# Is a state being unentangled equivalent to statistical independence for all pairs of subsystem observables?

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?

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 $$O_A,O_B$$. $$\tag*{\square}$$

• ...near where you introduce the $\rho_i$. Are they tacitly there and I'm missing them?
• @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 $O=0$. Commented Mar 9, 2023 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 $H_A$. Commented Mar 9, 2023 at 20:46