Mathematical Introduction: Using the Bloch sphere to represent a qubit

Required materials

• paper and pencil

Key idea

Let us consider a qubit represented by ${\displaystyle |\mathrm{\psi <!--\; \psi \; -->}⟩=\mathrm{\alpha <!--\; \alpha \; -->}|0⟩+\mathrm{\beta <!--\; \beta \; -->}|1⟩}$, where, in the most general case, ${\displaystyle \mathrm{\alpha <!--\; \alpha \; -->}}$ and ${\displaystyle \mathrm{\beta <!--\; \beta \; -->}}$ are complex numbers. At first sight, four parameters – the real and imaginary parts of ${\displaystyle \mathrm{\alpha <!--\; \alpha \; -->}}$, and the real and imaginary parts of ${\displaystyle \mathrm{\beta <!--\; \beta \; -->}}$– are needed to determine the state of the qubit.

The state of the qubit can also be written in an exponential form:

${\displaystyle |\mathrm{\psi <!--\; \psi \; -->}⟩=\mathrm{\alpha <!--\; \alpha \; -->}|0⟩+\mathrm{\beta <!--\; \beta \; -->}|1⟩=\left|\mathrm{\alpha <!--\; \alpha \; -->}\right|e^{i{\mathrm{\phi <!--\; \phi \; -->}}_{\mathrm{\alpha <!--\; \alpha \; -->}}}|0⟩+\left|\mathrm{\beta <!--\; \beta \; -->}\right|e^{i{\mathrm{\phi <!--\; \phi \; -->}}_{\mathrm{\beta <!--\; \beta \; -->}}}{|1⟩}^{}}$

In this equation, two real numbers (${\displaystyle \left|\mathrm{\alpha <!--\; \alpha \; -->}\right|}$ and ${\displaystyle \left|\mathrm{\beta <!--\; \beta \; -->}\right|}$) and two angles (${\displaystyle {\mathrm{\phi <!--\; \phi \; -->}}_{\mathrm{\alpha <!--\; \alpha \; -->}}}$ and ${\displaystyle {\mathrm{\phi <!--\; \phi \; -->}}_{\mathrm{\beta <!--\; \beta \; -->}}}$) are needed to describe the state of the qubit – so, still four parameters.

Eliminating two parameters

One can now factorise out ${\displaystyle e^{i{\mathrm{\phi <!--\; \phi \; -->}}_{\mathrm{\alpha <!--\; \alpha \; -->}}}}$:

${\displaystyle |\mathrm{\psi <!--\; \psi \; -->}⟩=e^{i{\mathrm{\phi <!--\; \phi \; -->}}_{\mathrm{\alpha <!--\; \alpha \; -->}}}(\left|\mathrm{\alpha <!--\; \alpha \; -->}\right||0⟩+\left|\mathrm{\beta <!--\; \beta \; -->}\right|e^{i({\mathrm{\phi <!--\; \phi \; -->}}_{\mathrm{\beta <!--\; \beta \; -->}}-<!--\; -\; -->{\mathrm{\phi <!--\; \phi \; -->}}_{\mathrm{\alpha <!--\; \alpha \; -->}})}|1⟩)}$

or after omitting the so-called global phase ${\displaystyle e^{i{\mathrm{\phi <!--\; \phi \; -->}}_{\mathrm{\alpha <!--\; \alpha \; -->}}}}$:

${\displaystyle |\mathrm{\psi <!--\; \psi \; -->}⟩=\left|\mathrm{\alpha <!--\; \alpha \; -->}\right||0⟩+\left|\mathrm{\beta <!--\; \beta \; -->}\right|e^{i\mathrm{\phi <!--\; \phi \; -->}}|1⟩}$

with ${\displaystyle \mathrm{\phi <!--\; \phi \; -->}={\mathrm{\phi <!--\; \phi \; -->}}_{\mathrm{\alpha <!--\; \alpha \; -->}}-<!--\; -\; -->{\mathrm{\phi <!--\; \phi \; -->}}_{\mathrm{\beta <!--\; \beta \; -->}}}$

Hence, measurable properties of the qubit state depend only on ${\displaystyle \left|\mathrm{\alpha <!--\; \alpha \; -->}\right|}$, ${\displaystyle \left|\mathrm{\beta <!--\; \beta \; -->}\right|}$ and the relative phase ${\displaystyle \mathrm{\phi <!--\; \phi \; -->}}$. The global phase does not affect the outcome of a measurement. This leaves us with three parameters needed to describe the qubit state.

→ We are left with three parameters due to the elimination of a global phase.

If one performs a measurement on a qubit described by the state ${\displaystyle |\mathrm{\psi <!--\; \psi \; -->}⟩=\mathrm{\alpha <!--\; \alpha \; -->}|0⟩+\mathrm{\beta <!--\; \beta \; -->}|1⟩}$, the probability of finding it in state ${\displaystyle |0⟩}$ is ${\displaystyle {\left|\mathrm{\alpha <!--\; \alpha \; -->}\right|}^2}$, and the probability of finding it in state ${\displaystyle |1⟩}$ is ${\displaystyle {\left|\mathrm{\beta <!--\; \beta \; -->}\right|}^2}$(see “Basics of Quantum Physics” [link]). These two probabilities sum up to 1:

${\displaystyle {\left|\mathrm{\alpha <!--\; \alpha \; -->}\right|}^2+{\left|\mathrm{\beta <!--\; \beta \; -->}\right|}^2=1}$

Let us now jump to a very similar equation we know from trigonometry. The equation is valid for any angle, but for reasons that will become clear later, we choose the angle ${\displaystyle \frac{\mathrm{\theta <!--\; \theta \; -->}}{2}}$.

${\displaystyle {\cos }^2<!--\; \; -->\frac{\mathrm{\theta <!--\; \theta \; -->}}{2}+{\sin }^2<!--\; \; -->\frac{\mathrm{\theta <!--\; \theta \; -->}}{2}=1}$

By comparing the two equations, one can write ${\displaystyle \left|\mathrm{\alpha <!--\; \alpha \; -->}\right|}$ and ${\displaystyle \left|\mathrm{\beta <!--\; \beta \; -->}\right|}$ as a function of the angle ${\displaystyle \mathrm{\theta <!--\; \theta \; -->}}$:

${\displaystyle \left|\mathrm{\alpha <!--\; \alpha \; -->}\right|=\cos <!--\; \; -->\frac{\mathrm{\theta <!--\; \theta \; -->}}{2}}$ and ${\displaystyle \left|\mathrm{\beta <!--\; \beta \; -->}\right|=\sin <!--\; \; -->\frac{\mathrm{\theta <!--\; \theta \; -->}}{2}}$

with ${\displaystyle \mathrm{\theta <!--\; \theta \; -->}\in <!--\; \in \; -->[0,\mathrm{\pi <!--\; \pi \; -->}]}$

Or, if we go back to the general description of a qubit state:

${\displaystyle |\mathrm{\psi <!--\; \psi \; -->}⟩=\mathrm{\alpha <!--\; \alpha \; -->}|0⟩+\mathrm{\beta <!--\; \beta \; -->}|1⟩=\cos <!--\; \; -->\frac{\mathrm{\theta <!--\; \theta \; -->}}{2}|0⟩+\sin <!--\; \; -->\frac{\mathrm{\theta <!--\; \theta \; -->}}{2}{e}^{i\mathrm{\phi <!--\; \phi \; -->}}|1⟩}$

The interval ${\displaystyle [0,\mathrm{\pi <!--\; \pi \; -->}]}$ reflects the fact that because of the normalisation relation ${\displaystyle {\left|\mathrm{\alpha <!--\; \alpha \; -->}\right|}^2+{\left|\mathrm{\beta <!--\; \beta \; -->}\right|}^2=1}$, ${\displaystyle \left|\mathrm{\alpha <!--\; \alpha \; -->}\right|}$ and ${\displaystyle \left|\mathrm{\beta <!--\; \beta \; -->}\right|}$ are real numbers between 0 and 1: ${\displaystyle \left|\mathrm{\alpha <!--\; \alpha \; -->}\right|,\left|\mathrm{\beta <!--\; \beta \; -->}\right|\in <!--\; \in \; -->[0,19]}$. ${\displaystyle \left|\mathrm{\alpha <!--\; \alpha \; -->}\right|}$ and ${\displaystyle \left|\mathrm{\beta <!--\; \beta \; -->}\right|}$ can be visualised as coordinates on a quarter circle in the first quadrant of the sphere (both numbers are nonnegative).

→ We are left with merely two parameters due to the normalisation relation ${\displaystyle {\left|\mathrm{\alpha <!--\; \alpha \; -->}\right|}^2+{\left|\mathrm{\beta <!--\; \beta \; -->}\right|}^2=1}$.

Conclusion

Combining these ideas and writing ${\displaystyle \mathrm{\alpha <!--\; \alpha \; -->}}$ and ${\displaystyle \mathrm{\beta <!--\; \beta \; -->}}$ as functions of ${\displaystyle \mathrm{\theta <!--\; \theta \; -->}}$ and ${\displaystyle \mathrm{\phi <!--\; \phi \; -->}}$ leads to the Bloch sphere description of a qubit state:

${\displaystyle |\mathrm{\psi <!--\; \psi \; -->}⟩=\cos <!--\; \; -->\frac{\mathrm{\theta <!--\; \theta \; -->}}{2}|0⟩+\sin <!--\; \; -->\frac{\mathrm{\theta <!--\; \theta \; -->}}{2}{e}^{i\mathrm{\phi <!--\; \phi \; -->}}|1⟩}$

with ${\displaystyle \mathrm{\phi <!--\; \phi \; -->}\in <!--\; \in \; -->[0,2\mathrm{\pi <!--\; \pi \; -->}]}$ and ${\displaystyle \mathrm{\theta <!--\; \theta \; -->}\in <!--\; \in \; -->[0,\mathrm{\pi <!--\; \pi \; -->}]}$.

On the Bloch sphere:

• ${\displaystyle \mathrm{\theta <!--\; \theta \; -->}}$(polar angle) spans from 0 to π;
• ${\displaystyle \mathrm{\phi <!--\; \phi \; -->}}$(azimuthal angle) spans from 0 to 2π.

→ Hence, a qubit state can be described by only two parameters: two angles.

Any pair of angles ${\displaystyle (\mathrm{\theta <!--\; \theta \; -->},\mathrm{\phi <!--\; \phi \; -->})}$ corresponds to exactly one point on the surface of the unit sphere, and therefore to one unique qubit state (up to a global phase).

Special cases

Task for students: Quiz

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