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Context of the question:

Schlosshauer (978-3-540-35773-4, p. 33) states:

"A useful intuitive way of quantifying the entanglement present in this state [(1)] is to consider the following question: How much can the observer learn about one system by measuring the other system?"

$$|\psi\rangle=\frac{1}{\sqrt{2}}\left(|\psi_1\rangle|\phi_1\rangle\pm|\psi_2\rangle|\phi_2\rangle\right)\tag{1}$$

Taken at face-value, the phrase above doesn't elaborate on what measurement (in what basis) is meant. So one could assume that entanglement strength of state (1) is a property that is defined "relative to two certain measurement bases" - one for measurement on the first and one for measurement on the second system.

Although I have the suspicion that that's not the case: Even if (1) is a product state one could learn "a lot" about system 1 when measuring system 2 for a certain pair of measurements (measurement bases). The difference seems to come down to the used measurement bases or rather - whether correlations appear in any measurement basis. Therefore, there seems to be some clarification regarding the measurement bases missing in Schlosshauers characterization.

I tried to improve the cited phrase.

A useful intuitive way of quantifying the entanglement present in this state [(1)] is to consider the following question: How much can the observer learn about one system by measuring the other system (regarding measurements in any bases)?

Unfortunately, this is not satisfying either since there is a pair of measurements where there is no correlation, even for the entangled state - namely if $|\psi\rangle=|00\rangle+|11\rangle$ and one measures $O=|+\rangle\langle +|-|-\rangle\langle -|$ on both systems.

So, a second try:

"A useful intuitive way of quantifying the entanglement present in this state [(1)] is to consider the following question: How much does a measurement on one system change measurement behaviour of the 2nd system?"

Question: Is entanglement of a bipartite system (if defined by correlation between measurements) a property that is defined relative to certain measurement bases? If not, then Schlosshauers explanation is somewhat misleading, since it doesn't mention measurement bases at all. How would a correction of said explanation look like then?

Update:

It seems to me that Schlosshauers description is correct - I was able to find it similar in the accepted answer to this question. Therefore, I am probably misunderstanding the phrase "How much can the observer learn about one system by measuring the other system?".

So far I took the view that it must be understood this way: If there is e.g. a measurement on the product state $|0\rangle_A\otimes|0\rangle_B$ that first measures in the $|0\rangle,|1\rangle$ basis on system A, you would always get the result "$|0\rangle$". If one measured system B after (in the same basis), you would always get the result $|0\rangle$". In this sense, you could learn something about system B by measuring A. But that is not how said phrase is meant. It is meant in this way: Measurement on system A influences measurement on system B. In other words: Measurement on system A changes the measurements statistics of a subsequent measurement on system B in comparison to solely measuring B (without measuring A first).

If somebody could doublecheck what I wrote in my update, I think my question has been solved! Thank you.

manuel459
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  • @TobiasFünke It's hard for me to extract but is this what you mean?: For bipartite states, measurement results are correlated (for any Observables) (in the stochastically defined sense) if and only if the bipartite state is entangled. And what Schlosshauer describes is correlation (for any Observables) (in the stochastically defined sense). So he misses the "for any Observables"? – manuel459 Apr 10 '23 at 19:32
  • If so, there still is that one pair of bases where there are no correlations. So is Schlosshauers take of using "how much one can learn" problematic in the first place? – manuel459 Apr 10 '23 at 19:44
  • Sorry, I don't really understand your comment (the question, too, as it seems and given only the context here also the statement of the book...). Entanglement is basis independent: Given a bipartition of your Hilbert space $H=H_1\otimes H_2$, whether or not a density matrix $\rho$ (say, for our discussion here a pure state) on $H$ is entangled is basis-independent. Now as shown in the linked post, for a pure $\rho$ entanglement is equivalent to a statement regarding the correlation of observables. – Tobias Fünke Apr 10 '23 at 19:59
  • Let me retry: Obviously Schlosshauer tries to explain the "strength" of entanglement with correlation. He therefore gives a rather intuitive explanation of correlation between the results of two measurements. If you take this explanation at face value, it is basis dependent. Therefore, Schlosshauers characterization must be wrong. Or is it? – manuel459 Apr 10 '23 at 21:22
  • And what I tried to say before is that I concluded from your answer in the question you linked that: For bipartite systems a state is entangled if and only if it is correlated (meaning, that measurement results depend on each other for any given two measurement bases). And I asked if that was a correct conclusion. – manuel459 Apr 10 '23 at 21:26
  • Going into detail: Schlosshauer uses "how much one can learn about system A by measuring system B" as a intuitive characterization of correlation. I state that the following is better suited: "How much measurement results of an observable $O_A$ depend on measurement results of $O_B$. Reason: See original question examples of product states where one can learn about system A by measuring B if one looks at certain observables. Question: Would you agree? – manuel459 Apr 10 '23 at 21:31
  • Sorry, I honestly do not understand what the book says (and hence I only commented in case it is related and helps). Other than that, the linked post says what it says: A pure state is entangled if and only if it is not a product state, which in turn is the case if and only if it is correlated - this means there exists at least one pair of local observables, such that the respective joint probability distribution does not factorize. Schlosshauer seems to have a quantifying concept of entanglement in mind (?), whereas this is just a yes-no-thing; at least in this form. – Tobias Fünke Apr 11 '23 at 04:47
  • Regarding your edit, see the no-communication theorem. I don't really understand your edit, but I have the impression this might be relevant...The measurement statistics of $B$ are not altered if $A$ performs a measurement (or does something locally at all). – Tobias Fünke Apr 11 '23 at 10:18

1 Answers1

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First, the definition of entanglement.

Let $\mathcal{H} \cong \mathcal{H}_1 \otimes \mathcal{H}_2$ be your composite Hilbert space (of two particles). A state $\lvert \psi \rangle \in \mathcal{H}$ is entangled precisely when $\lvert \psi \rangle \neq \lvert \phi \rangle_1 \otimes \lvert \varphi \rangle_2$ for $\lvert \phi \rangle_1 \in \mathcal{H}_1$ and $\lvert \varphi \rangle_2 \in \mathcal{H}_2$.

It is important to note that this definition of being entangled is utterly independent of the chosen basis for you Hilbert space. Rather, whether two systems are entangled depends on the tensor product factorization of the composite Hilbert space.

It is meant in this way: Measurement on system A influences measurement on system B. In other words: Measurement on system A changes the measurements statistics of a subsequent measurement on system B in comparison to solely measuring B (without measuring A first).

This seems fine if you are looking to translate the mathematical definition above into words. Though, I think it would be more precise to say "A and B are entangled if the measurement outcomes of A are quantum correlated with the measurement outcomes of B". Where quantum correlation is (tautologically) defined as the correlations allowed in the theory of quantum mechanics.

However, all this discussion has been about whether a bipartite state has the quality of being entangled or not. Nothing about quantitatively how entangled two systems represented by a state are. A common (I think) measure for how much a pure state is entangled is given by the von Neumann entropy (also called entanglement entropy, but entanglement entropy broadly refers to the measure being used to quantify entanglement). Let $\rho$ be a density matrix representing a pure state of a two particle system. Then, $\rho_A = \text{Tr}_B(\rho)$ is the reduced density matrix for system A. Then, the von Neumann entropy for $\rho_A$ is

$$S(\rho_A) = -\text{Tr}_A(\rho_A\log_2\rho_A)$$

which gives somewhat of a measure of how entangled the two systems represented by the state $\rho$ are. For why, cf. slide 8 and 9 here.

I think this von Neumann entropy measure of entanglement is what Schlosshauer has in mind (cf. section 2.4.3 in the Decoherence text you're working through). The von Neumann entropy is basis independent as can be verified. Hence, this measurement of how entangled two systems represented by a state are is basis independent.

Silly Goose
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  • Thank you for your answer. I explicitly tried to avoid the von neumann entropy to keep it as little technical as possible. Would you agree with the following as a characterization of how entangled 2 systems are: Strong entanglement <-> Measurement of A changes the measurements statistics of a subsequent measurement on system B "strongly" in comparison to solely measuring B (without measuring A first). – manuel459 Apr 11 '23 at 02:02
  • @manuel459 The problem with everyday language is that it is not precise and thus (very) often ambiguous. You should try to encapsulate such a statement in mathematical terms... – Tobias Fünke Apr 11 '23 at 04:48
  • I think something like that is what Schlosshauer has in mind. Since he calls the Bell states maximally entangled because the Bell states give perfect correlation when Alice makes a measurement in $z$ basis and so does Bob. But I think what's really meant is that the Bell states are maximally entangled because their reduced density matrices are maximally mixed. And if your reduced density matrices are all maximally mixed, you "know as little as possible" about a given subsystem. @manuel459 – Silly Goose Apr 11 '23 at 19:34