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I imagine the answer is yes since, if so, the definition of unentangled is rather non-obvious and yet it gives an operational way to check for statistical independence.

I am working with the standard (I think) definitions. I will use the vector representation of states (thereby limiting the discussion to pure states) though I'm sure whatever proof is supplied will be generalizable. Consider two systems with Hilbert spaces $\mathcal{H}_1$ and $\mathcal{H}_2$, as well as the corresponding composite system in $\mathcal{H}_1 \otimes \mathcal{H}_2$. A state $|\psi\rangle \in \mathcal{H}_1 \otimes \mathcal{H}_2$ is said to be unentangled if it is possible to write $|\psi\rangle$ in product form: $|\psi\rangle = |\psi^{(1)}\rangle \otimes |\psi^{(2)}\rangle$ for some $|\psi^{(i)}\rangle \in \mathcal{H}_i, i=1,2.$ A state is said to be uncorrelated if the condition of statistical independence is obeyed by the probability distributions associated with arbitrary observables on a particular subsystem, represented by operators of the form $A^{(1)} \otimes I$ and $I\otimes A^{(2)}$ (that is, if the joint pdf in the given state for arbitrary two observables factors into marginal pdfs for the individual observables).

That unentangled implies uncorrelated is clear, but I can't think of how to prove the converse. Is it true and, if so, can someone sketch the proof?

Tobias Fünke
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EE18
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1 Answers1

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Prelude: Consider a bipartite system of finite dimensions $H=H_A\otimes H_B$ and a state $\rho$ (pure or mixed) on $H$ with reduced density matrices $\rho_1$ and $\rho_2$ on $H_A$ and $H_B$, respectively.

Following Ref. 1, we say that $\rho$ state is uncorrelated if for all joint projective measurements represented by (hermitian) projection operators of the form $P_a \otimes P_b$ the associated joint probability distributions have no correlations, i.e. it holds that

$$\mathcal P_{\rho}(a,b) = \mathrm{Tr}\,\rho\, P_a \otimes P_b = \mathrm{Tr}\,\rho_1 \, P_a\, \mathrm{Tr}\, \rho_2 \, P_b = \mathcal P_{{\rho_1}}(a) \,\mathcal P_{{\rho_2 }}(b) \quad \tag{1} $$

for all hermitian projection operators $P_a, P_b$ on $H_A,H_B$. Define further for two hermitian operators $O_A,O_B$ on $H_A,H_B$ the following quantity:

$$ C_\rho(O_A,O_B):=\mathrm{Tr}\,\rho\,O_A\otimes O_B - \mathrm{Tr}\, \rho_1\, O_A \, \mathrm{Tr}\, \rho_2\, O_B \, \tag{2} \quad .$$

Theorem: The following statements are equivalent:

$(\mathrm i)$ $\rho=\rho_1 \otimes \rho_2$ $(\mathrm{ii})$ $C_\rho(O_A,O_B) = 0$ for all hermitian $O_A, O_B$ on $H_A, H_B$ $(\mathrm{iii})$ $\rho$ is uncorrelated.

Since a pure bipartite state is not entangled if and only if it is a product state, the theorem shows that in this case it holds that $\rho$ is not entangled if and only if it is uncorrelated.

Proof: There are several ways to prove this theorem, see e.g. also Ref. 1. To start, let us first prove the equivalence between $(\mathrm i)$ and $(\mathrm{ii})$.

To this end, note that $(2)$ can be rewritten as

$$C_\rho(O_A,O_B)=\mathrm{Tr}\, \left(\rho-\rho_1\otimes\rho_2 \right) \left(O_A\otimes O_B\right) \tag 3 \quad . $$

The above vanishes for all hermitian $O_A,O_B$ if and only if $$\mathrm{Tr}\, \left(\rho-\rho_1\otimes\rho_2 \right) O \tag 4 $$ vanishes for all hermitian operators $O$ on $H$; this follows from the fact that every hermitian operator $O$ on $H$ can be written as a linear combination of tensor products of hermitian operators, see e.g. this Math SE post. Vanishing of equation $(4)$ for all hermitian $O$ in turn is equivalent to $\rho-\rho_1\otimes\rho_2=0$.

To conclude the proof, we show the equivalence between $(\mathrm{ii})$ and $(\mathrm{iii})$. To do so, note that if $(2)$ vanishes identically, it must vanish in particular for all hermitian projection operators $O_A=P_a$ and $O_B=P_b$, which yields $(1)$, i.e. $\rho$ is uncorrelated. To prove the converse, just make use of the fact that every hermitian operator admits a spectral representation, which shows that if $(1)$ holds for all (orthogonal) projection operators $P_a$, $P_b$, then $(2)$ vanishes for all hermitian $O_A,O_B$. $\tag*{$\square$}$

References and further reading:

1.Quantum Information Meets Quantum Matter. Bei Zeng, Xie Chen, Duan-Lu Zhou, Xiao-Gang Wen. Springer. Section 1.3.3.

2.When is $\rho_{AB}=\rho_A\otimes \rho_B$ true?

Tobias Fünke
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  • Thank you for this proof (my apologies also for my errant attempt at an edit; I see now that the two statements I thought were the same are not actually so. Perhaps it is worthwhile to write that you are proving that three different statements are equivalent as it's not completely obvious, at least to the untrained eye.). The only line I don't follow is below (3) and beginning "Thus, (3) implies that...". How do you use the previous sentence here (do you use properties of the trace)? Also, I would have expected it to be necessary to write some "there exists" statements... – EE18 Mar 09 '23 at 20:36
  • ...near where you introduce the $\rho_i$. Are they tacitly there and I'm missing them? – EE18 Mar 09 '23 at 20:36
  • @EE18 Regarding your first comment/question: If you accept (as shown e.g. in the Math SE link) that $(3)$ vanishes for all hermitian $O_A,O_B$ if and only if it vanishes for all hermitian $O$ on $H$, then the argument is simple: If $\rho$ is a product state, i.e. $\rho=\rho_1 \otimes \rho_2$, then $C_\rho(O_A,O_B)=0$ for all (hermitian) $O_A,O_B$. Conversely, if it vanishes for all $O_A,O_B$, it vanishes for all hermitian $O$. There are several ways now; one is choosing $O=\rho-\rho_1\otimes\rho_2$ which gives $\mathrm{Tr}O^2=0$; but for our choice it holds that $O^2\geq O$ and thus $O=0$. – Tobias Fünke Mar 09 '23 at 20:43
  • Regarding your second question: I don't understand your comment here, could you elaborate? Do you know what reduced density matrices are? They exist for every bipartite state (pure or mixed). For example, in your notation: If $\rho$ is not entangled, then $\rho_i = |\psi^{(1)}\rangle\langle\psi^{(i)}|$. I think I've used the exact same notion of (stat.) independence as you did, since by the properties of reduced density matrices it holds that $\mathrm{Tr} \rho A\otimes \mathbb I^{(2)} = \mathrm{Tr}\rho_1 A$ and likewise for $\rho_2$. The second trace is understood as the trace on $H_A$. – Tobias Fünke Mar 09 '23 at 20:46
  • Ah, my bad regarding the second question. I didn't realize $\rho_i$ was the notation for the reduced density matrix in particular. I'd forgotten the fact that if a state factors then it (uniquely) factors into a product of the corresponding reduced density matrices. – EE18 Mar 09 '23 at 20:48
  • Regarding reduced density matrices, see e.g. this or this. Regarding my argument with the hermitian operators below eq. 3, see also this. Regarding your last comment: yes, exactly! if it factors, it factors in the reduce density matrices! – Tobias Fünke Mar 09 '23 at 20:49
  • Regarding my first question and your followup, and for the converse. What does it mean to write $O^2\geq O$ (for operators)? – EE18 Mar 09 '23 at 20:53
  • @EE18 Ah, sorry. It is a widely used notation for positive semi-definite operators and means the following (in our case here for everywhere defined bounded operators/finite-dimensions): For an operator $A$ we write $A\geq 0$ if for all $x\in H$ it holds that $\langle x,Ax\rangle \geq 0$. For a complex Hilbert space (the usual setting in QM), this implies in particular that $A$ is hermitian; see also Wikipedia. – Tobias Fünke Mar 09 '23 at 20:57
  • Got it. When you write $O^2\geq O$ presumably you mean $O^2\geq 0$ then, right? – EE18 Mar 09 '23 at 20:59
  • For this particular argument, see also this. The point is that $\Delta \rho:=\rho-\rho_1 \otimes \rho_2$ is, as a difference of density operators, hermitian. Thus $(\Delta \rho)^2 \geq 0$ (easy to prove). But the trace of a positive semi-definite operator is always larger or equal zero - with equal zero if and only if the operator is zero. As shown in the above link, $(\Delta \rho)^2=0 \Longrightarrow \Delta \rho=0$. And yes, sorry, I meant $O^2 \geq 0$. – Tobias Fünke Mar 09 '23 at 20:59
  • Lovely, thank you! This answers all of my questions. If it's possible, can you leave all of the comments up? They greatly help me understand your proof (for future reference). – EE18 Mar 09 '23 at 21:02
  • Sure. I hope I did not do any mistake (comments are hard to edit and read etc.). So you should really double-check everything. That being said, if you still have questions, let me know. BTW: It could be that some moderator puts all of this in the chat, but the link to the chat should be right here then. – Tobias Fünke Mar 09 '23 at 21:02
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    @EE18 I've edited the post to try to make the theorem more readable, as you've suggested. – Tobias Fünke Mar 10 '23 at 08:36
  • @EE18 in the light of your question, my comment regarding the factoring into the reduced density matrices is wrong, strictly speaking. I should've said: up to trivial re-arrangements. Sorry for the confusion. – Tobias Fünke Aug 04 '23 at 14:58
  • Just coming back to read this lovely proof. Can you go over one of the key steps for me one more time -- namely, you say that $(\rho-\rho_1\otimes\rho_2)^2$ is positive semi-definite as part of the proof (I believe). Is the argument that it's Hermitian, so that $\langle \psi|(\rho-\rho_1\otimes\rho_2)^2|\psi \rangle = (\langle \psi|\rho-\rho_1\otimes\rho_2)(\rho-\rho_1\otimes\rho_2|\psi \rangle) \geq 0$ by the positivity of the inner product/norm? – EE18 Aug 04 '23 at 19:04
  • @EE18 Yes, indeed! – Tobias Fünke Aug 04 '23 at 19:09
  • @EE18 I've edited the answer to make one point clearer (the phrasing was not good at this point). – Tobias Fünke Aug 04 '23 at 19:17
  • Thank you! (Very) last question, I promise -- this proof seems to require that the operator $\rho-\rho_1\otimes\rho_2$ be able to represent an observable (i.e. setting $\rho-\rho_1\otimes\rho_2$). Is this just something we have to accept (i.e. is it a tacit axiom that this is an observable)? – EE18 Aug 04 '23 at 20:04
  • @EE18 Why do you think so? It is only required that it is hermitian, which it is. Whether all hermitian operators (even ignoring superselection rules) are indeed physical observables is another topic... – Tobias Fünke Aug 04 '23 at 20:30
  • On March 9 you wrote "There are several ways now; one is choosing $O = \rho-\rho_1\otimes\rho_2$." I follow that that choice leads us to the required conclusion (using $(\rho-\rho_1\otimes\rho_2)^2$ positive semi-definite and that trace zero for a positive semi-definite operator implies that operator is 0), but I cannot see how any other choice of $O$ works? – EE18 Aug 04 '23 at 20:33
  • Let us continue this discussion in chat. – Tobias Fünke Aug 04 '23 at 20:35