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I have this general state $\lvert \psi \rangle_{AB} = \sum_{\alpha\beta}c_{\alpha\beta}\lvert\alpha \rangle \otimes \lvert \beta\rangle$ of two qbits. I want to write the state of the system after the measurement. The measurement is performed only on the first qbit. Let's say that the value found was $\alpha_1$, I wrote:

$\lvert\psi\rangle_{AB} \longrightarrow \lvert \psi_{\alpha1}\rangle$ = $\lvert \alpha_1\rangle \langle \alpha_1 \rvert \otimes1\sum_{\alpha\beta}c_{\alpha\beta}\lvert \alpha\rangle\otimes\lvert \beta \rangle$ = $\lvert \alpha_1\rangle\otimes\sum_\beta\lvert\beta\rangle$.

Is it right? Is this the state of the system after measurement?

This is a decomposed state, as it seems to me. The lecture note I was studying said that the state of the system after the measurement is always a decomposable state, and I should show that this is true and also that the second qbit is dependent on the result of the measurement due to the first. I cannot see that. If the result of my calculation is right, the second qbit is the combination $\sum_\beta\lvert \beta\rangle$, but was eliminated any possibility of a tensor product with any other state other than $\lvert \alpha_1 \rangle$. That is, anything that is not $\lvert \alpha_1\rangle$ in the first qbit is not possible anymore, but any result in the second qbit is still possible.

What is wrong with my conclusion, and why is that?

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You're right that the post-measurement state would be represented by the action of the projection operator $(\vert \alpha_1 \rangle\langle \alpha_1\vert \otimes \mathbb{I})$ on the pre-measurement state (up to a constant, of course). However, the resultant of this would not get rid of all the coefficients $c_{\alpha\beta}$. Rather (check this by explicitly writing the Kronecker delta function before you get rid of the summation over $\alpha$), it would give $\vert \alpha_1\rangle \otimes \sum_{\beta} c_{\alpha_1 \beta} \vert\beta\rangle$. Thus, the post-measurement state is a product state but the state of the $B$ subsystem is $\sum_{\beta} c_{\alpha_1 \beta} \vert\beta\rangle$ where the dependence of the coefficients on $\alpha_1$ makes it clear that the post-measurement state of the $B$ system depends on the particular state onto which the $A$ system collapses.

Notice that if $c_{\alpha\beta}$ were of the form $c_{\alpha} c_{\beta}$ then $\sum_{\beta}c_{\alpha_1\beta} = c_{\alpha_1}\sum_\beta c_{\beta}$ and thus the post-measurement state of system $B$ wouldn't depend on the outcome of the measurement on system $A$ because the prefactor $c_{\alpha_1}$ would be weeded out in normalization without any physical significance whatsoever. As you'd notice, this is simply reflecting the fact that the post-measurement state of one of the subsystems doesn't depend on the measurements done on the other subsystem in a product state but does depend on it in an entangled state.

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    $\begingroup$ Yeah, now I see. I was too fast getting rid off all of the coefficients. Now it is clear why it was not making sense to me. You are completely right, thank you very much for pointing out the mistake and for the very detailed answer. $\endgroup$
    – Dimitri
    Jun 10, 2020 at 2:41

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