4.2.2 One-time pad

Before discussing teleportation of quantum states, it is useful to first understand a simpler procedure for probabilistic bits, which is called one-time pad. This procedure allows Alice to encrypt a message and send it to Bob in such a way that only Bob can understand what the message is. That is, in case anyone else intercepts the message (such as their classmate Eve), they would have no idea what the actual message is. The fact that this is even possible is somewhat surprising. Indeed, what advantage does Bob have over Eve so that he can decode Alice’s message correctly while Eve has absolutely no clue about its contents?

The trick is for Alice and Bob to first meet in a coffee shop. Take two coins, glue them together with a chewing gum, flip the resulting ‘double coin’, and separate the two coins again. Alice and Bob each take one of the flipped coins. Now they share a pair of random bits that are described by the state from (3.5), namely

You can think of this as a shared secret! Moreover, only Alice and Bob know what this secret is – to find it out, they can simply look at their respective coins (thus measuring them). They will both see the same side, and each of the sides occurs with probability $1/2$. This is a very good secret, since Eve cannot predict it better than blindly guessing!

Now, let us see how Alice and Bob can make use of it. Suppoose that Alice has a secret message $m\in\{0,1\}$ that she wants to send to Bob. The overall state of all their bits is described by the state

where the first two bits ($m$ and the first half of $r$) belong to Alice and the third bit (the second half of $r$) belongs to Bob.

To send her message, Alice needs to look at her half of the shared random bit $r$ and

1. 1.
   ​

   if she sees $0$ then she sends $m$ over to Bob as it is,

2. 2.
   ​

   if she sees $1$ then she sends $\mathrm{NOT}(m)$ over to Bob.

Let’s imagine that Eve intercepts this message. What does she see? Irrespective of the value of $m$, she will see $0$ with probability $1/2$ and $1$ with probability $1/2$. This is because Alice inverts $m$ with probability $1/2$, which effectively randomizes $m$ in such a way that Eve sees it as a uniformly random bit.

But what about Bob? Doesn’t Alice’s message appear uniformly random to him as well? Luckily, Bob has the other half of the secret random bit they shared. While originally Alice’s message appears random to him, too, he can decode it by applying exactly the same procedure as Alice: look at his half of the shared random bit and

1. 1.
   ​

   if he sees $0$ then he takes Alice’s message as it is,

2. 2.
   ​

   if he sees $1$ then he applies a $\mathrm{NOT}$ operation to Alice’s message.

Overall, Alice’s message is either transmitted as it is or inverted twice, meaning that Bob always understands it correctly. However, from Eve’s perspective it has been inverted with probability $1/2$, meaning that what she sees is a uniformly random bit. Hence, this is a perfectly secure way for Alice and Bob to communicate!

Let us understand more formally what is going on here. When Alice inverts her first bit if her second bit is equal to $1$, this is the same as applying $\mathrm{CNOT}_{2\to 1}$ on her two bits. Next, Alice sends the first bit over to Bob. Then, when Bob decodes Alice’s message, what he does is to apply a $\mathrm{CNOT}_{3\to 1}$ operation (which he can do since, in addition to the third bit, he now also holds the first bit). The resulting state is

We can evaluate this as follows:

Thus, the message bit is back in its original state but is now with Bob.

An interesting aspect of the above one-time pad protocol is not only that it lets Alice send a deterministic message $[m]$ to Bob but even a probabilistic one. It follows by linearity that if Alice’s message is instead a probabilistic bit with distribution $p$, then the initial state is $p\otimes r$ and the final state is again $p\otimes r$, where this time $p$ is with Bob. However, from Eve’s perspective the transmitted message is still uniformly random. The surprising part about this is that by sending a uniformly random bit Alice manages to secretly transmit a probabilistic bit whose distribution she might not even be aware of.

This procedure is quite similar to quantum teleportation, where Alice can transmit a qubit state $\left|\psi\right\rangle$ to Bob by sending two (instead of one) uniformly random bit. For teleportation, they have to use a shared maximally entangled state $\left|\Phi^{+}\right\rangle$ instead of the shared random bit $r$. In both cases, the shared resource is measured and thus consumed during the procedure. We discuss this in the following section.