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Measuring a qubit and ending up with a bit feels a little like tossing out infinities in renormalization. Does neglecting the part of the wave function with a vanishing Hilbert space norm amount to renormalizing of Hilbert space?

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The question was inspired, if that's the right word, but the recent paper by Max Tegmark: arxiv.org/PS_cache/arxiv/pdf/1008/1008.1066v1.pdf – user1632 Jan 27 '11 at 22:56
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I'm not certain what "neglecting the part of the wave function with a vanishing Hilbert space norm" might mean precisely, and I'm a little scared to find out. But, as far as renormalization goes, it is a clean and precise process and in modern understanding it doesn't have to do with subtracting infinities. Rather it is about scale dependence of coupling constants, you can look at any QFT book which has a chapter on renormalization group. – user566 Jan 27 '11 at 23:08
No. What you are referring to is the "collapse" of the wavefunction from a quantum state vector to a classical value. – user346 Jan 28 '11 at 5:00
Measuring a qubit means we're entangling the qubit with the measuring apparatus and the environment. Decoherence leads to a subjective branching into many worlds. – QGR Jan 28 '11 at 9:36

2 Answers

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No, those are two very different processes (as far as I understand).

  1. Renormalization: When you are calculating vacuum expectation values, for instance $\langle \Omega\mid T(\phi(\mathbf{p})\phi(0))\mid \Omega\rangle$, you discover that these values are infinite. However, you can interpret this infinity, in a consistent manner, as the value of this correlation function at other momentum $\mathbf{p}^{\prime}$ and a finite part that relates the correlation function at the two different momenta. Nothing is really lost in the renormalization procedure, it is just a matter of how to introduce a measured quantity (the correlation function at this other momentum) into the theory.

  2. Measurement: The measurement concerns a certain state $\mid \psi\rangle$ coupled to the a measurement device. Originally, before being coupled, the pure state has entropy equal to zero. Later, by the time evolution of the coupled system, the system being measured has, after tracing over the measurement device states, entropy larger than zero. The difference is the information lost by the system in the process. So, something is lost in the measurement process, contrary to renormalization.

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up vote 1 down vote

I am maybe a bit uncertain what you are asking, but from what I understand the answer would be no. Renormalization is a procedure for absorbing infinities in an interacting field theory. A quantum bit is really just a state, but referred to in information theoretic terms. The two physics are not directly related as such. In a measurement if one considers it as a collapse there is a new normalization (renormalization?) of the system state, which is just the state vector which pertains to the measurement outcome.

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