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

Strange "spin" is NOT a real thing

What are "real" elementary particles ?

- What are Left-handed and right-handed ?
- S-matrix and cross section.
- Calculation of amplitude in QED reaction.

First we explain about "left-handed" and "right-handed" spinors.

Can we visualize these states ??

As shown in this page, u(p) and v(p ) of Dirac's spinors are

*(Eq.1)*

And

*(Eq.2)*

where

*(Eq.3)*

Here we use Pauli matrices of

*(Eq.4)*

And γ matrices are

*(Eq.5)*

*(Eq.6)*

From Eq.3, Eq,4, Eq.5 and Eq.6, the numerator of Eq.1 (of u_{1}(p) ) is

*(Eq.7)*

where we use (-1, +1, +1, +1) version.

So u_{1}(p) of Eq.1 is

*(Eq.8)*

In the same way, u_{2}(p) of Eq.1 is

*(Eq.9)*

v_{1}(p) of Eq.1 is

*(Eq.10)*

v_{2}(p) of Eq.1 is

*(Eq.11)*

Here we use the γ matrix of

*(Eq.12)*

When this γ^{5} acts on u_{1}(p),

*(Eq.13)*

When the mass m is zero, and the particle moves in the z direction,

*(Eq.14)*

Substituting Eq.14 into the last line of Eq.13, and paying attention to σ3

*(Eq.15)*

From Eq.15,

*(Eq.16)*

If we define

*(Eq.17)*

The wavefunction can be divided like

*(Eq.18)*

Using the relations of

*(Eq.19)*

and

*(Eq.20)*

Dirac equation is

*(Eq.21)*

Eq.21 means when the mass of fermion is zero, only left-handed (or right-handed ) particle is left.

(When the fermions get mass, the left-handed and right-handed particles are mixed, they say. )

But unfortunately, we **can not visualize** these left-handed (or right-handed ) states.

To get a cross section in QED. we need to calculate the probability amplitude in Feynman diagram.

First I advise you these are very "**mathematical**" than physical.

In this web page we use the notations and relations of this page and this page.

*(Ap.1)*

The wavepacket representing some state can be expressed as

*(Eq.22)*

where the sum of all probabilities adds up to 1,

*(Eq.23)*

where we use the relations and | k > of Ap.1.

The probability (= a square of amplitude ) of the transition ( A+B → 1+2 ) is

*(Eq.24)*

where iT includes the **interaction term** of S-matrix.

S-matrix is an unitary operator, which causes transition during some time.

So, we can define

*(Eq.25)*

The delta functions of **energy and momentum conservation** are always included in Eq.25.

(This is calculated later.)

*(Fig.1)*

As shown in Fig.1, the wavepacket of B is supposed to be uniformly distributed in impact parameter b (= two dimensional plane).

So the number of scattering event (N) and the cross section (σ) are

*(Eq.26)*

where we integrate by b in some plane.

And n_{B} means the number density of B.

Using Eq.22, Eq.24, Eq.25, and Eq.26, the (differential) cross section is

*(Eq.27)*

where exponential function of b means de Broglie's waves (= probability amplitude ) of b.

Eq.27 includes six momentum integration of A ( or B ) ( 3 × dk_{A}, and 3 × dk_{A}* ).

First we integrate, as follows,

*(Eq.28)*

Here we use the next formula of delta function.

*(Eq.29)*

When f(k) is

*(Eq.30)*

f '(k) becomes

*(Eq.31)*

As a result, if we consider E = mc^{2} approximately and use Eq.29 and Eq.31, Eq.28 becomes

*(Eq.32)*

where velocity v = p/m, and the momentum p = ħ k.

Next in Eq.27, we do the integration of

*(Eq.33)*

And we do the next integration using the results of Eq.33.

*(Eq.34)*

And in Eq.27, we do the next integrations of

*(Eq.35)*

From Eq.32, Eq.33, Eq.34, and Eq.35, Eq.27 becomes

*(Eq.36)*

Considering Eq.23, Eq.36 becomes

*(Eq.37)*

And then we integrate by momentum (wave number k) of the final state 2, as follows,

*(Eq.38)*

The integration of dk^{3} of the final state 1 can be expressed using a solid angle, as follows,

*(Eq.39)*

Using the formula of Eq.29,

*(Eq.40)*

Here we suppose the center of mass frame,

*(Eq.41)*

where E_{cm} is the total energy.

From Eq.41, Eq.40 becomes

*(Eq.42)*

From Eq,38, Eq.39, Eq.40, and Eq.42, Eq.37 becomes

*(Eq.43)*

Using Eq.41,

*(Eq.44)*

Using the approximation of

*(Eq.45)*

where | v_{A} - v_{B} | is 2c, approximately.

Using Eq.45, Eq.44 is

*(Eq.46)*

This is the differential cross section in relation to a solid angle.

Here we confirm the **dimention** of Eq.46 is a square of meter.

Basically, the dimension of Dirac delta function is a **reciprocal** of the number in it, as follows,

*(Eq.47)*

The dimensions of the wave number (k), energy (E), light speed (c), and Planck constant (h) are

*(Eq.48)*

So the dimension of **M** of Eq.25 is

*(Eq.49)*

where the exponential function of S matrix has no dimension.

And from Ap.1, one pair of |k> is the dimension of E/k^{3}.

So the dimension of Eq.46 is

*(Eq.50)*

As a result, the dimension of differential cross section is **a square of meter**.

Next we calculate the amplitude of the most important reaction of QED.

In Fig.2, the high energy electron (e-) and positron (e+) crash into each other and changes into virtual photon.

And then the virtual photon (= photon propagator) changes into muon (μ-) and antimuon(μ+).

This calculation process applies to the generation of quark and antiquark, too.

Of course, these final products are unstable, so they changes into the stable electron, γ rays, and neutrino soon.

About Feynman diagram and concrete calculation methods, see also this page.

Here we suppose the momentums of the initial electron and positron are P_{A} and P_{B}.

(So their wave numbers are k_{A} and k_{B}.)

And the momentums of the final products muon and antimuon are P_{1} and P_{2}.

Using Dirac wavefunction of

*(Eq.51)*

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

*(Eq.52)*

And the photon propagator of

*(Eq.53)*

The probability amplitude of Fig.2 is

*(Eq.54)*

where we use < k | and | k > of Ap.1.

From Eq.25, M is

*(Eq.55)*

And a squared matrix element is

*(Eq.56)*

Using the trace technique of Eq.49-Eq.51 of this page. Eq.56 is

*(Eq.57)*

where we use the relations of u and v in Ap.1.

And the final states can take all spin states, but the inital states can take one of up and down spins.

So 1/2 × 1/2 = 1/4 is added. (Though this is a little difficult to imagine.)

From Eq.55, Eq.57 is

*(Eq.58)*

Here we use the next trace formulas of gamma matrices.

*(Eq.59)*

This homepage uses (-1, 1, 1, 1) version metric tensor.

The trace of odd γ matrices is zero, so

*(Eq.60)*

In the similar way,

*(Eq.61)*

Here the mass of the electron is supposed to be zero ( **m _{e} = 0** ), and do the calculations such as

and

and

We can get

*(Eq.63)*

Here the momentums and energies of the electron (A) and positron (B) are supposed to be

*(Eq.64)*

And those of muon (1) and antimuon (2) is

*(Eq.65)*

where the angle between momentum of muon and z axis is θ, as follows,

*(Eq.66)*

The momentum and energy of the virtual photon is the sum of the electron and positron, so from Eq.64,

*(Eq.67)*

where k is the wave number of photon.

From Eq.64 to Eq.66

*(Eq.68)*

Using Eq.67 and Eq.68, Eq.63 is

*(Eq.69)*

Here we define

*(Eq.70)*

Using Eq.70, Eq.44 is

*(Eq.71)*

Substituting Eq.69 into Eq.71

*(Eq.72)*

where we use the fine structure constant α

This is a little complicated.

But the important point is that the **cross section is proportial to each charge**, which can be said also in Rutherford scattering.

As you see in the trace of γ matrices, these are only "mathematical" things. ( NOT physical objects at all.)

2011/12/25 updated. Feel free to link to this site.