# Least Action

Nontrivializing triviality..and vice versa.

## Cadabra – a Computer Algebra System for field theory

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 problems

(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.

Written by Vivek

November 8, 2010 at 12:35 AM

## Coulomb’s Law in (n+1) dimensions

The pairwise interaction energy in $n$ spatial dimensions is given by

$E = -\int\frac{d^n k}{(2\pi)^{n}}\frac{e^{i\vec{k}\cdot\vec{r}}}{k^2+m^2}$

To evaluate this $n$-dimensional Fourier integral, we must employ hyperspherical coordinates. The volume element is

$dV = k^{n-1}\sin^{n-2}\phi_1\sin^{n-3}\phi_2\ldots\sin\phi_{n-2}\,dk d\phi_1\ldots d\phi_{n-1}$

The angles $\phi_{1}, \ldots, \phi_{n-2}$ range from $0$ to $\pi$ and $\phi_{n-1}$ ranges from $0$ to $2\pi$. Writing $\vec{k}\cdot\vec{r} = kr\cos\phi_1$, the integral can be written as

$E = -\frac{(2\pi)^{d} C}{(2\pi)^{n}}\int_{0}^{\infty}\frac{dk\,k^{n-1}}{k^2+m^2}\int_{0}^{\pi}d\phi_1\,e^{ikr\cos\phi_1}\sin^{n-2}\phi_1$

The constant $C$ is equal to the product of integrals of the form $\int_{0}^{\pi} \sin^{n-k}\phi\,d\phi$. The number of such integrals depends on the dimension $n$. In two and three dimensions, $C = 1$. In more than 3 dimensions, there are $n-3$ such terms, for $k = 2, 3, \ldots, n-2$ (the integral over $\phi_{n-1}$ produces a $2\pi$, which we’ve already factored out). The value of $d$ is $0$ for 2 dimensions, and is $1$ for 3 and higher dimensions.

The integral over $\phi_1$ produces a regularized confluent hypergeometric function $_{2}F_3(a_1,a_2;b_1,b_2;\alpha)$. Specifically, for $n \geq 2$, the integral over $\phi_1$ produces

$\frac{\sqrt{\pi}e^{-ikr}\Gamma\left(\frac{n-1}{2}\right)}{\Gamma\left(\frac{n}{2}\right)}\left[_{2}F_{3}\left(\frac{n-1}{4},\frac{n+1}{4};\frac{1}{2},\frac{n-1}{2},\frac{n}{2};-k^2 r^2\right) + (ikr) _{2}F_{3}\left(\frac{n+1}{4},\frac{n+3}{4};\frac{3}{2},\frac{n+1}{2},\frac{n}{2};-k^2 r^2\right)\right]$

(As a check, for $n =2$, this becomes $\pi J_{0}(kr)$, which is what we obtained for 2 spatial dimensions in a previous post.)

The resulting integration over $k$ 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.

Written by Vivek

October 21, 2010 at 6:39 PM

## Feynman Diagrams using Adobe Illustrator

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 $\LaTeX$ file, which Jaxodraw actually uses to export diagrams into $\LaTeX$ 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.

Written by Vivek

October 17, 2010 at 11:53 AM

## FeynArts on Mathematica in Windows

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:

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!

Written by Vivek

October 15, 2010 at 7:08 PM

## Gaussian Integrals and Wick Contractions

Theorem 1: For nonzero real $a$,

$\langle x^{2n}\rangle = \frac{\int_{-\infty}^{\infty}dx\,e^{-\frac{1}{2}ax^2}x^{2n}}{\int_{-\infty}^{\infty}dx\,e^{-\frac{1}{2}ax^2}} = \frac{1}{a^{n}}(2n-1)(2n-3)\cdots 5\cdot 3\cdot 1$

Theorem 2: For a real symmetric $N\times N$ matrix $A$, a real $N \times 1$ vector $x$ and a real $N\times 1$ vector $J$, we have

$\int_{-\infty}^{\infty}\dotsi\int_{-\infty}^{\infty}\,dx_1\cdots dx_N\, e^{-\frac{1}{2}x^{T}Ax + J^{T}x} = \left(\frac{(2\pi)^N}{\det(A)}\right)^{1/2}e^{\frac{1}{2}J^{T}A^{-1}J}$

Proof:

Define $O$ to be the real orthogonal matrix which diagonalizes $A$ to its diagonal form $D$. That is,

$A = O^{T} D O$

where $O^{T}O = OO^{T} = I$. Further, let us choose $O$ to be a special orthogonal matrix, so that $\det(O) = +1$. Also, define $Y = OX$. Now,

$dy_1 \cdots dy_n = |\det(O)| dx_1 \cdots dx_n = dx_1 \cdots dx_n$

The argument of the exponential is

$-\frac{1}{2}X^{T}AX + J^{T}X = -\frac{1}{2}Y^{T}O(O^{-1}DO)(O^{T}Y) + J^{T}O^{T} Y$

which equals

$-\frac{1}{2}Y^{T}DY + J^{T}O^{T}Y$

The first term is

$-\frac{1}{2}Y^{T}DY = -\frac{1}{2}\sum_{i}y_{i}D_{ii}y_{i}$

Let $C^{T} = J^{T}O^{T}$ (that is, $C = OJ$). So the second term is

$J^{T}O^{T}Y = C^{T}Y$

So the argument of the exponential becomes

$-\frac{1}{2}\sum_{i}y_{i}D_{ii}y_{i} + \sum_{i}C_{i}Y_{i}$

which can be written as

$-\frac{1}{2}\sum_{i}\left[D_{ii}\left(y_{i}-\frac{C_i}{D_{ii}}\right)^2 - \frac{C_i^2}{D_{ii}}\right]$

The integrand becomes

$e^{-\frac{1}{2}\sum_{i}\left[D_{ii}\left(y_{i}-\frac{C_i}{D_{ii}}\right)^2 - \frac{C_i^2}{D_{ii}}\right]} = \prod_{i}e^{-\frac{1}{2}D_{ii}\left(y_{i}-\frac{C_i}{D_{ii}}\right)^2}e^{\frac{1}{2}C_{i}D_{ii}^{-1}C_{i}}$

Hence the quantity to be evaluated is

$\left\{\int_{-\infty}^{\infty}\dotsi\int_{-\infty}^{\infty}dy_1 \cdots dy_n \prod_{i}e^{-\frac{1}{2}D_{ii}\left(y_{i}-\frac{C_i}{D_{ii}}\right)^2}\right\}e^{\frac{1}{2}\sum_{i}C_{i}D_{ii}^{-1}C_{i}}$

The quantity in curly brackets involves $N$ definite Gaussian integrals, the $i^{th}$ term of which is

$\sqrt{\frac{2\pi}{D_{ii}}}$

using the identity $\int_{-\infty}^{\infty} dt\,e^{-\frac{1}{2}at^2} = \sqrt{\frac{2\pi}{a}}$.

So the continued product is

$\left(\frac{(2\pi)^N}{\det(D)}\right)^{1/2} = \left(\frac{(2\pi)^N}{\det(A)}\right)^{1/2}$

whereas the exponential factor sticking outside is

$e^{\frac{1}{2}\sum_{i}C_{i}D_{ii}^{-1}C_{i}} = e^{\frac{1}{2}C^{T}D^{-1}C} = e^{\frac{1}{2}(J^{T}O^{T}OA^{-1}O^{-1}OJ)} = e^{\frac{1}{2}J^{T}A^{-1}J}$

This completes the proof.

Theorem 3: $\langle x_{i} x_{j} \cdots x_{k}x_{l}\rangle = \sum_{\mbox{Wick}}(A^{-1})_{ab}\cdots(A^{-1})_{cd}$

Proof: If $m$ is the number of terms in $\langle x_{i} x_{j} \cdots x_{k}x_{l}\rangle$, differentiate the identity proved in Theorem 2 above $m$ times with respect to each of $J_i, J_j, \ldots, J_k, J_l$.

As an example, differentiate the RHS of the identity in Theorem 2 with respect to $J_{i}$. The derivative is

$\frac{d}{dJ_{i}}\exp\left\{\frac{1}{2}\sum_{m,n}J_{m}(A^{-1})_{mn}J_{n}\right\} = \frac{1}{2}\left(\sum_{m,n}(\delta_{mi}(A^{-1})_{mn}J_{n} + J_{m}(A^{-1})_{mn}\delta_{ni})\right)\exp\left\{\frac{1}{2}\sum_{m,n}J_{m}(A^{-1})_{mn}J_{n}\right\}$

the prefactor of the exponential is

$\frac{1}{2}(\sum_{n}(A^{-1})_{in}J_{n} + \sum_{m}J_{m}(A^{-1})_{mi})$

which becomes

$\sum_{j}(A^{-1})_{ij}J_{j}$

using the fact that $A$ is symmetric. Repeating this exercise, we see that bringing down $x_{i}$ involves differentiating with respect to $J_{i}$, and effectively introduces a matrix element of $A^{-1}$. It is obvious by induction that the sum must run over all possible permutations of matrix indices of $A^{-1}$. That is, the sum must run over every possible permutation $(a, b), \ldots (c, d)$ of pairing the indices $i, j, \ldots k, l$. This completes the “proof”.

Written by Vivek

October 9, 2010 at 9:45 PM

## Klein Gordon Field Part II

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 $(\Box + m^2)$, and not on the source $J(x)$. Note also, that $x$ is short for $x_{\mu}$, the position 4 vector. In our notation $\bf{x}$ will denote the position 3 vector.

The first step in our analysis is to solve for the propagator. To do, we write $G(x-x')$ in terms of its Fourier transform $G(p)$:

$G(x-x') = \int \frac{d^{4}p}{(2\pi)^4} e^{-ip\cdot(x-x')} G(p)$

where the different components of $p^{\mu}$ 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 $p^{\mu}p_{\mu} - m^2 = 0$. Let us operate on the above equation by $(\Box_{x} + m^2)$. This gives us

$(\Box_{x} + m^2) G(x-x') = \int \frac{d^{4}p}{(2\pi)^4} e^{-ip\cdot(x-x')}(-p^2 + m^2) G(p)$

Using the integral representation of the Dirac delta distribution, we get

$G(p) = \frac{1}{p^2-m^2} = \frac{1}{(p^{0})^2-|{\bf{p}}|^2-m^2} = \frac{1}{(p^{0})^2-E_{p}}$

where we have used the energy momentum relation in the last two equalities. Putting this back into the analysis equation, we get

$G(x-x') = \int \frac{d^{4}p}{(2\pi)^4} e^{-ip\cdot(x-x')}G(p)$

Do you see a problem with this expression?

The integration over $p^0$ cannot be performed as the integrand has poles at $p^{0} = \pm E_{p}$. This is where Feynman’s genius comes in…we’ll take a look at it in the next blog post!

Written by Vivek

June 22, 2009 at 11:13 AM

## Klein Gordon Field Part I

I’m trying to gain a deeper insight into the Green’s function for the Klein Gordon field, which is used to define the propagator. Peskin and Schroeder (P&S) have provided partial motivational details, and I am trying to complete the argument. Posts that will follow in a similar fashion will be primarily for new students of quantum field theory, like myself, and not for experts or students who have already worked with QFT. So if you belong to the latter category, you should find nothing new here.

For newcomers to quantum field theory, the Klein Gordon Field is a solution to Klein Gordon’s equation:

$(\Box + m^2)\phi = J(x)$

where as usual, $\Box$ is the d’Alembertian operator and m is the mass of the scalar field (for now, if we consider a real Klein Gordon field, this is just the pass of the particle excitation mode of the field). Note that the “free” Klein Gordon field is one for which the source term J(x) is zero. A general solution to the above equation can be written in terms of the Green’s function for the field, defined by

$(\Box_{x} + m^2)G(x-x') = -\delta^{4}(x-x')$

where $\Box_{x}$ operates only on terms involving $x$ (and not on $x'$), and the right hand side is our usual Minkowski space Dirac delta distribution. As stated, the Green’s function provides a concise way of writing a general solution to Klein Gordon equation with a source term:

$\phi(x) = \phi_{0}(x) - \int d^{4}x' G(x-x')J(x')$

where $\phi_{0}(x)$ is any solution to the free Klein Gordon equation, and can be chosen appropriately to satisfy boundary conditions. The interesting feature of the Klein Gordon formalism in QFT is that it provides a window to several tools and viewpoints that are thereafter used freely, for instance the concepts of particle/antiparticle as a natural outcome of second quantization, Feynman diagrams for scattering amplitudes, propagators isomorphic to Green’s functions and the notion of causality. The free Klein Gordon equation of course does not include interactions and is admittedly “simple”, but as is the case in almost all of science, simple toy models are quite powerful in conveying the features and limitations of a theory, pushing it almost to the fringe. In this framework, we can even gain a glimpse into some of the many infinities plaguing field theory.

In the next blog post, we will look at how the Green’s function nicely comes out of the above formalism with a minimal amount of mathematics. Then, we will discuss a small aspect of Feynman’s brilliance in taking it forward to connect it to the physics of the Klein Gordon field. Have fun!

Written by Vivek

June 22, 2009 at 10:06 AM