Angles on a sphere for quantum computing

Two angles name a point

A point on a sphere of radius 1 is fixed by two angles. The polar angle $\theta$ is measured down from the north pole, so $0 \le \theta \le \pi$ : $\theta = 0$ is the north pole, $\theta = \pi/2$ the equator, $\theta = \pi$ the south pole. The azimuth $\varphi$ is measured around the equator, counter-clockwise from the $+x$ axis, so $0 \le \varphi < 2\pi$ . The point's coordinates are

x = \sin\theta\cos\varphi, \quad y = \sin\theta\sin\varphi, \quad z = \cos\theta

The height $z = \cos\theta$ depends on $\theta$ alone, so a fixed $\theta$ is a circle of latitude.

Check yourself

Which point has $\theta = \pi/2$ and $\varphi = \pi/2$ ?

A $(1, 0, 0)$ , on the equator at $+x$
B $(0, 0, -1)$ , the south pole
C $(0, 0, 1)$ , the north pole
D $(0, 1, 0)$ , on the equator at $+y$

Show the answer

Answer: D. $(0, 1, 0)$ , on the equator at $+y$

$(\sin\tfrac{\pi}{2}\cos\tfrac{\pi}{2},\ \sin\tfrac{\pi}{2}\sin\tfrac{\pi}{2},\ \cos\tfrac{\pi}{2}) = (0, 1, 0)$ .

Why not the others

A. $+x$ is $\varphi = 0$ . A quarter turn counter-clockwise, $\varphi = \pi/2$ , reaches $+y$ .
B. The south pole has $\theta = \pi$ . Here $\sin\theta = 1$ and $\cos\theta = 0$ , so the point is on the equator.
C. The north pole has $\theta = 0$ . With $\theta = \pi/2$ the point is on the equator, and $\varphi = \pi/2$ turns it a quarter turn from $+x$ to $+y$ .

Check yourself

What is the height $z$ of the point with $\theta = 2\pi/3$ (any $\varphi$ )?

$z = \;?$

Hint

$z = \cos\theta$ .

Show the answer

Answer: $z = -\frac{1}{2}$

$z = \cos(2\pi/3) = -1/2$ : 120° from the north pole is 30° below the equator.

Common slips

$\frac{1}{2}$ : $\cos(2\pi/3)$ is negative: $\theta = 120°$ is past the equator, in the southern half. $z = -1/2$ .

The 2:1 angle mapping

The Bloch sphere writes a qubit state as $\cos\frac{\theta}{2}\ket0 + e^{i\varphi}\sin\frac{\theta}{2}\ket1$ . Why half the polar angle? The two sizes are non-negative and their squares add to 1, so they are the cosine and sine of one angle between 0 and 90°: 0 for $\ket0$ , 90° for $\ket1$ . A polar angle runs from 0 to 180°, pole to pole. Calling the amplitude angle $\theta/2$ stretches the first range onto the second, so $\ket0$ and $\ket1$ land on opposite poles and every state gets a point on the whole sphere. The price: every angle between states doubles on the sphere. That is why orthogonal states end up at opposite points.

Check yourself

A state is $\cos 45°\ket0 + \sin 45°\ket1$ . What is its polar angle $\theta$ on the Bloch sphere?

A $22.5°$
B $90°$ : it sits on the equator
C $45°$

Show the answer

Answer: B. $90°$ : it sits on the equator

$\theta/2 = 45°$ , so $\theta = 90°$ . Equal amplitudes put the state on the equator, as $P(0) = 1/2$ requires.

Why not the others

A. That halves instead of doubling. The amplitudes contain $\theta/2 = 45°$ , so $\theta = 90°$ .
C. 45° is the angle in the amplitudes, which is $\theta/2$ . Double it: $\theta = 90°$ . The state is $\ket{+}$ , on the equator.

Two identities connect the half-angle to the sphere:

\cos^2\frac{\theta}{2} = \frac{1 + \cos\theta}{2}

\sin^2\frac{\theta}{2} = \frac{1 - \cos\theta}{2}

Since the height is $z = \cos\theta$ , ${P(0) = (1 + z)/2}$ : the north pole ( $z = 1$ ) gives $P(0) = 1$ , the equator gives $1/2$ and the south pole gives 0.

Worked example

From amplitudes to a point

Locate $\cos\tfrac{\pi}{8}\ket0 + \sin\tfrac{\pi}{8}\ket1$ on the sphere.

Step 1
Match it to $\cos\frac{\theta}{2}\ket0 + e^{i\varphi}\sin\frac{\theta}{2}\ket1$ : $\theta/2 = \pi/8$ , so $\theta = \pi/4$ . There is no phase factor, so $\varphi = 0$ .
Step 2
Coordinates: $(\sin\tfrac{\pi}{4}, 0, \cos\tfrac{\pi}{4}) = (\tfrac{\sqrt2}{2}, 0, \tfrac{\sqrt2}{2})$ , halfway between the north pole and $+x$ .
Your turn
Use ${P(0) = (1 + z)/2}$ to find $P(0)$ for this state.
$P(0) = \;?$
Hint
1. Here $z = \sqrt2/2$ .
Show the answer
Answer: $P(0) = 0.854$
$(1 + \tfrac{\sqrt2}{2})/2 = \tfrac{2+\sqrt2}{4} \approx 0.854$ , which equals $\cos^2(\pi/8)$ .
Common slips
- $1.707$ : That is $1 + z$ , which is more than 1. Divide by 2: $\tfrac{2 + \sqrt2}{4}$ .
- $\frac{1}{\sqrt{2}}$ : That is $z$ itself. Use ${P(0) = (1 + z)/2}$ .

Check yourself

A state's point has height $z = -\tfrac12$ . What is $P(0)$ ?

$P(0) = \;?$

Hint

$P(0) = \cos^2(\theta/2) = (1 + \cos\theta)/2$ , and $z = \cos\theta$ .

Show the answer

Answer: $P(0) = \frac{1}{4}$

$P(0) = (1 + z)/2 = (1 - \tfrac12)/2 = \tfrac14$ .

Common slips

$-\frac{1}{2}$ : $z$ is a coordinate, not a probability. Use ${P(0) = (1 + z)/2}$ .
$\frac{3}{4}$ : That is $(1 - z)/2 = P(1)$ . For $P(0)$ use $(1 + z)/2$ .

Go further

Free resources for when you want more than this refresher covers.

Spherical coordinate systemWikipedia · Article
Compares the physics and maths conventions, which swap the names θ and φ, and relates the angles to latitude and longitude.
Precalculus 2e: Double-angle, half-angle, and reduction formulasOpenStax · Free book
Derivations and exercises for the half-angle identities that turn θ/2 into a height on the sphere.
Bloch sphere representation of quantum states for a spin ½ particleQuVis, University of St Andrews · Interactive
Set the polar and azimuthal angles with sliders and read off the state, written with spin-up and spin-down kets.

Lessons that use it

Lesson 3: The Bloch sphereTwo angles for every qubit state, and why orthogonal means opposite