Given a pure state $|\psi\rangle_{AB}$ on a joint system $AB$, we can consider the reduced density operator $\sigma_A = Tr_B(|\psi \rangle \langle \psi|)$ on $A$ and subsequently purify this state into $|\phi\rangle_{RA}$ using a new register $R$ (ie $\sigma_A = Tr_R(|\phi \rangle \langle \phi|)$. My question is: what is the joint system on $RAB$? Is it the product state $|\phi \rangle \langle \phi|_{RA} \otimes \sigma_{B}$ where $\sigma_B = Tr_A(|\psi \rangle \langle \psi|)$.
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$\begingroup$ Have you tried any examples? $\endgroup$– DaftWullieJul 1, 2020 at 10:25
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2 Answers
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Let's write the state in terms of its Schmidt decomposition. $$ |\psi\rangle_{AB}=\sum_i\alpha_i|u_i\rangle_A|v_i\rangle_B, $$ ($|u_i\rangle,|v_i\rangle$ are orthonormal bases). Then $$ \rho_A=\sum|\alpha_i|^2|u_i\rangle\langle u_i| $$ and the purification that you would attempt would map $|u_i\rangle\mapsto |u_i\rangle_A|w_i\rangle_R$ (state $\sum_i\alpha_i|u_i\rangle|w_i\rangle$). Hence, you can see that the overall state would become $$ \sum_i\alpha_i|u_i\rangle_A|v_i\rangle_B|w_i\rangle_R, $$ which is GHZ-like, and certainly not separable unless $|\psi_{AB}\rangle$ was separable.
answered Jul 1, 2020 at 10:30
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$\begingroup$ Thank you. I'm a bit puzzled by the fact that if we trace out B in $|\psi\rangle_{ABR}$ it doesn't come back to the purification of $\rho_A$ (a pure state on RA). I guess I had a wrong intuition about what purification is. $\endgroup$ Jul 1, 2020 at 10:45
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$\begingroup$ The issue is that the purification starts just from $\rho_A$ and completely ignores the existence of the $B$ system. If you wanted a global state $|\psi\rangle_{ABR}$ to trace out $B$ returning a pure state, then, as you said $|\psi\rangle_{ABR}$ would have to be $AR|B$ separable. But the purification unitary $U_{AR}$ only acts on $A$ and $R$. So it cannot change the entanglement across the $AR|B$ partition. So the initial $|\psi\rangle_{AB}$ would have needed to be separable. $\endgroup$ Jul 1, 2020 at 10:58
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$\begingroup$ (I realise I've rather randomly introduced this "purification unitary". Hopefully it helps to convey what I'm trying to say without any further specifics.) $\endgroup$ Jul 1, 2020 at 11:00
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$\begingroup$ @DaftWullie Apologies for commenting on this old question. If you don't mind me asking Wullie, how can $\sum_i\alpha_i|u_i\rangle|w_i\rangle$ even exist here? If A is the reduced state, and a mixed one, then it's already entangled with B. So how can it be purified by R as you have shown if we also consider B, and get the GHZ-like state at the end? Doesn't monogamy of entanglement prevent that? $\endgroup$ Sep 30, 2020 at 21:02
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1$\begingroup$ @GaussStrife Depends what you want it for. Personally, I don't see a context where you would want to purify onto a reference system - B is already that purification. But that's what the question was, so.... $\endgroup$ Oct 1, 2020 at 12:41
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One way to paraphrase this question is to say that it hypothesizes a state that will not exist in general and then asks what that state is. There is no state of RAB in general that is consistent with both RA and AB being pure; this can only happen when RAB is in a pure product state $|\psi_R\rangle|\psi_A\rangle|\psi_B\rangle$.
For example, it could be that R, A, and B are all qubits, and we start with AB maximally entangled. This means that A is completely mixed, and purifying A to RA is again maximally entangled. However, there is no state of RAB that is consistent with both RA and AB being maximally entangled, which is a phenomenon commonly known as the monogamy of entanglement.
So what specifically is going wrong in the question? Well, we can certainly consider the state of A in isolation, and consider a purification of this state in RA, but there is no way to do this that is consistent with the existence of B and the state of AB being pure. Purification is not a physical process, for instance, where some quantum channel takes as input the system A and outputs RA that purifies A.
