# Bloch Sphere

> The Bloch sphere is a round picture of a qubit: every possible state is one point on its surface, and the top and bottom points are 0 and 1.

Page: https://formulabro.yunusemrevurgun.com/quantum-mechanics/bloch-sphere/  
Category: Quantum Mechanics  
Published: 2026-10-22  
Updated: 2026-10-22

## Formula

$$
|\psi\rangle = \cos\frac{\theta}{2}|0\rangle + e^{i\varphi}\sin\frac{\theta}{2}|1\rangle, \qquad P(0) = \cos^2\frac{\theta}{2}
$$

Read it out loud: "Ket psi equals cosine of theta over two, times ket zero, plus e to the power i phi, times sine of theta over two, times ket one. The chance of measuring zero is cosine squared of theta over two. Theta is the angle from the top point of the sphere, and phi is the angle that turns around the vertical axis. Theta is said THAY-tuh, and phi is said FYE. The fraction theta over two means half of theta."

## Before you start

- **Angles: degrees, radians and $\pi$**: An angle can be measured in degrees, where a full turn is 360 degrees, or in radians, where a full turn is $2\pi$ radians. The Greek letter $\pi$, said pie, is about 3.14159. So $\pi$ radians is 180 degrees, and 1 radian is about 57.3 degrees. Many calculators and programming functions expect radians unless told otherwise.
- **Cosine, sine and their squares**: Cosine and sine are ratios that come from a right triangle. For a point on a circle of radius 1, cosine gives the horizontal position and sine gives the vertical position. At 0 degrees the cosine is 1 and the sine is 0. At 90 degrees the cosine is 0 and the sine is 1. The small raised 2 in $\cos^2 x$ means the cosine multiplied by itself, so $\cos^2 30^\circ = 0.866^2 = 0.75$.
- **Half an angle, $\frac{\theta}{2}$**: The fraction $\frac{\theta}{2}$ means half of the angle $\theta$. If $\theta = 60^\circ$, then $\frac{\theta}{2} = 30^\circ$. The formula uses half angles so that the trip from the top pole to the bottom pole is an angle that runs from 0 to $\pi$ radians.
- **Greek letters $\theta$, $\varphi$ and $\psi$**: Greek letters are extra names, because the Latin alphabet runs out. $\theta$ is theta, said THAY-tuh, and is used for an angle. $\varphi$ is phi, said FYE, and is a second angle. $\psi$ is psi, said SIGH, and is the usual name for a wave function or a qubit state.
- **Complex numbers and $e^{i\varphi}$**: The symbol $i$ is the imaginary unit, with $i^2 = -1$. The number $e$, about 2.718, is a fixed number that appears in growth and turning. Together, $e^{i\varphi} = \cos\varphi + i\sin\varphi$. This is a complex number with size 1 that sits on a circle, and the angle $\varphi$ tells you how far it has turned. Multiplying by it turns a number without changing its size.
- **Probability, $P(0)$**: The probability of an outcome is the chance that it happens, written as a number from 0 to 1. $P(0)$ means the probability of measuring 0. The probabilities of all outcomes add up to 1. Here the two outcomes are 0 and 1, so $P(0) + P(1) = 1$. A probability of 0.75 means 75 percent, which is three times in four.
- **Poles, the equator and the sphere**: A pole is the top or bottom point of a sphere, the two points furthest apart along a vertical line. The equator is the ring halfway between them. A sphere of radius 1 is the set of all points exactly 1 unit away from its center. The Bloch sphere is such a sphere, with its poles named $|0\rangle$ and $|1\rangle$.

More notation help: https://formulabro.yunusemrevurgun.com/math-basics/

## What each symbol means

- $|\psi\rangle$: The state of one qubit, written as a ket. On the Bloch sphere each state is one point. A phase that multiplies both parts of the state by the same turn is left out, because it has no effect on any measurement. The state has no unit, because it is a recipe made of amplitudes.
- $\theta$: The polar angle, measured from the top pole $|0\rangle$ down to the bottom pole $|1\rangle$. It is an angle in radians, or in degrees in the example. It runs from 0 to $\pi$ radians, which is 0 to 180 degrees. A value of 60 degrees is one third of the way from the top pole to the bottom pole, measured as an angle.
- $\varphi$: The azimuthal angle, the turn around the vertical axis, like longitude on a globe. It is an angle in radians or degrees. It runs from 0 to $2\pi$ radians, which is 0 to 360 degrees. In the example it is 90 degrees, a quarter turn around the axis.
- $|0\rangle$: The basis state at the top pole, where $\theta = 0$. It is one of the two standard results of a measurement, and it has no unit. On the sphere it is the top point, where the state gives 0 with certainty.
- $|1\rangle$: The basis state at the bottom pole, where $\theta = \pi$. It is the other standard result of a measurement, and it has no unit. On the sphere it is the bottom point, where the state gives 1 with certainty.
- $\cos\frac{\theta}{2}$: The amplitude of $|0\rangle$. It is a pure number with no unit, and for the angles on this page it runs from 1 down to 0. Its squared size is the chance of measuring 0. At $\theta = 60^\circ$ it equals $\cos 30^\circ \approx 0.866$.
- $e^{i\varphi}$: A phase factor, a complex number with size exactly 1. It has no unit. It turns the amplitude of $|1\rangle$ by the angle $\varphi$ without stretching it, so it leaves the chance of each measurement along the vertical axis unchanged.
- $\sin\frac{\theta}{2}$: The size of the amplitude of $|1\rangle$. It is a pure number with no unit. Its squared size is the chance of measuring 1. At $\theta = 60^\circ$ it equals $\sin 30^\circ = 0.5$.
- $P(0)$: The probability of measuring 0, a number from 0 to 1 with no unit. It equals the squared cosine of half the angle. At $\theta = 60^\circ$ it is 0.75, or 75 percent.

## The formula piece by piece

- $\theta$: Start with the angle $\theta$, which says how far the point has moved from the top pole toward the bottom pole. Zero means the top pole, which is the state $|0\rangle$, and $\pi$ radians means the bottom pole, which is the state $|1\rangle$.
- $\frac{\theta}{2}$: Half of that angle. Halving means that the whole trip from pole to pole is covered by angles from 0 to $\frac{\pi}{2}$, which is a quarter of a circle, so the cosine and sine below change smoothly along that quarter circle.
- $\cos\frac{\theta}{2}$: The amplitude of the top basis state. It starts at 1 when the point is at the top pole and falls to 0 at the bottom pole. As the point moves down the sphere, the amplitude of $|0\rangle$ shrinks.
- $e^{i\varphi}$: The phase factor, a number with size 1 that turns the other amplitude by the angle $\varphi$. It moves the point around the sphere, but it does not change the chances of the two basic results.
- $P(0) = \cos^2\frac{\theta}{2}$: The chance of measuring 0. Squaring the cosine of the half angle turns the amplitude into a probability, from 1 at the top pole down to 0 at the bottom pole.
- $|\psi\rangle = \cos\frac{\theta}{2}|0\rangle + e^{i\varphi}\sin\frac{\theta}{2}|1\rangle, \qquad P(0) = \cos^2\frac{\theta}{2}$: Every single-qubit state is one point on the sphere, located by its two angles, and the chance of measuring 0 is the squared cosine of half of the first angle.

## What the formula is doing, and why

**The picture.** Every single-qubit state can be drawn as a point on a sphere of radius 1, called the Bloch sphere. It is named after the physicist Felix Bloch. Two angles locate the point. The angle $\theta$ measures how far the point is from the top pole, and $\varphi$ measures how far it has turned around the vertical axis. A phase that multiplies the whole state by the same turn has no physical effect, so the picture leaves it out.

**The poles and the equator.** At $\theta = 0$ the formula gives $|0\rangle$ at the top pole, and at $\theta = \pi$ it gives $|1\rangle$ at the bottom pole. Points on the equator, where $\theta = \frac{\pi}{2}$, are even mixes. Each one gives 0 with probability $\cos^2\frac{\pi}{4} = 0.5$. Their angle $\varphi$ tells you where around the equator they sit.

**Why half angles?** The chance of 0 is $\cos^2\frac{\theta}{2}$, and the chance of 1 is $\sin^2\frac{\theta}{2}$. The two add up to 1. The height of the point above the equator is $z = \cos\theta$, and this equals $P(0) - P(1)$. Using the half angle in the amplitudes lets the angle $\theta$ run once from the top pole to the bottom pole, from 0 to $\pi$, so each state on the sphere is counted once.

**What the second angle does.** The angle $\varphi$ does not change $P(0)$ or $P(1)$, because $|e^{i\varphi}|^2 = 1$. It does change the state. It shows up when you measure along a different axis, such as the horizontal axis of the sphere.

**Gates are turns.** A quantum gate, which is an operation on one qubit, acts as a rotation of this sphere. A gate that swaps 0 and 1, often called the X gate, is a half turn that moves the top pole to the bottom. That is why the sphere is such a useful picture for single-qubit operations.

## An everyday picture

Think of a globe. A point on the globe is located by its angle from the North Pole and by its longitude, the angle around the axis. The North Pole is $|0\rangle$ and the South Pole is $|1\rangle$. Points near the top are mostly 0, points near the bottom are mostly 1, and points on the equator are an even mix. The angle from the pole plays the role of $\theta$, and longitude plays the role of $\varphi$.

## Worked example: A qubit at 60 degrees

A qubit has $\theta = 60^\circ$ and $\varphi = 90^\circ$. We want the chance of measuring 0 and the chance of measuring 1. The full trip from the top pole to the bottom pole is 180 degrees, so $\theta = 60^\circ$ is one third of the way down. The numbers are chosen for this example.

1. **Find the half angle.** The formulas use $\frac{\theta}{2}$, so we halve the angle: $\frac{60^\circ}{2} = 30^\circ$. Halving first is what the formulas ask for, because the chances are built from half of the angle from pole to pole.
2. **Find the amplitude of 0.** The amplitude of $|0\rangle$ is $\cos 30^\circ = \frac{\sqrt{3}}{2} \approx 0.866$. The value $\frac{\sqrt{3}}{2}$ is a standard result: $\sqrt{3}$ is about 1.732, and half of that is 0.866. A calculator gives the same number.
3. **Square it to get the chance of 0.** Squaring turns the amplitude into a probability: $P(0) = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4} = 0.75$. Squaring the top and the bottom gives $\frac{3}{4}$, which is 75 percent.
4. **Find the chance of 1.** The chance of 1 is $\sin^2 30^\circ = \left(\frac{1}{2}\right)^2 = \frac{1}{4} = 0.25$. It also equals $1 - 0.75$, because the two chances must add up to 1. That is a useful check on the arithmetic.
5. **Check the phase factor.** The angle $\varphi = 90^\circ$ gives $e^{i\varphi} = \cos 90^\circ + i\sin 90^\circ = 0 + i = i$. The factor $i$ has size 1, so it turns the amplitude of 1 without stretching it, and it does not change either chance.
6. **Check the height.** The height of the point above the equator is $z = \cos 60^\circ = 0.5$. This equals $0.75 - 0.25 = 0.5$, which matches the rule that the height is the difference between the two chances.

**Answer:** The chance of measuring 0 is **75 percent**, and the chance of measuring 1 is **25 percent**. The point sits closer to the top pole, and its height is $z = 0.5$, which equals $0.75 - 0.25$. In 100 runs we would expect about 75 results of 0.

## Common mistakes

- Using $\cos^2\theta$ instead of $\cos^2\frac{\theta}{2}$. The probability uses the half angle. With $\theta = 60^\circ$ the wrong formula gives 0.25 instead of 0.75.
- Feeding degrees into a function that expects radians. Python's math.cos expects radians, so convert with math.radians first. Passing 30 directly gives the cosine of 30 radians, about 0.154, not 0.866.
- Thinking $\varphi$ changes the measured odds of 0 and 1. For measurements along the vertical axis it does not, because $|e^{i\varphi}| = 1$. It matters for measurements along other axes.
- Treating the sphere as ordinary three-dimensional space. Its points are quantum states, not locations in a room, even though the picture is drawn in three dimensions.

## Where you will meet it

- Quantum computing, where each single-qubit gate is a rotation of the sphere, which makes the effect of the gate easy to picture.
- Describing noise in a qubit: noise pulls the point toward the inside of the sphere, which records a loss of quantum coherence, the steady phase relation between 0 and 1.
- Nuclear magnetic resonance, where the same picture describes how the spins of atomic nuclei turn under radio-frequency pulses.
- Designing control pulses that rotate a qubit from $|0\rangle$ to $|1\rangle$ in superconducting circuits and trapped-ion hardware.

## Questions people ask

### Why is the sphere's radius 1?

Because valid amplitudes satisfy $|\alpha|^2 + |\beta|^2 = 1$. Every valid qubit state fits on the sphere of radius 1, and every point on that sphere is a valid state.

### Is the Bloch sphere a real sphere in space?

No. It is an abstract picture of the state space of one qubit. A point on it is a quantum state, not a place in the room, although the picture is drawn in three dimensions.

### Why does the overall phase not matter?

An overall phase multiplies both amplitudes by the same turn of size 1. The probabilities depend only on squared sizes, so every measurement result stays the same. Only the relative phase between $|0\rangle$ and $|1\rangle$ is physical.

## Related formulas

- [Qubit State](https://formulabro.yunusemrevurgun.com/quantum-mechanics/qubit-state/)
- [Born Rule](https://formulabro.yunusemrevurgun.com/quantum-mechanics/born-rule/)
- [Wave Function Normalization](https://formulabro.yunusemrevurgun.com/quantum-mechanics/wave-function-normalization/)

## Sources

- [Bloch sphere, Wikipedia](https://en.wikipedia.org/wiki/Bloch_sphere)
- [Qubit, Wikipedia](https://en.wikipedia.org/wiki/Qubit)

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