Top page (correct Bohr model including the two-electron atoms)

Anomalous magnetic moment in QED.

Special relativity is wrong.

- Propagator of Klein-Gordon field.
- Propagator of Dirac field .
- Propagators of photons and strange " zero " component ?

*(Fig.1) Internal lines = propagator.*

Propagator is a main tool in calculating various values in quantum electrodynamics.

QED propagators were introduced by changing classical Lorentz force into relativistic interaction.

This means if special relativity is wrong, these interactions using propagators are wrong.

In this section, we explain feynman propagators in various fields ( scalar, Dirac, and electromagnatic fields.)

To understand propagators, we need to understand the basic quantum field theory first.

So first read this page (if you have not read ).

The plain wave solution of real Klein-Gordon field is

*(Eq.1)*

Caution: here we use (-1, 1, 1, 1) version metric tensor.

If you want to use (1, -1, -1, -1) version, change notation like ( kx → -kx, k^{2} → -k^{2} ).

The **meanings of them are completely the same**. (Only notations are different.)

( See also appendix. )

Propagator ( Green function ) of the scalar field is expressed as

*(Eq.2)*

where θ(t-t') means **Heaviside step function** of

*(Eq.3)*

This is used for time-ordered.

In the step function (= θ ) of Eq.3, when (t-t') is positive, θ becomes "1", and when t-t' is negative, θ becomes zero.

So Eq.2 means that the scalar particles are *created first* in the vacuum and *later annihilated*.

( Eq.1 includes "creation" ( *a ^{†}* ) and "annihilation" (

Of course, as shown in Eq.1, propagator contains

( So the propagator itself is NOT a real thing. But QED uses this concept to get values. )

Using complex integral, step function can be expressed as

*(Eq.4)*

and

*(Eq.5)*

where we make the exponential function parts the same.

--------------------------------------------------

First, we prove Eq.4.

If you want to skip this proof, please proceed to the next section.

To prove Eq.4, we have to use Cauchy's residue theorem.

In the residue theorem, only the coefficient of 1/(z-a) is left by the complex integration.

*(Eq.6)*

and

*(Eq.7)*

where integral is counterclockwise closed contour.

And when the closed contour doesn't contain "pole", Eq.7 becomes zero.

*(Eq.4)*

In Eq.4, when t-t' is positive ( t-t' > 0 ), the exponential parts of Eq.4 becomes zero in the region of ω = -i∞, as follows,

*(Eq.8)*

So we can treat the integration from -∞ to +∞ along the real line as the **complex integration** of Fig.2.

*(Fig.2) Step function ( complex integration ) of Eq.4. ( t-t' > 0 )*

As shown in Fig.2, "pole" exists in the minus imaginary area, so it doesn't become zero.

Expanding Eq.4, the coefficient c_{-1} becomes " 1 ", as follows,

*(Eq.9)*

Fig.2 is **clockwise** closed contour, so the result becomes opposite ( 1 → -1 ).

*(Eq.10)*

As a result, when t-t' > 0, the step function θ becomes "1",

*(Eq.11)*

When t-t' is negative ( t-t' < 0 ), the exponential part of Eq.4 becomes zero, as ω → i∞,

*(Eq.12)*

So the integration of Eq.4 is equal to the **counterclockwise** closed contour in the complex field,

*(Fig.3)*

As shown in Fig.3, this closed contour doesn't contain a pole.

So the step function θ becomes zero, when t-t' is negative.

Next we think about the step function of Eq.5.

*(Eq.5)*

*(Fig.4) Step function of Eq.5.*

In Eq.5, "pole" exists in the **plus** imaginary area, as shown in Fig.4.

In the same way as Eq.4, we can get the results of

*(Eq.13)*

---------------------------------------------

*(Eq.1) Klein-Gordon wavefunction.*

The creation and annihilation operators of Eq.1 satisfy the following commutation relation.

*(Eq.14)*

Using two different wavefunctions of Eq.1, we calculate

*(Eq.15)*

Due to the relation between the vacuum of both sides and annihilation operators,
only aa^{†} term is left

*(Eq.16)*

Using the commutation relation of Eq.14, the result of Eq.15 becomes

*(Eq.17)*

In the same way, exchanging x and x' of Eq.17,

*(Eq.18)*

In Eq.18, ω_{k} is an *even function* of k_{i}, as follows,

*(Eq.19)*

So if k_{i} values are determined, ω_{k} is also determined, and it's always *positive*.

Changing the sign of 1-3 component k_{i} ( k → -k ) in Eq.18, the exponential function of momentum ( e^{ikx} ) parts become the same as Eq.17, as follows,

*(Eq.20)*

Using Eq.4 and Eq.17, the first term of Eq.2 is

*(Eq.21)*

From Eq.5 and Eq.18 (Eq.20), the second term of Eq.2 is

*(Eq.22)*

Changing ω to k^{0}, as follows,

*(Eq.23)*

As a result, the **Klein-Gordon propagator** of Eq.2 is the sum of Eq.21 and Eq.22, as follows,

*(Eq.24)*

This is what is called "**propagator**" (= Green's function ) of scalar field.

And the scalar propagator is a *basis* for other fields, so it is very important.

In QED, Dirac propagator is mainly used. So next we explain Dirac field.

As shown on this page, Dirac's solutions are,

*(Eq.25)*

And its conjugate transpose (× γ^{0}) is

*(Eq.26)*

And the particle ( c, c^{†} ) and antiparticle ( d, d^{†} ) satisfy the **anticommutation** relations of

*(Eq.27)*

Other combinations become zero at the right hand sides.

**Dirac propagator** is expressed as

*(Eq.28)*

Dirac field satisfies *anticommutation* relation, so the "**minus**" sign is added in the second term of Eq.28.

Considering annihilation by the vacuum at both ends, only "cc^{†}" is left.

*(Eq.29)*

From Eq.29, we get

*(Eq.30)*

where anticommutation of Eq.27 is used.

The important point is that matrix parts (= u ) of Eq.30 can not be numbers, as follows,

*(Eq.31)*

To put it simply, the two matrices u **only stand in line** in Eq.31.

The state of Eq.31 can be expressed as ,

*(Eq.32)*

If two matrices u of Eq.32 only stand in line (not united into number),
when we multiply Eq.32 by another u(k) from right side,

*(Eq.33)*

where we use the relation of

*(Eq.34)*

About the relations of "u", see this page.

Eq.33 is equal to

*(Eq.35)*

Eq.35 is equal to Eq.5-12 of this page, which matrix u(k) needs to satisfy.

As a result, Eq.32 is proved.

In the same way, we can define

*(Eq.36)*

Adding another v(k) from the right side of Eq.36, we get

*(Eq.37)*

Eq.37 is equal to Eq.5-13 of this page.

Using Eq.32, Eq.30 becomes

*(Eq.38)*

In calculating the second term of Dirac propagator (Eq.28), " c^{†} c " term is annihilated.

*(Eq.39)*

where " d d^{†} " term is left.

*(Eq.40)*

As you notice, the two matrices part v(k) of Eq.40 is **a row matrix × a column matrix**.

So Eq.40 is completely different from Eq.30.

To use Dirac's propagator, we have to make them the same kind thing "artificially".

Using Eq.36, Eq.40 becomes

*(Eq.41)*

Note that coefficient p_{μ} of Eq.41 originally means Dirac's *differential operators*.

So if the exponential function changes (at the final form ), they also change as

*(Eq.42)*

So basically, Dirac propagator resembles Klein-Gordon propagator of Eq.24 except for their coefficients, as follows,

*(Eq.43) Klein-Gordon propagator.*

*(Eq.44) Dirac propagator.*

Comparing Eq.43 and Eq.44, Dirac propagator is

*(Eq.45)*

Eq.45 is Dirac's propagator, which is often used in QED.

As shown on this page, Lagrangian of the electromagnetic field in the vacuum can be expressed as

*(Eq.46)*

As shown in Eq.46, if μ and ν are the same, F_{μν} becomes zero.

So the zero component of the **canonical momentum** is zero, as follows,

*(Eq.47)*

But as is explained later, if the zero component of photon can not be defined, the propagator calculation becomes **impossible**.

So we have to produce the zero component canonical momentum "compulsorily".

We define Lagrangian which includes the zero component, as follows,

*(Eq.48)*

This is called " **Lagrangian of Feynman gauge** ".

( Here we use usual SI unit, so magnetic permeability μ_{0} is added. )

The second term of Eq.48 vanishes if we consider Lorentz gauge condition of

*(Eq.49)*

So the cannonical momentum of the zero component is, ( x^{0} = ct )

*(Eq.50)*

where Lagrangian involved is

*(Eq.51)*

And the canonical momentums of other components ( j = 1, 2, 3 ) are

*(Eq.52)*

where Lagrangian involved is

*(Eq.53)*

μ and ν of Eq.48 contain all 0 - 3 components.

So the coefficient 1/4 (and 1/2) is cancelled out.

Like de Broglie's relation, we can define the **commutation** relations of

*(Eq.54)*

If so,

*(Eq.55)*

where **g ^{μν}** appears.

Other combinations are all zero.

Using Eq.55, we can prove Eq.54, as follows,

*(Eq.56)*

and

*(Eq.57)*

Here (-1,1,1,1) version metric tensor g is used.

See this page.

As shown on this page, the solution of A_{μ} (x), which satisfies Maxwell equation is

*(Eq.58)*

The time derivative of Eq.58 is

*(Eq.59)*

And to satisfy Eq.55, the following relations must be valid.

*(Eq.60)*

Other combinations are all zero.

As shown in Eq.60, only the zero component of photon particles has "negative" energy ( = "**negative**" number ). ( g_{00} = -1 )

So this zero component photon is **NOT** a real thing.

Of course, if we consider Lorentz gauge condition of Eq.49, this zero component photon can be removed.

This strange property is caused by the **minus sign** of the second term in Eq.48.

( This minus sign is artificially introduced. )

Why we define the minus sign in Eq.48 (and Eq.60) ?

In fact, if we don't define the minus sign of Eq.60, we can NOT calculate the g-factor.

In other words, to calculate g-factor, the strange relation of Eq.60 is introduced.

This is explained later.

Next we prove Eq.55 using the relations of Eq.60.

Substituting Eq.58 and Eq.59 into Eq.55,

*(Eq.61)*

where

*(Eq.62)*

where time t is the same.

From Eq.61 and Eq.62 we have

*(Eq.63)*

Eq.55 is gotten.

Eq.58 resembles Eq.1.

So propagators of photons are almost same as those of Klein-Gordon field of Eq.24, except for *zero photon mass*

So the **photon propagators** are

*(Eq.64)*

So only in the zero component photon, the sign becomes **opposite**.

And photons are **vectors**, so propagators between photons of different components are *zero*, as shown in Eq.64.

Here we explain Hamiltonian of Coulomb gauge briefly.

When the scalar potential (= A^{0} ) is zero,

*(Eq.65)*

Hamiltonian is

*(Eq.66)*

Of course, in this webpage, we use Lorentz gauge, so we have to consider "minus" energy of the zero component photons in addition to this.

2013/1/20 updated. Feel free to link to this site.