Modeling a measurement as unitary

[atyy]

Another thought. If the question is how can the state of the detector can be prepared, then it doesn't have to be unitary. One way of state preparation is to make a projective measurement, since at the end of the measurement the wave function collapses into a known definite state. Within Copenhagen unitary evolution only applies between measurements, and there is no need for state preparation to be unitary, since measurement can be a form of state preparation.

 

[PeterDonis]

Within Copenhagen unitary evolution only applies between measurements, and there is no need for state preparation to be unitary, since measurement can be a form of state preparation.

Good point; I should have clarified that I am trying to find out whether the measurement process, or more generally any state preparation process, *can* be modeled as unitary, because if it can't, and that could be proven, it would seem to throw any interpretation of QM that assumes it can (such as the MWI) out of court.

 

[The_Duck]

I'm not arguing that it can't be modeled that way. I'm arguing that such a model, taken at face value, is not unitary.

As I mentioned, Bohm in his QM textbook works out in some detail what happens when you pass a spin-1/2 particle through a Stern-Gerlach apparatus. The initial spin state is a superposition of spin-up and spin-down. The initial spatial state is a wave packet. The final state is of course a superposition of a wave packet with spin up and a spatially separated wave packet with spin down. This nicely corresponds with your simplified model of a measurement:

$|\uparrow\rangle |R\rangle \to |\uparrow\rangle |U\rangle$

$|\downarrow\rangle |R\rangle \to |\downarrow\rangle |D\rangle$

$\frac{1}{\sqrt 2}(|\uparrow\rangle + |\downarrow\rangle) |R\rangle \to \frac{1}{\sqrt 2}(|\uparrow\rangle |U\rangle + |\downarrow\rangle |D\rangle)$

Here $|\uparrow\rangle$ and $|\downarrow\rangle$ are spin states, $|R\rangle$ is the spatial state representing the initial wave packet, and $|D\rangle$ and $|U\rangle$ are the two possible final-state wave packets. The spatial state is the "measuring apparatus" and becomes entangled with the spin state.

You work this out just by integrating Schrödinger's equation for a spin-1/2 particle in the appropriate magnetic field, so the process is manifestly completely unitary. This is a nice simple model of measurement of essentially the kind you were considering in the OP. The Hamiltonian doesn't even need to be time-dependent.

In your original post you worried that if the final states $|\uparrow\rangle|U\rangle$ and $|\downarrow\rangle|D\rangle$ are unchanged under time evolution then time evolution is not unitary. You're right--if these states were invariant under time evolution then unitarity would be violated. Since we known the Schrödinger equation produces unitary evolution, we can conclude that these two states are not invariant under time evolution. In the case of the Stern-Gerlach experiment, this is abundantly clear: $|U\rangle$ and $|D\rangle$ are moving wave packets and therefore clearly not invariant under time evolution.

 

[atyy]

Good point; I should have clarified that I am trying to find out whether the measurement process, or more generally any state preparation process, *can* be modeled as unitary, because if it can't, and that could be proven, it would seem to throw any interpretation of QM that assumes it can (such as the MWI) out of court.

State preparation can be obtained by measurement in which the wave function collapses, followed by choosing those cases in which the collapse was into the desired state. So if MWI can replace collapse with unitary evolution, it will make all stages of measurement (preparation, interaction, collapse) unitary. The MWI versions that seem closest to solving the problem incorporate decoherence, in which it is the system, apparatus and environment as a whole that evolve unitarily.

 

[PeterDonis]

$|U\rangle$ and $|D\rangle$ are moving wave packets and therefore clearly not invariant under time evolution.

By "moving wave packets" I assume you mean something like coherent states?

 

[The_Duck]

By "moving wave packets" I assume you mean something like coherent states?

I mean a wave packet like $\psi(x) = \exp(i k x - x^2/2\sigma^2)$.

 

[atyy]

Although I think it is only collapse that is necessarily non-unitary, and since measurement and collapse can be used as state preparation, I don't think MWI has to solve the problem of state preparation separately from the non-unitarity of collapse. But since we discussed time-dependent Hamiltonians above, here are some references about the extent to which unitary operations with time-dependent Hamiltonians enable one to move between any two states.
http://arxiv.org/abs/quant-ph/0106128
http://arxiv.org/abs/quant-ph/0108114

 

[stevendaryl]

Although I think it is only collapse that is necessarily non-unitary, and since measurement and collapse can be used as state preparation, I don't think MWI has to solve the problem of state preparation separately from the non-unitarity of collapse. But since we discussed time-dependent Hamiltonians above, here are some references about the extent to which unitary operations with time-dependent Hamiltonians enable one to move between any two states.
http://arxiv.org/abs/quant-ph/0106128
http://arxiv.org/abs/quant-ph/0108114

I don't understand the comments about MWI. I thought the whole point of MWI was to get rid of collapse, so there is nothing non-unitary going on.

 

[atyy]

I don't understand the comments about MWI. I thought the whole point of MWI was to get rid of collapse, so there is nothing non-unitary going on.

Yes. What I'm saying is that if MWI is able to convincingly get rid of collapse, then it doesn't have to solve state preparation as a separate problem.

The other point I was making is that although collapse can prepare a state, state preparation within Copenhagen is not necessarily non-unitary, but collapse is necessarily non-unitary.