I've been studying quantum mechanics and quantum field theory for a few years now and one question continues to bother me.
The Schrödinger picture allows for an evolving state, which evolves through a unitary, reversible evolution (Schrödinger’s equation, represented by the Hamiltonian operator) and an irreversible evolution (wave function collapse, represented by a projection operator).
The Heisenberg picture holds the states constant and evolves the operators instead. It provides an equivalent representation of the unitary evolution on operators, but I haven't yet seen an equivalent Heisenberg representation of wave function collapse. Is there any accepted explanation for how to represent state collapse in the Heisenberg picture?
Thanks!
Answer
Well, I think you said the answer yourself when you used the words "projection operator." In the Heisenberg picture the operators get projected down to a subspace at the time of the collapse. In other words, the operator 'collapses' by picking up a projection piece that kills the unphysical part of the state.
Forget about pictures for a second, the physical thing is the full matrix element
⟨ψ,t1|U(t1,t2)O(t2)U(t2,t1)|ψ,t1⟩
Knowledge of the hamiltonian is buried inside of the time evolution operator U.
The Schrodinger picture amounts to grouping the U with the state so that |ψ(t)⟩=U(t,t∗)|ψ(t∗)⟩, the Heisenberg picture amounts to grouping the U with the operator so that O(t)=U(t,t∗)O(t∗)U(t∗,t). This is clearly an artificial split and nothing can ever depend on your choice of picture: if you express things in terms of the full matrix element the difference between the pictures always amounts to a different way of grouping terms.
How do we describe collapse? There is some special time tc, the collapse time, at which something non-unitary happens. We cannot use U to evolve past tc.
Or in other words, the relationship U(t2,t1)=U(t2,tc)U(tc,t1)
So let's say we want to evaluate the physical matrix element, we have to include this projection operator
⟨ψ,t1|U(t1,tc)N∗cPcU(tc,t2)O(t2)U(t2,tc)NcPcU(tc,t1)|ψ,t1⟩
So again we have a choice of how we group things. We could group things in a Schrodinger way so that
|ψ(tc+ϵ)⟩=NcU(tc+ϵ,tc)PcU(tc,t1)|ψ,t1⟩=NcPc|ψ,tc⟩+O(ϵ)
This is the 'state collapse.' At tc the state changes so that it is projected down onto a subspace.
Or, we could group things in a Heisenberg way, so that
O(tc+ϵ)=U(tc+ϵ,tc)N∗cPcU(tc,t1)O(t1)U(t1,tc)NcPcU(tc,tc+ϵ)=|Nc|2PcO(tc)Pc+O(ϵ)
This is "the operator being projected onto a subspace." The state is the same, but the operator now includes a projection piece that cancels out the part of the state that is no longer physical.
EDIT # 1: I previously said U(t2,t1)=U(t2,tc)PcU(tc,t1), which is incorrect. The basic point still stands but the math was technically wrong.
Whoops! I was right the first time. Thanks to Bruce Connor for making me rethink through this point. I was confused because I thought wanted the transformation rule PcUPc, which is how you would project the time evolution operator to the collapsed subspace. But that is not what we want here: the time evolution operator is special. The point is that you project down to the subspace (say a position eigenstate) at tc, then you evolve normally from there. In particular you are allowed to evolve out of the subspace. For example, after we observe a particle at position x the particle is allowed to evolve a probability to be at x′. You don't want to force the evolution to stay in the subspace, that's what the second Pc would have done.
EDIT # 2: Sorry for all the edits, this is a little more subtle to get exactly right than I originally thought. You aren't just projecting the state down to a subspace, you are projecting the state and then rescaling it so that it has the correct normalization.
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