I know two particles in a Bell state cannot be written as a product state as they are entangled. But what if I had a classically correlated state$$\rho = \frac{1}{2}(|11\rangle\langle 11| + |00\rangle\langle00|)$$How do I write this as a product state?
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$\begingroup$ I don't think either density matrix is pure (both are |1><1|+|0><0|) does this mean the state is actually an entangled state? Since it cannot be written as a product state? $\endgroup$– DJamesCommented Jan 26 at 22:38
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$\begingroup$ Sorry, my previous comment was wrong (what is correct is that if at least one reduced density matrix is pure, then indeed $\rho=\rho_1\otimes \rho_2$). Regarding your specific question: Why don't you write $\rho=\rho_1\otimes \rho_2$ and see where it leads to? From a more general perspective, I think you should read a bit about mixed entangled/separable states. $\endgroup$– Tobias FünkeCommented Jan 26 at 22:48
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$\begingroup$ The problem is I don't know the product form. I only know the separable form. I thought a separable state can always be written in a product form but maybe this is not true. $\endgroup$– DJamesCommented Jan 26 at 22:51
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$\begingroup$ Well, for your $\rho$ it is very easy to check if the product form holds true... OTOH, it depends on what you mean by product state. If you mean that it is the tensor product of its two reduced density matrices, then it is indeed straightforward here. If not, please explain. $\endgroup$– Tobias FünkeCommented Jan 26 at 22:53
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1$\begingroup$ You might want to convert this into an answer, perhaps with 1-2 calculation steps. It might help potential future readers with a similar question/problem. $\endgroup$– Tobias FünkeCommented Jan 26 at 23:01
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3 Answers
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After a discussion in comments, it turns out this cannot be written as a product state.
The reduced density matrix of each subsystem is $\frac{1}{2}(|0\rangle\langle 0| + |1\rangle\langle1|)$ and so $$\rho_1\otimes\rho_2 = \frac{1}{4}(|00\rangle\langle00|+|11\rangle\langle11| + |01\rangle\langle01| + |10\rangle\langle10|)\neq\rho$$
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$\begingroup$ Good. From a more general perspective, note that if both reduced density matrices are mixed, there are infinitely many bipartite density matrices (which might differ in their separability), having these two as reduced density matrices. The product state is of course one of them (but tautologically there is only one such state). OTOH, as already explained, if at least one reduced DM is pure, it is the product state. See e.g. my answer here and the links therein $\endgroup$ Commented Jan 26 at 23:10
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As you write yourself, the state is classically correlated. Thus, it cannot be of product form: A product does not have any kind of correlations, whether quantum or classical.
answered Jan 26 at 23:37
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"Product state" here refers to a pure state. The given state, however, is mixed (as can be seen immediately from its eigenvalues). Thus, it cannot be written as a single pure state, be it a product state or not.
answered Jan 26 at 23:23
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$\begingroup$ As I said in the comments, it of course depends what you mean by the term "product state". But in my experience it refers to mixed states as well in general. Yes, if you restrict to pure states, everything is trivial. $\endgroup$ Commented Jan 26 at 23:24
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1$\begingroup$ j@TobiasFünke ... and if you allow for mixed states, this is clearly a mixture of product states, which is as unentangled as it gets. In the context of the question (which starts by talking about the Bell state, which cannot be written as a product state), this seems to be the natural interpretation. -- Let me offer an alternative answer. $\endgroup$ Commented Jan 26 at 23:35
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