I went through several books regarding zero dimensional path integral formulation. Can anyone please explain it to me what is that more physically. I read in the book that the formulation in 0+0D made the Feynman path integral calculation much easier. Why are we even considering zero dimension concept?

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    $\begingroup$ The zero dimensional path integral is like a playground, where you can exploreore complicated topics (gauge theory, topological QFT, supersymmetry, etc.), without getting bogged down in the messiness that higher-dimensional QFTs offer. I'm not sure if it has physical meaning or applications, it's just a simplified version of a harder physical theory that still lets you study the important features. $\endgroup$ – Bob Knighton Nov 3 '17 at 13:52
  • $\begingroup$ Means is it just for the sake of mathematical calculation? What I understood was that the whatever frame we would be considering for these fields are reduced to zero dimension. Please make it further clear. Like zero dimension means here as 0 in (spacetime) or 0 in (space -time) $\endgroup$ – Jyotsna Sharma Nov 3 '17 at 14:01
  • $\begingroup$ For 0+1 dimensional case you can look at very pedagogical paper found at authors.library.caltech.edu/8383/1/BOOejp07b.pdf. I don't quite understand what you mean by 0+0 dimensional field theory. Do you consider discrete state system evolving in discrete time as 0+0 dimensional field theory? $\endgroup$ – Sunyam Nov 3 '17 at 14:40

Zero-dimensional quantum field theory is exactly like a standard quantum field theory, except that the background spacetime is exactly one point.

Consider, for a moment, a $d$-dimensional QFT defined by a partition functional (in the Euclidean signature)


where $\phi$ is a placeholder for all fields and $J$ is a placeholder for all of the currents. Varying $\mathcal{Z}$ with respect to $J$ gives a way to calculate correlation functions.

Now, a zero-dimensional QFT is formulated in the same light, but, since the fields don't vary with position (since there is no position with which to vary), the path integral simply becomes a regular integral:


In this light, $\mathcal{Z}$ is a function of a variable $J$ and $S$ is a function of $\phi$ and $J$, not a functional.

Why is this even a useful thing to define? The biggest use is being able to study properties of QFTs without the extra complications of having to work with multiple dimensions. For instance, consider a zero-dimensional QFT, whose action is given by


This is just a zero-dimensional analogue of $\phi^4$ theory that we're all used to studying. Let's say I want to compute the four-point function


Expanding the exponential in both the numerator and denominator, we find that $\langle\phi^4\rangle$ is a perturbative expansion in $\lambda$. If I did a little more work, I could easily discover the diagrammatic picture of this expansion! We just found a diagrammatic expansion with minimal work. Whoo!

This is also a good playground for considering more complicated QFT topics, like gauge theory, fermionic path integrals, resummation, and even supersymmetry (see the following wonderful set of notes by David Skinner).

This is even useful outside of physics. A mathematician might see the power in this very quickly. By using a zero-dimensional path integral, we can find a way to enumerate the number of graphs with $k$ external lines, $n$ vertices, and with specific rules for how many edges can hit a node. Zero-dimensional path integrals also have geometric importance, as some supersymmetric theories have cohomologies corresponding to those of interesting topological spaces.

All in all, zero dimensional QFTs are just very useful playgrounds for exploring aspects of more complicated QFTs, and also have utility in pure mathematics.

I hope this helps!

  • $\begingroup$ What would the partition function expansion be in zero field theory? $\endgroup$ – fielder Dec 9 '19 at 12:57
  • $\begingroup$ @fielder I'm not exactly sure what you're asking. The perturbation expansion of the partition function is just a Taylor series in whatever coupling you have, just as in standard field theory. $\endgroup$ – Bob Knighton Dec 9 '19 at 14:48

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