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:

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 \((V^\star)^{\otimes 2}\). 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 \((A^{-1}) \in (V^\star)^{\otimes 2}\) 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 \(\Sigma_n \in (V^\star)^{\otimes n}\) 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.