I'm trying to recently understand the Copenhagen interpretation. We all know due to Noether's theorem energy is translational invariance in time.

However, in quantum mechanics the experimentalist can have the same initial configuration ("physical system" + "Apparatus") and do a measurement and get different outcomes. Hence, while energy (as a number) is conserved Nother's theorem obviously does not apply (during the measurement).

Now, the Copenhagenist seems to say the measuring instrument is some kind of semi-classical apparatus and the system is purely quantum - the measurement is an artefact of this.


How does the Copenhagenist justify this apparent discrepancy (violation of translational invariance in time)?

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    $\begingroup$ Have you ever heard of Ehrenfest's theorem? $\endgroup$ – BioPhysicist Sep 12 '19 at 14:18
  • $\begingroup$ Yes, however, for the Copenhagenist if semi (quantum/)classical mechanics and unitary evolution is there is to everything ... Then :/ $\endgroup$ – More Anonymous Sep 12 '19 at 14:20
  • $\begingroup$ More explicitly, during unitary evolution atleast if I have the same initial state after time $t$ I have the same final state. In classical mechanics if I have the same initial configuration then after time $t$ I have the same final state. Over here same refers time translational invariance. Given this any semi-classical theory should agree on time translational invariance. $\endgroup$ – More Anonymous Sep 12 '19 at 14:29
  • $\begingroup$ FWIW, Noether's original theorem applies to a classical Lagrangian system, not quantum mechanics per se. $\endgroup$ – Qmechanic Sep 12 '19 at 15:48
  • $\begingroup$ @Qmechanic I see. However people do use Noether's theorem in QM and at least during the unitary evolution the original interpretation is correct. $\endgroup$ – More Anonymous Sep 12 '19 at 15:50

The reason you cannot apply Noether's theorem reasoning to this scenario is because the dynamics, given by the Hamiltonian, are not transitionally invariant in time. If the system is governed by the free Hamiltonian $H_{sys}$ that is time independent then it is fine. However, when you are measuring the system you have to apply an external interaction to do the measurement. The external iteration term $H_{int}$ in the Hamiltonian is switched on temporarily to do the measurement and then it is switched off. The total Hamiltonain is therefore time dependent $$H(t)=H_{sys}+\alpha (t)H_{int}$$ $\alpha (t)$ is a stet function in time.

When you are changing the dynamics externally to the system, you cannot expect of the energy to be conserved. Noether's theorem is only valid if the (time translation) symmetry actually holds.

  • $\begingroup$ I explicitly write: same initial configuration "physical system" + "Apparatus" ... In which case Noether's theorem should apply to the whole system (please update your answer in light of this comment) $\endgroup$ – More Anonymous Sep 12 '19 at 18:28
  • $\begingroup$ @MoreAnonymous The validity of Noether's theorem depends on the dynamics, initial configuration has nothing to do with it. The whole system ("physical system" + "Apparatus") has a time dependent Hamiltonian, thus, for the whole system Noether's theorem does not hold, thus, the energy of the whole system is not conserved. $\endgroup$ – oleg Sep 12 '19 at 19:44
  • $\begingroup$ @oleg I can do the double slit experiment with a measuring instrument stopping the interference pattern and use a battery put the room as a closed system ... Then does your objection still hold? (If I consider the battery part of the apparatus?) $\endgroup$ – More Anonymous Sep 12 '19 at 19:52

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