Communication Cost of Simulating Bell Correlations

What classical resources are required to simulate quantum correlations? For the simplest and most important case of local projective measurements on an entangled Bell pair state, we show that exact simulation is possible using local hidden variables augmented by just one bit of classical communication. Certain quantum teleportation experiments, which teleport a single qubit, therefore admit a local hidden variables model.

Figure 1

The protocol: The shared unit vectors ${\widehat{\lambda }}_1$ and ${\widehat{\lambda }}_2$ described in the text divide the Bloch sphere into four quadrants, as shown. Alice and Bob’s actions depend on which quadrant their respective measurement axes lie in, and in Bob’s case, the bit he receives from Alice. (a) Alice’s output: if $\widehat{a}$ lies in the shaded region, Alice outputs $-1$; in the unshaded region, she outputs $+1$. (b) The communication: Alice sends $c=+1$ if her measurement axis lies in the N or S quadrants, and $-1$ otherwise. (c) Bob’s output: this depends on the bit received from Alice. The shading is as for (a).

Figure 2

Construction used to evaluate Eq. (2): (a) We first integrate over ${\widehat{\lambda }}_2$, taking $\widehat{b}$ to point along the positive $z$ axis [18]. Observe that $\mathrm{s}\mathrm{g}\mathrm{n}[\widehat{b}·({\widehat{\lambda }}_2-{\widehat{\lambda }}_1)]$ is positive in the top spherical cap (shaded) and negative otherwise. The area of the top spherical cap is $A_{+}=2\pi {\int }_0^t\sin \theta d\theta =2\pi (1-\cos t)$ where $\cos t=\widehat{b}·{\widehat{\lambda }}_1$, hence $\int d{\widehat{\lambda }}_2\mathrm{s}\mathrm{g}\mathrm{n}[\widehat{b}·({\widehat{\lambda }}_2-{\widehat{\lambda }}_1)]=A_{+}-(4\pi -A_{+})=-4\pi \cos t=-4\pi \widehat{b}·{\widehat{\lambda }}_1$. (b) We now take $\widehat{a}$ to point along the positive $z$ axis [19], set $\widehat{b}=(\sin r,0,\cos r)$, and integrate over ${\widehat{\lambda }}_1$, obtaining $\int d{\widehat{\lambda }}_1\mathrm{s}\mathrm{g}\mathrm{n}(\widehat{a}·{\widehat{\lambda }}_1)\widehat{b}·{\widehat{\lambda }}_1={\int }_0^{\pi }\sin \theta d\theta {\int }_0^{2\pi }d\varphi \mathrm{s}\mathrm{g}\mathrm{n}(\cos \theta )(\cos \theta \cos r+\sin \theta \cos \varphi \sin r)=2\pi \cos r=2\pi \widehat{a}·\widehat{b}$.