A Place of Ideas

Thoughts on Feynman Diagrams

I'll readily admit I don't know much about quantum field theory. I plan to learn it sometime, but that time has not yet come. But despite that, I think I've noticed something interesting.

Recap

To my understanding, Feynman diagrams are a description of a complicated infinite series, whose sum somehow describes something physical.

We begin with some input and output edges, each labeled with a momentum. We connect these edges into networks in all possible ways; each network becoming a term in the sum.

For example, for a three-particle interaction with two external wires, both labelled \(p\), the first few diagrams might look like this.

To compute these terms, we do the following:

It's probably a bit more complicated for particles with spin, but I don't know the details.

So, in the above example, labelling the external edges \(p\) and \(q\) and using coupling constant \(\lambda\), the example diagrams become the following terms. (I may be missing some constant factors.)

\[\begin{align*} & \frac{\delta(q-p)}{p^2 - m^2} \\ &\frac{1}{2} \int \int \frac {\lambda^2 \delta(r+s-p) \delta(q-r-s) d^4 r d^4 s} {(p^2 - m^2)(q^2 - m^2)(r^2 - m^2)(s^2 - m^2)} \\ &\frac{1}{6} \int \int \int \frac {\lambda^2 \delta(q-p) \delta(r+s+t) \delta(-r-s-t) d^4 r d^4 s d^4 t} {(p^2 - m^2)(r^2 - m^2)(s^2 - m^2)(t^2 - m^2)} \\ &\frac{1}{4} \int \int \frac {\lambda^2 \delta(r+(-r)-p) \delta(q-s-(-s)) d^4 r d^4 s} {(p^2 - m^2)(q^2 - m^2)(r^2 - m^2)(s^2 - m^2)} \end{align*}\]

Simplifying, these are \(\delta(q-p)\) times:

\[\begin{align*} & \frac{1}{p^2 - m^2} \\ &\frac{\lambda^2}{2(p^2 - m^2)^2} \int \frac {d^4 r} {(r^2 - m^2)((p-r)^2 - m^2)} \\ &\frac{\delta(0)\lambda^2}{6(p^2 - m^2)} \int \int \frac {d^4 r d^4 s} {(r^2 - m^2)(s^2 - m^2)((-r-s)^2 - m^2)} \\ &\frac{\delta(p)\lambda^2}{4m^4} \left(\int \frac{d^4 r}{(r^2 - m^2)}\right)^2 \end{align*}\]

A lot of these are just infinity, but supposedly there are ways to deal with that.

Abstracting the Problem

This setup reminded me of something. Tensor networks, a graphical alternative to tensor index notation.

So think of these momenta as indices. This makes our vector space something like \(V = \mathbb{R}^{\mathbb{R}^4}\).

Looking at the Feynman calculation, the propagator becomes an element of \((V^*)^{\otimes 2}\). The vertex contribution becomes a tensor in \(V^{\otimes (\text{vertex degree})}\). And the integrals just become tensor contraction!

So let's forget the integrals, and think purely in terms of abstract tensors!

Let \((A^{-1}) \in (V^*)^{\otimes 2}\) be the propagator, and \(\lambda \in V^{\otimes 3}\) be the vertex contribution. (Why \(A\) inverse? You'll see soon.) Then the sum from before is:

\[\begin{align*} & (A^{-1})_{ij} \\ &+ \tfrac{1}{2} \lambda^{klm} \lambda^{nop} (A^{-1})_{ik} (A^{-1})_{ln} (A^{-1})_{mo} (A^{-1})_{pj} \\ &+ \tfrac{1}{6} \lambda^{klm} \lambda^{nop} (A^{-1})_{ij} (A^{-1})_{kn} (A^{-1})_{lo} (A^{-1})_{mp} \\ &+ \tfrac{1}{4} \lambda^{klm} \lambda^{nop} (A^{-1})_{ik} (A^{-1})_{lm} (A^{-1})_{no} (A^{-1})_{pj} \\ &+ \text{terms of order \(\lambda^4\) and higher} \end{align*}\]

Most of the infinities in the sum have disappeared. But I figure they aren't truly gone; just hiding in the infinite dimensionality of \(V\).

One more thing. The above formulation is limited to three-particle interactions. Let's generalize to \(n\)-particle interactions.

To do this, we simply replace \(\lambda \in V^{\otimes 3}\) with the sequence \(\lambda_n \in V^{\otimes n}\).

Derivatives

Here's the fun part. Let \(\Sigma_n \in (V^*)^{\otimes n}\) be the sum over Feynman diagrams for \(n\) external wires. Then:

\[\begin{align*} \Sigma_{n+k} &= \frac{\partial}{\partial\lambda_k} \Sigma_n \\ \Sigma_{n+2} &= \left(2\frac{\partial}{\partial A} + A^{-1}\right) \Sigma_n \end{align*}\]

The derivative of a Feynman sum is another Feynman sum!

In particular, \(\Sigma_n = \frac{\partial^n}{\partial^n\lambda_1}\). So we can simplify:

\[\begin{align*} \frac{\partial^k}{\partial^k\lambda_1} \Sigma_0 &= \frac{\partial}{\partial\lambda_k} \Sigma_0 \\ \frac{\partial^2}{\partial^2\lambda_1} \Sigma_0 &= \left(2\frac{\partial}{\partial A} + A^{-1}\right) \Sigma_0 \end{align*}\]

And rearrange:

\[\begin{align*} \frac{\partial\Sigma_0}{\partial\lambda_k} &= \frac{\partial^k\Sigma_0}{\partial^k\lambda_1} \\ 2\frac{\partial\Sigma_0}{\partial A} &= \frac{\partial^2\Sigma_0}{\partial^2\lambda_1} - A^{-1}\Sigma_0 \end{align*}\]

We now have formulas for all the derivatives of \(\Sigma_0\), in terms of the \(\lambda_1\)-derivatives.

Symmetry

We can extract one more equation from a symmetry of the system. If in \(\Sigma_0\), we multiply \(\lambda_n\) by \(t^n\) and \(A\) by \(t^2\), all the \(t\)s cancel. So:

\[\begin{align*} \Sigma_0(\lambda,A) &= \Sigma_0((t^n\lambda_n)_{n\in\mathbb{N}},t^2A) \\ 0 &= \left(\frac{\mathrm{d}}{\mathrm{d}t} \Sigma_0((t^n\lambda_n)_{n\in\mathbb{N}},t^2A) \right)_{t=1} \\ &= \frac{\partial\Sigma_0}{\partial A^{ij}} 2A^{ij} + \sum_n \frac{\partial\Sigma_0}{\partial(\lambda_n)^{i_1 \dots i_n}} n(\lambda_n)^{i_1 \dots i_n} \\ &= \frac{\partial^2\Sigma_0}{\partial (\lambda_1)^i \partial (\lambda_1)^j} A^{ij} - \Sigma_0\dim V + \sum_n n \frac{\partial^n\Sigma_0}{\partial(\lambda_1)^{i_1} \dots \partial(\lambda_1)^{i_n}} (\lambda_n)^{i_1 \dots i_n} \end{align*}\]

And that's a differential equation for \(\Sigma_0\)! Specifically, it's a homogeneous linear differential equation, whose order is the maximum number of particles that can be involved in a single interaction.