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Actually, unitary pseudo-collapse?
Von Neuman said quantum mechanics proceeds by two processes: unitary evolution and nonunitary reduction, also now called projection, collapse and splitting.

Collapse is non-unitary by definition, and the impossibility of unitary collapse is also a basic mathematical theorem. But we no longer believe, as Bohr did, in two Kingdoms, one classical and one quantum. Instead, we now talk about the emergence of the pseudo-classical. Similarly, I suggest, we can now talk about a pseudo-collapse, which is pseudo-unitary.

In a collapse two things happen: First, the matrix diagonalizes, and then, it becomes one dimensional, ie one element of the main diagonal becomes 1 and all the others become 0. This second part is the nonunitary projection part. (In a slight variation, we have two corresponding matrices, labelled system and apparatus, and the coresponding "pointer state" in the apparatus goes to 1, and all the other pointer states go to 0.) Now along comes decoherence, and we have three matrices, S,A, and E: System, Apparatus and Environment. Actually, these three matices are submatrices, ie diagonal blocks in one bi gger matrix. You can still diagonalize everything by unitary operations. But you still can't project it to a one dimensional definite outcome by unitary means. However, I think you can make the S block one dimensional by unitary operations if you push the off diagonal terms into the S-E quadrant. Is this correct?

This seems almost obvious to me, but I can't find confirmation. In fact, as I searched, I found lots of discussion on the diagonalization process, but almost nothing about the selection step, which is the real crux of the process, as far as `I am concerned. Why is this? Can someone point me to a good discussion of the selection process? And confirm that my pseudo-unitary pseudo-collapse is mathematically possible?

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I cannot understand what for a WF collapse can still be needed today. The Gell-Mann Griffiths Omnes school (Consistent Histories) has shown that not even a medium is necessary for decoherence. Decoherence makes ordinary logic work well in QM, is only impractical to compute. I think they solved the problem but I also may be wrong, did not followed the field. –  Lupercus Sep 5 '12 at 1:53

3 Answers 3


Rigorous adherence to the liturgical rituals of the "Church of the Larger Hilbert Space" is feasible in principle yet exponentially inefficient in practice.


One way to answer this question is by reference to a feasible numerical computation.

So fire-up MatLab; specify the dynamical system as (say) $n\sim 10$ interacting qubits; specify any desired Hamiltonian; choose some starting energy $E$; then integrate a dynamical trajectory $\psi(\,t\,|\,E\,)$.

Now perform the following exercise:

Exercise I  Solely from operator expectation values associated to an $n$-qubit unitary trajectory $\psi(\,t\,|\,E\,)$, estimate the single-qubit Bloch relaxation parameters $T_1(E)$ and $T_2(E)$.

If you are more ambitious:

Exercise II (extra credit)  Follow in the footsteps of Urey, Onsager, Dirac, Feynman (etc.) and estimate the thermodynamic transport coefficients of larger systems of interacting qubits, again wholly by reference to the unitary dynamical trajectory $\psi(\,t\,|\,E\,)$.

Then we have the following assertion

Assertion  Without loss of accuracy in estimating $T_1(E)$ and $T_2(E)$, all but a fraction $\mathcal{O}(e^{-n})$ of the unitary dynamical trajectory $\psi(\,t\,|\,E\,)$ can be discarded.

The portion of the unitary trajectory that can be discarded — for the practical purpose of estimating thermodynamic parameters — is of course associated to qubits in "cat" states.

That is the practical reason why the "Church of the Larger Hilbert Space" is not popular among systems engineers … its liturgical rituals are exponentially inefficient!

Alternate doctrines

Efficient computational recipes for estimating thermodynamic parameters are extant in the literature; these — necessarily non-unitary — quantum dynamical recipes are reviewed in an answer to the question "Reversing gravitational decoherence."

Open questions

We have seen that rigorous adherence to the liturgical rituals of the "Church of the Larger Hilbert Space" is feasible in principle yet exponentially inefficient in practice. It is natural to wonder:  "Does Nature herself embrace the strictly unitary yet exponentially inefficient liturgical rituals of the Larger Hilbert Space? Or does she resort to computationally efficient yet non-unitary dynamical trajectories similar to those of mortal engineers?"

These questions are open.

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Thanks for your answer. I am very interested in a more efficient way of doing QM. I have not been able to do your suggested exercises, or look up all of your references, but I am still working on them. –  Jim Graber Sep 8 '12 at 13:01

I believe "a good discussion of the selection process" can be found in . However, I prefer just to reject collapse, based, e.g., on Schlosshauer's analysis of experimental data: "no positive experimental evidence exists for physical state-vector collapse;" (M. Schlosshauer, Annals of Physics, 321 (2006) 112-149). Collapse is an approximation in the best case. A more general reason to reject collapse - it contradicts unitary evolution.

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Thanks for good reference. –  Jim Graber Sep 8 '12 at 12:52
Glad you liked it, but which one of the two?:-) –  akhmeteli Sep 8 '12 at 14:50
Actually both of them, but I had already known about Schlosshauer. The other was new to me. –  Jim Graber Sep 9 '12 at 14:32
You may also wish to look at their earlier work - it is older, but much, much shorter. It was eye-opening to me :-) –  akhmeteli Sep 9 '12 at 14:59

Yes, this is correct. Let $\rho$ be the initial state of the system S and let $|0>$ be the initial state of the apparatus. Take $P_m$ to be a set of rank-1 projectors you measure, with $m$ denoting the outcome. Then $\rho \otimes |0> \mapsto \sum_m P_m \rho \otimes |m>$, which we can denote as $U$, can be extended to a unitary operation for any initial state $\rho$. This is because for any two pure initial states $|\psi>, |\phi>$ we have that $(<\phi| \otimes <0|) U^\dagger U (|\psi> \otimes |0>) = <\phi|\psi>$. In words, $U$ acts as a unitary operation on this particular subspace. For a proof that there exists a unitary extension to the entire Hilbert space, see for example Nielsen and Chuang, Quantum Computation and Quantum Information.

At the conclusion of this process, the system state is of the form $\sum_m p_m |m><m|$, which is what I think you meant by "block one-dimensional", unless I misunderstood your question.

Also, a good review of decoherence is this article by Zurek. If this is what interests you, I recommend checking this out

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Thanks for Zurek reference. I was surprised he discussed "selective loss of information". I thought that no information loss was the whole point of unitarity. –  Jim Graber Sep 8 '12 at 12:54
I was also able to briefly look at a copy of Nielsen and Chuang. i looked through the index and table of contents, but did not locate the extension proof. Can you give me a more detailed pointer? TIA. –  Jim Graber Sep 8 '12 at 12:57
It's pretty early in the book, after they talk about the measurement postulate, page 95. You need to do exercise 2.67, just underneath, to understand why the assertions are true. But where he talks about loss of information he means loss of information in a subspace. Unitarity does mean that information is preserved, but if you only see a part of the evolving system, the effective evolution might no longer be unitary. –  SMeznaric Sep 8 '12 at 18:41
I editted the answer to change the state, as it said before that the joint state becomes mixed and it should have said the system state becomes mixed block one dimensional. If you have the environment in addition to the ancilla you can make the system-apparatus state block-one dimensional as well in the same way. –  SMeznaric Sep 8 '12 at 20:53
I will be able to look at N&C again on Monday, and I will check out pp 95+. Thanks for the pointer. I think your point about loss of information on a subspace probably answers my question. –  Jim Graber Sep 9 '12 at 14:41

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