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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. Ref. 1 and Ref. 2. 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.States, Effects, and Operations Fundamental Notions of Quantum Theory: Lectures in Mathematical Physics at the University of Texas at Austin. Karl Kraus. Springer Berlin Heidelberg, 1983. Chapter 4.*

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

Tobias Fünke
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