I have been studying decoherence in quantum mechanics (not in qft, and don't know how it is described there) and renormalization in QFT and statistical field theory, I found at first a similarity between the two procedures: on one side decoherence tells us to trace over the degrees of freedom we don't monitor, in some way intrinsically unknown to recover a classical picture picture from quantum mechanics, on the other side by renormalizing we also integrate over "our ignorance" but this time, the U.V. physics or high energy modes to get the infra-red physics we observe. Beyond the technical similarity (taking a trace, for discrete Kadanoff-Wilson transformations) it feels that in both cases we are forced to do these procedures because we start from a wrong picture where we separate the free object (purely quantum in the first case, with bare parameters in the second one) and then calculate the effects of the interactions, that are responsible from what we, observers, see, classical and infra-red physics. This where it comes to me that some interested links between the two concepts may exist or be pointed out (and also wonder what decoherence becomes in QFT).

I still see one huge asymmetry between the two, decoherence is dynamical, it has a typical time of decay, where renormalization is static.

I hope I could explain my interrogation clearly, and some interesting comments will come.

  • $\begingroup$ On tracing over some degrees of freedom from a "total" density matrix, you get an "effective" state. If you did this on a state of 2 entangled qubits (say a Bell state), you'd get a single qubit in a mixed state. If you consider a field theory (with regulators, etc) you could write it as a bunch of harmonic oscillators at each position in space (or alternatively each momentum). Then you could trace over some of them (say the high momentum modes, or maybe those corresponding to a specific volume in space). That will give you a mixed state and you would have performed RG. (contd..) $\endgroup$
    – Siva
    May 17, 2013 at 8:40
  • $\begingroup$ (..contd) In the case of decoherence, the degrees of freedom you've traced over belong to the environment, and not to your system of consideration. So, operationally they're kinda similar but conceptually quite different. $\endgroup$
    – Siva
    May 17, 2013 at 8:41

1 Answer 1


In general, decoherence and renormalization are two different things.

Decoherence is loss of quantum correlations due to lack of information (either on the interaction, or - on states of some particles). The most typical example is when you loss a particle - then you need to trace out with respect to their degrees of freedom, effectively changing all entanglement in (classical) randomness.

Renormalization is a procedure, when you construct (or deal with) an effective theory, disregarding some degrees of freedom only to threat them as a single object (e.g. instead of four particles only one "effective particle"). Typically you don't "loss" particles, but treat collections of then as a new one. However, the procedure is lossy - i.e. you lose some properties; in particular, from a pure state you can go to mixed ones.

However, typically you rather still hold an approximate pure state, rather than the exact one which (after neglecting some degrees of freedom) is mixed; see e.g.:


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