13.8.1: Qubits in the Bloch sphere

Consider a qubit in an arbitrary superposition

\[|\;\psi\rangle=\alpha\;|\;0\rangle+\beta\;|\;1\rangle . \tag{13.47} \]

If \(\alpha\) and \(\beta\) were real, equation 13.17 would define a circle of radius one, and we would picture the qubit as a point in the boundary of the circle. However, \(\alpha\) and \(\beta\) are complex numbers, so we will have to work a little bit harder to derive a similar intuition.

Every complex number can be represented by a phase and a magnitude, so we can rewrite \(\alpha\) and \(\beta\) as:

\[\alpha=A e^{i a} \quad \beta=B e^{i b} \tag{13.48} \]

from the normalization of the kets (Equation 13.17), we can derive that

\[\begin{align*}
1 &=|\alpha|^{2}+|\beta|^{2} \\
&=A^{2}+B^{2}, \tag{13.49}
\end{align*} \nonumber \]

and this now is the equation of a circle centered at the origin, so both \(A\) and \(B\) can be rewritten in terms of an angle\(^5\) \(\theta\)/2.

\[A=\cos \frac{\theta}{2} \quad B=\sin \frac{\theta}{2}. \tag{13.50} \]

let us introduce this result in the equation 13.47 of the original superposition:

\[|\;\psi\rangle=\cos \frac{\theta}{2} e^{i a}|\;0\rangle+\sin \frac{\theta}{2} e^{i b}|\;1\rangle. \tag{13.51} \]

we can still do one more thing, take \(e^{ia}\) out as a common factor

\[\begin{align*}
|\;\psi\rangle &=e^{i a}\left(\cos \frac{\theta}{2}\;|\;0\rangle+\sin \frac{\theta}{2} e^{i(b-a)}\;|\;1\rangle\right) \\
&=e^{i a}\left(\cos \frac{\theta}{2}\;|\;0\rangle+\sin \frac{\theta}{2} e^{i \varphi}\;|\;1\rangle\right) \tag{13.52}
\end{align*} \nonumber \]

where we have renamed \(\varphi = b − a\). If we ignore the global phase factor(\(e^{ia}\)), the two angles \(\theta\) and \(\varphi\) define a point in a unit sphere. This sphere is called the Bloch Sphere, and is shown in Figure 13.2.

Each point in its surface represents one possible superposition of the states \(|\;0 \rangle\) and \(|\; 1 \rangle\). For example, consider the qubit in the state \(|\; \eta\rangle = \frac{1}{\sqrt{2}} (|\; 0\rangle + |\; 1\rangle\)), we can compare this state with equation 13.52, and conclude that then \(\theta\) /2 = \(\pi\)/4, \(\varphi\) = 0, so that the qubit \(|\; \eta\rangle\) is represented by a vector parallel to the x-axis of the Bloch Sphere.

Figure 13.2: Geometrical representation of a Qubit: Bloch Sphere

When we introduced operators we said that they transformed the superposition state of qubits while preserving their normalization. In the context of the Bloch Sphere, this means that operators move dots around the unit sphere, i.e., define trajectories.

Back to equation 13.52, we still need to consider what happens to the global phase factor eia that we had ignored. This factor would seem to imply that the dot in the Bloch sphere can rotate over itself by an angle a. However, as we are ultimately interested in the probability of each state (because it is the state and not the superpositions what we measure), we should see what happens to this factor as we take the square of the dot products. For example let us examine the probability that measuring the qubit from equation 13.52 yields \(|\; 1\rangle\) as an answer

\[\begin{align*}
|\langle 1 \mid \psi\rangle|^{2} &=\left|\langle 1|\cdot e^{i a}\left(\cos \frac{\theta}{2}|0\rangle+\sin \frac{\theta}{2} e^{i \varphi}|1\rangle\right)\right|^{2}\\
&=\left|e^{i a}\right|^{2} \times\left|\cos \frac{\theta}{2}\langle 1 \mid 0\rangle+\sin \frac{\theta}{2} e^{i \varphi}\langle 1 \mid 1\rangle\right|^{2} \\
&=1 \times\left|0+\sin \frac{\theta}{2} e^{i \varphi} \times 1\right|^{2} \\
&=\left|\sin \frac{\theta}{2} e^{i \varphi}\right|^{2}. \tag{13.53}
\end{align*} \nonumber \]

We see that the global phase factor squares to one, and so plays no role in the calculation of the probability. It is often argued that global phase factors disappear when computing probabilities, and so, are not measurable.

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\(^5\)The choice of the angle as \(\theta\)/2 instead of \(\theta\) is a technical detail for us, it is adequate for using the spin of an electron as a qubit, if instead, we used the polarization of the photon, then the adequate choice would be \(\theta\). This is related to fermions and bosons of which you may have heard, but whose nature is irrelevant to us here.