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In decoherence theory, the basic situation is the following (I illustrate with two level system for simplicity). I want to measure a system $S$ by the mean of an apparatus $A$. Around it there is the environment $E$.

I assume the system is initially in $|\psi_S \rangle = a |0 \rangle + b |1\rangle$, the apparatus and environment in a state $|0\rangle$

The first step if the pre-measurement: system and apparatus get quantum correlated:

$$|\psi_S \rangle |0\rangle |0\rangle \rightarrow \left( a|00\rangle + b|11\rangle \right) |0\rangle $$

The second step is the correlation with the environment:

$$\left( a|00\rangle + b|11\rangle \right) |0\rangle \rightarrow a|000\rangle + b|111\rangle$$

Tracing out the environment, we find the good mixed density matrix for system-apparatus:

$$\rho_{SA}=|a|^2 |00\rangle \langle 00| + |b|^2 |11 \rangle \langle 11 |$$

My question

In decoherence theory, as far as I understood there is no need for a collapse: all the measurement can be done based on unitary dynamics as I showed. However what disturbs me is that in practice, the experimentalist will either find $|00\rangle$ or $|11\rangle$ (with probabilities $|a|^2$ or $|b|^2$).

But if there were no collapse: what does the final global state becomes ? Is it still in $a|000\rangle + b|111\rangle$ ? I would find it weird because we now know that $SA$ is in $|00\rangle$.

Then is it now in $|000\rangle$ ? If it is the case it means a collapse occured on the system $SAE$.

I am confused.


About the bounty: In the comment it appears that indeed decoherence theory doesn't say anything about when I will have read the outcome (if I find $|00\rangle$, what will the environment state be).

However what confuses me is that taking a system composed of three photons, we know for sure that if the two first are in $|00\rangle$ the last one will be in $|0\rangle$ if the initial state was $a |000\rangle+b|111\rangle$. The entanglement will be completly broken basically. And it seems to be a missing explanation from decoherence.

What I want to be 100% sure about is that indeed decoherence doesn't explain this. I would like to see a clear source in which this specific issue is stated, and where it is said it does'nt solve it. In many sources we can see that it doesn't solve the measurement problem but it is very vague. If it means that it doesn't explain the origin of probability only it doesn't mean it would'nt solve this problem.

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    $\begingroup$ Decoherence doesn't solve the measurement problem. It doesn't help predict which outcome will occur when something is measured, and it doesn't alleviate our inability to derive Born's rule. But it does help us identify which observable (if any) has actually been measured, and with what quality, so that we know when Born's rule can safely be applied. See @PeterShor's answer to a related question. $\endgroup$ – Chiral Anomaly Jun 6 at 23:40
  • $\begingroup$ @ChiralAnomaly thank you for your comment. Tell me if I am right in my reformulation. Decoherence theory gives us a state for $SAE$ after the measurement. We see that the reduced density matrix on $SA$ has the shape it should. However we still do not really know how to interpret all this. Basically if I found $|00\rangle$ on $SA$, we don't really know what is means about the state $SAE$. Maybe more directly: it gives the good statistical behavior of a measurement process (i.e in term of density matrix). But once one actually reads the result, decoherence fails to explain what happens next $\endgroup$ – StarBucK Jun 6 at 23:46
  • $\begingroup$ But if we wanted to restrict the physics to unread measurements, it is a satisfying answer. $\endgroup$ – StarBucK Jun 6 at 23:48
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    $\begingroup$ Yes, those are the right ideas. By the way, for the theory to have any real content, we need a good model of the environment as a very complicated system so that the entanglement in SAE is practically irreversible. That's the key to inferring which observable was measured. $\endgroup$ – Chiral Anomaly Jun 6 at 23:52
  • $\begingroup$ @ChiralAnomaly Great ! Thanks $\endgroup$ – StarBucK Jun 6 at 23:54
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Let us denote the System + Apparatus' Hilbert space as $\mathcal{H_{SA}}$ and the Environment's Hilbert space as $\mathcal{H_E}$. Your SAE state is an entangled state which lives in the Hilbert space $\mathcal{H_{SA}} \otimes \mathcal{H_E}$. The experimentalist performs her operations on SA after tracing out the environment, where she now works with an effective theory of SA whose Hilbert space is just $\mathcal{H_{SA}}$, while the Hilbert space $\mathcal{H_E}$ ceases to exist in her description. This means that she has destroyed any entanglement between SA and E, and is left with your reduced density matrix on SA. As a result, she cannot conclude that the environment is the state $| 0\rangle$ anymore because the operation of tracing over has now reduced her Hilbert space from $\mathcal{H_{SA}} \otimes \mathcal{H_E}$ to $\mathcal{H_{SA}}$. The notion of a "global state" living on $\mathcal{H_{SA}} \otimes \mathcal{H_E}$ does not exist anymore. The experimentalist can now perform a measurement on SA and find $| 00\rangle$ or $| 11\rangle$. The measurement of the environment can now give a totally uncorrelated value as indicated from the initially entangled state, depending on its reduced density matrix.

Another way to see this is from the path integral perspective, where the transitional amplitudes are now given by "influence functional". Here you have integrated out the environment in SAE path integral, and thus are left with "influence phases" in the integral due to the SA-E interaction. Finally you are only concerned with amplitudes between in and out states living in $\mathcal{H_{SA}}$.

See Section 4 of this review, where the author talks about a similar SAE state as in the question. The preceding sections of this paper might be useful too, as well as the references therein. For the influence functional, see this pioneering paper.

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  • $\begingroup$ Thank you for your answer. I agree that the experimentalist will not know the state of the environment which is why he focuses on $SA$ as you said. However if we want a global theory, the state of the environment still exists even if we do not have access to it. So I do not agree when you say "the global state doesn't exist anymore". It exists, it is just that the experimentalist doesn't have access to it. Then, with this global vision it seems a collapse is still required in the description: decoherence doesn't remove the need for it. You still need it at the very end. $\endgroup$ – StarBucK Jun 21 at 11:32
  • $\begingroup$ I already read the paper you refer (Zurek). However, this question of is there indeed a collapse at the very end for the global state is not (as far as I read) explained explicitly in this document (but maybe I missed the part where he talks about it as the document is quite big). It is the kind of resource I am looking for ! Thanks ! $\endgroup$ – StarBucK Jun 21 at 11:33
  • $\begingroup$ @StarBucK My main point here is that if the experimentalist performs the measurements on SA after tracing out the environment (as asked in OP), this effectively destroys the entanglement between the SA and E Hilbert spaces. As a consequence, your environment can be in any state depending on its reduced density matrix. This is different from performing measurements on SAE Hilbert space where the entanglement is untouched, and where you will get $|0\rangle$ for the environment as well. $\endgroup$ – Bruce Lee Jun 22 at 14:36
  • $\begingroup$ @StarBucK Have a look at Section 4 of Zurek, which is quite insightful. The main problem in that section is phrased as you have posed your problem. The author doesn't address your question exactly, however IMO it is a very good analysis of the SAE problem. $\endgroup$ – Bruce Lee Jun 22 at 15:53
  • $\begingroup$ I'm sorry but I don't see where my question is answered in this document. It explain something different which is how the environment plays a role to make decoherence and why it solves basis ambiguity. But it doesn't explain what happens after the measurement is read, neither that decoherence actually doesn't really answer this question so that you still need a collapse at the very end. @Chiral Anomaly told me that indeed in a way you still need a collapse when you read the measurement. My point is that I would like a good reference in which it is clearly stated bc it is a very important point $\endgroup$ – StarBucK Jun 27 at 12:53

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