# Calculating the reduced density matrix of two separate experiments

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 $$\beta$$.

• 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? Nov 30, 2020 at 20:37
• 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). Nov 30, 2020 at 20:58
• I think it'd be better if you explained why you're unsure, otherwise the bottom problem isn't solved. Nov 30, 2020 at 21:01
• 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. Nov 30, 2020 at 21:13
• I'll write this into an answer, otherwise moderators will (justly) complain. Nov 30, 2020 at 21:15

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).

• 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 $Tr[\lvert \phi_{1}\rangle\langle \phi_{1}\rvert \otimes \dotsb \hat{O} \lvert \phi_{k}\rangle\langle \phi_{k}\rvert \dotsb \otimes \lvert \phi_{m}\rangle\langle \phi_{m}\rvert]$
– DJA
Nov 30, 2020 at 22:47
• Yes – and you may notice that your expression can be simplified even further Nov 30, 2020 at 22:49
• To just the operator itself right?
– DJA
Nov 30, 2020 at 22:50
• right! The trace of a tensor product is the product of the traces. Nov 30, 2020 at 22:51
• Your help has been invaluable to my understanding!! Thanks so much!!!
– DJA
Nov 30, 2020 at 22:52