Bloch Sphere¶
A state can be pictured as a point on a sphere.
From a List of Numbers to a Picture¶
On the previous page, we wrote a state as two amplitudes. We also found that a sign is a real difference in the state, yet it is easy to miss in a list of numbers.
How can we picture a state, including its sign, instead of reading a list of numbers?
We have already met a picture like this. The qubit section of What Is Quantum Computing? showed a sphere with two sliders. Here we use the sphere as a tool for reading a state: where a point sits tells us the probabilities and the phase. It is a picture of the numbers you already know, not a new object.
A Map of States¶
The Bloch sphere is a globe-like picture of the states of one qubit. Every state is one point on it.
- Poles — |0⟩ at the north pole (certain 0) and |1⟩ at the south pole (certain 1).
- Equator — |+⟩, |−⟩, |+i⟩, and |−i⟩, each equally likely to give 0 or 1.
- Point — one state. A point is not an outcome.
Start with |0⟩ at the top and |+⟩ on the equator. Here are all six named points, with the two angles that place them. The polar angle is measured down from the north pole, and the phase angle is measured around the equator.
Is the qubit pointing up?
No. The sphere is a picture of the state. North is not a direction in the room; it is where the state |0⟩ is drawn.
Sphere Explorer¶
Move the point and watch what happens. Start from a preset, or set the two angles yourself. The sphere, the description of the state, and the probability bars each sit in their own place.
1. Choose a preset, or move the point yourself
Top: certain 0. Equator: 50/50. Bottom: certain 1. The dashed line from the point to the vertical axis shows its height.
Polar angle (0° to 180°, down from the north pole)
Phase angle (0° to 360°, around the equator)
A phase of 0° means the two amplitudes have the same sign, like |+⟩. A phase of 180° means opposite signs, like |−⟩.
2. The state at this point
Intermediate readouts
3. The probabilities
The numbers you see are the size of each amplitude and the phase angle. The sphere shows the same state as a point, and the bars show what that state will do when it is measured in the Z basis.
Height Decides the Probabilities¶
On the previous page, the probabilities came from squaring the amplitudes. On the sphere, the same information is the height of the point.
Between the poles the probabilities slide smoothly. The dashed line in the explorer runs from the point straight across to the vertical axis. Where it lands is the height, and the height sets the bars.
Check your understanding. First, predict before the bars appear:
Then decide which pairs of points give the same Z-basis probabilities:
Now try one yourself, using the explorer above:
Make outcome 1 happen 25% of the time.
Around the Equator: Phase as an Angle¶
Slide the phase angle in the explorer while the polar angle stays at 90°. The point travels around the equator. Its height does not change, so the bars do not change. The state does.
Phase is the angle around the equator. Compare |+⟩ and |−⟩. The faded diamond is |+⟩ and the solid point is |−⟩:
They sit at the same height, on opposite sides of the sphere. This is the same-counts, different-states idea from the previous pages, now as a picture. The real-amplitude states you built on the previous page all sit on one circle of the sphere, where the phase is 0° or 180°.
If the height is the same, in what way are |+⟩ and |−⟩ different?
They sit at different places around the equator. The Z-basis counts cannot tell them apart. Later pages show how to tell them apart. Nothing on this page does.
Two things to keep in mind. The sphere pictures a pure state of one qubit, up to a shared overall phase. And a later page shows an operation that turns a phase difference into a difference you can see.
Beyond the Core¶
The formal picture¶
The formula. The two angles give the state:
So the amplitude of \(|0\rangle\) is \(\cos(\theta/2)\), and the amplitude of \(|1\rangle\) is \(e^{i\varphi}\sin(\theta/2)\). The sizes are the two numbers the main view showed. The angle \(\varphi\) is the relative phase from the previous page. The probabilities are \(\cos^2(\theta/2)\) and \(\sin^2(\theta/2)\), and \(\varphi\) does not appear in them. That is why moving around the equator changes no bar.
The Bloch vector. The point has three coordinates:
The height is \(z\), and \(z = |\alpha|^2 - |\beta|^2 = P(0) - P(1)\). Since \(P(0) + P(1) = 1\), this gives
The explorer now shows the written form, \((x, y, z)\), and \(P(0)\) beside the bars. Check that \(P(0)\) from the formula matches the bar.
Global phase. Multiply both amplitudes by the same phase factor and the point does not move. The global phase shift in the explorer changes the written amplitudes, while the point and the bars stay put. This is why the sphere shows a state only up to a shared overall phase: the picture fixes the |0⟩ amplitude to be real and not negative.
Opposite points are orthogonal. Two points on opposite sides of the sphere are orthogonal states: their inner product is zero. Your current point and its opposite:
The limits of the picture. The sphere is one qubit in a pure state. Two qubits do not fit on one sphere, and a mixed state does not sit on the sphere's surface. We name these limits here and treat them later.
Naming the formalism¶
A mixed state is described by a density matrix. On the Bloch picture it sits inside the ball rather than on its surface, and a point at the centre is as uncertain as a state can be. A later treatment makes these ideas precise.
From Picture to Motion¶
Compare what you can do now with what is still unknown.
A state can start at one point and be needed at another:
A point can sit anywhere on the sphere, but we do not yet know how to get it from one place to another.
Takeaway¶
A qubit's state can be pictured as a point on a sphere. Height sets the probabilities and the angle around the equator carries the phase, and the picture shows one qubit in a pure state only.
How can we move a state from one point to another?
Next: Quantum Gates →