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