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In QFT a polynomial (of degree >2) in the fields is said to be an interaction term, Ex.: $\lambda\phi^4$.

Question

Is it possible to give an interpretation to terms like $\frac{1}{\phi^n}$? (for $n\in\mathbb{N}$)

Cheers

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  • $\begingroup$ Make a lattice, and do a Monte-Carlo, and then all the issues of renormalization go away at finite lattice spacing, and you can understand all the field potential terms immediately. $\endgroup$
    – Ron Maimon
    Nov 8, 2012 at 18:35
  • $\begingroup$ @RonMaimon: But then you have to explain why your results are independent of the lattice chosen, and the renormalization issues reappear in full complexity. $\endgroup$ Nov 8, 2012 at 18:50
  • $\begingroup$ @ArnoldNeumaier: Yes, of course, but they are obvious then. $\endgroup$
    – Ron Maimon
    Nov 8, 2012 at 19:16

2 Answers 2

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In principle, yes, but only if the expectation value of $\phi$ is nonzero, so $\phi$ would immediately be shifted. Moreover, the result would be badly nonrenormalizable, so nobody is using such terms.

An important case of a nonpolynomial interaction that received considerable attention in 1+1D is the interaction $\sin\phi(x)$ of the sine-Gordon model http://en.wikipedia.org/wiki/Sine-Gordon#Quantum_version

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  • $\begingroup$ Could you in principle have something like $1/(1-\phi)$ which then would just result in a simple power series expansion? $\endgroup$
    – Lagerbaer
    Nov 8, 2012 at 18:18
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    $\begingroup$ @Lagerbaer: yes, though the renormalizability issue remains. An important case of nonpolynomial interactions that received considerable attention in 1+1D is the sine-Gordon model en.wikipedia.org/wiki/Sine-Gordon#Quantum_version $\endgroup$ Nov 8, 2012 at 18:29
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As Arnold Neumaier points out, the particle interpretation is not very good.

Another problem: The interaction $V(\phi,x) = 1/\phi^n(x)$ isn't stable if $n$ is odd. This energy isn't bounded below. You can try to fix this by setting $V(\phi,x) = 1/|\phi|^n(x)$, but there's still a kind of stability problem. The minimum of this potential is at $\phi(x) = \infty$, so if you start in the naive vacuum, you'll generate a huge expectation value.

If you put this theory on a lattice, you'll see the magic of effective field theory in action: The expectation values of observables whose support is large relative to the lattice spacing will be governed by an effective field theory with polynomial interactions. You'll find you could have computed these expectation values just as well by assuming that the lattice theory was renormalizable, with coefficients gotten by matching the input and output of renormalization flow.

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