Top page ( Quantum mechanics and Einstein relativity are wrong )

QED anomalous magnetic moment or g-2 factor uses wrong math

Relativistic quantum field theory is unrealistic.

- Conversion of Notations from ( -1, +1, +1, +1 ) to ( +1, -1, -1, -1 ) versions.
- Calculation of Eq.85 of QED g-factor.
- Calculation of Eq.88 of QED g-factor.

On this web site, we use ( -1, +1, +1, +1) or (-+++) version of metric tensor.

So only the sign of the zero component of the four-vector changes by the index positions.

If you want to change to ( +1, -1, -1, -1 ) or (+---) version of metric tensor or signature, do as shown in Eq.1, Eq.2 and Fig.1-.

*(Eq.1)*

( x^{μ}, p^{μ}, and k^{μ} = four-vectors )

And the matrices are

*(Eq.2)*

Basically, the **meaning** of the equations is **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 - → +) when encountering the letters with subindex like p_{μ} → -p_{μ}, and p^{2} (= p^{μ}p_{μ} ) → -p^{2}.

*(Fig.1) time (= t or x ^{0} ) and space (= x^{j} or x,y,z ) variables*

This website uses the metric signatrue of (-1,1,1,1) or (-+++) where only the sign of zero component vector (= time ) with subindex changes (= ct = x^{0} = -x_{0}, where x_{0} = -ct ), while the space-component vectors do not change ( x^{j} = x_{j}, j = 1,2,3 ) their signs regardless of whether subindex or superscript in the spacetime four vectors ( this p.3, this p.3 ).

If you want to switch from (-1,1,1,1) or (-+++) into (1,-1-1-1) or (+---) version of metric signature, change the sign only of the space coordinate or 1-3 components ( x^{j} = -x_{j} j = 1,2,3 ) instead of the time variable or zero-component of four vectors.

*(Fig.2) "c" is light speed.*

Fig.2 is the relativistic energy (= E = cp^{0} ) and momentum (= p^{1}, p^{2}, p^{3} or p_{x}, p_{y}, p_{z} ).

If you want to switch from this website's (-1,+1,+1,+1) or (-+++) version of metric signature into (1,-1,-1,-1) or (+---) version, change only the sign of letters with subscript like p_{μ} → - p_{μ}

*(Fig.3)*

Using Fig.2, relativistic energy-momentum-mass relation is expressed like Fig.3.

In this notation rule, when the same letter is used twice in one term like p_{μ}p^{μ} or p^{2}, it means the sum of all μ = 0,1,2,3 components (= even when the summation simbol Σ is not added ) like the upper figure ( this p.9, this p.3-4, this p.2 ).

*(Fig.4)*

In the derivative (= ∇ or ∂ ) or differential operators, when the index is at the lower position (= subindex or subscript ), it means the common normal differential operator.

When the index position is at the higher position (= superscript ), the sign of only the zero component (= time ) differential operator switches into the negative, while the signs of all other space derivatives remain the same irrespective of the index position in this website (-1,1,1,1) version of metric signature.

In the (1,-1-1-1) version of metric signature, the sign of the differential operator with superscript besomes the opposite where only space differential operators become negative, while the sign of the zero-component time differential operator does not change irrespective of the index position ( this p.2 ).

*(Fig.5)*

*(Fig.6)*

Relativistic energy (= E = ℏω where ω is relativistic fictitious angular frequency, ℏ = h/2π where h is Planck constant ) and momentum (= p ) can be expressed as differential operators together with the exponential function of the relativistic wavefunctions ( this p.2 ), based on de Broglie wavelength theory.

Here we use "four-vector potential (= this vector potential is an invisible fictitious potential introduced only for nonphysical theoretical calculations )" like current density, as follows,

*(Fig.7)*

In Fig.7, "A" is vector (magnetic) potential allegedly used for expressing the magnetic field B, and "φ" is *scalar (electric) potential* allegedly used for expressing the electric field E.

*(Fig.8) On this website (-1,1,1,1), only the zero-component changes its sign in its subindex.*

Relativistic Dirac equation uses nonphysical γ matrices like

*(Fig.9)*

where I means 2 × 2 **identity matrix**.

*(Fig.10)*

where σ_{j} mean 2 × 2 Pauli matrices of

*(Fig.11)*

γ^{0}γ^{0} = I ( = 4× 4, identity matrix ),

*(Fig.12)*

And γ matrices satisfy **anticommutation** relations like,

*(Fig.13)*

*(Fig.14)*

*(Fig.15)*

*(Fig.16)*

Using Fig.2 and Fig.4,
the relativistic Dirac equation can be expressed like Fig.16.

(+1,-1,-1,-1) version of Dirac equation is in this p.20, this p.2-3.

*(Fig.17)*

Using the metric tensor (= g_{μν} ), the relativistic momentum and energy can be expressed like Fig.17.

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

*(Eq.4)*

In Eq.4, "q" is an external virtual photon, "p" is an incoming electron, and "k" is the internal virtual electron's propagator.

Here we use " l " of Eq.80 on this page,

*(Eq.5)*

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

*(Eq.6)*

As shown on this page (Eq.80), the denominator D^{3} contains *l ^{2}*.

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 "finite" value. (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
this.

Using Eq.5, the second term of Eq.3 ( last equation ) becomes

*(Eq.10)*

One term of Eq.10-right-side equation can be expressed as,

*(Eq.11)*

where "a" of left is changed into "b" (= originally, the left and right sides of γ^{μ} are independent from each other in the variables "a" of γ^{a}k_{a}, so the left side "a" 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 ( last equation ) 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, γ^{0}γ^{0} = 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)*

*(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 ( this p.3-(13) ) of

*(Eq.17)*

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

Adding g_{μν} to both sides of Eq.17,

*(Eq.17-2)*

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 (= originally, independent from the left-side a ).

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 and Fig.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 g^{μc}g^{bc} = 1.

And the second term of Eq.18 becomes

*(Eq.22)*

where γ^{a}γ^{b} = -g^{ab}.

And Eq.17, Eq.16, and Eq.17-3 are used.

Summing the first term of Eq.3 (last equation), Eq.21, and Eq.22,

*(Eq.22-2)*

As a result, we can prove Eq.3.

*(Fig.18) Summary of Eq.3 = Eq.10.*

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)*

*(Eq.24)*

The momentum and energy are conserved, as follows,

*(Eq.25)*

In Eq.25, p and p' are "incoming" and "outgoing" electrons, respectively.

The important point is that "q" becomes **virtual** photon which violates relativity, if Eq.25 is 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 γ^{μ} 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" (= anomalous magnetic moment value can be freely changed, which is happening in this QED calculation ).

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 q^{2} = 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.19) Summary of Eq.23*

Next we sum up the above values to prove Eq.23.

First we calculate the values outside the γ^{μ} 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, p^{μ} temrs are

*(Eq.38)*

From Eq.37 and Eq.38, the **sum of p ^{μ} temrs** are

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

where we use p' = p + q.

From Eq.37 and Eq.40, the **sum of q ^{μ} terms** are

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 m ^{2} terms** in the γ is

From Eq.33 (= one of Eq.31 ), **q ^{2} term** in the γ is

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

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 d^{3}k , 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 d^{4}p to Eq.4-10, and integrating Eq.4-10 with respect to p^{0}, we get the equation of

*(Eq.4-11)*

where θ means "step function".

Eq.4-11 contanins the integration of d^{4}p and Eq.4-1, so this is Lorentz invariant.

( Of course, integral region must be from minus infinity to plus infinity to make d^{4}p 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.

2022/8/27 updated. Feel free to link to this site.