Appendix of QED

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Conversion of Notations from ( -1, +1, +1, +1 ) to ( +1, -1, -1, -1 ) versions.

On this web site, we use ( -1, +1, +1, +1) version of metric tensor.
So only the zero component of the four-vector changes by the index positions.
If you want to change to ( +1, -1, -1, -1 ) version of metric tensor, do as shown in Eq.1 and Eq.2.

(Eq.1)

( xμ, pμ, and kμ = four-vectors )

And the matrices are
(Eq.2)

Basically, the meanings of the equations themselves are completely the same in both ( -1, +1, +1, +1 ) and ( +1, -1, -1, -1 ) versions.
Only the notations are different in them.
So change only the signs ( + → -, or - → +) at proper places, if you change versions.

Calculation of Eq.85 of QED g-factor.

Here we prove conversion of the equations in Eq.85 of this page.

(Eq.3) = (Eq.85)

where
(Eq.4)

In Eq.4, "q" is photon, "p" is electron, and "k" is in the electron's propagator.
Here we use " l " of Eq.80 on this page,
(Eq.5)

Using Eq.5, the last terms of Eq.3 ( third line ) can be expressed as
(Eq.6)

As shown on this page (Eq.80), the denominator D3 contains l2.
So the denominator is a even function of l.
This means that if the numerator is an odd function, integration becomes zero, as follows,
(Eq.7)

Actually, Eq.7 takes the power of infinity.
Remember that the new variable l is made by shifting k by a "definite" constant. (See Eq.5.)
If the integral region of k is NOT from -∞ to +∞, this l is neither odd nor even function.
So the convenient tool of Eq.7 depends on infinity.

Using Eq.7, the last term of Eq.6 becomes zero, so the third line of Eq.3 is
(Eq.8)

Using Eq.4, the first line of Eq.3 becomes
(Eq.9)

As a result, Eq.8 proves to be equal to the last three terms of Eq.9.
As you may notice, this QED method is wrong math, as shown in Fig.10' and Fig.10'' in this page.

Using Eq.5, the second term of Eq.3 ( second line ) becomes
(Eq.10)

One of Eq.10 ( right side ) can be expressed as,
(Eq.11)

where "a" of left is changed into "b".
So Eq.11 proves to be equal to the remaining term of Eq.9.

This means the remaing terms of Eq.3 ( second line ) must be zero.
( So these "zero" terms are added intentionally. )
Other terms of Eq.10 is
(Eq.12)

Eq.12 is an odd function of l, so becomes zero by integration.

In the same way, the following term of Eq.10 becomes zero.
(Eq.13)

The last term left in Eq.10 is
(Eq.14)

Here we use the formula of
(Eq.15)

Eq.15 can be proved using the relation of
(Eq.16)

Eq.16 means
(Eq.H-1)

For example, γ0γ0 = I ( = 4× 4, identity matrix ),
(Eq.H-2)

And γ matrices satisfy anticommutation relation like,
(Eq.H-3)

Next we prove Eq.15. For exmaple, when μ of Eq.15 is zero,
(Eq.H-4)

Using the relations of Eq.16 and Eq.H-1, the left side of Eq.H-4 is
(Eq.H-5)

where the anticommutation relation is used.

Eq.H-5 is equal to
(Eq.H-6)

where
(Eq.H-7)

As a result, we can prove Eq.H-4.
( In the same way, Eq.15 proves to be true. )

And we also use the formula of
(Eq.17)

When μ is different from ν, it means "odd" functions, so Eq.17 becomes zero by integration.

Adding gμν to both sides of Eq.17,
(Eq.17-2)

where we use
(Eq.17-3)

So Eq.17 is right.

Using Eq.15, Eq.14 becomes
(Eq.18)

where we change the index "a" of the right γ and k into b.

We can change like ( k → l ) of the term which contains one l and one k, as follows,
(Eq.20)

Because, the difference between l and k vanishes due to the odd function property of another l.

So the first term of Eq.18 is (using Eq.17),
(Eq.21)

where summing 0-3 components with respect to c ( and b).
And only the case of μ = b = c is left, so gμcgbc = 1.

And the second term of Eq.18 becomes
(Eq.22)

where γaγb = -gab.
And Eq.17 and Eq.17-3 are used.

Summing the first term of Eq.3 (second line), Eq.21, and Eq.22,
(Eq.22-2)

As a result, we can prove Eq.3.

(Fig.1) Summary of Eq.3 = Eq.10.

Calculation of Eq.88 of QED g-factor.

Next we try to prove Eq.85 (= Eq.3 ) is equal to Eq.88 (= Eq.23 ) of this page.

(Eq.23) = (Eq.88)

Here we need to use the following relations which Dirac equation satisfies. (See this page (Eq.5-12).)
(Eq.5-12)

So,
(Eq.24)

The momentum and energy are conserved, as follows,
(Eq.25)

In Eq.25, p and p' are "in" and "out" electrons, respectively.
The important point is that "q" becomes virtual photon which violates relativity, if Eq.5-12 and Eq.25 are satisfied.
From Eq.24 and E.25,
(Eq.26)

Using Eq.24 and Eq.25, the second term of Eq.23 (1) becomes,
(Eq.27)

(Eq.15)

Using Eq.15 and Eq.24, the following term, which is included in the last line of Eq.27, changes into
(Eq.28)

And another term, which is included in the last line of Eq.27, becomes
(Eq.29)

As you notice, they try to move p to the right of γμ and p' to the left, and change them into "-m".
So if the virtual photon "q" is zero, p' = p, we don't need to move them, when we try to change them into "-m".
And another term, which is included in the last line of Eq.27 is
(Eq.30)

The last one term of Eq.27 is
(Eq.31)

Using Eq.15 and Eq.24, one term included in Eq.31 is
(Eq.32)

And using Eq.25, another term of Eq.31 is
(Eq.33)

where we use the relations of
(Eq.34)

And another mass shell condition is
(Eq.35)

From Eq.35,
(Eq.36)

As I explain later, we do q2 = 0, so, pq = 0, in this case, virtual photon q becomes zero ( q = 0 ).
In this case, g-factor cannot be fixed at a single value.
Summing all calculation values, we can prove Eq.23.
( Using p' = p + q, and x + y + z = 1, and of course, we need to add the remaining terms of Eq.23.)

(Fig.2) Summary of Eq.23

Next we sum up the above values to prove Eq.23.
First we calculate the values outside the γμ term (= vertex ).
From the second line of Eq.23-1, we pick up the terms which don't include γ,
(Eq.37)

From Eq.28 and Eq.31, pμ temrs are
(Eq.38)

From Eq.37 and Eq.38, the sum of pμ temrs are
(Eq.39)

From Eq.29 and Eq.32 (= one of Eq.31 ), the sum of p'μ terms are
(Eq.40)

where we use p' = p + q.

From Eq.37 and Eq.40, the sum of qμ terms are
(Eq.41)

where we use x + y + z = 1.

From Eq.39, Eq.40 and Eq.41, we can get the second line of Eq.23-2.

Next, from Eq.28, Eq.29, Eq.30, Eq.32 (= one of Eq.31), Eq.33 (= one of Eq.31), and the second line of Eq.23-1, the sum of m2 terms in the γ is
(Eq.42)

From Eq.33 (= one of Eq.31 ), q2 term in the γ is
(Eq.43)

As a result, the first line of Eq.23-2 is gotten.

Lorentz invariance in propagator.

This section is moved from Eq.4-7 (Eq.4-4) to Eq.4-11 of this page.

(Eq.4-4)

For example, when we calculate the next propagator, which means the transmission of the particle,
(Eq.4-7)

where Eq.4-19 is used.
(Eq.4-19)

Two square roots of ω in Eq.4-4 change to
(Eq.4-8)

After integrating with respect to one d3k , so only another k is left here.

We use the following formula in delta function.
(Eq.4-9)

Taking the formula of Eq.4-9, we can get
(Eq.4-10)

where we use Eq.4-1.
(Eq.4-1)

Adding d4p to Eq.4-10, and integrating Eq.4-10 with respect to p0, we get the equation of
(Eq.4-11)

where θ means "step function".

Eq.4-11 contanins the integration of d4p and Eq.4-1, so this is Lorentz invariant.
( Of course, integral region must be from minus infinity to plus infinity to make d4p Lorentz invariant. )
So Eq.4-8 is Lorentz-invariant, too.
As you notice, we can change the coefficients such as c and h freely, keeping Lorentz invariance.
So we can adjust these values "artificially" to get the correct energy values.

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