5. Entanglement

The proof of Theorem 4 (http://planetmath.org/3distributeddynamicalsystems#Thmthm4) showed the structure presheaf has non-unique descent, reflecting the fact that measuring devices do not necessarily reduce to products of subdevices. Similarly, as we will see, measurements do not in general decompose into independent submeasurements. Entanglement, $\gamma$, quantifies how far a measurement diverges in bits from the product of its submeasurements. It turns out that $\gamma >0$ is necessary for a system to generate more information than the sum of its components: non-unique descent thus provides “room at the top” to build systems that perform more precise measurements collectively than the sum of their components.

Entanglement has no direct relation to quantum entanglement. The name was chosen because of a formal resemblance between the two quantities, see Supplementary Information of [1].

Definition 10.

Entanglement over partition $𝒫=\{M_1\mathrm{\dots }{M}_m\}$ of $src({𝔪}_{𝐃})$ is

$$\gamma ({𝔪}_{𝐃},𝒫,d_{out})=H[{𝔪}_{𝐃}^{\mathrm{♮}}\circ d_{out}\parallel \underset{i=1}{\overset{m}{\otimes }}{\pi }_j\circ {𝔪}_j^{\mathrm{♮}}\circ d_{out}]$$

where ${\pi }_j:𝒱S^{𝐃}\to 𝒱S^{M_j}$ and ${𝔪}_j=\{(k,l)\in {𝔪}_{𝐃}|k\in M_j\}$.

Projecting via ${\pi }_j$ marginalizes onto the subspace $𝒱S^{M_j}$. Entanglement thus compares the measurement performed by the entire system with submeasurements over the decomposition of the source occasions into partition $𝒫$.

Theorem 9 (effective information decomposes additively when entanglement is zero).

$$\gamma ({𝔪}_{𝐃},𝒫,d_{out})=0⟹ei({𝔪}_{𝐃},d_{out})=\sum _{i=1}^{m}ei({𝔪}_j,d_{out}).$$

Proof: Follows from the observations that (i) $H[p\parallel p_1\otimes p_2]=0$ if and only if $p=p_1\otimes p_2$; (ii) $H[p_1\otimes p_2\parallel q_1\otimes q_2]=H[p_1\parallel q_1]+H[p_2\parallel q_2]$; and (iii) the uniform distribution on $𝐃$ is a tensor of uniform distributions on subsystems of $𝐃$. $\mathrm{■}$

The theorem shows the relationship between effective information and entanglement. If a system generates more information “than it should” (meaning, more than the sum of its subsystems), then the measurements it generates are entangled. Alternatively, only indecomposable measurements can be more precise than the sum of their submeasurements.

We conclude with some detailed computations for $X\times Y\stackrel{𝑔}{\to }Z$, Diagram (11) (http://planetmath.org/4measurement#id2). Let $𝒫=\{X|Y\}$.

Theorem 10 (entanglement and effective information for $g\mathrm{:}X\mathrm{\times }Y\mathrm{\to }Z$).

$\gamma ({𝔪}_{XY},𝒫,{\delta }_z)$ $={\displaystyle \sum _{(x,y)\in g^{-1}(z)}}{\displaystyle \frac{1}{|g^{-1}(z)|}}{\log }_2{\displaystyle \frac{|g^{-1}(z)|}{|g_{x\times Y}^{-1}(z)|\cdot |g_{X\times Y}^{-1}(z)|}}$

$=ei({𝔪}_{XY},{\delta }_z)-ei({𝔪}_{X\bullet },{\delta }_z)-ei({𝔪}_{\bullet Y},{\delta }_z).$

Proof: The first equality follows from Propositions 5 (http://planetmath.org/4measurement#Thmthm5) and 6 (http://planetmath.org/4measurement#Thmthm6)

$$\gamma ({𝔪}_{XY},𝒫,{\delta }_z)=\sum _{(x,y)\in g^{-1}(z)}=\sum _{(x,y)\in g^{-1}(z)}\frac{1}{|g^{-1}(z)|}{\log }_2\left[\frac{1}{|g^{-1}(z)|}\cdot \frac{|g^{-1}(z)|}{|g_{x\times Y}^{-1}(z)|}\frac{|g^{-1}(z)|}{|g_{X\times Y}^{-1}(z)|}\right].$$

From the same propositions it follows that $ei({𝔪}_{XY},{\delta }_z)-ei({𝔪}_{X\bullet },{\delta }_z)-ei({𝔪}_{\bullet Y},{\delta }_z)$ equals

$${\log }_2\frac{|X|\cdot |Y|}{|g^{-1}(x)|}-\sum _x\frac{|g_{x\times Y}^{-1}(z)|}{|g^{-1}(z)|}{\log }_2\frac{|X|\cdot |g_{x\times Y}^{-1}(z)|}{|g^{-1}(z)|}-\sum _y\frac{|g_{X\times y}^{-1}(z)|}{|g^{-1}(z)|}{\log }_2\frac{|Y|\cdot |g_{X\times y}^{-1}(z)|}{|g^{-1}(z)|}$$

$$={\log }_2\frac{1}{g^{-1}(z)}-\sum _{(x,y)\in g^{-1}(z)}\frac{1}{|g^{-1}(z)|}\cdot {\log }_2\frac{|g_{X\times y}^{-1}(z)|}{|g^{-1}(z)|}\cdot \frac{|g_{x\times Y}^{-1}(z)|}{|g^{-1}(z)|}.$$

Entanglement quantifies how far the size of the pre-image of $g^{-1}(z)$ deviates from the sizes of its $X\times y$ and $x\times Y$ slices as $x$ and $y$ are varied. $\mathrm{■}$

By Corollary 8 (http://planetmath.org/4measurement#Thmthm8) entanglement also equals $ei({𝔪}_{X\bullet }\to {𝔪}_{XY},{\delta }_z)-ei({𝔪}_{\bullet Y},{\delta }_z)$. In Diagram (11) (http://planetmath.org/4measurement#id2) entanglement is the vertical arrow minus both arrows at the bottom, or the difference between opposing diagonal arrows. Note that the diagonal arrows from left to right are constructed by adding edge $v_Y\to v_Z$ to the null system and the subsystem ${𝔪}_{X\bullet }=\{v_X\to v_Z\}$ respectively. Entanglement is the difference between the information generated by the diagonal arrows. It quantifies the difference between the information $\{v_Y\to v_Z\}$ generates in two different contexts.

Corollary 11 (characterization of disentangled set-valued functions).

Function $X\mathrm{\times }Y\stackrel{𝑔}{\mathrm{\to }}Z$ performs a disentangled measurement when outputting $z$ iff

$$g^{-1}(z)=g_{x\times Y}^{-1}(z)\times g_{X\times y}^{-1}(z)$$

for any $x\mathrm{,}y$ such that $g\mathit{}\mathrm{(}x\mathrm{,}y\mathrm{)}\mathrm{=}z$.

Proof: By Theorem 10 entanglement is zero iff

$$|g^{-1}(z)|=|g_{x\times Y}^{-1}(z)|\cdot |g_{X\times y}^{-1}(z)|$$

for any $x,y$ such that $g(x,y)=z$. This implies the desired result since $g^{-1}(z)\hookrightarrow g_{x\times Y}^{-1}(z)\times g_{X\times y}^{-1}(z)$. $\mathrm{■}$

Thus, the measurement generated by $g$ is disentangled iff its pre-image $g^{-1}(z)$ satisfies a strong geometric “rectangularity” constraint: that the pre-image decomposes into the product of its $x\times Y$ and $X\times y$ slices for all pairs of slices intersecting $g^{-1}(z)$. The categorizations performed within a disentangled measuring device have nothing to do with each other, so that the device is best considered as two (or more) distinct devices that happen to have been grouped together for the purposes of performing a computation.

Example 4.

An XOR-gate $g:X\times Y\to Z$ outputting 0 generates an entangled measurement. The pre-image is $g^{-1}(0)=\{00,11\}$ so the XOR-gate generates 1 bit of information about occasions $v_X$ and $v_Y$. However, the bit is not localizable. The measurement generates no information about occasion $v_X$ taken singly, since its output could have been 0 or 1 with equal probability; and similarly for $v_Y$.

Finally, and unsurprisingly, a function is completely disentangled across all its measurements iff it is a product of two simpler functions:

Corollary 12 (completely disentangled functions are products).

If $X\mathrm{\times }Y\stackrel{𝑔}{\mathrm{\to }}Z$ is surjective, then
$\gamma \mathit{}\mathrm{(}{\mathrm{m}}_{X\mathit{}Y}\mathrm{,}\mathrm{P}\mathrm{,}{\delta }_z\mathrm{)}\mathrm{=}\mathrm{0}$ for all $z\mathrm{\in }Z$ iff $g$ decomposes into $X\mathrm{\times }Y\stackrel{g_{\mathrm{1}}\mathrm{\times }{g}_{\mathrm{2}}}{\mathrm{\to }}{Z}_{\mathrm{1}}\mathrm{\times }{Z}_{\mathrm{2}}\mathrm{=}Z$ for $X\stackrel{g_{\mathrm{1}}}{\mathrm{\to }}{Z}_{\mathrm{1}}$ and $Y\stackrel{g_{\mathrm{2}}}{\mathrm{\to }}{Z}_{\mathrm{2}}$.

Proof: The reverse implication is trivial. In the forward direction, note that $Z=\{g^{-1}(z)|z\in Z\}$ and, by Corollary 11, each pre-image has product structure $g^{-1}(z)=g_{x\times Y}^{-1}(Z)\times g_{X\times Y}^{-1}(z)$. Let $Z_1=\{g_{X\times y}^{-1}|y\in Y\text{ and }z\in Z\}$ and similarly for $Z_2$. Define

$$g_1:X\to Z_1:x\mapsto \text{the unique element of form }{g}_{X\times y}^{-1}(z)\text{ containing it,}$$

and similarly for $g_2$.

Title	5. Entanglement
Canonical name	5Entanglement
Date of creation	2014-04-22 23:06:40
Last modified on	2014-04-22 23:06:40
Owner	rspuzio (6075)
Last modified by	rspuzio (6075)
Numerical id	4
Author	rspuzio (6075)
Entry type	Feature