In textbooks it's stated that one should renormalize the field strengths in addition to coarse graining and rescaling by a factor of for example $z$. For a scalar $\phi^4$ theory they choose $z$ to be such that the coefficient of the $|\nabla \phi |^2$ remains unchanged. Books reason that this choice of $z$ results in a RG flow with 2 relevant directions which is convenient to describe a ferromagnetic phase transition.

About the choice of $z$ I have seen three different opinions:

1-Textbooks claim that one can choose any other values for $z$ and he will get different coarse grained Hamiltonians with different critical behaviors.

2-In this question the accepted answer argues that different values of $z$ just give us different but "equivalent" RG flows.

3-For each theory the value of $z$ is uniquely determined.

Personally the second opinion is false as it's evident even in the Gaussian model that different $z$ s will give different RG flows. But the first statement also seems peculiar to me, since I think that one shouldn't renormalize the field variables if he attempts to find the effective Hamiltonian of the system when the system is observed over larger length scales (if you view an image from a larger distance, your eyes don't enhance the intensity of light which comes from the image). Another problem with the first statement is that why should two systems with exactly the same statistical field theories have different critical behaviors?

And if the third statement is true, what're the criteria according to which we should define $z$ ? Because apparently people define $z$ in different contexts with different reasons. For example in treating the non-linear sigma model it's not important that how many relevant parameters the coarse grained Hamiltonian will have.


1 Answer 1


The fact that the critical behavior of the system can depend on the choice of the wave-function renormalization $Z$ is completely false.

The critical behavior is a physical property of the system (it can be measured experimentally, via the correlation functions, for example), and as such, it is independent of the way one does a calculation, using RG, Monte-Carlo or anything else. So the critical behavior does not depend on the definition of $Z$. (Of course, when one does explicit calculations, which entails approximations, then the result might depend on the way the calculation, and thus the definition of $Z$. But that's a disease of the approximation, not a general property of the RG flow.)

Now, one thing that can depend on the definition of $Z$ is the flow of the coupling constants, and thus of the Hamiltonian. But this is not in contradiction with what I wrote above, since the coarse-grained Hamiltonian is not physical (in the sense that it cannot be measured directly). The standard definition of $Z$ is that it is chosen so that the flow of the Hamiltonian goes to a fixed point if the system is critical. But once more, one should be careful. Having a fixed point in the flow implies scale invariance (in the physical quantities, like the correlation functions), but not having a fixed point does not imply that the system is not critical. Indeed, if one chooses, for instance, a "wrong" definition of $Z$, then there won't be a fixed point, but the system is still critical. The flow is equivalent, but more difficult to interpret, since it does have a fixed point, even though the system is scale invariant.

So in summary, 1) is wrong, the critical behavior is always the same (even though the coarse-grained Hamiltonian will be different); 2) is correct; 3) is wrong, although there is usually a natural definition of $Z$ depending on the problem one is studying.

  • $\begingroup$ Your answer seems completely convincing. But is there any references which explicitly show that difference values of $z$ don't change anything about the behavior of the system? Is it related to the fact that the wave function renormalization is an invertible transformation -in contrast to coarse graining which is not invertible and some information about the system are lost after this procedure- ? $\endgroup$
    – Hossein
    Sep 17, 2016 at 19:29
  • $\begingroup$ I don't really have any reference on that particular point unfortunately... But think of it that way: the RG is just a smart way to compute a trace, so the way you do it does not change the physical properties/observables. Hence the way you do RG does not change the physics, it just help you get the correct behavior more or less easily. (Note that the fact that you lose information after a RG transformation is not stricto sensus true, even though one reads it everywhere. There are ways to keep (in principle) all the information. But you can't do it in practice. But that's another question.) $\endgroup$
    – Adam
    Sep 17, 2016 at 19:37

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