation of the wavefunction is as a probability amplitude
-the probability of finding a particle described by a wavefunction Ψ in a region V is given by:
P(V) = ∫ |Ψ(x)|² d³x

Question 17

Probability Density Current

Definition

Answer 17

probability density only changes in a region if:
ji = -1\2mi (Ψ∂iΨ - Ψ∂iΨ)
-the probability density current flows through the surface of that region

Question 18

Probability Density Current

Derivation

Answer 18

-multiply Schrodinger equation by Ψ* (1)
-take conjugate of Schrodinger equation, then multiply by Ψ (2)
-subtract: (1) - (2)
=>
∂/∂t(|Ψ|²) = -1\2mi ∂i (Ψ∂iΨ - Ψ∂iΨ)
=>
ji = -1\2mi (Ψ∂iΨ - Ψ∂iΨ)

Question 19

Quantum Mechanics of Charged Particles

Answer 19

• charged particles in the presence of an electromagnetic field requires a particular form of the Schrodinger equation
• in this case want a Hamiltonian that gives the Lorentz Force Law from the Hamilton equations

Question 20

Hamilton Equations

Answer 20

x’ = ∂H/∂pi
p’i = - ∂H/∂xi
-these equation are completely general

Question 21

Lorentz Force Law

Answer 21

Fi = Q (Ei + εijk x’j Bk)

Question 22

Hamiltonian From the Lorentz Force Law

Answer 22

-write the Lorentz Force Law in terms of electromagnetic potentials
-swap εijk for deltas
-sub in dA/dt
=>
Fi = Q(-∂iV - dA/dt + x’j∂iAj)

Question 23

Electromagnetic Form of the Schrodinger Equation

Answer 23

i ∂Ψ/∂t = δij(pi-QAi)(pj-QAj)/2m Ψ + VQΨ

-recognise that the derivatives act on A as well as Ψ

Question 24

Classical Angular Momentum

Answer 24

Li = εijk rj pk

Question 25

Quantum Angular Momentum Operator Components

Answer 25

Li^ = -i εijk rj ∂k

Question 26

Angular Momentum Commutation Relations

Answer 26

[L1^,L2^] = iL3^
[L2^,L3^] = iL1^
[L3^,L1^] = iL2^

Question 27

Total Angular Momentum Operator

Answer 27

L²_^ = L1²^ + L2²^ + L3²^

-it can be shown that this commutes with Li^

Question 28

Angular Momentum Eigenvalues

Answer 28

• boundary conditions on on the eigenstates of L_^ and Li^ impose certain allowed values / quantum numbers, specifically:
• L²_^ has eigenvalues l(l+1) for l∈ℕ
• Li^ has eigenvalues m∈Z such that -l≤m≤l

Question 29

Angular Momentum Operator

Simultaneous Eigenstates

Answer 29

-can be in simultaneous eigenstate of L_^ and Li^ but NOT e.g. L1^ and L2^

Question 30

Intrinsic Angular Momentum

Answer 30

• intrinsic angular momentum is spin
• the boundary condition argument does not apply in this case since spin is an intrinsic property
• introduce spin operators, Si^ and S^_² that obey the same commutation relationships as the angular momentum operators

Question 31

Spin Operator Eigenvalues

Answer 31

S^_²ψ = s²ψ
e.g.
Si^ψ = msψ

Question 32

Ladder Operators for Spin

Answer 32

-consider a simultaneous eigenstate of S3^ and S^_², ψ, with eigenvalues ms and s²
-construct ladder operator:
S±^ = S1^ ± iS2^

Question 33

Spin Ladder Operators

Commutation Relations

Answer 33

1) [S3^,S±^] = ± S±^
2) [S+,S-] = 2 S3^
3) [S±^,S^²] = 0
- (1) => these are ladder operators for S3^: S+ ψ is an S3^ eigenstate with eigenvalue ms+1, similarly acting with S-^ gives eigenvalue ms-1
- (3) => fact that S±^ commute with S^² tells us that when we change S3^ value we still have S^_²eigenstate with the same value, i.e. changing third component of angular momentum you don’t change overall angular momentum

Question 34

Total Angular Momentum and S+ & S-

Answer 34

-for given total spin, S±^ gives us all possibilities of S3^
S^_² = S1^² + S2^²+ S3^²
= 1/2 (S+S- + S-S+) + S3^²
= S3^ + S-^S+^ + S3^²
-OR
= -S3^ + S-^S+^ + S3^²

Question 35

Angular Momentum

ψmax

Answer 35

-let ψmax be the state with the maximum possible ms for given S
-know that:
S3^S+^ ψmax= (mmax + 1)S+ ψmax
-this must hold but also have a maximum therefore S+^ψmax must =0
-S+^ annihilates ψmax
-similarly S-^ annihilates ψmin

Question 36

Relation Between Total Spin and Maximum Eigenvalue for One Spin Component

Answer 36

S^_² ψmax 
= (S3^ + S-^S+^ + S3^²) ψmax 
= (mmax + mmax²) ψmax
= mmax ( 1+mmax)  ψmax
=>
s = √[mmax(mmax+1)]
-AND
s = √[mmin(mmin-1)]

Question 37

Allowed Values of Spin

Answer 37

-putting the two expressions for s in terms of mmax and mmin together =>
mmin = ±mmax
-and m's differ by an integer amount, so:
mmax = {0,1/2,1,3/2,3,...}
-spin statistics theorem
=>
integer spin = bosons
half integer spin = fermions

Question 38

Schrodinger Equation for an Electron

Answer 38

-an electron is spin 1/2 so need two degrees of freedom, spin up or spin down (1/2,-1.2)
-we promote ψ to a spinor-valued wave function with two components for the two spin states
-with two component spinors, can write 2x2 representation of spin operators:
si = 1/2 σi
-where σi are the Pauli matrices
-these ideas can be incorporated straight into the Schrodinger Equation to form the Pauli Equation, a non-relativistic equation for a spin 1/2 charged particle