Archive for the ‘Quantum Field Theory’ Category
Although their importance and applications are widely recognized in theoretical physics, differential forms are not part of the standard curriculum in physics courses, except for the rare mention in a general relativity course, e.g. if you have used the book by Sean Carroll. In this blog post, we describe the construction of the field strength tensor for Yang Mills Theory, using differential forms. The derivation (which is based on the one given by Tony Zee in his QFT book) is elegant, and mathematically less tedious than the more conventional derivation based on the matrix gauge field. It also illustrates the power of the differential forms approach.
To begin with, let me define a normalized matrix gauge field which is times the usual matrix gauge field. So in the equations that appear below, the covariant derivative has no lurking factors of . This greatly simplifies the algebra for us, as we don’t have to keep track of conjugations and sign changes associated with . The covariant derivative is thus in terms of this new gauge field. The matrix 1-form is
So, . But since , only the antisymmetric part of the product survives and hence we can write
We want to construct an appropriate 2-form from this 1-form A. Now, if d denotes the exterior derivative, then dA is a 2-form, as is A^2. These are the only two forms we can construct from A. So, must be a linear combination of these two forms. This is a very simple, and neat argument!
Now, the transformation law for the gauge potential is
where is a 0 form (so ). Applying d to the transformation law, we get
where the negative sign in the third term comes from moving the 1-form d past the 1-form A. Squaring the transformation law yields
Now, $UU^\dagger = 1$, so applying d again to both sides we get . So, we can write the square transformation law as
whereas if we recall the expression for the transformation of , it was just
Clearly if we merely add and , the last 3 terms on the RHS of each cancel out, and we get
which is the expected transformation law for a field strength of the form :
The differential form approach uses compact notation that suppresses the Lorentz index as well as the group index , and gives us a fleeting glimpse into the connection between gauge theory and fibre bundles.
For a gentle yet semi-rigorous introduction to differential forms, the reader is referred to the book on General Relativity by Sean Carroll.
This seems very exciting:
Cadabra is a computer algebra system (CAS) designed specifically for the solution of problems encountered in field theory. It has extensive functionality for tensor computer algebra, tensor polynomial simplification including multi-term symmetries, fermions and anti-commuting variables, Clifford algebras and Fierz transformations, implicit coordinate dependence, multiple index types and many more. The input format is a subset of TeX. Both a command-line and a graphical interface are available.
There are two interesting papers on this. The first is a semi-technical overview, and the other (hep-th/0701238) is a more comprehensive one geared towards an audience familiar with various problems in modern field theory. The abstract of the second paper reads:
Introducing Cadabra: a symbolic computer algebra system for field theory problemsAuthors: Kasper Peeters(Submitted on 25 Jan 2007 (v1), last revised 14 Jun 2007 (this version, v2))
Abstract: Cadabra is a new computer algebra system designed specifically for the solution of problems encountered in field theory. It has extensive functionality for tensor polynomial simplification taking care of Bianchi and Schouten identities, for fermions and anti-commuting variables, Clifford algebras and Fierz transformations, implicit coordinate dependence, multiple index types and many other field theory related concepts. The input format is a subset of TeX and thus easy to learn. Both a command-line and a graphical interface are available. The present paper is an introduction to the program using several concrete problems from gravity, supergravity and quantum field theory.
Source = http://arxiv.org/abs/hep-th/0701238
The pairwise interaction energy in spatial dimensions is given by
To evaluate this -dimensional Fourier integral, we must employ hyperspherical coordinates. The volume element is
The angles range from to and ranges from to . Writing , the integral can be written as
The constant is equal to the product of integrals of the form . The number of such integrals depends on the dimension . In two and three dimensions, . In more than 3 dimensions, there are such terms, for (the integral over produces a , which we’ve already factored out). The value of is for 2 dimensions, and is for 3 and higher dimensions.
The integral over produces a regularized confluent hypergeometric function . Specifically, for , the integral over produces
(As a check, for , this becomes , which is what we obtained for 2 spatial dimensions in a previous post.)
The resulting integration over is rather tricky. At this point, I don’t know if its possible to do it by hand, so I am using Mathematica to perform it.
Edit: a few hours later..
So, it seems this integral does not have a closed form representation, or at least not one that Mathematica can find.
I recently became aware that Professor Daniel Schroeder used Adobe Illustrator to make Feynman Diagrams in Chapter 9 of his Thermal Physics book. A google search yielded two videos by a Youtube user named AjabberWok. I’m taking the liberty of embedding them here. If you would rather view them in a new window, just click on the videos to to be taken to the appropriate youtube link.
In a previous post, I briefly outlined the installation of FeynArts, an open source Mathematica package for creating Feynman diagrams. There are also other tools such as the open source JaxoDraw program, and the axodraw file, which Jaxodraw actually uses to export diagrams into lineart. I personally endorse the use of open source programs such as FeynArts and Jaxodraw, but for those of you who already have access to Adobe Illustrator, you might find these videos quite useful.
As much as I hate to admit it, some of us are constrained to use Mathematica on Windows for some reason or the other. If you happen to be part of this ‘some of us’ subset (at least temporarily, in my case) and are eager to get your hands on FeynArts, a nifty package for Mathematica which lets you make and play with Feynman diagrams, you’re going to have some trouble installing it on the current version of Mathematica. If you are using Windows, I suggest downloading the tarball and unzipping its contents to:
C:\Users\<your Windows user name>\AppData\Roaming\Mathematica\Applications\
The important thing to note is that the tarball actually unpacks into a new directory (FeynArts-3.5 for now). You need to ensure that the contents of this directory are placed in the above directory, so that FeynArts-3.5 is not a subfolder of the above folder. In short, this is what my folder looks like:
The next step is to fire up Mathematica and select “Install” from the File Menu. Select “Package” under the “Type of item to Install” tab and navigate to the above directory using the “File” option under the Source tab. Select FeynArts35.m and type “FeynArts” in the Install Name field. Finally, under “Default Installation Directory”, check “Your user Mathematica base directory”.
This last step is important: if you check the “System-wide Mathematica base directory” option, you’ll end up with the $FeynArtsDir variable pointing to “C:\Program Files\Wolfram Research\Mathematica\7.0\AddOns\Applications”. In this case, FeynArts will work only if you dump the contents of “C:\Users\<your Windows user name>\AppData\Roaming\Mathematica\Applications\” into “C:\Program Files\Wolfram Research\Mathematica\7.0\AddOns\Applications”, something you shouldn’t do as this makes the Application folder messy.
A workaround should be to change the Directory variable in Setup.m for FeynArts, to point to the folder of your choice. I tried this, but ran into other path-related problems. Anyway, so long as it works, who cares!
Theorem 1: For nonzero real ,
Theorem 2: For a real symmetric matrix , a real vector and a real vector , we have
Define to be the real orthogonal matrix which diagonalizes to its diagonal form . That is,
where . Further, let us choose to be a special orthogonal matrix, so that . Also, define . Now,
The argument of the exponential is
The first term is
Let (that is, ). So the second term is
So the argument of the exponential becomes
which can be written as
The integrand becomes
Hence the quantity to be evaluated is
The quantity in curly brackets involves definite Gaussian integrals, the term of which is
using the identity .
So the continued product is
whereas the exponential factor sticking outside is
This completes the proof.
Proof: If is the number of terms in , differentiate the identity proved in Theorem 2 above times with respect to each of .
As an example, differentiate the RHS of the identity in Theorem 2 with respect to . The derivative is
the prefactor of the exponential is
using the fact that is symmetric. Repeating this exercise, we see that bringing down involves differentiating with respect to , and effectively introduces a matrix element of . It is obvious by induction that the sum must run over all possible permutations of matrix indices of . That is, the sum must run over every possible permutation of pairing the indices . This completes the “proof”.
In the last post, we looked at how a general solution to the sourced KG equation can be written in terms of a source-free KG solution and the Green’s function of the KG field. The important thing to note here is that the Green’s function depends only on the structure of the equation, specifically, on the differential operator , and not on the source . Note also, that is short for , the position 4 vector. In our notation will denote the position 3 vector.
The first step in our analysis is to solve for the propagator. To do, we write in terms of its Fourier transform :
where the different components of should be treated as independent, unrelated by the energy momentum relation. Right now, we’re only at the level of mathematics, so we do not enforce . Let us operate on the above equation by . This gives us
Using the integral representation of the Dirac delta distribution, we get
where we have used the energy momentum relation in the last two equalities. Putting this back into the analysis equation, we get
Do you see a problem with this expression?
The integration over cannot be performed as the integrand has poles at . This is where Feynman’s genius comes in…we’ll take a look at it in the next blog post!