Bloch Sphere Visualizer
Bloch Sphere Visualizer plots a qubit state on a 3D sphere using θ and φ angles, computing Bloch coordinates and state components for quantum computing studies.
Formulas Used in Bloch Sphere Visualizer
The visualizer uses the following formulas for a single qubit state:
Qubit State:
\\[ |\psi\rangle = \cos\left(\frac{\theta}{2}\right) |0\rangle + e^{i\phi} \sin\left(\frac{\theta}{2}\right) |1\rangle \\]Bloch Vector Coordinates:
\\[ x = \sin\theta \cos\phi, \quad y = \sin\theta \sin\phi, \quad z = \cos\theta \\]State Components:
\\[ \alpha = \cos\left(\frac{\theta}{2}\right), \quad \beta = e^{i\phi} \sin\left(\frac{\theta}{2}\right) \\] \\[ \beta_{\text{re}} = \sin\left(\frac{\theta}{2}\right) \cos\phi, \quad \beta_{\text{im}} = \sin\left(\frac{\theta}{2}\right) \sin\phi \\]Where:
- \\( \theta \\): Polar angle (0 to \\( \pi \\))
- \\( \phi \\): Azimuthal angle (0 to \\( 2\pi \\))
- \\( \alpha \\): Coefficient of \\( |0\rangle \\)
- \\( \beta \\): Coefficient of \\( |1\rangle \\)
- \\( x, y, z \\): Bloch sphere coordinates
Example Calculations
Example 1: Superposition State (θ = 90°, φ = 0°)
Input: θ = 90°, φ = 0°
\\[
|\psi\rangle = \cos\left(\frac{90^\circ}{2}\right) |0\rangle + e^{i \cdot 0^\circ} \sin\left(\frac{90^\circ}{2}\right) |1\rangle = \frac{1}{\sqrt{2}} |0\rangle + \frac{1}{\sqrt{2}} |1\rangle
\\]
\\[
x = \sin 90^\circ \cos 0^\circ = 1, \quad y = \sin 90^\circ \sin 0^\circ = 0, \quad z = \cos 90^\circ = 0
\\]
\\[
\alpha = \cos\left(\frac{90^\circ}{2}\right) \approx 0.707, \quad \beta_{\text{re}} = \sin\left(\frac{90^\circ}{2}\right) \cos 0^\circ \approx 0.707, \quad \beta_{\text{im}} = \sin\left(\frac{90^\circ}{2}\right) \sin 0^\circ = 0
\\]
Result: Bloch Coordinates: (1, 0, 0), State: [0.707, 0.707 + 0i]
Example 2: Spin-Up State (θ = 0°, φ = 0°)
Input: θ = 0°, φ = 0°
\\[
|\psi\rangle = \cos\left(\frac{0^\circ}{2}\right) |0\rangle + e^{i \cdot 0^\circ} \sin\left(\frac{0^\circ}{2}\right) |1\rangle = 1 |0\rangle + 0 |1\rangle
\\]
\\[
x = \sin 0^\circ \cos 0^\circ = 0, \quad y = \sin 0^\circ \sin 0^\circ = 0, \quad z = \cos 0^\circ = 1
\\]
\\[
\alpha = \cos\left(\frac{0^\circ}{2}\right) = 1, \quad \beta_{\text{re}} = \sin\left(\frac{0^\circ}{2}\right) \cos 0^\circ = 0, \quad \beta_{\text{im}} = \sin\left(\frac{0^\circ}{2}\right) \sin 0^\circ = 0
\\]
Result: Bloch Coordinates: (0, 0, 1), State: [1, 0 + 0i]
Example 3: General State (θ = 120°, φ = 45°)
Input: θ = 120°, φ = 45°
\\[
|\psi\rangle = \cos\left(\frac{120^\circ}{2}\right) |0\rangle + e^{i \cdot 45^\circ} \sin\left(\frac{120^\circ}{2}\right) |1\rangle \approx 0.5 |0\rangle + (0.707 e^{i \cdot 45^\circ}) |1\rangle
\\]
\\[
x = \sin 120^\circ \cos 45^\circ \approx 0.612, \quad y = \sin 120^\circ \sin 45^\circ \approx 0.612, \quad z = \cos 120^\circ = -0.5
\\]
\\[
\alpha = \cos\left(\frac{120^\circ}{2}\right) \approx 0.5, \quad \beta_{\text{re}} = \sin\left(\frac{120^\circ}{2}\right) \cos 45^\circ \approx 0.612, \quad \beta_{\text{im}} = \sin\left(\frac{120^\circ}{2}\right) \sin 45^\circ \approx 0.612
\\]
Result: Bloch Coordinates: (0.612, 0.612, -0.5), State: [0.5, 0.612 + 0.612i]