QUANTUM 03
Picturing any single-qubit state as a point on a sphere
Sinclair College - Dr. Mike Libassi
TODAY
01 · TWO ANGLES
Normalization and global phase remove two of the four numbers in a pair of complex amplitudes, leaving exactly two angles.
02 · READING THE SPHERE
|0⟩. Measures 0 with certainty.
|1⟩. Measures 1 with certainty.
Every 50/50 superposition lives here.
Relative phase. Invisible to a 0-or-1 measurement.
03 · LANDMARKS
| State | Position | Vector | Made by |
|---|---|---|---|
| |0⟩ | north pole · +z | [1, 0] | start state |
| |1⟩ | south pole · −z | [0, 1] | X on |0⟩ |
| |+⟩ | equator · +x | 1/√2 [1, 1] | H on |0⟩ |
| |−⟩ | equator · −x | 1/√2 [1, −1] | H on |1⟩ |
| |+i⟩ | equator · +y | 1/√2 [1, i] | S on |+⟩ |
| |−i⟩ | equator · −y | 1/√2 [1, −i] | S† on |+⟩ |
03 · LANDMARKS
Perfectly distinguishable states, like |0⟩ and |1⟩ or |+⟩ and |−⟩.
Measuring one state in the other's basis is a pure 50/50 coin flip.
Noisy, partly random states drift inward. The center is completely random.
04 · LIMITS AND SECURITY
Entangled qubits have no individual arrows, so the sphere can't draw them.
A phase on the whole state has no physical effect and isn't shown.
Z axis = rectilinear basis, X axis = diagonal basis, 90° apart.
An eavesdropper who measures along the wrong axis gets a random answer and knocks the qubit onto the other axis.
WRAP-UP
Any single qubit is a point on the sphere, set by θ and φ
Latitude sets the odds of 0 vs. 1; longitude is the hidden phase
Opposite points are distinguishable; perpendicular axes are coin flips
The picture works for one qubit only, not entangled systems
DISCUSSION
If φ can't be seen by measuring 0 or 1, how could you ever detect it?
Hint: what does a Hadamard gate do to |+⟩ compared with |−⟩?