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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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? – DJames Jan 26 '24 at 22:38
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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. – Tobias Fünke Jan 26 '24 at 22:48
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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. – DJames Jan 26 '24 at 22:51
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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. – Tobias Fünke Jan 26 '24 at 22:53
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Can you tell me how? I tried a change of basis but did not get a product state out of it. – DJames Jan 26 '24 at 22:56
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I mean you have calculated both reduced density matrices...Now compute their tensor product and just compare to the expression you have for $\rho$. To be clear: I have not said it is possible. I want you to realize, however, that it is straightforward to check, at least for this particular case... – Tobias Fünke Jan 26 '24 at 22:56
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Ok, I have done this and I see that the product of the reduced density matrices does not match the original state. Therefore I will conclude it cannot be written as a product state – DJames Jan 26 '24 at 22:59
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1You 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. – Tobias Fünke Jan 26 '24 at 23:01
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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 $$\rho_1\otimes\rho_2 = \frac{1}{4}(|00\rangle\langle00|+|11\rangle\langle11| + |01\rangle\langle01| + |10\rangle\langle10|)\neq\rho$$
DJames
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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 – Tobias Fünke Jan 26 '24 at 23:10
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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.
Norbert Schuch
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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. – Tobias Fünke Jan 26 '24 at 23:24
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1j@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. – Norbert Schuch Jan 26 '24 at 23:35
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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.
Norbert Schuch
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