3.1.8 Correlated distributions

Some two-bit distributions are not product distributions – they cannot be written as in Eq. 3.22 or Eq. 3.24, no matter what values of $q_{a}$ and $r_{b}$ you choose. We will say that a distribution is correlated if it is not a product distribution. One example of a correlated distribution is

To see that this is not a product distribution, suppose we measure one of the bits. The outcome $a$ will be completely random, i.e., either $0$ or $1$ with probability 50% each. However, once we know the outcome $a$, the state of the remaining bit is completely determined – measuring it would yield the same outcome with probability 100%. Hence the state of the remaining bit is $b=a$, which depends on the outcome $a$ of the measurement performed on the other bit. We saw in 3.5 that this cannot be the case for a product distribution. Thus, we have proved that Eq. 3.28 describes a correlated state. In fact, the two bits are perfectly correlated since both measurement outcomes are completely random but always identical ($a=b$). Because of this property we say that Eq. 3.28 describes a pair of perfectly correlated random bits.

Correlated distributions arise naturally through some kind of interaction. For example, suppose you toss a fair coin, write the outcome on a piece of paper, hide the paper in an envelope, and pass the envelope to a friend. From the perspective of your friend (who knows the preparation procedure but not what is written on the paper inside the envelope), the state of your coin (bit 1) and of the piece of paper inside their envelope (bit 2) is described by

$$\frac{1}{2}\vbox{\hbox{\includegraphics[width=30.0pt]{figs/0.png}}}\otimes\vbox{\hbox{\includegraphics[width=30.0pt]{figs/envelope-heads-en.pdf}}}+\frac{1}{2}\vbox{\hbox{\includegraphics[width=30.0pt]{figs/1.png}}}\otimes\vbox{\hbox{\includegraphics[width=30.0pt]{figs/envelope-tails-en.pdf}}}$$

which is nothing but an amusing way of writing the two-bit state in Eq. 3.28.

How can we create correlated states in Quirky? Local operations alone are not enough since those can only create product states. However, we can use the controlled-NOT operation to make the two bits interact, as in the following Quirky computation:

Why does this work? You can figure this out in the following exercise:

To check whether an arbitrary two-bit distribution

corresponds to a product or a correlated state, you can simply compute the following quantity:

If $\Delta(p)=0$ then $p$ is a product distribution; otherwise it is correlated. For example, for the state in Eq. 3.28 we find that

which confirms that it is indeed correlated and not a product state. If you are curious why this simple condition works, you can read the explanation below. But since it is not essential for understanding the rest, you can also feel free to skip it.

To prove that $\Delta(p)=0$ is equivalent to $p$ being a product state, we need to show two things. First, let us show that if $p$ is a product state then $\Delta(p)=0$. Indeed, if $p=q\otimes r$ then

How about the converse claim – could we have $\Delta(p)=0$ even when $p$ is not a product state? It turns out that this is not possible! To prove this, let us assume that $\Delta(p)=0$ and show that $p=q\otimes r$, for some probabilistic bits $q$ and $r$. Let us choose these two bits as follows:

Note from Fig. 3.3 that $q$ and $r$ are simply the distributions of outcomes that we would obtain if we measured the first or the second bit, respectively. Let us now verify that this choice of $q$ and $r$ indeed produces the state $p$:

Here we first used Eq. 3.26, then we multiplied out the product, next we used $\Delta(p)=0$ to replace $\boldsymbol{p_{01}p_{10}}$ by $\boldsymbol{p_{00}p_{11}}$ (see Eq. 3.29) , and finally we simplified $p_{00}+p_{01}+p_{10}+p_{11}=1$, since $p$ is a probability distribution. The three terms that we abbreviated by “…” can be treated similarly. Can you fill in the details and verify one of them yourself?

We saw before in 3.5 that $p$ cannot be a product state if measuring the first bit “disturbs” the state of the second bit. In fact, the converse is also true, as you can show in the following homework.