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Piece of Information 1:

This text states that since correlation and entanglement are equivalent for pure states, one could use a quantum-correlation measure as a quantifier of entanglement. For quantum correlation, one could possibly(?) use the so called correlation-function defined as

$$K=E(O_A\otimes O_B)_{\psi}-E(O_A\otimes I)_{\psi}\cdot E(I\otimes O_A)_{\psi}.\tag{cf}$$

E stands for the expected value of a quantum measurement on $|\psi\rangle$, given by the respective obersvables in brackets. Coincidentally, this is quite similar to the "quantum mutual information" (see here) that uses entropies when considering mixed states - I can't find any publication that elaborates on this connection though or mentions (cf) in that exact shape.

Vocabulary: If $K=0$ for some observables, the measurement results of two measurements on both parts of the system are independend of each other.

Thought 1:

A pure bipartite state could therefore be called entangled if one can find $O_A$, $O_B$ so that $K\neq 0$. Reason being, that for pure product states obviously $K=0$ for any $O_A$, $O_B$.

  • Question 1: Do you agree? Is that something one would find in literature? The problem probably is that it only works for pure bipartite states.

Piece of Information 2: 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?"

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

The states used to define $|\psi\rangle$ in (cf) are not necessarily mutually orthgonal!

Thought 2:

I guess you would agree that Schlosshauer (in the background) could use (or even uses) the correlation function to quantify the "amount of" entanglement of (1). Reason being that if $K=1$ or $K=-1$ for some $O_A$, $O_B$, this is just the mathematical way of saying that "one could learn something about system B by measuring system A (with the concrete $O_A$, $O_B$ in mind)". So he seems to use "the farther K from 0 (the more correlated the measurement results), the more entangled is $|\psi\rangle$" - at least, if one ascribes Schlosshauer to use said mathematical background.

Question 2: Do you agree? (so far, intuitively)

Obviously, as one can see, $K$ depends on $O_A$, $O_B$. And Schlosshauer doesn't make exactly clear, if $O_A$, $O_B$ are fixed or how they are treated. For a proper definition of "the amount of entanglement" using (cf) one has to make clear, how $O_A$, $O_B$ are handled or implemented in the definition. What's more, Schlosshauers intuitive approach must have a definition at the bottom, where it is clear how $O_A$, $O_B$ are considered. There are two options in my opinion:

  1. Schlosshauer could mean "the farthest value $K$ that can be from $0$ (therefore "trying all $O_A$, $O_B$) quantifies the entanglement. The farther this value from $0$, the more entangled is $|\psi\rangle$.

  2. Obviously, one could alternatively just use $O_A=|\psi_1\rangle\langle\psi_1|-|\psi^{\perp}_1\rangle\langle\psi^{\perp}_1|$ and $O_B=|\phi_1\rangle\langle\phi_1|-|\phi^{\perp}_1\rangle\langle\phi^{\perp}_1|$.

Question 3: Given that Question 1 can be answered with "yes". What is the correct mathematical background for Schlosshauers intuitive saying? 1. or 2.?

Update: I tried to verify 2) and found a problem, that also exists for 1 (I think), because obviously the measurement basis in 2) is "optimal". Well, given $|\phi_1\rangle$ and $|\phi_2\rangle$ are orthogonal, Schlosshauer states that by measuring System B one knows exactly what state system A will be in after the measurement. However, if I calculate $K$ for observables in 2) in that case ($|\phi^{\perp}_1|$ then being $|\phi_2$) I arrive at $K=1-|\langle\psi_1|\psi_2\rangle|^2$. This result doesn't match with Schlosshauers phrase since my result means that "only if the states of the first system are orthogonal, one can know what state system A is in by measuring system B".

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    $\begingroup$ Schlosshauer's statement is, IMHO, way too vague. With all due respect, why do you bother so much about one single sentence in one book? Is this, again: vague, concept later explained or referred to in the book? Have you checked e.g. other sources? $\endgroup$ Commented Apr 14, 2023 at 6:28
  • $\begingroup$ @TobiasFünke: He keeps coming back to it when he talks about reduced density matrices and decoherence. There, he finds that the off diagonals depend on the orthgonality of $|\phi_1\rangle$ and $|\phi_2\rangle$ (he also uses a bipartite state of the form $|\psi\rangle$). This is connected to the quote that intuitively defines entanglement strength: In using and discussing the quote, he finds that the more orthogonal $|\phi_1\rangle$ and $|\phi_2\rangle$ are, the more entangled the state is. So he infers: the more entanglement in $|\psi\rangle$, the less off diagonals (the more decoherence). $\endgroup$
    – manuel459
    Commented Apr 14, 2023 at 13:25
  • $\begingroup$ @TobiasFünke I tried to find other sources doing the same as Schlosshauer but the closest I could find was the concept of"quantum mutual information" which is similar to (cf) but for mixed states. Nothing I could find restricted its view to bipartite states. If Schlosshauers statement is way too vague - is at least my mathematical variant correct in your opinion (using "2.")? In words it would mean: "A state is strongly entangled if the correlation of measurement results of measurements on both parts of the system can be very far from 0 (the absolute value of the correlation can be high)." $\endgroup$
    – manuel459
    Commented Apr 14, 2023 at 13:29
  • $\begingroup$ Edit: My variant would fit with Schlosshauers version if $|\psi_1\rangle$, $|\psi_2\rangle$ are orthonormal in Schlosshauers elaboration, but they aren't. Maybe that is a mistake there? $\endgroup$
    – manuel459
    Commented Apr 14, 2023 at 13:44
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    $\begingroup$ Certainly, the quantum mutual information, for a pure state, is a measure (even the measure) of entanglement. $\endgroup$ Commented Apr 20, 2023 at 14:42

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About the update

Your update makes all sense, because if B could know what state system A is by just measuring his system very often is related to the problem of to distinguish quantum states:

"Alice chooses a state $\left | \psi_i\right>$ from some fixed set of states $\{\left|\psi_j\right>|1\leq j\leq n\}$ known to both parties. She gives the state to Bob, whose task it is to identify the index $i$ of the state Alice has given him." -Nielsen and Chuang, pg. 86. 10th anniversary edition.

Bob can reliably distinguish the states if and only if they are orthogonal. If they are not, the best he can do is to identify what not is the state

"However, it is possible for him to perform a measurement which distinguishes the states some of the time, but never makes an error of mis-identification" - Nielsen and Chuang, pg. 92. 10th anniversary edition.

Look that this is true regard how many times Bob measures A, i.e., how many copies Alice send to him, because we can't say she sends the same state every time (remember the partial state of Alice is a misture of two non-orthogonal states). If Bob could build a clone machine, it would be possible to him to do local tomography and reliably distinguish the states of Alice, but this is not possible. By converse, if he could distinguish the non-orthogonal states of Alice, he can build a clone machine (Nielsen and Chuang, pg. 531. 10th anniversary edition.).

We are in a little modified version of the distinguishing problem, where the system of Bob is entangled with the Alice's system, but nothing changes, since these results are general. Schlosshauer probably refers to orthogonal states; if not, he is missing the no-cloning theorem.

About the questions

When people talk about pure bipartite states, often they are considering the Schmidt decomposition. So given a state $\left|\Psi \right>$, let's consider

$$ \left|\Psi \right> = \sum_j \lambda_j \left|\phi_j^A \right>\left|\phi_j^B \right>, \quad \lambda_j \geq 0. $$

Now, if we consider the correlation function for the state in this form, we get

$$ K = \sum_{j,k} \lambda_j \lambda_k \left < \phi_j^A \right| O_A \left |\phi_k^A\right> \left < \phi_j^B \right| O_B \left |\phi_k^B\right> - \left(\sum_j \lambda_j^2 \left < \phi_j^A \right| O_A \left |\phi_j^A\right> \right ) \left (\sum_j \lambda_j^2 \left < \phi_k^B \right| O_B \left |\phi_j^B\right> \right). $$

Now, it comes the answer of your first question: the state is a product state if and only if just one $\lambda_j$, say $\lambda_1$, is different from zero. But if it happens, (by normalization $\lambda_1 = 1$), then

$$ K = \left < \phi_1^A \right| O_A \left |\phi_1^A\right> \left < \phi_1^B \right| O_B \left |\phi_1^B\right> - \left < \phi_1^A \right| O_A \left |\phi_1^A\right> \left < \phi_1^B \right| O_B \left |\phi_1^B\right> =0 , \quad \forall O_A,O_B. $$

So

If the state $\left | \Psi \right>$ is a product state, then $K=0$ for every $O_A,O_B$.

and we can say that

If the state $\left | \Psi \right>$ is entangled, then $K\neq0$ for some $O_A,O_B$.

Now, to learn what happens to entangled states, let's consider first a special kind of entangled states. If we say $\lambda_1 = \sqrt{1-\epsilon}$, $\lambda_2 = \sqrt{\epsilon}$ and $\lambda_j = 0, j>2$, then

$$ \left| \Psi \right> = \sqrt{1-\epsilon} \left|\phi_1^A \right>\left|\phi_1^B \right> + \sqrt{\epsilon} \left|\phi_2^A \right>\left|\phi_2^B \right>, \quad \epsilon \in [0,1] $$

For small $\epsilon$, we can say this is state is "almost product". The correlation function now will be

$$ K = \epsilon (1-\epsilon) (O_A^{11}+O_A^{22})(O_B^{11}+O_B^{22}) + \sqrt{\epsilon(1-\epsilon)}(O_A^{12}O_B^{12}+O_A^{21}O_B^{21}) $$ where $O_A^{ij} = \left < \phi_i^A \right| O_A \left |\phi_j^A\right>$ and the same for Bob observable. Now we see that $K$ depends on $\epsilon$, that is our entanglement parameter, but also depends if $O_A$ and $O_B$ have support over the spaces spanned by $\{\left |\phi_1^A\right>,\left |\phi_2^A\right> \}$ and $\{\left |\phi_1^B\right>,\left |\phi_2^B\right> \}$. For example, if $O_A = \left | \phi_3^A\right >\left < \phi_3^A\right |$ and $O_B = \left | \phi_3^B\right >\left < \phi_3^B\right |$, clearly $K=0$ despite $\left|\Psi\right>$ being entangled or not. Now it comes the answer of your second question: Yes, we can use the correlation function to quantify entanglement but, as you noticed, we must consider all observables $O_A,O_B$, or at least take the optimal ones.

By the expression we obtained for $K$, it's not hard to search for optimal observables. You probably noticed that we must consider observables with support in the spaces spanned by $\{\left |\phi_1^A\right>,\left |\phi_2^A\right> \}$ and $\{\left |\phi_1^B\right>,\left |\phi_2^B\right> \}$. We can consider $O_A = \left |\phi_1^A\right>\left <\phi_2^A\right| + \left |\phi_2^A\right>\left <\phi_1^A\right|$ and $O_B = \left |\phi_1^B\right>\left <\phi_2^B\right| + \left |\phi_2^B\right>\left <\phi_1^B\right|$ so we get $$ K = \sqrt{\epsilon(1-\epsilon)} $$ which is the max. value for each $\epsilon$, since $\sqrt{\epsilon(1-\epsilon)} \geq \epsilon(1-\epsilon)$ for $\epsilon \in [0,1]$. It's not hard to check that $K$ is maximal if and only if $\epsilon = 1/2$, which is the greatest entanglement we can have for the kind of states we considered (effectively two qubit states).

General pure bipartite states

The general case appears to be not so hard to deal. We can consider $$ O_A = \sum_{j\neq k} \left|\phi_j^A\right>\left<\phi_k^A\right| $$ and the same for B, so we get $$ K = \sum_{j\neq k} \lambda_j \lambda_k = \left (\sum_j \lambda_j \right)^2 - 1 $$ So to maximize $K$, we must maximize the function inside the parenthesis. By Lagrange multiplier method, we must consider the function $$ \mathcal L = \sum_j \lambda_j + \mu \left (\sum_j \lambda_j^2 - 1\right) $$ $$ \frac{\partial \mathcal L}{\partial \lambda_j} = 1-2\mu \lambda_j = 0 $$ $$ \lambda_j = \frac{1}{2\mu}, \forall j. $$ If the state $\left | \Psi \right >$ has $r$ non-zero Schmidt coefficients, then $$ \sum_j \lambda_j^2 = \frac{1}{4\mu^2}\sum_j 1 = \frac{r}{4\mu^2} = 1 $$ which implies that $$ \mu = \frac{\sqrt{r}}{2} \rightarrow \lambda_j = \frac{1}{\sqrt{r}}. $$

in the special case where the dimension of Hilbert spaces of A and B are equal $d_A=d_B =d$ and the shared state $\left | \Psi \right>$ has $d$ non-zero Schmidt coefficients, we get $$ \lambda_j = \frac{1}{\sqrt{d}} \rightarrow \left | \Psi \right> = \frac{1}{\sqrt{d}} \sum_{j=1}^d \left|\phi_j^A\right> \left| \phi_j^B\right> $$

which is the maximal entangled state. So

Considering the optimal observables $O_A,O_B$, the correlation function $K$ is maximal if and only if $\left |\Psi \right>$ is the maximal entangled state.

In summary, the correlation function could be a good entanglement quantifier for pure bipartite states if we consider optimal observables $O_A,O_B$. If we not consider the optimal ones, we can be tricky and find observables without correlation for a given entangled state.

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  • $\begingroup$ Thank you for your very clear and founded answer. Regarding my update: I think what is meant with "an observer can know what state A is in by measuring B" is that "an observer can know what state A is in by measuring B very often" (law of larg numbers, in the sense that there are many copies). So the confusion still exists. While Schlosshauer says "an observer can know what state system A is in by measuring B very often" i only find this to be true if the states of system A are orthogonal. I would appreciate if you could update your answer considering this. Thank you again very much! $\endgroup$
    – manuel459
    Commented Apr 18, 2023 at 16:49

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