2.1.1 Probabilities versus amplitudes

Quantum bits are very similar to probabilistic bits. There are only two major differences:

1. 1.
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   probabilities are replaced by amplitudes (which can also be negative),

2. 2.
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   amplitudes are squared during the measurement (while probabilities are not).

We will explain these differences in more detail shortly, but let us first describe the possible states of a qubit. Recall how we used the two sides of a coin to denote the two possible deterministic states of a probabilistic bit (see Fig. 1.1)? In quantum computing, these two states are commonly denoted by $\left|0\right\rangle$ and $\left|1\right\rangle$ to distinguish them from the classical bits $[0]$ and $[1]$. Just like with probabilistic bits, a general qubit state $\left|\psi\right\rangle$ is then a linear combination or superposition of these two deterministic states:

Here, the greek letter $\psi$ (pronounce “psi”) is the name of the qubit state (just like we named the probabilistic bit $p$). The brackets $\left|\cdot\right\rangle$ form a so-called “ket” that indicates that we are dealing with a quantum state. For comparison, recall from Eq. 1.7 that an arbitrary probabilistic bit $p$ can be written as

Note that Eq. 2.1 looks identical to this, except the probabilities $p_{0}$ and $p_{1}$ are replaced by the amplitudes $\psi_{0}$ and $\psi_{1}$, and the classical notation $[0]$ and $[1]$ is replaced by the quantum notation $\left|0\right\rangle$ and $\left|1\right\rangle$! However, there is one major difference: while the probabilities in Eq. 2.2 are subject to

the amplitudes are subject to

In particular, this implies $\psi_{0}^{2}\leq 1$ and $\psi_{1}^{2}\leq 1$, and hence $\psi_{0},\psi_{1}\in[-1,1]$. In contrast, the constraints from Eq. 2.3 on probabilities imply that $p_{0},p_{1}\in[0,1]$. The crucial difference is that amplitudes are actually allowed to be negative while probabilities are not!⁹⁹ 9 In fact, amplitudes are even allowed to be so-called complex numbers. We will not need them in this course, but you’re encouraged to browse the web to learn more about this.

Just like with probabilistic bits, it is convenient to represent qubit states by vectors. In complete analogy with Eq. 1.6, we represent the deterministic qubit states $\left|0\right\rangle$ and $\left|1\right\rangle$ by the two basis vectors:

A general quantum state $\left|\psi\right\rangle$ from Eq. 2.1 is then represented as