the world particle content all the particles are spin ½ fermions!
TRANSCRIPT
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The WorldParticle content
All the particles are spin ½ fermions!
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Schrodinger Wave Equation
He started with the energy-momentum relation for a particle
he made the quantum mechanical replacement:
How about a relativistic particle?
Expecting them to act on plane waves
ipxrpiiEt eee
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The Quantum mechanical replacement can be made in a covariant form. Just remember the plane wave can be written in a covariant form:
As a wave equation, it does not work.It doesn’t have a conserved probability density.It has negative energy solutions.
ipxrpiiEt eee
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0222 cmp
0p E
There are two solutions for each 3 momentum p (one for +E and one for –E )
ipxxpitip eaeax 0
)(Plane wave solutions for KG Eq.
ipxipxipx eameapeap 2220
2220 mpp
p
xpiiEtxpiiEt
p
ipxipx ebeaebeax
)(
Time dependence can be determined.
It has negative energy solutions.
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Expansion of the KG Field by plane:
p
xpiiEtxpiiEt
p
ipxipx ebeaebeax
)(
p
ipxipx eaeax
)(
If Φ is a real function, the coefficients are related:
The proper way to interpret KG equation is it is not a Wavefunction Equation but actually a Field equation just like Maxwell’s Equations.
Plane wave solutions just corresponds to Plane Waves.
It’s natural for plane waves to contain negative frequency components.
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Dirac
This result is too beautiful to be false; it is more important to have beauty in one's equations than to have them fit experiment.
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Blaming the negative energy problem on the second time derivative of KG Eq., Dirac set out to find a first order differential equation.This Eq. still needs to give the proper energy momentum relation. So Dirac propose to factor the relation!For example, in the rest frame:
mct
i
Made the replacement
First order diff. Eq.
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and we need:
Now put in 3-momenta:
Suppose the momentum relation can be factored into linear combinations of p’s:
Expand the right hand side:
We get
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Dirac propose it could be true for matrices.
It’s easier to see by writing out explicitly:
Oops! What!
00110 0110 or
No numbers can accomplish this!
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2 by 2 Pauli Matrices come very close
1i jiji 0,
ijji ,
10
01
0
0
01
10321
i
i
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Dirac find it’s possible for 4 by 4 matrices
We need:
that is
He found a set of solutions:
Dirac Matrices
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Dirac find it’s possible for 4 by 4 matrices
Pick the first order factor:Make the replacement and put in the wave function:
If γ’s are 4 by 4 matrices, Ψ must be a 4 component column:
03210
mczyxt
i
It consists of 4 Equations.
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The above could be done for 2 by 2 matrices if there is no mass.Massless fermion contains only half the degrees of freedom.
A pure left-handed or a pure right-handed must be massless.
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Now put in 3-momenta:
Suppose the momentum relation can be factored into linear combinations of p’s:
Expand the right hand side:
pppp k
k
Now β, γ do not need to be the same.
3211 k 321 ---1 k
We can choose:
交叉項抵銷 ppk
k
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Massless fermions contain only half the degrees of freedom.
3211 k 321 ---1 k
0 p 0
p
0
0
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Expansion of a solution by plane wave solutions for KG Eq.
p
xpiiEtxpiiEt
p
ipxipx ebeaebeax
)(
p
ipxipx eaeax
)(
If Φ is a real function, the coefficients are related:
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Multiply on the left with
mcp
0222 ucmp 0222 cmp 0p E
There are again two sets of solutions for each 3 momentum p (one for +E and one for –E )
ueaueax xpitipipx 0
)(
2
1
2
1
B
B
A
A
B
A
u
u
u
u
u
uu
Plane wave solutions for Dirac Eq.
Negative Energy again!First order equation doesn’t escape it!
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How about u?
0p 0p0p
0p
0p
0p
0p
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You may think these are two conditions, but no.
We need:
Multiply the first by
So one of the above is not independent if 0222 cmp
0p 0p
20p
0p
0p
0p
0p 20p
22 pp
1
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We need:
or 0p0p
How many solutions for every p?
Go to the rest frame!
)0,(or)0,(
mmp
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0p
0p
0p
0p
)0,(
mp
020
00
B
A
u
u
m
0Bu uA is arbitrary
1
0,
0
1Au
u has two solutions corresponding to spin up and spin down in the rest frame.
Go to the rest frame!
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0p
0p
0p
0p
)0,(
mp
000
02
B
A
u
um
0Au uB is arbitrary
1
0,
0
1Bu
Two solution (spin down and spin up antiparticle)
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It’s not hard to find four independent solutions.
There are four solutions for each 3 momentum p (two for particle and two for antiparticle)
or
-
We got two positive and two negative energy solutions!Negative energy is still here!In fact, they are antiparticles.
0p0p
0p
0p
0p
0p
0p
0p
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Electron solutions:
ueauea rpiiEtipx
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Positron solutions: u to v
0),(4,3 pEumcp
0),(2,1 pEvmcp
veaveauea rpiiEtipxrpiiEt
)2,1()4,3(
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Expansion of a solution by plane wave solutions for KG Eq.
p
xpiiEtxpiiEt
p
ipxipx ebeaebeax
)(
p
ipxipx evbeuax
)(
Expansion of a solution by plane wave solutions for Dirac Eq.
220 mpEp
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4-columns4-rows
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Find the Green Function of Dirac Eq.
')',()',( 4 xxIxxmcGxxGi
Now the Green Function G is a 4 ˣ 4 matrix
')',()',( 4 xxIxxmcGxxGi
How about internal lines?
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')',()',( 4 xxIixxmcGxxGi
)(
~
2)'( )'(
4
4
pGepd
xxG xxip
ipGmp )(~
)'(4
44
2)'( xxipe
pdxx
mp
ipG
)(~
Using the Fourier Transformation
22 mp
mpi
Fermion Propagator
p
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4 ˣ 4 matrices
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Bilinear Covariants
Ψ transforms under Lorentz Transformation:
Interaction vertices must be Lorentz invariant.
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The weak vertices of leptons coupling with W
e
We
ee ?
eWeg
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Bilinear Covariants
Ψ transforms under Lorentz Transformation:
Interaction vertices must be Lorentz invariant.
How do we build invariants from two Ψ’s ?
A first guess:
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Maybe you need to change some of the signs:
It turns out to be right!
We can define a new adjoint spinor:
is invariant!
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In fact all bilinears can be classified according to their behavior under Lorentz Transformation:
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A
p
ipxipx evbeuax
)(
)( 1pe
0)( 11 upeau p
p
ipxipx euaevbx
)(
0u
)(0 313133peuuuau ppppp
Feynman Rules for external lines
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Photons:
It’s easier using potentials:
forms a four vector.
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4-vector again
Charge conservation
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Now the deep part:
E and B are observable, but A’s are not!
A can be changed by a gauge transformation without changing E and B the observable:
So we can use this freedom to choose a gauge, a condition for A:
J
cA
4
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For free photons:
0 A
Almost like 4 KG Eq.
ipxeaxA )(
Energy-Momentum Relation
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Polarization needs to satisfy Lorentz Condition:
We can further choose
Lorentz Condition does not kill all the freedom:
then Coulomb Guage
The photon is transversely polarized.
For p in the z direction:
For every p that satisfy
there are two solutions!
Massless spin 1 particle has two degrees of freedom.
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A
),( q
0)( i
Feynman Rules for external photon lines
2,1,
)()()(ip
ipxipp
ipxipp eaeaxA
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Gauge Invariance
Classically, E and B are observable, but A’s are not!A can be changed by a gauge transformation without changing E and B the observable:
But in Qunatum Mechanics, it is A that appear in wave equation:
),(' xtAA Transformation parameter λ is a
function of spacetime.
AAAAAA ''
In a EM field, charged particle couple directly with A.
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Classically it’s force that affects particles. EM force is written in E, B.But in Hamiltonian formalism, H is written in terms of A.
txetxAc
ep
mH ,,
2
12
Quantum Mechanics or wave equation is written by quantizing the Hamiltonian formalism:
eAie
mti
2
2
1
Is there still gauge invariance?
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B does not exist outside.
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eAie
mti
2
2
1
Gauge invariance in Quantum Mechanics:
In QM, there is an additional Phase factor invariance:
),(),( txetx i
It is quite a surprise this phase invariance is linked to EM gauge invariance when the phase is time dependent.
),(),( ),( txetx txie
),( txAA
This space-time dependent phase transformation is not an invariance of QM unless it’s coupled with EM gauge transformation!
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),(),( ),( txetx txie
),( txAA
),(),(),(),(),( txietxietxietxietxie eeeee
Derivatives of wave function doesn’t transform like wave function itself.
),(),(),( txietxietxie eieeieieAeieA
ieAeieA txie ),(
2
2
1
mti
Wave Equation is not invariant!
But if we put in A and link the two transformations:
This “derivative” transforms like wave function.
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AieeAie txie
),(
ie
teie
ttxie ),(
eAie
mti
2
2
1
2
2
1Aie
mie
ti
2
2
1Aie
meie
tei ieie
In space and time components:
The wave equation:
can be written as
It is invariant!
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),(),( ),( txetx txie
),( txAA
ieAeieAD txie ),(
Your theory would be easily invariant.
This combination will be called “Gauge Transformation”It’s a localized phase transformation.
Write your theory with this “Covariant Derivative”.
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There is a duality between E and B.
Without charge, Maxwell is invariant under:
Maybe there exist magnetic charges: monopole
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Magnetic Monopole
0 A
The curl of B is non-zero. The vector potential does not exist.
03 AxdadB
If A exists,
there can be no monopole.
But quantum mechanics can not do without A.
Maybe magnetic monopole is incompatible with QM.
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But Dirac did find a Monopole solution:
sin
cos1
r
gA
rr ArA
rrrA
r
A
r
AA
rrA
1
sin
11
sinsin
1
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Dirac Monopole
sin
cos1
r
gA
It is singular at θ = π. Dirac String
It can be thought of as an infinitely thin solenoid that confines magnetic field lines into the monopole.
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sin
cos1
r
gA
Dirac String doesn’t seem to observe the
symmetry
But a monopole is rotationally symmetric.
In fact we can also choose the string to go upwards (or any direction):
sin
cos1'
r
gA
They are related by a gauge
transformation!
It has to!
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Charge Quantization
sin
cos1
r
gA
Since the position of the string is arbitrary, it’s unphysical.
4eg
Since the string is unphysical. 14 iege
g
ne
2
BadeAade
Using any charge particle, we can perform a Aharonov like interference around the string. The effects of the string to the phase is just like a thin solenoid:
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Finally…. Feynman Rules for QED
e
e
e
e
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4-columns4-rows
4 ˣ 4 matrices
4 ˣ 4 matrices
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1 ˣ 1 in Dirac index
Dirac index flow,from left to right!
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kp
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'kp
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using
The first term vanish!
Photon polarization has no time component.
0 ump
0),( spump
The third term vanish!
Numerator simplification
In the Lab frame of e
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Denominator simplification
in low energy
Assuming low energy limit:
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0k
2nd term:
k1
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00 00
Amplitude squared
Finally Amplitude:
00
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02 kkk
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旋轉帶電粒子所產生之磁偶極
磁偶極矩與角動量成正比
Lm
epr
m
eervr
vr
eriiA
222222
Lm
e 2
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Sm
egS
m
es
2
帶電粒子自旋形成的磁偶極
Anomalous magnetic moment
12theory 108.89.11596521871
2
1 g
12experi 103.44.11596521881
2
1 g