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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
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
The proper way to interpret KG equation is it is actually a field equation just like Maxwell’s Equations.
Consider we try to solve this eq as a field equation with a source.
)(2 xjm
We can solve it by Green Function.
')',()',( 2 xxxxmxxG
G is the solution for a point-like source at x’.
By superposition, we can get a solution for source j.
)'()',(')()( 40 xjxxGxdxx
Green Function for KG Equation:
')',()',( 42 xxxxGmxxG
By translation invariance, G is only a function of coordinate difference:
)'()',( xxGxxG
The Equation becomes algebraic after a Fourier transformation.
)(
~
2)'( )'(
4
4
pGepd
xxG xxip
1)(~22 pGmp
)'(4
44
2)'( xxipe
pdxx
22
1)(
~
mppG
This is the propagator!
'x x
Green function is the effect at x of a source at x’.
That is exactly what is represented in this diagram.
KG Propagation
The tricky part is actually the boundary condition.
B
A
B
A
For those amplitude where time 1 is ahead of time 2, propagation is from 1 to 2.
C1
2
B
A
B
A
For those amplitude where time 2 is ahead of time 1, propagation is from 2 to 1.
C
1
2
is actually the sum of the above two diagrams!
To accomplish this, 22
1)(
~
mppG
imp
pG
22
1)(
~
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.
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
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!
Dirac find it’s possible for 4 by 4 matrices
We need:
that is
He found a set of solutions:
Dirac Matrices
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.
The above could be done for 2 by 2 matrices if there is no mass.
Massless fermion contains only half the degrees of freedom.
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
0222 cmp
0p E
There are two solutions for each 3 momentum p (one for +E and one for –E )
xpitipipx eaeax
0
)(
Plane wave solutions for KG Eq.
ipxipxipx eameapeap 2220
2220 mpp
p
xpiiEtxpiiEt
p
ipxipx eaeaeaeax
)(
Expansion of a solution by plane wave solutions for KG Eq.
p
xpiiEtxpiiEt
p
ipxipx ebeaebeax
)(
p
xpiiEtxpiiEt
p
ipxipx eaeaeaeax
)(
If Φ is a real function, the coefficients are related:
Multiply on the left with
mcp
0222 ucmp 0222 cmp 0p E
There are 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.
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
0p
0p
0p
0p
)0,(
mp
020
00
B
A
u
u
m
0Bu uA is arbitrary
1
0,
0
1Au
Two solution (spin up spin down)
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)
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
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.
Bilinear Covariants
Ψ transforms under Lorentz Transformation:
Interaction vertices must be Lorentz invariant.
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:
Maybe you need to change some of the signs:
It turns out to be right!
We can define a new adjoint spinor:
is invariant!
A
p
ipxipx evbeuax
)(
)( 1pe
0)( 11 upeau p
p
ipxipx euaevbx
)(
0u
)(0 313133peuuuau ppppp
Feynman Rules for external lines
Find the Green Function of Dirac Eq.
')',()',( 4 xxIxxmcGxxGi
')',()',( 42 xxxxGmxxG
Now the Green Function G is a 4 ˣ 4 matrix
')',()',( 4 xxIxxmcGxxGi
How about internal lines?
')',()',( 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
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
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.
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.
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?
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!
),(),( ),( 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.
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!
),(),( ),( 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”.
There is a duality between E and B.
Without charge, Maxwell is invariant under:
Maybe there exist magnetic charges: monopole
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.
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.
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!
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:
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