2.1.2 Qubit as a circle

Figure 2.1: Qubit state |ψ⁢(θ)⟩\left|\psi(\theta)\right\rangle as a point on the unit circle.

Notice how Eq. 2.4 for qubit amplitudes is reminiscent of the equation x2+y2=1x^{2}+y^{2}=1 of a circle. Let us work out this correspondence in more detail because it will help us to visualize quantum bits and to understand them on a more intuitive level.

A convenient way to parametrize the qubit amplitudes is by letting

ψ0=cos⁡θ,ψ1=sin⁡θ\psi_{0}=\cos\theta,\qquad\psi_{1}=\sin\theta

for some angle θ∈[0,2⁢π)\theta\in[0,2\pi). In fact, it will often be convenient to allow the angle θ\theta to be an arbitrary real number (which is fine provided we keep in mind that any two angles that differ by 2⁢π2\pi result in the same amplitudes). Since cos2⁡θ+sin2⁡θ=1\cos^{2}\theta+\sin^{2}\theta=1, we are guaranteed to automatically satisfy Eq. 2.4. With this choice, a general qubit state looks as follows:

|ψ⁢(θ)⟩=cos⁡θ⁢|0⟩+sin⁡θ⁢|1⟩=(cos⁡θsin⁡θ).\left|\psi(\theta)\right\rangle=\cos\theta\left|0\right\rangle+\sin\theta\left% |1\right\rangle=\begin{pmatrix}\cos\theta\\ \sin\theta\end{pmatrix}. (2.5)

One can visualize this as a unit vector in two dimensions that starts at the origin of the plane and has angle θ\theta with the horizontal |0⟩\left|0\right\rangle axis (see Fig. 2.1). In particular, |0⟩=|ψ⁢(0)⟩\left|0\right\rangle=\left|\psi(0)\right\rangle and |1⟩=|ψ⁢(π2)⟩\left|1\right\rangle=\left|\psi(\frac{\pi}{2})\right\rangle. The set of all qubit states then correspond to a unit circle centered at the origin. In contrast, recall from Fig. 1.2 that the set of all states of a probabilistic bit is a line segment connecting the points (10)\bigl{(}\begin{smallmatrix}1\\ 0\end{smallmatrix}\bigr{)} and (01)\bigl{(}\begin{smallmatrix}0\\ 1\end{smallmatrix}\bigr{)} located on the two coordinate axes. The two sets are compared in Fig. 2.2.

Figure 2.2: State spaces of a probabilistic bit (blue) and quantum bit (red).
Exercise 2.1 (States on the circle).

Consider the following two states of a qubit:

|+⟩=|0⟩+|1⟩2,|−⟩=|0⟩−|1⟩2.\left|+\right\rangle=\frac{\left|0\right\rangle+\left|1\right\rangle}{\sqrt{2}% },\qquad\left|-\right\rangle=\frac{\left|0\right\rangle-\left|1\right\rangle}{% \sqrt{2}}.

Where do these two states lie on the circle? What angles θ\theta do they correspond to?

Solution. Note that
|+⟩=12⁢(11)=|ψ⁢(π/4)⟩,|−⟩=12⁢(1−1)=|ψ⁢(−π/4)⟩.\displaystyle\left|+\right\rangle=\frac{1}{\sqrt{2}}\begin{pmatrix}1\\ 1\end{pmatrix}=\left|\psi(\pi/4)\right\rangle,\qquad\left|-\right\rangle=\frac% {1}{\sqrt{2}}\begin{pmatrix}1\\ -1\end{pmatrix}=\left|\psi(-\pi/4)\right\rangle.
Thus the angles are θ=±π/4\theta=\pm\pi/4, and the two states are located 45 degrees upwards and downwards from |0⟩\left|0\right\rangle, respectively.