## Covariant gauge transformation of Dirac equation to get the spin adjusted Klein-Gordon equation.

Posted by peeterjoot on September 3, 2011

# Motivation.

In section 36.4 of [1] is a covariant treatment of the gauge transformed Dirac equation, with the end goal of finding the Klein-Gordon relation with the spin terms required for electromagnetic phenomena.

Two typos to start things off, and (36.63-64) should be respectively

Other than this there are no typos till the end where a factor of two is lost, and we should have

It’s slightly tempting to re-derive this, with the inclusion of the and factors. However, a bit of play using the Geometric Algebra (GA) operators discussed in [2] proves more productive. This text introduces a purely GA formalism for the Dirac equation which I’ll not use here. Instead I’ll use the more conventional Feynman slash notation so that a four vector with coordinates is written with its basis as

# Geometric Algebra notation.

We require the GA dot, and wedge operators

In contrast to the matrix notation, the product of two identical gamma matrices is written as the unit scalar value (or grade zero product), instead of a using an explicit four by four identity matrix representation. We similarly label the product of different basis elements, for example as a grade two element, or bivector. Thus the dot product is the grade zero term of the multivector product , and the wedge product is the grade zero term of the same. More generally, for a product of multivectors and the grade selection operator is defined as

This is an abstract notation encoding the instructions to select just the grade elements of the multivector product if they exist. In this notation, the product of vectors splits into scalar and bivector terms which may also be expressed as the dot and wedge products

# Gauge transforming the Dirac equation.

Our electron equation is

After a gauge transformation

this is

We left multiply by the conjugate quantity , and our task is now to reduce operator equation

We have two contributions for the scalar parts, one is the product of scalars, and the other is the dot product of the vectors

The grade one (four vector) components sum to zero since those are only the scalar times four vector portions and have opposing signs

Only the vector products can contribute to the grade two portion of the multivector product, so retain only the wedges between those

Here braces have been used to denote the range of operation of our differential operator . With , our momentum operator is

allowing us to write

For the electromagnetic field bivector lets write

So the grade two terms of 3.12 are

Assembling all results we have

# Some checks

There are two tasks that remain. One is to verify that this reproduces the result expressed in terms of the gauge covariant derivative . The other is to verify that this also reproduces the earlier result with an explicit split into energy and momentum operators.

## Gauge covariant derivative form

With our dot product takes the form

We thus write

differing from the text only by the inclusion of and factors. Since we have

we have

or

With , this reproduces 1.3, the (corrected) result (36.78) from the text.

## Verifying the space time split into energy and spatial momentum operators.

Temporarily working with we have

With and , this is

Restoring and , and multiplying throughout by , 4.21 becomes

Now let’s expand the product in terms of the matrices used in the text previously. That is

Since

and

So

For the magnetic terms we have

For , considering the case (since it is zero otherwise), we have

So

Thus the space time split of the field is

This recovers (36.22) from the text (once one adds back in the and and fixes the sign error in the electric field term).

# A last word on covariant gauge transformations.

Usually discussions of gauge invariance involve a Lagrangian. Such a beastie for the Dirac equation has not yet been discussed yet in the text. Suppose we just apply the phase transformation to the Dirac equation of the same form that we use with the Klein-Gordon Lagrangian

Observe that if the dimensions of are those of the vector potential , then is dimensionless.

Now lets transform the Dirac equation, utilizing the freedom to adjust the phase arbitrarily

With the usual assignment

this becomes

The bigger question of why 5.28 is the usual transformation remains. I don’t know the answer for that one. It is common in Lagrangian transformation discussions for the Klien-Gordon equation, and I’m guessing it’s something better justified in places I’ve not yet read.

So in the limit where , the Dirac equation is transformed as

and we can say that our momentum operator transforms

The gauge covariant derivative is then just what we get when we factor out an from the gauge transformed momentum operator

So with

or

we define

# References

[1] BR Desai. *Quantum mechanics with basic field theory*. Cambridge University Press, 2009.

[2] C. Doran and A.N. Lasenby. *Geometric algebra for physicists*. Cambridge University Press New York, Cambridge, UK, 1st edition, 2003.

## Leave a Reply