0
$\begingroup$

Thank you all for so much help as I work through Zee's QFT book. Here I finally have a question about physics instead of math.

Zee makes several comments that the main thing left to do in QFT is find a way to solve

$$ Z(J)=\int\!D\varphi\,\exp\!\left[ i\int\!d^4x\,\frac{1}{2}\big[ (\partial\varphi)^2-m^2\varphi^2 \big] -\frac{\lambda}{4!}\varphi^4+J\varphi \right] .$$

Since we cannot solve it directly, even by the magical process for "discretizing and undiscretizing" infinite dimensional path integrals, Zee makes a study of two simplifed cases. He presents a "baby problem"

$$ Z_B(J)=\int_{-\infty}^\infty\!dq\,\exp\!\left[ -\frac{1}{2}m^2q^2-\frac{\lambda}{4!}q^4+Jq \right], $$

and a "child problem"

\begin{align*} Z_C(J)&=\int_{-\infty}^\infty\int_{-\infty}^\infty\dots\int_{-\infty}^\infty\!dq_1\,dq_2\,\dots dq_N\,\exp\!\left[ -\frac{1}{2}\,q\cdot A\cdot q-\frac{\lambda}{4!}q^4+J\cdot q \right]\\ &=\iint\dots\int\!dq_1\,dq_2\,\dots dq_N\,\exp\!\left[ -\frac{1}{2}\sum_{m,n=1}^N q_m A_{mn} q_n-\frac{\lambda}{4!}\sum_{k=1}^N q_k^4+\sum_{l=1}^N J_l q_l \right]. \end{align*}

Analytically, I see the difference is that $J$ and $\varphi$ are functions of a continuous variable in $Z$, a discrete variable in $Z_C$, and they are not functions of anything in $Z_B$.

Q1: What would be the physical significances of $Z_B$ and $Z_C$ respectively?

Q2: What does it mean for $q$ not to be a function of anything?

For $q$ to be a function of a discrete variable, I see that $q$ describes, essentially, the vertical displacement position of a set of oscillators located at some number of discrete $\vec x_k$. If it's not a function of anything, does that mean it is a single oscillator? How could it not be a function of time at least? Is the location of the single oscillator coded into $q$ itself so $q$ means "the oscillator at some $\vec x_0$?" If you can go into detail about the physics we are looking at here, that would be very helpful.

$\endgroup$
1
  • 1
    $\begingroup$ Many mathematical tools used in QFT and Zee's textbook are of no physical significance whatsoever. The math behind QFT is notoriously involved, so a lot of time is usually spent on toy models so that the reader can get familiar with the mathematical tricks. There are only few QFT models with significance in high energy physics – QED, the Standard Model, etc. $\endgroup$ – Prof. Legolasov Oct 22 '20 at 6:40
2
$\begingroup$

Q1: What would be the physical significances of 𝑍𝐵 and 𝑍𝐶 respectively?

$Z_B$ doesn't really have a physical meaning in the sense of representing an interesting physical system. It is a mathematical toy problem that is easier to think about than the full path integral. If you really want to assign a physical meaning to it, it would amount to integrating over the value of a field at one point in spacetime.

$Z_C$ could represent any number of different systems, really. You can think of it as a general form of a discretized lattice representation of a quantum field theory of an interacting scalar field. In 0+1 dimensions, this could represent the position of a particle in a potential like $V(x)=m^2 x^2 + \lambda x^4$ (identifying $\phi$ with $x$). The variable $q$ would be identified with the position of the particle and the index would represent time. In 3+1 dimensions, this could represent a discretized version of a scalar field theory. In this case the value of $q$ represents the value of the scalar field, and the index represents different pionts in spacetime.

The full path integral is a continuum limit of $Z_C$, and exactly what field theory you get (how many dimensions, etc) will depend on how the lattice is constructed and how the limit is taken.

Q2: What does it mean for 𝑞 not to be a function of anything?

I wouldn't get too hung up on this. I think the 0-dimensional baby problem is just that, a baby problem that's easy to get your head around, but is not a very good model of an interesting physical system.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.