The Bloch sphere · Kets & Gates

1 / 12 · Learn

Before you start

Lesson 2 described a qubit state by the sizes of its amplitudes and one relative phase. The Bloch sphere draws both as a single point, and turns every measurement basis into an axis.

By the end of this lesson you can:

convert between amplitudes and the Bloch angles $\theta$ and $\varphi$ , removing a global phase first;
place $\ket0$ , $\ket1$ , $\ket{\pm}$ and $\ket{\pm i}$ on the sphere;
explain why orthogonal states sit at opposite points;
use the fact that every state on one circle of latitude has the same $P(0)$ .

Check yourself

A warm-up: make a guess before reading on. The lesson comes back to it.

$\ket0$ and $\ket1$ are orthogonal: $\braket{0}{1} = 0$ . On the Bloch sphere, how far apart are the points of two orthogonal states?

AIt depends on the phases of the two states.
B45°
C180°: they sit at opposite points.
D90°: perpendicular, like the vectors.

Show the answer

Answer: C. 180°: they sit at opposite points.

Orthogonal states are always antipodal: $\ket0$ and $\ket1$ are the two poles, $\ket{+}$ and $\ket{-}$ are opposite points on the equator. This lesson shows why.

Why not the others

A. For every orthogonal pair the answer is the same: opposite points, 180° apart. The phases decide where the pair sits, not how far apart they are.
B. That halves the angle instead of doubling it. The sphere plots $\theta$ , twice the angle inside the amplitudes, so orthogonal states (90° apart as vectors) are 180° apart on the sphere.
D. A natural guess, since orthogonal vectors are perpendicular. But the sphere doubles angles: the amplitudes use $\theta/2$ while the sphere plots $\theta$ . Orthogonal states sit 180° apart, like $\ket0$ and $\ket1$ at the two poles. 90° on the sphere means an overlap of $1/2$ , as for $\ket0$ and $\ket{+}$ .

Refresher: Angles on a sphere This lesson uses spherical coordinates. Open this for $\theta$ , $\varphi$ and the half-angle.

2 / 12 · Learn

Two angles are enough

A state $\alpha\ket0 + \beta\ket1$ has two complex amplitudes, so four real numbers. Normalization, ${\abs\alpha^2 + \abs\beta^2 = 1}$ , fixes one. A global phase changes nothing (Lesson 2), so you can always make $\alpha$ real and non-negative, which fixes another. Two real numbers are left, and they can be written as two angles, in what these lessons call the Bloch form:

\ket\psi = \cos\frac{\theta}{2}\ket0 + e^{i\varphi}\sin\frac{\theta}{2}\ket1

Here $0 \le \theta \le \pi$ and $0 \le \varphi < 2\pi$ , and $\varphi$ is the relative phase from Lesson 2. The amplitudes use half of $\theta$ , so in the 0/1 basis $P(0) = \cos^2(\theta/2)$ . Why the half? That comes later in this lesson.

Check yourself

Read off the Bloch angles of $\cos\tfrac{\pi}{8}\ket0 + e^{i\pi/4}\sin\tfrac{\pi}{8}\ket1$ .

A $\theta = \pi/4$ , $\varphi = \pi/4$
B $\theta = \pi/8$ , $\varphi = \pi/8$
C $\theta = \pi/8$ , $\varphi = \pi/4$
D $\theta = \pi/4$ , $\varphi = \pi/2$

Hint

Line the state up with the Bloch form. The cosine contains $\theta/2$ , not $\theta$ ; the exponential contains $\varphi$ itself.

Show the answer

Answer: A. $\theta = \pi/4$ , $\varphi = \pi/4$

Match the state to the Bloch form: $\theta/2 = \pi/8$ gives $\theta = \pi/4$ , and $e^{i\varphi} = e^{i\pi/4}$ gives $\varphi = \pi/4$ .

Why not the others

B. The angle in the cosine is $\theta/2$ , so $\theta = \pi/4$ . And $\varphi$ is read from $e^{i\varphi} = e^{i\pi/4}$ directly, so $\varphi = \pi/4$ .
C. A common slip is to read the angle straight off the cosine. The amplitudes contain $\theta/2$ , so $\theta/2 = \pi/8$ and $\theta = \pi/4$ . The sphere angle is always twice the angle you see in the amplitudes.
D. Only $\theta$ is halved in the Bloch form. The phase factor is $e^{i\varphi}$ itself, so $\varphi = \pi/4$ .

Check yourself

A state has Bloch angle $\theta = \pi/3$ (any $\varphi$ ). You measure it in the 0/1 basis. What is $P(0)$ ?

$P(0) = \;?$

Hints

Which angle appears inside the amplitude of $\ket0$ ?
The $\ket0$ amplitude is $\cos(\theta/2)$ ; square it.

Show the answer

Answer: $P(0) = \frac{3}{4}$

$P(0) = \cos^2(\theta/2) = \cos^2(\pi/6) = (\sqrt3/2)^2 = 3/4$ , whatever $\varphi$ is.

Common slips

$\frac{1}{4}$ : That is $\cos^2\theta$ . The amplitude of $\ket0$ is $\cos(\theta/2) = \cos(\pi/6) = \sqrt3/2$ , so $P(0) = 3/4$ .
$\frac{\sqrt{3}}{2}$ : That is the amplitude $\cos(\pi/6)$ . A probability is the amplitude squared: $(\sqrt3/2)^2 = 3/4$ .
$\frac{1}{2}$ : That is $\cos\theta$ . The probability is $\cos^2(\theta/2) = \cos^2(\pi/6) = 3/4$ .

3 / 12 · Learn

Landmarks on the sphere

The angles name a point on a sphere of radius 1. $\theta$ is measured down from the north pole; $\varphi$ is measured around the equator, counter-clockwise from the $+x$ axis. The landmarks:

$\ket0$ at the north pole ( $\theta = 0$ ) and $\ket1$ at the south pole ( $\theta = \pi$ );
$\ket{+}$ and $\ket{-}$ on the equator at $+x$ ( $\varphi = 0$ ) and $-x$ ( $\varphi = \pi$ );
$\ket{+i}$ and $\ket{-i}$ on the equator at $+y$ ( ${\varphi = \pi/2}$ ) and $-y$ ( $\varphi = 3\pi/2$ ).

At a pole $\varphi$ does nothing: at the south pole the state is $e^{i\varphi}\ket1$ , a global phase times $\ket1$ .

Check yourself

Where on the sphere is $\ket{-} = \tfrac{1}{\sqrt2}(\ket0 - \ket1)$ ?

AAt the centre, halfway between $\ket0$ and $\ket1$ .
BAt the south pole, because of the minus sign.
COn the equator at $-x$ : $\theta = \pi/2$ , $\varphi = \pi$ .
DOn the equator at $+x$ , the same point as $\ket{+}$ : the probabilities match.

Show the answer

Answer: C. On the equator at $-x$ : $\theta = \pi/2$ , $\varphi = \pi$ .

$\ket{-}$ has $\theta = \pi/2$ (equal sizes) and $\varphi = \pi$ (the factor $-1 = e^{i\pi}$ ), so it sits on the equator at $-x$ , opposite $\ket{+}$ .

Why not the others

A. Every state $\cos\tfrac{\theta}{2}\ket0 + e^{i\varphi}\sin\tfrac{\theta}{2}\ket1$ is a point on the surface: the arrow always has length 1. Points inside the ball describe mixtures, such as a coin choosing $\ket0$ or $\ket1$ .
B. The south pole is $\ket1$ , where $P(1) = 1$ . $\ket{-}$ gives 0 and 1 equally often, so $\cos(\theta/2) = 1/\sqrt2$ , $\theta = \pi/2$ : the equator. The minus sign is the relative phase $\varphi = \pi$ , which picks the $-x$ side.
D. Equal 0/1 probabilities put both states on the equator, but the relative phase picks the longitude. $-1 = e^{i\pi}$ , so $\varphi = \pi$ : the opposite side, $-x$ .

Check yourself

Match each state to its Bloch angles.

Match each of these…

$\ket1$
$\ket{+i} = \tfrac{1}{\sqrt2}(\ket0 + i\ket1)$
$\ket{-i} = \tfrac{1}{\sqrt2}(\ket0 - i\ket1)$
$\ket{-}$

…with one of these

$\theta = \pi/2$ , $\varphi = \pi/2$
$\theta = \pi$ , any $\varphi$
$\theta = \pi/2$ , $\varphi = 3\pi/2$
$\theta = \pi/2$ , $\varphi = \pi$

Hints

Equal sizes put a state on the equator; $\ket1$ is a pole.
Write the $\ket1$ factor as $e^{i\varphi}$ : $i = e^{i\pi/2}$ , $-1 = e^{i\pi}$ , $-i = e^{3i\pi/2}$ .

Show the answer

$\ket1$	$\theta = \pi$ , any $\varphi$
$\ket{+i} = \tfrac{1}{\sqrt2}(\ket0 + i\ket1)$	$\theta = \pi/2$ , $\varphi = \pi/2$
$\ket{-i} = \tfrac{1}{\sqrt2}(\ket0 - i\ket1)$	$\theta = \pi/2$ , $\varphi = 3\pi/2$
$\ket{-}$	$\theta = \pi/2$ , $\varphi = \pi$

$\ket1$ is the south pole. The other three have equal sizes ( $\theta = \pi/2$ ) and $\ket1$ factors $i = e^{i\pi/2}$ , $-i = e^{3i\pi/2}$ and $-1 = e^{i\pi}$ .

4 / 12 · Learn

The half-angle trap

Take $\cos\tfrac{\pi}{6}\ket0 + \sin\tfrac{\pi}{6}\ket1 = \tfrac{\sqrt3}{2}\ket0 + \tfrac12\ket1$ . The angle inside the amplitudes is $\pi/6 = 30°$ , but that angle is $\theta/2$ . On the sphere the state sits at $\theta = \pi/3 = 60°$ from the north pole, tilted toward $+x$ (its $\varphi$ is 0). A quick check uses its height on the sphere, ${z = \cos\theta = 1/2}$ . Since ${\cos^2(\theta/2) = (1 + \cos\theta)/2}$ (see the sphere refresher), ${P(0) = (1 + z)/2 = 3/4}$ , which matches $(\sqrt3/2)^2$ .

Check yourself

Place the state $\cos\tfrac{\pi}{3}\ket0 + \sin\tfrac{\pi}{3}\ket1 = \tfrac12\ket0 + \tfrac{\sqrt3}{2}\ket1$ on the sphere. Use the sliders or drag the tip, then check.

Set the state on the Bloch sphere.

Hints

Is $\pi/3$ the angle $\theta$ , or $\theta/2$ ?
The amplitudes contain $\theta/2$ , so $\theta = 2\pi/3 = 120°$ .
Set $\theta$ to 120° and $\varphi$ to 0.

Show the answer

Answer: $\frac{1}{2}|0\rangle + \frac{\sqrt{3}}{2}|1\rangle$

$\theta/2 = \pi/3$ , so $\theta = 2\pi/3 = 120°$ : past the equator toward $\ket1$ , on the $+x$ side ( $\varphi = 0$ ). That fits $P(1) = 3/4 > 1/2$ .

Check yourself

Write the state with Bloch angles ${\theta = 2\pi/3}$ and $\varphi = \pi/4$ : enter the amplitudes of $\ket0$ and $\ket1$ .

Hints

Write the two amplitudes as $\cos(\theta/2)$ and $e^{i\varphi}\sin(\theta/2)$ .
Here $\theta/2 = \pi/3$ .

Show the answer

Answer: $\frac{1}{2}|0\rangle + (\frac{\sqrt{6}}{4}+\frac{\sqrt{6}}{4}i)|1\rangle$

$\cos(\pi/3) = \tfrac12$ and $e^{i\pi/4}\sin(\pi/3) = e^{i\pi/4}\tfrac{\sqrt3}{2}$ , so the state is $\tfrac12\ket0 + e^{i\pi/4}\tfrac{\sqrt3}{2}\ket1$ , with $P(0) = 1/4$ .

5 / 12 · Learn

Orthogonal means opposite

Now, why the half? The two sizes are $\cos t$ and $\sin t$ for one angle $t$ from 0 to 90°: ${t = 0}$ is $\ket0$ , ${t = 90°}$ is $\ket1$ . As arrows in the plane, $\ket0$ and $\ket1$ are perpendicular. A sphere's polar angle runs from 0 to 180°, pole to pole, and the Bloch form stretches one range onto the other by setting $t = \theta/2$ . Every angle doubles, so perpendicular arrows become opposite points. This holds for any orthogonal pair. The state orthogonal to $(\theta, \varphi)$ is $\sin\tfrac{\theta}{2}\ket0 - e^{i\varphi}\cos\tfrac{\theta}{2}\ket1$ , the antipode ${(\pi - \theta,\ \varphi + \pi)}$ . Check: the inner product is $\cos\tfrac{\theta}{2}\sin\tfrac{\theta}{2} - \sin\tfrac{\theta}{2}\cos\tfrac{\theta}{2} = 0$ , because the bra conjugates $e^{i\varphi}$ and the phases cancel.

As vectors

$\ket0$ and $\ket1$ are perpendicular: 90° apart. $\ket0$ and $\ket{+}$ are 45° apart.

On the Bloch sphere

$\ket0$ and $\ket1$ are opposite: 180° apart. $\ket0$ and $\ket{+}$ are 90° apart.

The state $\cos t\ket0 + \sin t\ket1$ sits at angle $t$ on the left and at angle $2t$ from the north pole on the right. Every angle doubles.

Check yourself

A state sits at $\theta = 2\pi/3$ , $\varphi = \pi/4$ . Where is the state orthogonal to it?

A $\theta = \pi/3$ , $\varphi = \pi/4$
B $\theta = 2\pi/3$ , $\varphi = 5\pi/4$
C $\theta = \pi/3$ , $\varphi = 5\pi/4$
D $\theta = \pi/6$ , $\varphi = \pi/4$ : 90° further round.

Hints

How far apart are orthogonal states on the sphere?
Orthogonal states are antipodal. The antipode of $(\theta, \varphi)$ is $(\pi - \theta, \varphi + \pi)$ .
Here $\pi - \theta = \pi/3$ .

Show the answer

Answer: C. $\theta = \pi/3$ , $\varphi = 5\pi/4$

The antipode of $(\theta, \varphi)$ is $(\pi - \theta, \varphi + \pi)$ : here $(\pi/3, 5\pi/4)$ .

Why not the others

A. Reflecting through the equator swaps $P(0)$ and $P(1)$ , but this point is not opposite: the antipode is also on the other side in $\varphi$ , at $\varphi + \pi = 5\pi/4$ .
B. Turning $\varphi$ by $\pi$ alone keeps the same latitude, so the same $P(0)$ ; an orthogonal state must have the opposite probabilities. You also need $\theta \to \pi - \theta$ .
D. That point is 90° away, which on the sphere means an overlap of $1/2$ , not 0. Orthogonal states are 180° apart: $(\pi - \theta, \varphi + \pi) = (\pi/3, 5\pi/4)$ .

So what does 90° on the sphere mean? $\ket0$ and $\ket{+}$ are 90° apart, and $\abs{\braket{0}{+}}^2 = 1/2$ : measured in the 0/1 basis, $\ket{+}$ gives 0 half the time. In general, two states $\Theta$ apart on the sphere have overlap $\abs{\braket{a}{b}}^2 = \cos^2(\Theta/2)$ : 1 at 0°, $1/2$ at 90°, 0 at 180°. The half-angle again. (The inner-product refresher covers $\braket{a}{b}$ .)

Check yourself

$\ket{+}$ and $\ket{+i}$ sit 90° apart on the equator, at $+x$ and $+y$ . You measure $\ket{+i}$ in the ± basis. What is $P(+)$ ?

$P(+) = \;?$

Hints

How far apart are the two states on the sphere?
Two states $\Theta$ apart have overlap $\cos^2(\Theta/2)$ .

Show the answer

Answer: $P(+) = \frac{1}{2}$

$P(+) = \cos^2(90°/2) = 1/2$ . You found the same number in Lesson 2 from $\abs{1 + i}^2/4$ .

Common slips

$0$ : That would make them orthogonal. Orthogonal states are 180° apart; at 90° the overlap is $\cos^2(45°) = 1/2$ .
$\frac{1}{\sqrt{2}}$ : That is $\cos 45° = \abs{\braket{+}{+i}}$ , the size of the inner product. The probability is its square: $1/2$ .

6 / 12 · Learn

Same odds, one circle

$P(0) = \cos^2(\theta/2)$ depends on $\theta$ alone. So every state on one circle of latitude (fixed $\theta$ ) has the same 0/1 statistics, and moving around that circle changes only $\varphi$ , the relative phase, which the 0/1 basis cannot see (Lesson 2). Fixing $P(0)$ fixes a latitude, not a point. In terms of the height ${z = \cos\theta}$ , ${P(0) = (1 + z)/2}$ : each circle of latitude is one height.

Check yourself

Select every state with $P(0) = 1/4$ in the 0/1 basis.

Select all that apply.

A $\tfrac{1}{\sqrt2}(\ket0 + e^{i\pi/4}\ket1)$
B $\tfrac12\ket0 + \tfrac{\sqrt3}{2}\ket1$
C $\tfrac12\ket0 + e^{i\pi/3}\tfrac{\sqrt3}{2}\ket1$
D $\tfrac{\sqrt3}{2}\ket0 + \tfrac12\ket1$
E $\tfrac12\ket0 - \tfrac{\sqrt3}{2}\ket1$

Show the answer

Answers: B. $\tfrac12\ket0 + \tfrac{\sqrt3}{2}\ket1$ ; C. $\tfrac12\ket0 + e^{i\pi/3}\tfrac{\sqrt3}{2}\ket1$ ; E. $\tfrac12\ket0 - \tfrac{\sqrt3}{2}\ket1$

The first three all have $\abs\alpha = 1/2$ , so $\theta = 2\pi/3$ and $P(0) = 1/4$ . They differ only in $\varphi$ (0, $\pi$ and $\pi/3$ ): three points on one circle of latitude.

Why not the others

A. $\tfrac{1}{\sqrt2}(\ket0 + e^{i\pi/4}\ket1)$ is on the equator, with $P(0) = 1/2$ .
D. $\tfrac{\sqrt3}{2}\ket0 + \tfrac12\ket1$ has $P(0) = 3/4$ : it lies on the mirror-image circle in the northern half, $\theta = \pi/3$ .

7 / 12 · Explore

Measuring along another axis

Predict, then see

The state $\tfrac{\sqrt3}{2}\ket0 + \tfrac12\ket1$ sits at $\theta = 60°$ , $\varphi = 0$ : tilted from the north pole toward $+x$ . The ± basis is the $x$ axis of the sphere, with $\ket{+}$ at $+x$ and $\ket{-}$ at $-x$ . You measure this state in the ± basis. About how often do you get $+$ ?

AAbout 50%, since the state is not on the $x$ axis
BAbout 75%, the same as $P(0)$
CAbout 87%
DAbout 93%

Show what happens

What happens: D. About 93%

Compare the readout with your prediction, and note how $P(+)$ changed as you moved the state. Then pick, below, the explanation that fits.

Why not the others

A. Off the axis does not mean 50/50. Only states 90° from $+x$ give 50/50 in the ± basis; this one is 30° from $+x$ .
B. 75% is $P(0)$ , which uses the angle to $+z$ . The state is 60° from $+z$ but only 30° from $+x$ , so it is closer to $\ket{+}$ than to $\ket0$ .
C. That is $\cos 30°$ , using the angle to $+x$ without halving and squaring. The probability is $\cos^2(30°/2) = \cos^2 15° \approx 0.933$ .

Check yourself

Which explanation fits what you saw: $P(+) \approx 93\%$ for the state at $\theta = 60°$ , $\varphi = 0$ ?

AWhat counts is the angle to $+x$ , and $P(+)$ is its cosine, $\cos 30° \approx 87\%$ .
BWhat counts is $\theta$ , the angle from the north pole, just as for $P(0)$ .
CWhat counts is $\varphi$ alone: $\varphi = 0$ points at $+x$ , so $+$ is nearly certain.
DWhat counts is the angle between the state and $+x$ , the $\ket{+}$ end of the axis: 30°, so $P(+) = \cos^2(30°/2)$ .

Show the answer

Answer: D. What counts is the angle between the state and $+x$ , the $\ket{+}$ end of the axis: 30°, so $P(+) = \cos^2(30°/2)$ .

The state is 30° from $+x$ , so $P(+) = \cos^2(30°/2) = \cos^2 15° = \tfrac{2+\sqrt3}{4} \approx 93\%$ , the number you computed in Lesson 2's worked example. On the sphere every basis is an axis, and the outcome odds depend only on the angle between the state and the ends of that axis. 75% is $P(0)$ , which uses the angle to $+z$ instead.

Why not the others

A. The angle to $+x$ is right, but the odds use the half-angle, squared: $\cos^2(30°/2) \approx 93\%$ . The plain cosine fails at 90°: $\ket0$ is 90° from $+x$ , $\cos 90° = 0$ , yet $\ket0$ gives $+$ half the time.
B. $\theta$ sets the 0/1 odds because $\ket0$ and $\ket1$ sit on the $z$ axis. The ± basis is the $x$ axis, so the angle that counts is the one to $+x$ : 30° here, not 60°. Using $\theta$ would give 75%, which the readout does not show.
C. $\varphi = 0$ puts the state on the $+x$ side, but how close it gets to $+x$ also depends on $\theta$ . At the north pole, $\theta = 0$ , the state is $\ket0$ , and it gives $+$ only half the time whatever $\varphi$ is.

A formula for any axis

If the state's point on the sphere is the unit vector $\vec r$ and the measurement axis is $\hat n$ , the outcome at $+\hat n$ has probability $(1 + \hat n\cdot\vec r)/2 = \cos^2(\Theta/2)$ , where $\Theta$ is the angle between them. For $\hat n = \hat z$ this is $P(0) = (1 + z)/2$ ; for the ± basis, $\hat n = \hat x$ . The readout shows $\hat n\cdot\vec r$ as you move the state.

Check yourself

A state sits 60° from the $+x$ axis of the sphere. You measure it in the ± basis. What is $P(+)$ ?

$P(+) = \;?$

Hints

Which angle matters for a ± measurement?
Two points $\Theta$ apart give $\cos^2(\Theta/2)$ .

Show the answer

Answer: $P(+) = \frac{3}{4}$

$P(+) = \cos^2(60°/2) = \cos^2 30° = 3/4$ , or equally $(1 + \cos 60°)/2 = 3/4$ .

Common slips

$\frac{1}{4}$ : That is $\cos^2 60°$ . Use the half-angle: $\cos^2(60°/2) = \cos^2 30° = 3/4$ .
$\frac{1}{2}$ : That is $\cos 60°$ . The probability is $(1 + \cos 60°)/2 = 3/4$ .

8 / 12 · Worked example

Removing a global phase

Worked example

When the $\ket0$ amplitude is not real

Find $\theta$ and $\varphi$ for $\ket\psi = \tfrac{i}{\sqrt2}\ket0 + \tfrac{1}{\sqrt2}\ket1$ . The Bloch form needs the $\ket0$ amplitude real and non-negative, so first remove a global phase.

Step 1
Factor out the phase of the $\ket0$ amplitude, $i = e^{i\pi/2}$ , using ${1/i = -i}$ :
$\ket\psi = i\left(\tfrac{1}{\sqrt2}\ket0 - \tfrac{i}{\sqrt2}\ket1\right)$
Step 2
Drop the global factor $i$ : it changes no prediction. Now ${\cos(\theta/2) = 1/\sqrt2}$ , so ${\theta/2 = \pi/4}$ and ${\theta = \pi/2}$ . The state is on the equator.
Your turn
The $\ket1$ amplitude is now $-\tfrac{i}{\sqrt2} = e^{i\varphi}\sin\tfrac{\pi}{4}$ . What is $\varphi$ ?
1. A $\varphi = \pi/2$
2. B $\varphi = 3\pi/2$ (the same direction as $-\pi/2$ )
3. C $\varphi = \pi$
4. D $\varphi = 0$
Show the answer
Answer: B. $\varphi = 3\pi/2$ (the same direction as $-\pi/2$ )
$-i = e^{3i\pi/2}$ , so $\varphi = 3\pi/2$ , which is the same direction as $-\pi/2$ .
Why not the others
- A. $e^{i\pi/2} = i$ , but the amplitude has the factor $-i$ . $-i = e^{-i\pi/2} = e^{3i\pi/2}$ .
- C. $e^{i\pi} = -1$ , not $-i$ . A factor of $-i$ is a quarter turn clockwise: $\varphi = 3\pi/2$ .
- D. That reads the original $\ket1$ amplitude, $1/\sqrt2$ , before the global phase was removed. Phase is relative: after factoring out $i$ , the $\ket1$ amplitude carries $-i$ .
Your turn
So where is $\ket\psi$ on the sphere?
1. AOn the equator at $+y$ , the point of $\ket{+i}$
2. BAt the north pole
3. COn the equator at $+x$ , the point of $\ket{+}$
4. DOn the equator at $-y$ , the point of $\ket{-i}$
Show the answer
Answer: D. On the equator at $-y$ , the point of $\ket{-i}$
$\theta = \pi/2$ and $\varphi = 3\pi/2$ : the $-y$ point, which is $\ket{-i}$ . Indeed $\ket\psi = i\ket{-i}$ .
Why not the others
- A. That is $\varphi = \pi/2$ . Here $\varphi = 3\pi/2$ , the opposite side of the equator.
- B. $\theta = \pi/2$ puts the state on the equator, not at a pole.
- C. Equal sizes do put it on the equator, but the $i$ on $\ket0$ is a relative phase: after removing the global phase the relative phase is $3\pi/2$ , not 0.

Here is one more worked example. This time the first amplitude is negative.

Another worked example

Second example

Take $\ket\psi = -\tfrac12\ket0 + \tfrac{\sqrt3}{2}\ket1$ . Factor out $-1 = e^{i\pi}$ : $\ket\psi = -\left(\tfrac12\ket0 - \tfrac{\sqrt3}{2}\ket1\right)$ .
Now ${\cos(\theta/2) = \tfrac12}$ , so ${\theta/2 = \pi/3}$ and ${\theta = 2\pi/3}$ .
The $\ket1$ amplitude is ${-\tfrac{\sqrt3}{2} = e^{i\pi}\sin\tfrac{\pi}{3}}$ , so $\varphi = \pi$ . The state sits 120° from the north pole, on the $-x$ side.

Check yourself

Now on your own. Find the Bloch angle $\varphi$ of $i\tfrac{\sqrt3}{2}\ket0 + i\tfrac12\ket1$ . Give a value from 0 up to $2\pi$ , such as pi/2.

$\varphi = \;?$

Hints

In the Bloch form, which amplitude must be real and non-negative?
Factor the $\ket0$ phase, $i = e^{i\pi/2}$ , out of the whole state; that global phase can be dropped.
Dividing by $e^{i\gamma}$ subtracts $\gamma$ from the phase of the $\ket1$ amplitude.

Show the answer

Answer: $\varphi = 0$

Factor the phase of the $\ket0$ amplitude, $\pi/2$ , out of the whole state and drop it. The $\ket1$ amplitude's phase becomes $\pi/2 - \pi/2$ , so $\varphi = 0$ (taken between 0 and $2\pi$ ). Also $\theta = \pi/3$ .

Common slips

$1.571$ : That reads $\varphi$ from the $\ket1$ factor alone. The $\ket0$ amplitude carries a phase too, and only the difference counts: factor the $\ket0$ phase out of the whole state first, and the $\ket1$ phase changes with it.

9 / 12 · Practice

Practice

Check yourself

A state has $P(0) = 0.8$ in the 0/1 basis. What is its Bloch angle $\theta$ , in degrees? The nearest whole degree is fine.

$\theta = \;?$

Hints

Which formula links $P(0)$ and $\theta$ ?
$P(0) = \cos^2(\theta/2)$ , so $\cos(\theta/2) = \sqrt{P(0)}$ .
Find $\theta/2 = \arccos\sqrt{0.8} \approx 26.6°$ , then double it.

Show the answer

Answer: $\theta = 53.13$

$\cos(\theta/2) = \sqrt{0.8} \approx 0.894$ , so $\theta/2 \approx 26.6°$ and $\theta \approx 53.1°$ . Check: $\cos\theta = 2(0.8) - 1 = 0.6$ .

Common slips

$26.565$ : That is $\theta/2$ , the angle in the amplitude. Double it: $\theta \approx 53.1°$ .
$36.87$ : That solves $\cos\theta = 0.8$ . The relation is $P(0) = \cos^2(\theta/2)$ , or equally $\cos\theta = 2P(0) - 1$ .

Check yourself

What are the Bloch angles of $\tfrac12\ket0 + e^{5i\pi/4}\tfrac{\sqrt3}{2}\ket1$ ?

A $\theta = \pi/2$ , $\varphi = 5\pi/4$
B $\theta = \pi/3$ , $\varphi = 5\pi/4$
C $\theta = 2\pi/3$ , $\varphi = 5\pi/4$
D $\theta = 2\pi/3$ , $\varphi = 3\pi/4$

Hints

Is the $\ket0$ amplitude already real and non-negative?
$\cos(\theta/2)$ is the $\ket0$ amplitude; the factor on $\ket1$ is $e^{i\varphi}$ .
Here $\theta/2 = \pi/3$ .

Show the answer

Answer: C. $\theta = 2\pi/3$ , $\varphi = 5\pi/4$

$\cos(\theta/2) = \tfrac12$ gives $\theta = 2\pi/3$ , and the factor $e^{5i\pi/4}$ gives $\varphi = 5\pi/4$ .

The $\ket0$ amplitude $\tfrac12$ is $\cos(\theta/2)$ , so $\theta/2 = \pi/3$ and $\theta = 2\pi/3$ . The factor on $\ket1$ is $e^{5i\pi/4}$ , so $\varphi = 5\pi/4$ .

Why not the others

A. $\theta = \pi/2$ is the equator, where both amplitudes have size $1/\sqrt2$ . Here the $\ket0$ amplitude is $\tfrac12$ , so $\theta = 2\pi/3$ .
B. $\pi/3$ is the angle in the amplitudes, $\theta/2$ . Double it: $\theta = 2\pi/3$ .
D. The factor on $\ket1$ is $e^{5i\pi/4}$ , so $\varphi = 5\pi/4$ . Read the angle of that factor directly.

Check yourself

A state sits at $\theta = 5\pi/6$ , $\varphi = 3\pi/2$ . Where is the state orthogonal to it?

A $\theta = 5\pi/6$ , $\varphi = \pi/2$
B $\theta = \pi/3$ , $\varphi = 3\pi/2$
C $\theta = \pi/6$ , $\varphi = \pi/2$
D $\theta = \pi/6$ , $\varphi = 3\pi/2$

Hints

How far apart are orthogonal states on the sphere?
Orthogonal states are antipodal. The antipode of $(\theta, \varphi)$ is $(\pi - \theta, \varphi + \pi)$ .
Here $\pi - \theta = \pi/6$ .

Show the answer

Answer: C. $\theta = \pi/6$ , $\varphi = \pi/2$

The antipode of $(5\pi/6, 3\pi/2)$ is $(\pi/6, \pi/2)$ .

Orthogonal states are antipodal: $\theta \to \pi - \theta = \pi/6$ and $\varphi \to \varphi + \pi = \pi/2$ (taken between 0 and $2\pi$ ).

Why not the others

A. Turning $\varphi$ by $\pi$ keeps the latitude, so $P(0)$ stays the same. An orthogonal state has $P(0)$ and $P(1)$ swapped.
B. That point is 90° away on the sphere, where the overlap is $1/2$ (here it is 50%). Orthogonal states are 180° apart.
D. Reflecting through the equator is not enough: the antipode is also on the other side in $\varphi$ .

10 / 12 · Explain

Perpendicular, yet opposite

Explain it

$\ket0$ and $\ket1$ are perpendicular vectors, yet on the Bloch sphere they sit at opposite poles, 180° apart. Why?

Show a model answer

The amplitudes in the Bloch form use the half-angle $\theta/2$ , but the sphere plots $\theta$ itself. A state $\cos t\ket0 + \sin t\ket1$ is at angle $t$ as a vector and at $2t$ on the sphere, so every angle doubles. $\ket0$ has $\theta/2 = 0$ and $\ket1$ has $\theta/2 = \pi/2$ , 90° apart like the vectors, so $\theta = 0$ and $\theta = \pi$ : opposite poles. The same holds for every orthogonal pair, which always sits at antipodes. By the same doubling, 90° on the sphere means an overlap of $1/2$ , as for $\ket0$ and $\ket{+}$ .

A good answer covers

The amplitudes use the half-angle $\theta/2$ , while the sphere plots $\theta$ .
So angles double: the 90° between orthogonal vectors becomes 180° on the sphere.
Gives an example, such as $\ket{+}$ and $\ket{-}$ at $\pm x$ , or $\ket0$ and $\ket{+}$ 90° apart with overlap $1/2$ .

11 / 12 · Mastery check

Mastery check

Check yourself

This was your opening guess. $\ket0$ and $\ket1$ are orthogonal: $\braket{0}{1} = 0$ . On the Bloch sphere, how far apart are the points of two orthogonal states?

AIt depends on the phases of the two states.
B45°
C90°: perpendicular, like the vectors.
D180°: they sit at opposite points.

Show the answer

Answer: D. 180°: they sit at opposite points.

Orthogonal states are antipodal, 180° apart.

The amplitudes use $\theta/2$ and the sphere plots $\theta$ , so the 90° between orthogonal vectors doubles to 180°: $\ket0$ and $\ket1$ are the poles, $\ket{+}$ and $\ket{-}$ are opposite points of the equator.

Why not the others

A. For every orthogonal pair the answer is the same: opposite points, 180° apart. The phases decide where the pair sits, not how far apart they are.
B. That halves the angle instead of doubling it. The sphere plots $\theta$ , twice the angle inside the amplitudes, so orthogonal states (90° apart as vectors) are 180° apart on the sphere.
C. A natural guess, since orthogonal vectors are perpendicular. But the sphere doubles angles: the amplitudes use $\theta/2$ while the sphere plots $\theta$ . Orthogonal states sit 180° apart, like $\ket0$ and $\ket1$ at the two poles. 90° on the sphere means an overlap of $1/2$ , as for $\ket0$ and $\ket{+}$ .

Check yourself

Where is $\cos\tfrac{\pi}{12}\ket0 + \sin\tfrac{\pi}{12}\ket1$ on the sphere? Give its angle $\theta$ from the north pole, in degrees.

$\theta = \;?$

Show the answer

Answer: $\theta = 30$

The amplitudes contain $\theta/2 = \pi/12 = 15°$ , so $\theta = 30°$ .

Common slips

$15$ : 15° is $\pi/12$ , the angle in the amplitudes: that is $\theta/2$ . The sphere angle is twice it, 30°.
$150$ : That measures from the south pole. $\theta$ starts at $\ket0$ , the north pole.

Check yourself

Two states sit at $(\theta, \varphi) = (3\pi/4, 0)$ and $(3\pi/4, \pi)$ . Which statement is true?

AThe same state: only $\theta$ affects any measurement.
BSame 0/1 statistics, but they are different states.
CDifferent 0/1 statistics, because they are different points.
DOrthogonal states, because their $\varphi$ values are 180° apart.

Show the answer

Answer: B. Same 0/1 statistics, but they are different states.

Same $\theta$ means the same $P(0)$ ; different $\varphi$ means a different relative phase, so a different state.

Both have $\theta = 3\pi/4$ , so $P(0) = \tfrac{2-\sqrt2}{4}$ for both. Their $\varphi$ values differ, so they differ by a relative phase. In the ± basis $P(+)$ is 85.4% for one and 14.6% for the other.

Why not the others

A. Only $\theta$ affects the 0/1 statistics. $\varphi$ is a relative phase. In the ± basis $P(+)$ is 85.4% for one and 14.6% for the other.
C. $P(0) = \cos^2(\theta/2)$ depends on $\theta$ only. Both are on the same circle of latitude, so both give $P(0) = \tfrac{2-\sqrt2}{4}$ .
D. Orthogonal states are antipodal: $\theta \to \pi - \theta$ as well as $\varphi \to \varphi + \pi$ . These two share a latitude; their overlap is 50%, not 0.

Check yourself

Which state is orthogonal to $\ket\psi = \tfrac{\sqrt3}{2}\ket0 + e^{3i\pi/4}\tfrac12\ket1$ ?

A $-\tfrac{\sqrt3}{2}\ket0 - e^{3i\pi/4}\tfrac12\ket1$
B $\tfrac{\sqrt3}{2}\ket0 - e^{3i\pi/4}\tfrac12\ket1$
C $\tfrac12\ket0 + e^{3i\pi/4}\tfrac{\sqrt3}{2}\ket1$
D $\tfrac12\ket0 - e^{3i\pi/4}\tfrac{\sqrt3}{2}\ket1$

Show the answer

Answer: D. $\tfrac12\ket0 - e^{3i\pi/4}\tfrac{\sqrt3}{2}\ket1$

The partner of $c\ket0 + e^{i\varphi}s\ket1$ is $s\ket0 - e^{i\varphi}c\ket1$ : $\theta \to \pi - \theta$ swaps the sizes, and $\varphi \to \varphi + \pi$ adds the minus sign.

Swap the sizes and add a minus sign to the $\ket1$ term: $\tfrac12\ket0 - e^{3i\pi/4}\tfrac{\sqrt3}{2}\ket1$ . Its inner product with $\ket\psi$ is $\tfrac{\sqrt3}{2}\cdot\tfrac12 - \tfrac12\cdot\tfrac{\sqrt3}{2} = 0$ : the bra conjugates the factor $e^{3i\pi/4}$ on the $\ket1$ term of $\ket\psi$ , and that cancels the same factor in the partner.

Why not the others

A. This is $-\ket\psi$ , the same state with a global phase. Its overlap with $\ket\psi$ is 1.
B. This turns $\varphi$ by $\pi$ but keeps $\theta$ , so it has the same $P(0)$ as $\ket\psi$ ; its overlap with $\ket\psi$ is 25%. An orthogonal state must have $P(0)$ and $P(1)$ swapped.
C. This reflects $\ket\psi$ through the equator but keeps $\varphi$ ; its overlap with $\ket\psi$ is 75%. The antipode also needs $\varphi + \pi$ : a minus sign on the $\ket1$ term.

Check yourself

A state has Bloch angles $\theta = \pi/2$ , $\varphi = \pi/2$ . You measure it in the 0/1 basis. What is $P(1)$ ?

$P(1) = \;?$

Show the answer

Answer: $P(1) = \frac{1}{2}$

$P(1) = \sin^2(\theta/2) = \sin^2(\pi/4) = \tfrac12$ ; $\varphi$ does not matter in the 0/1 basis.

Common slips

$1$ : That is $\sin^2\theta$ . The $\ket1$ amplitude is $e^{i\varphi}\sin(\theta/2)$ , so $P(1) = \sin^2(\theta/2) = \tfrac12$ .

12 / 12 · Recap

Recap

From memory

Before reading the summary, write down what you remember. Pulling it back from memory is what makes it last.

How do you find $\theta$ and $\varphi$ for a state $\alpha\ket0 + \beta\ket1$ ?
Show the key points
- Remove a global phase so the $\ket0$ amplitude is real and non-negative.
- $\cos(\theta/2) = \abs\alpha$ : the amplitudes contain the half-angle, so double it to get $\theta$ .
- $\varphi$ is the phase of the $\ket1$ amplitude after that step, taken between 0 and $2\pi$ .
Why are orthogonal states at opposite points of the sphere?
Show the key points
- The amplitudes use $\theta/2$ but the sphere plots $\theta$ , so angles double.
- The 90° between orthogonal vectors becomes 180° on the sphere.
- The partner of $(\theta, \varphi)$ is $(\pi - \theta, \varphi + \pi)$ ; 90° on the sphere means an overlap of $1/2$ .
What do all the states on one circle of latitude have in common, and how do they differ?
Show the key points
- They share $\theta$ , so they share $P(0) = \cos^2(\theta/2)$ .
- They differ only in the relative phase $\varphi$ , which the 0/1 basis cannot see.
- They are still different states: a measurement along another axis, such as the ± basis, tells them apart.

Key ideas

Up to a global phase, every qubit state can be put in the Bloch form $\cos\tfrac{\theta}{2}\ket0 + e^{i\varphi}\sin\tfrac{\theta}{2}\ket1$ : a point on a sphere of radius 1.
$\ket0$ and $\ket1$ are the poles; $\ket{\pm}$ sit on the equator at $\pm x$ and $\ket{\pm i}$ at $\pm y$ .
The amplitudes use the half-angle $\theta/2$ , so angles double on the sphere: orthogonal states are antipodal, and 90° apart means an overlap of $1/2$ .
A circle of latitude is a fixed 0/1-basis probability $P(0) = \cos^2(\theta/2)$ ; moving around it changes only the relative phase $\varphi$ .
Every basis is an axis of the sphere. The odds of each outcome depend only on the angle between the state and that end of the axis: $\cos^2(\Theta/2)$ .

Go further

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

Bloch sphereWikipedia · Article
Writes the sphere with the Pauli matrices and shows that mixed, partly random states fill the inside of the ball.
Time development of two-level quantum states in the Bloch sphereQuVis, University of St Andrews · Interactive
Switch on a magnetic field and watch a spin state turn around the sphere over time, a motion this course doesn’t cover.

Math refreshers

The toolkit topics this lesson leans on.

Angles on a sphereSpherical coordinates, and why the Bloch sphere uses θ/2
Inner products and bases⟨φ|ψ⟩, orthogonality, and |⟨φ|ψ⟩|² as a probability

Before you start

Two angles are enough

Landmarks on the sphere

The half-angle trap

Orthogonal means opposite

Same odds, one circle

Measuring along another axis

Removing a global phase

When the ∣0⟩\ket0∣0⟩ amplitude is not real

Second example

Practice

Perpendicular, yet opposite

Mastery check

Recap

Key ideas

Go further

Math refreshers

When the $\ket0$ amplitude is not real