The QED Lagrangian density is

where

and

is the gauge covariant derivative and ( is the elementary charge, which is in atomic units, i.e. the electron has a charge )

is the electromagnetic field tensor. It’s astonishing, that this simple Lagrangian can account for all phenomena from macroscopic scales down to something like . So it’s not a surprise that Feynman, Schwinger and Tomonaga received the 1965 Nobel Prize in Physics for such a fantastic achievement.

Plugging this Lagrangian into the Euler-Lagrange equation of motion for a field, we get:

The first equation is the Dirac equation in the electromagnetic field and the second equation is a set of Maxwell equations () with a source , which is a 4-current comming from the Dirac equation.

The fields and are quantized. The first approximation is that we take as a wavefunction, that is, it is a classical 4-component field. It can be shown that this corresponds to taking the tree diagrams in the perturbation theory.

We multiply the Dirac equation by from left to get:

and we make the following substitutions (it’s just a formalism, nothing more): , , , to get

or:

This can be written as:

where the Hamiltonian is given by:

or introducing the electrostatic potential and writing the momentum as a vector (see the appendix for all the details regarding signs):

The right hand side of the Maxwell equations is the 4-current, so it’s given by:

Now we make the substitution , which states, that we separate the largest oscillations of the wavefunction and we get

We use the identity to get:

The constant factor in front of the Lagrangian is of course irrelevant, so we drop it and then we take the limit (neglecting the last term) and we get

After integration by parts we arrive at the Lagrangian for the Schrödinger equation:

The Dirac equation implies the Klein-Gordon equation:

Note however, the in the true Klein-Gordon equation is just a scalar, but here we get a 4-component spinor. Now:

We rewrite :

The nonrelativistic limit can also be applied directly to the Klein-Gordon equation:

Taking the limit we again recover the Schrödinger equation:

we rewrite the right hand side a little bit:

Using (see the appendix for details):

we get the usual form of the Schrödinger equation for the vector potential:

A little easier derivation:

and letting we get the Schrödinger equation:

We want to solve the equation:

(1)

with , where is time-independent part whose eigenvalue problem has been solved:

and is a small time-dependent perturbation. form a complete basis, so we can express in this basis:

(2)

Substituting this into (1), we get:

so:

Choosing some particular state of the Hamiltonian, we multiply the equation from the left by :

where . Using :

we integrate from to :

Let the initial wavefunction at time be some particular state of the unperturbed Hamiltonian, then and we get:

(3)

This is the equation that we will use for the perturbation theory.

In the zeroth order of the perturbation theory, we set and we get:

In the first order of the perturbation theory, we take the solution obtained in the zeroth order and substitute into the right hand side of (3):

In the second order, we take the last solution, substitute into the right hand side of (3) again:

And so on for higher orders of the perturbation theory — more terms will arise on the right hand side of the last formula, so this is our main formula for calculating the coefficients.

As a special case, if doesn’t depend on time, the coefficients simplify, so we calculate them in this section explicitly. Let’s take

so at the time the Hamiltonian is unperturbed and we are interested in the time , when the Hamiltonian becomes (the coefficients will still depend on the variable) and we do the limit (this corresponds to smoothly applying the perturbation at the time negative infinity).

Let’s calculate :

Taking the limit :

Substituting this into (2) evaluated for :

The sum is over all , similarly for the other sum. Let’s also calculate the energy:

To evaluate this, we use the fact that and :

Where we have neglected the higher order terms, so we can identify the corrections to the energy coming from the particular orders of the perturbation theory:

The incoming plane wave state is a solution of

with . E.g.

We want to solve:

The solution of this is:

where

is the Green function for the Schrödinger equation. is not unique, it contains both outgoing and ingoing waves. As shown below, one can distinguish between these two by adding a small into the denominator, that moves the poles of the Green functions above and below the -axis:

Both and are well-defined and unique. One can calculate both Green functions explicitly:

Assuming , we can taylor expand :

and simplify the result even further:

Note: both functions may be divided by the factor due to the momentum integration.

Let’s get back to the solution of the Schrödinger equation:

It contains the solution on both sides of the equation, so we express it explicitly:

and multiply by :

where is the transition matrix:

Then the final solution is:

and in a coordinate representation:

Plugging the representation of the Green function for in:

where the scattering amplitude is:

Where is the final momentum.

The differential cross section is defined as the probability to observe the scattered particle in a given state per solid angle, e.g. the scattered flux per unit of solid angle per incident flux:

where we used and

Let’s write the explicit formula for the transition matrix:

The Born approximation is just the first term:

We have

where the ground state of the noninteracting Hamiltonian is:

and the ground state of the interacting Hamiltonian is:

Then:

We can also write

where

Let’s write several common expressions for the ground state energy:

The last expression incorporates the dependence of explicitly. The vacuum amplitude is sometimes denoted by :

The two point (interacting) Green (or correlation) function is:

The limit of is tacitly assumed to make this formula well defined (sometimes the other way of writing the same limit is used). Another way of writing the formula above for the Green function in QM is:

Last type of similar expressions to consider is the scattering amplitude:

where the initial state is let’s say a boson+fermion and the final state a boson+antifermion:

This is just an example, the and states can contain any number of (arbitrary) particles.

The evolution operator is dimensionless:

So:

where is an arbitrary mass scale. Length unit is , so then

For the particular forms of the Lagrangians above we get:

so , and we get

Example: what is the dimension of in ? Answer:

In order to get the above units from the SI units, one has to do the following identification:

The SI units of the above quantities are:

The SI units are useful for checking that the , and constants are at correct places in the expression.

In general, the covariant and contravariant vectors and tensors work just like in special (and general) relativity. We use the metric (e.g. signature -2, but it’s possible to also use the metric with signature +2). The four potential is given by:

where is the electrostatic potential. Whenever we write , the components of it are given by the upper indices, e.g. . The components with lower indices can be calculated using the metric tensor, so it depends on the signature convention:

In our case we got and (if we used the other signature convention, then the sign of would differ and would stay the same). The length (squared) of the vector is:

where .

The position 4-vector is (in any metric):

Gradient is defined as (in any metric):

the upper indices depend on the signature, e.g. for -2:

and +2:

The d’Alembert operator is:

the 4-velocity is (in any metric):

where is the proper time, and is the velocity in the coordinate time . In the metric with signature +2:

With signature -2 we get . The 4-momentum is (in any metric)

where is the rest mass. The fluid-density 4-current is (in any metric):

where is the fluid density at rest. For example the vanishing 4-divergence (the continuity equation) is written as (in any metric):

Momentum () and energy () is combined into 4-momentum as

For the signature we get and .

For we get (signature -2):

comparing those two we get the following useful relations (valid in any metric):

the following relations are also useful:

For the signature we get:

So for example the Klein-Gordon equation:

can be for signature written as:

and for as:

Note: for the signature +2, we would get and .

For the minimal coupling we get:

and for the lower indices:

We can also use the formula:

and rewrite the expansion using spherical harmonics: