Flashcards in Relativistic Kinematics, Classical Electromagnetism and Quantum Mechanics Deck (38)

Loading flashcards...

1

##
Invariant Mass Calculations

Overview

###
-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

2

##
Invariant Mass

Definition

### W² = (ΣEi)² - (Σpi_)²

3

## Wavefunction of X Particle

###
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

4

## Breit-Wigner Formula

###
ϱ(E) = R / [(E-mx)² + Γ²/4]

-Γ is the full width half maximum

5

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

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

6

## Derive the Maxwell Equation

###
-combine the field strength tensor:

Fμν = ∂μAν - ∂νAμ

-and

∂μFμν = Jν

=>

∂μ∂μAν - ∂μ∂νAμ = Jν

7

## Derive the Charge Continuity Equation

###
-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

8

## Derive Charge Conservation

###
-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

9

##
Gauge Invariance

Definition

###
-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'

10

##
Gauge Invariance

Fixing the Gauge

###
-in practice, we must choose some condition that A satisfies in order to 'fix the gauge'

-i.e. a condition that uniquely defines A

11

## Coulomb Gauge

### ∂iAi = 0

12

## Lorenz Condition

###
-not technically a gauge as it only restricts A and doesn't uniquely define it

∂μAμ = 0

-equivalent to:

∂ . A = 0

13

## Maxwell Equation Under the Lorenz Transformation

### ∂²Aμ = Jμ

14

##
Quantum Mechanics

Wave Mechanics Formulation

###
-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

15

## Where does the time dependent Schrodinger equation come from?

###
-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

16

## Probability Amplitude

###
-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

17

##
Probability Density Current

Definition

###
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

18

##
Probability Density Current

Derivation

###
-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*)

19

## Quantum Mechanics of Charged Particles

###
-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

20

## Hamilton Equations

###
x' = ∂H/∂pi

p'i = - ∂H/∂xi

-these equation are completely general

21

## Lorentz Force Law

### Fi = Q (Ei + εijk x'j Bk)

22

## Hamiltonian From the Lorentz Force Law

###
-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)

23

## Electromagnetic Form of the Schrodinger Equation

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

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

24

## Classical Angular Momentum

### Li = εijk rj pk

25

## Quantum Angular Momentum Operator Components

### Li^ = -i εijk rj ∂k

26

## Angular Momentum Commutation Relations

###
[L1^,L2^] = iL3^

[L2^,L3^] = iL1^

[L3^,L1^] = iL2^

27

## Total Angular Momentum Operator

###
L²_^ = L1²^ + L2²^ + L3²^

-it can be shown that this commutes with Li^

28

## Angular Momentum Eigenvalues

###
-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

29

##
Angular Momentum Operator

Simultaneous Eigenstates

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

30

## Intrinsic Angular Momentum

###
-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

31

## Spin Operator Eigenvalues

###
S^_²ψ = s²ψ

e.g.

Si^ψ = msψ

32

## Ladder Operators for Spin

###
-consider a simultaneous eigenstate of S3^ and S^_², ψ, with eigenvalues ms and s²

-construct ladder operator:

S±^ = S1^ ± iS2^

33

##
Spin Ladder Operators

Commutation Relations

###
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

34

## Total Angular Momentum and S+ & S-

###
-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

35

##
Angular Momentum

ψmax

###
-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

36

## Relation Between Total Spin and Maximum Eigenvalue for One Spin Component

###
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)]

37

## Allowed Values of Spin

###
-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

38