The EPR paradox steps in or, the quantum daemon bites its own tail.

The famous EPR (Einstein-Podolsky-Rosen) paradox describes a rather counterintuitive quality of quantum physics: the quantum state space of a set of several particles is the direct product of their individual spaces, and not the direct sum as one would expect. As a consequence, if A and B have separate particles a and b, which happen to be in an entangled state, then A may be able to gain information on B's measurement of b by measuring a.

It is possible for A to exploit this in order to cheat. Here's A's cheating scheme:

When Alice measures her qubit with respect to the horizontal basis, she gets either a zero or a one with 50% chance. However, this projects the entire state space of the entangled pair to either $\left\vert \rightarrow \rightarrow \right\rangle$ or $\left\vert \uparrow \uparrow \right\rangle$ (since the $\left\vert \rightarrow \uparrow \right\rangle$ and $\left\vert \uparrow \rightarrow \right\rangle$ components are zero from the start), which means that Bob will obtain the same result⁵. This is exactly the EPR paradox (table 4).

Table 4: Alice cheats on BB84
qubit pair A's basis $\widetilde{b_{i}}$ new state $b_{i}\prime$ B's basis P(0) P(1)

$\frac{1}{\sqrt{2}}\left\vert \rightarrow \rightarrow \right\rangle$ $\{\rightarrow ,\uparrow \}$ 0 $\left\vert \rightarrow \rightarrow \right\rangle$ 0 $\{\rightarrow ,\uparrow \}$ 1 0

+ 1 $\{\nearrow \nwarrow \}$ $\frac{1}{2}$ $\frac{1}{2}$

$\frac{1}{\sqrt{2}}\left\vert \uparrow \uparrow \right\rangle$ $\{\rightarrow ,\uparrow \}$ 1 $\left\vert \uparrow \uparrow \right\rangle$ 0 $\{\rightarrow ,\uparrow \}$ 0 1

= 1 $\{\nearrow \nwarrow \}$ $\frac{1}{2}$ $\frac{1}{2}$

$\frac{1}{\sqrt{2}}\left\vert \nearrow \nearrow \right\rangle$ $\{\nearrow ,\nwarrow \}$ 0 $\left\vert \nearrow \nearrow \right\rangle$ 0 $\{\rightarrow ,\uparrow \}$ $\frac{1}{2}$ $\frac{1}{2}$

+ 1 $\{\nearrow \nwarrow \}$ 1 0

$\frac{1}{\sqrt{2}}\left\vert \nwarrow \nwarrow \right\rangle$ $\{\nearrow ,\nwarrow \}$ 1 $\left\vert \nwarrow \nwarrow \right\rangle$ 0 $\{\rightarrow ,\uparrow \}$ $\frac{1}{2}$ $\frac{1}{2}$

1 $\{\nearrow \nwarrow \}$ 0 1

**Table 4:** Alice cheats on BB84
qubit pair	A's basis	$\widetilde{b_{i}}$	new state	$b_{i}\prime$	B's basis	P(0)	P(1)
$\frac{1}{\sqrt{2}}\left\vert \rightarrow \rightarrow \right\rangle$	$\{\rightarrow ,\uparrow \}$	0	$\left\vert \rightarrow \rightarrow \right\rangle$	0	$\{\rightarrow ,\uparrow \}$	1	0
+				1	$\{\nearrow \nwarrow \}$	$\frac{1}{2}$	$\frac{1}{2}$
$\frac{1}{\sqrt{2}}\left\vert \uparrow \uparrow \right\rangle$	$\{\rightarrow ,\uparrow \}$	1	$\left\vert \uparrow \uparrow \right\rangle$	0	$\{\rightarrow ,\uparrow \}$	0	1
=				1	$\{\nearrow \nwarrow \}$	$\frac{1}{2}$	$\frac{1}{2}$
$\frac{1}{\sqrt{2}}\left\vert \nearrow \nearrow \right\rangle$	$\{\nearrow ,\nwarrow \}$	0	$\left\vert \nearrow \nearrow \right\rangle$	0	$\{\rightarrow ,\uparrow \}$	$\frac{1}{2}$	$\frac{1}{2}$
+				1	$\{\nearrow \nwarrow \}$	1	0
$\frac{1}{\sqrt{2}}\left\vert \nwarrow \nwarrow \right\rangle$	$\{\nearrow ,\nwarrow \}$	1	$\left\vert \nwarrow \nwarrow \right\rangle$	0	$\{\rightarrow ,\uparrow \}$	$\frac{1}{2}$	$\frac{1}{2}$
				1	$\{\nearrow \nwarrow \}$	0	1

With this procedure Alice makes sure that whenever $\widetilde{b_{i}}=b_{i},$ Bob used the same basis to do his measurement, so he will get an identical result. And in the opposite case, as the angle between state and basis is $\frac{\pi }{4}$ , he obtains a 50%-50% distribution as expected.