2

So I was recently modelling something that turned out to be basically Wigner's friend.

I saw there were some differences (in the Wiki page) in how it was modelled: Namely, that Wigner's friend somehow managed to perform a measurement without interacting with the system! (I mean if there was an interaction term in the Hamiltonian it would be much better. On the other hand, the interaction term would destroy the tensor product! The interaction term becomes important during the time of the measurement in the Heisenberg picture).

Does this point mute the calculations on the Wikipedia page?

Qmechanic
  • 201,751
More Anonymous
  • 1,309
  • Let me ask you something, i measure an electron's spin in x direction without telling you. My measurement will force the electron to come out of superposition state. However i won't tell you this, you, as a second observer will consider the electron in superposition or not before your measurement? – Paradoxy Jun 20 '19 at 21:11
  • @Paradoxy the way I think of it is the following: the system + the measuring apparatus definitely obey unitary evolution... On the other hand your observation is also valid that a measurement occurred. However, the contradiction only comes when one expects there to be no interaction Hamiltonian. One has to use perturbation theory to relate both the wave functions. To find more detailed math on my opinion please see the link "something" – More Anonymous Jun 20 '19 at 21:30
  • One has to use perturbation theory to relate both the wave functions. To elaborate this point further .. when u have an interacting Hamiltonian one needs to use perturbation theory to relate the wave function that you measure and the wave function that I measure since they are different Hamiltonians – More Anonymous Jun 20 '19 at 21:52

0 Answers0