How much can I learn about correlations between two quits by measuring the sum of their values? What is the best way to formalize such a question?

Below is my original, longer formulation of the same question in spin language. I have an ensemble of pairs of spin-$1/2$ particles. All I can do is to measure total $S_z$ for each pair, each time learning one of the three standard outcomes: $+1$, $0$ or $-1$. Based on the knowledge of this probability distribution $P(S_z)$, what can be said about the density matrix of my ensemble? In particular, what is best measure for correlations between the two particles in these circumstances?

I understand that I won't be able to distinguish whether the correlations are classical or necessarily quantum-mechanical, but struggle to come up with a measure better than a classical correlation coefficient $$r=\frac{\langle s_1 s_2 \rangle-\langle s_1 \rangle \langle s_2 \rangle}{\sqrt{\langle (s_1- \langle s_1 \rangle)^2 \rangle \langle (s_2- \langle s_2 \rangle)^2 \rangle }}$$ where $s_1$ and $s_2$ are the $\pm 1/2$ spin projections (or $0$, $1$ qubit values) of the 1st and the 2nd particle, respectively, so that my only observable can be expressed as $S_z=s_1+s_2$. I then minimize $r$ over all possible $\langle s_1 \rangle$ and $\langle s_1 \rangle$ consistent with measured $P(S_z)$ and declare that the outcome is the minimal possible correlation in the pair. Is there a more elegant way?

UPDATE: Classically, if we take the definition of zero correlations as $\langle s_1 s_2 \rangle =\langle s_1 \rangle \langle s_2 \rangle$, then it implies that $s_1 = +1/2$ and $s_2=+1/2$ with some independent probabilities $p_1$ and $p_2$, so that the sum is distributed bimomially $P(+1)=p_1 p_2$ and $P(-1)=(1-p_1)(1-p_2)$. These two equations have a solution with $0\leq p_1, p_2 \leq 1$ only if $$P(+1) < 1-2 \sqrt{P(-1)}+P(-1)$$

Graphically, this condition is satisfied within the region bounded by a black line in the plot below. The rest of the available probability space is the region of excess correlations ($r_{\text{min}}>0$, depicted in red). enter image description here

Updated question: is there a quantum loophole in such analysis? I suspect not but would really like to come up with a shorter/cleaner argument.

Update no. 2 My selected spin/qubit degree of freedom is one of the many variables characterizing the state of the two particles. Can we be sure there exists no pure product state that can lead to $r_{\text{min}}>0$? See a related active question on fermionic entanglement of two particles.

Update no. 3 The work that motivated this question has been published in Nature Nanotechnology; the open access version is arXiv:1404.0030.

  • $\begingroup$ Will the special case of a pure (but otherwise arbitrarily entangled) composite system be of interest to you. $\endgroup$ Apr 25, 2013 at 12:40
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    $\begingroup$ The idea is to use the classification given in arxiv.org/abs/quant-ph/0006068 (Geometry of entangled states) by Marek Kuś Karol Žyczkowski. The manifolds of equal entanglement of the pure two-qubit system fall into three strata parameterized by a single parameter $\theta$. The nongeneric orbits are $\mathbb{R}P^3$ and $S^2 \times S^2 $ correspond to $\theta = \frac{\pi}{2}, 0 $ respectively. The generic orbits correspond to $0 < \theta <\frac{\pi}{2}$ are five dimensional twisted $S^2$ bundles over $\mathbb{R}P^3$. $\endgroup$ Apr 25, 2013 at 14:18
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    $\begingroup$ I expect an experimental point (uncertainty ellipse) deep in the correlated region & I want to be able to claim something like "if this is a pure state, then it should be at least ... [number] of [most appropriate entanglement measure] at 95% confidence. $\endgroup$
    – Slaviks
    Apr 25, 2013 at 16:24
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    $\begingroup$ Do you have good reason to think that your state is pure, or close to pure ? And do you have any way to measure the spin in another direction ? $\endgroup$ Apr 25, 2013 at 16:36
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    $\begingroup$ Yes, there are good reasons to expect high-fidelity pure state although present set of data does not prove it (because only $S_z$ is measured so there is always an option to justify correlations classically). We'll see how this develops and I'll post a link to the paper once it is out. $\endgroup$
    – Slaviks
    Apr 25, 2013 at 16:42

2 Answers 2


Edited to add the second part

Edited again, for part 3 and 4

$\newcommand\ket[1]{\left|#1\right>} \newcommand\bra[1]{\left<#1\right|} $

1. Absence of Quantum Loophole

You can easily see that there is no "quantum loophole" in your argument by writing explicitly any pure separable state. With your notations, we have : $$ \left(\sqrt{p_1}\ket\uparrow+e^{i\phi_1}\sqrt{1-p_1}\ket{\downarrow}\right) \left(\sqrt{p_2}\ket\uparrow+e^{i\phi_2}\sqrt{1-p_2}\ket{\downarrow}\right) $$ This expression obviously corresponds to all possible separable pure state and gives the uncorrelated distribution you describe.

Of course, as @Norbert Schuch said, if you allow for mixed states, you have access to whole region, which is the same as saying that you allow classical correlations.

To detect quantum correlations, as you are probably aware, you will need to have information about another direction, like $S_x$.

2. Lower bound of entanglement of a pure state

If you are sure that your state $\ket\psi$ is pure, you can have a lower bound of its entanglement. Let $P_\pm$ and $P_0$ be the three spin probabilities. The state you have can be written $$\ket{\psi}=\sqrt{P_+}\ket{\uparrow\uparrow} + e^{i\phi}\sqrt{P_-}\ket{\downarrow\downarrow} +\sqrt{P_0}(\alpha\ket{\uparrow\downarrow}+\beta\ket{\downarrow\uparrow}) \text{ with } |\alpha|^2+|\beta|^2=1$$ Since our measurement does not distinguish what happens in the $S_Z=0$ subspace, we probably have no information about the entanglement there. So we can take the fidelity $F$ with a maximally entangled correlated state for entanglement measure. We have $$ F:=\max_{\theta}\left|\left<\psi\right|\frac{\ket{\uparrow\uparrow}+e^{i\theta}\ket{\downarrow\downarrow}}{\sqrt2}\right|^2=\frac{\left(\sqrt{P_+}+\sqrt{P_-}\right)^2}2 $$ You cannot prove entanglement when $F=\tfrac12$. This corresponds to the zero correlation line :-)

Of course, all this is only valid for pure states, and could also correspond to classically correlated states.

3. Lower bound, take 2

Since the state is pure, its entanglement is fully characterized by the eigenvalues $\lambda_\pm$ of its partial trace $\rho=Tr_B \ket\psi \bra\psi$. They are the solutions of $\lambda^2-\lambda+\det\rho$ and are given by $\lambda_\pm=\tfrac12 \pm \sqrt{\tfrac14 -\det\rho}.$ We have the Schmidt decomposition
$\ket{\psi}=\sqrt{\lambda_+}\ket{0}\otimes\ket{0}+\sqrt{\lambda_-}\ket{1}\otimes\ket{1}$, where the $\ket{i}$s are the bases which diagonalise $ Tr_B \ket\psi \bra\psi$ and $ Tr_A \ket\psi \bra\psi$.

Before computing $\det\rho$, one can see that $0\le\det\rho\le\tfrac14$. When $\det\rho=0$, $\lambda_+= 1$, $\lambda_-=0$ and $\ket\psi$ is separable ; when $\det\rho=\tfrac14$, $\lambda_+=\lambda_-=\tfrac12$ and $\ket\psi$ is maximally entangled. The bigger $\det\rho$ is, the more $\ket\psi$ is entangled. More quantitatively, the entropy of entanglement of this state is $h(\lambda_-)$ where $h(\cdot)$ is the binary entropy function and the negativity of the state is $\det\rho$.

\begin{align} \det\rho &=\left(P_++P_0|\alpha|^2\right)\left(P_-+P_0|\beta|^2\right) -\left| \beta\sqrt{P_0P_+}+\alpha^*e^{i\phi}\sqrt{P_0P_-}\right|^2 \\ &=P_+P_- +|P_0\alpha\beta|^2 +P_0\left( |\alpha|^2P_-+ |\beta|^2P_+ - |\beta|^2P_ +- |\alpha|^2P_- - 2 Re\left( \beta \alpha^*e^{i\phi}\sqrt{P_+P_-} \right)\right)\\ &=\left| \sqrt{P_+P_-} + \beta \alpha^*e^{i\phi} P_0 \right|^2 \end{align} We want a lower bound on the entanglement. We therefore chose the free parameters $\alpha$, $\beta$, and $\phi$ in order to minimize $\det\rho$. Since we have $|\alpha\beta|\le\tfrac12$, $\min\det\rho=0$ if $ \sqrt{P_+P_-}\le \tfrac12 P_0$. This corresponds to the separability limit established earlier.

If we have $|\alpha\beta|\le\tfrac12$, then $$ \min\det\rho=\left( \sqrt{P_+P_-}- \tfrac12 P_0\right)^2 =\tfrac14 \left[\left(\sqrt{ P_+}+ \sqrt{ P_-}\right)^2 - 1\right]^2 = \left(F-\tfrac12\right)^2. $$ For these minimally entangled states, $\lambda_\pm=\frac12\pm\sqrt{F(1-F)}$.

4. Bonus : an upper bound on the entanglement

One can also look at an upper bound on the entanglement with the same approach: $$ \max\det\rho= \left| \sqrt{P_+P_-} + \tfrac12 P_0 \right|^2 = \left(G-\tfrac12\right)^2 \text{ with } G=\tfrac12\left( \sqrt{ P_+} - \sqrt{ P_-}\right)^2. $$ For these "maximally" entangled states, $\lambda_\pm=\frac12\pm\sqrt{G(1-G)}$.

A good way to put all this on a graph is to take $\sqrt{P_\pm}$ as coordinates. The possible measurement are in the top right quarter of a disk and the bound described here are constant along 45° straight lines.

  • $\begingroup$ If you have a non-trivial bound on the purity of your state, you can also compute a bound on $ F $. $\endgroup$ Apr 25, 2013 at 18:17
  • $\begingroup$ I have to admit I don't find the argument why to neglect the Sz=0 subspace very convincing. $\endgroup$ Apr 26, 2013 at 9:16
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    $\begingroup$ @NorbertSchuch: You don't have to buy this argument, since we only need a lower bound. However, one can make this more rigorous by looking into the fidelity with a general maximally entangled state of the form $\newcommand\ket[1]{\left|#1\right\rangle} I\otimes U(\ket{\uparrow\uparrow}+\ket{\downarrow\downarrow})/\sqrt2$. The latter form is not too complicated and allows to also have an upper bound on the possible entanglement. $\endgroup$ Apr 26, 2013 at 9:56
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    $\begingroup$ @Slaviks: Using the expression of $\left|\psi\right>$, one can show that if $F>1/2$, the Schmidt coefficients of the least entangled pure state are $\sqrt{(1\pm2\sqrt{F(F-1)})/2}$. This allows to easily compute the minimal Entropy of Entaglement $\endgroup$ Apr 26, 2013 at 13:16
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    $\begingroup$ @Norbert : I've added a more detailed and more complete answer, without mystery ;-) $\endgroup$ Apr 28, 2013 at 16:55

Computation of the $S_z$ probability distribution for each of the manifolds of equal entanglement:

Remark: Notations and references from Kuś and Žyczkowski are used.

Case 1: The separable case: The state vector is parametrized as (equation: 24)

$w = \begin{bmatrix} \cos \alpha \cos \beta e^{i \chi_1},& \cos \alpha \sin\beta e^{i \chi_2}, &\sin \alpha \cos\beta e^{-i \chi_1}, & \sin \alpha \sin \beta e^{-i \chi_1} \end{bmatrix}^{t}$

Here is the computation of the the probability distribution of $S_z$ for the most general pure two qubit state vector:

The total spin projector operator is given by

$S_z = \frac{1}{2} \sigma_3 \otimes 1_2 + 1_2 \otimes \frac{1}{2}\sigma_3$

The characteristic function of the probability distribution is given by:

$ \Phi_{S_z}(t) = w^{\dagger} e^{i t S_z} w =P(S_z = 1) e^{it} + P(S_z = 0)+ P(S_z = -1) e^{-it}$

From which the probability distribution can be read:

$P(S_z = 1) = \cos^2 \alpha\cos^2 \beta $

$P(S_z = 0) = \cos^2 \alpha\sin^2 \beta+ \sin^2 \alpha\cos^2 \beta $

$P(S_z = -1) = \sin^2 \alpha\sin^2 \beta $

Case 2: The maximally entangled case: The state vector is parameterized as (equation: 23)

$w = \frac{1}{\sqrt{2}}\begin{bmatrix} \cos \alpha e^{i \chi_1},& \sin \alpha e^{i \chi_2}, &\sin \alpha e^{-i \chi_1}, & \cos \alpha e^{-i \chi_1} \end{bmatrix}^{t}$

In this case:

$P(S_z = 1) = \frac{1}{2}\cos^2 \alpha$

$P(S_z = 0) = \sin^2 \alpha$

$P(S_z = -1) = \frac{1}{2}\cos^2 \alpha $

Case 3: The generic case: The state vector has been worked out:

$w = \frac{1}{\sqrt{1+\sin^2\eta}} \begin{bmatrix} (\cos \alpha \cos \beta -\sin\eta e^{i\phi}\sin\alpha \sin\beta) e^{i \chi_1},& (\cos \alpha \sin\beta + \sin\eta e^{i\phi} \sin\alpha \cos \beta) e^{i \chi_2}, \\&(\sin \alpha \cos\beta + \sin\eta e^{i\phi} \cos\alpha \sin \beta) e^{-i \chi_1}, & (\sin \alpha \sin \beta -\sin\eta e^{i\phi} \cos\alpha\cos\beta) e^{-i \chi_2} \end{bmatrix}^{t}$

Please, observe that this state vector interpolates between the separable case at $\eta =0$ and the maximally entangled case at $\eta =\frac{\pi}{2} $ (in this case it depends on a single combination $\alpha , \beta, \phi$ making two of them redundant.

This state vector describes the generic orbit for $0<\eta< \frac{\pi}{2}$.

The angle $\eta$ is related to the angle $\theta$ used in the article by the relation:

$ \sin\theta = \frac{2\sin\eta}{\sqrt{1+\sin^2\eta}}$

In this case, we obtain:

$P(S_z = 1) = \frac{1}{1+\sin^2\eta} ( \cos^2 \alpha\cos^2 \beta + \sin^2 \alpha\sin^2 \beta \sin^2 \eta - 2\cos\alpha\cos \beta \sin \alpha\sin \beta \sin \eta \cos \phi)$

$P(S_z = 0) = \frac{1}{1+\sin^2\eta} ( \cos^2 \alpha\sin^2 \beta (1+\sin^2\eta)+ \sin^2 \alpha\cos^2 \beta (1+\sin^2\eta) + 4\cos\alpha\cos \beta \sin \alpha\sin \beta \sin \eta \cos \phi)$

$P(S_z = -1) = \frac{1}{1+\sin^2\eta} (\sin^2 \alpha\sin^2 \beta + \cos^2 \alpha\cos^2 \beta \sin^2 \eta - 2\cos\alpha\cos \beta \sin \alpha\sin \beta \sin \eta \cos \phi)$

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    $\begingroup$ This is a very useful answer as well, make the decision on the bounty awkwardly difficult. At this stag I'm not yet sure which we will finally use, but the answer by @Frédéric Grosshans is somewhat more diverse and bring in a number of useful contexts. Thanks to both of you for the clear ideas and the effort for write them up properly! $\endgroup$
    – Slaviks
    Apr 30, 2013 at 10:49

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