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Consider setting up two experiments $\alpha $ and $\beta$.

In experimental setup $\alpha$ we can prepare $n$ possible pure states $\{| \psi_1 \rangle, | \psi_2 \rangle, ... | \psi_n \rangle\}$ with associated probabilities $\{p_1,p_2,...p_n\}$.

In experimental setup $\beta$ we prepare $m$ non-interacting systems. Each system prepared in its corresponding lower energy states $\{| \phi_1 \rangle, | \phi_2 \rangle, ... | \phi_n \rangle\}$.

My question is how would one deduce the density matrix operators for the quantum states prepared in each experiment. Is it simply a matter of noting that one can denote the density operator as a sum of a set of projectors as follows: $\rho = \sum_j p_j |j\rangle \langle j|.$

So then I may kind of assume that the density matrix operator for states prepared in $\alpha $ will be $\rho_{\alpha} = \sum_j p_i |\psi_i\rangle \langle \psi_i|.$

But I am unsure if this is correct I am also unsure how to construct the equivalent for experiment $\beta$.

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  • $\begingroup$ I'm not sure I understand "we can prepare $m$ non-interacting systems prepared..." in the second set up. You know that all $m$ systems have been prepared, each in a pure state? or you know that one of them only has been prepared, but you don't know which? $\endgroup$
    – pglpm
    Nov 30, 2020 at 20:37
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    $\begingroup$ Thank you for the more precise edit of your post. The density matrix for the first setup is correct. For the second setup, consider that you have a joint system with joint state $\lvert\phi_{i_1},\dotsc, \phi_{i_m}\rangle$, where $\phi_{i_k}$, with $i_k \in \{1,\dotsc,n\}$, is the state of the $k$th system. Then the density matrix is simply $\lvert\phi_{i_1},\dotsc, \phi_{i_m}\rangle\langle\phi_{i_1},\dotsc, \phi_{i_m}\rvert$ – that is, it's the density matrix that represents a pure state (of the joint system). $\endgroup$
    – pglpm
    Nov 30, 2020 at 20:58
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    $\begingroup$ I think it'd be better if you explained why you're unsure, otherwise the bottom problem isn't solved. $\endgroup$
    – pglpm
    Nov 30, 2020 at 21:01
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    $\begingroup$ Great! By the way, note that "non-interacting" is a superfluous specification here. It's a property of the dynamics of a joint system, rather than its state. It means that its evolution operator has the form $U_1 \otimes \dotsb \otimes U_m$, but it says nothing about its initial state. $\endgroup$
    – pglpm
    Nov 30, 2020 at 21:13
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    $\begingroup$ I'll write this into an answer, otherwise moderators will (justly) complain. $\endgroup$
    – pglpm
    Nov 30, 2020 at 21:15

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With some more thought you arrive at the answer for the second part of your question, since you have the correct understanding of the first part.

Setup $\alpha$: we know one system has been prepared in one of the $n$ possible pure states $\{\lvert\psi_1\rangle,\dotsc,\lvert\psi_n\rangle\}$. We don't know which, but we assign probabilities $p_1,\dotsc,p_n$ to the $n$ possibilities. Then the density matrix representing our knowledge of this setup is, as you correctly wrote, $\sum_i p_i \lvert\psi_i\rangle\langle\psi_i\rvert$.

Setup $\beta$: we know that $m$ systems have been prepared in $m$ pure states (possibly identical for some of them), each from among the set $\{\lvert\phi_1\rangle,\dotsc,\lvert\phi_n\rangle\}$. We also know the preparation state of each system, say $\lvert\phi_{i_k}\rangle$ for the $k$th system.

This means that their joint system has been prepared in the pure state $\lvert \phi_{i_1},\dotsc,\phi_{i_m}\rangle$, which is equivalent to the density matrix $\lvert \phi_{i_1},\dotsc,\phi_{i_m}\rangle\langle \phi_{i_1},\dotsc,\phi_{i_m}\rvert \equiv \lvert \phi_{i_1}\rangle\langle \phi_{i_1}\rvert \otimes \dotsb \otimes \lvert \phi_{i_m}\rangle\langle \phi_{i_m}\rvert$. (We're assuming that there are no bosonic or fermionic symmetrization complications.)

Saying that they're non-interacting is unimportant for the specification of the state (kinematics): that qualification refers to the dynamics of the joint system, meaning that it has the form $U_1 \otimes \dotsb \otimes U_m$, where $U_k$ is the evolution operator acting on the $k$th system.

Just to add a reference, my favourite book about these topics: Bengtsson, Życzkowski: Geometry of Quantum States: An Introduction to Quantum Entanglement (2nd ed., Cambridge 2017).

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  • $\begingroup$ Quick question that is related to system $\beta$. If I consider a hermitian operator $\hat{O}$ that describes a physical observable associated with the kth system of $\beta$ would the expectation value of this operator simply be $Tr[\lvert \phi_{1}\rangle\langle \phi_{1}\rvert \otimes \dotsb \hat{O} \lvert \phi_{k}\rangle\langle \phi_{k}\rvert \dotsb \otimes \lvert \phi_{m}\rangle\langle \phi_{m}\rvert]$ $\endgroup$
    – DJA
    Nov 30, 2020 at 22:47
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    $\begingroup$ Yes – and you may notice that your expression can be simplified even further $\endgroup$
    – pglpm
    Nov 30, 2020 at 22:49
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    $\begingroup$ To just the operator itself right? $\endgroup$
    – DJA
    Nov 30, 2020 at 22:50
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    $\begingroup$ right! The trace of a tensor product is the product of the traces. $\endgroup$
    – pglpm
    Nov 30, 2020 at 22:51
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    $\begingroup$ Your help has been invaluable to my understanding!! Thanks so much!!! $\endgroup$
    – DJA
    Nov 30, 2020 at 22:52

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