# Why is "this" particular decomposition/regouping of states observed?

This question is motivated from this as well as the regrouping of terms in Shor code.

The setup is not important, but it is being provided for context.

$$|\psi\rangle=\alpha|0\rangle+\beta|1\rangle$$

$$|\psi_2\rangle = \frac{1}{2}\left[ \alpha(|0\rangle+|1\rangle)(|00\rangle+|11\rangle)+\beta(|0\rangle-|1\rangle)(|10\rangle+|01\rangle) \right].$$

Which can be rewritten as:

$$|\psi_2\rangle = \frac{1}{2}\left[ |00\rangle(\alpha|0\rangle + \beta|1\rangle) + |01\rangle(\alpha|1\rangle + \beta|0\rangle) + \\|10\rangle(\alpha|0\rangle - \beta|1\rangle) + |11\rangle(\alpha|1\rangle - \beta|0\rangle) \right].$$

As we can see, all that needs to be done is measure the first two qubits and we can determine the third qubit's state

Why exactly is this regrouping of terms lead to what we observe in the third qubit's state? I have asked this question earlier, but there were no satisfactory answers. To be clear, I am looking for a QM-based explanation as to why this occurs, not the mathematical justification of why this happens (because it's easy to see that the terms can be regrouped in many ways) Does this have anything to do with non-locality? If so, how?

Let's forget about grouping the terms and write the state of $$\psi_2$$ as a superposition of basis states. We have, $$|\psi_2\rangle = \frac{1}{2}\left[ \alpha(|0\rangle+|1\rangle)(|00\rangle+|11\rangle)+\beta(|0\rangle-|1\rangle)(|10\rangle+|01\rangle) \right]$$ So, $$|\psi_2\rangle = \frac{1}{2}\left[ \alpha|000\rangle+\alpha|100\rangle+\alpha|011\rangle+\alpha|111\rangle+\beta|010\rangle-\beta|110\rangle+\beta|001\rangle-\beta|101\rangle\right]$$
Now, assume that we measured the first two qubits and the outcome, say, $$00$$. Then state will collapse to $$\alpha|000\rangle+\beta|001\rangle$$. Which means, the state of the third qubit after measurement is $$\alpha|0\rangle+\beta|1\rangle$$.