Part 1: Measurement Probabilities
(a) Measurement in the computational basis
For ∣ ψ ⟩ = 1 2 ∣ 0 ⟩ + 1 2 ∣ 1 ⟩ |\psi\rangle = \frac{1}{\sqrt{2}}|0\rangle + \frac{1}{\sqrt{2}}|1\rangle ∣ ψ ⟩ = 2 1 ∣0 ⟩ + 2 1 ∣1 ⟩
P ψ ( 0 ) = ∣ ⟨ 0 ∣ ψ ⟩ ∣ 2 = ∣ 1 2 ∣ 2 = 1 2 P_\psi(0) = \left|\langle 0|\psi\rangle\right|^2 = \left|\frac{1}{\sqrt{2}}\right|^2 = \frac{1}{2} P ψ ( 0 ) = ∣ ⟨ 0∣ ψ ⟩ ∣ 2 = ∣ ∣ 2 1 ∣ ∣ 2 = 2 1
P ψ ( 1 ) = ∣ ⟨ 1 ∣ ψ ⟩ ∣ 2 = ∣ 1 2 ∣ 2 = 1 2 P_\psi(1) = \left|\langle 1|\psi\rangle\right|^2 = \left|\frac{1}{\sqrt{2}}\right|^2 = \frac{1}{2} P ψ ( 1 ) = ∣ ⟨ 1∣ ψ ⟩ ∣ 2 = ∣ ∣ 2 1 ∣ ∣ 2 = 2 1 For ∣ ψ ′ ⟩ = e i π / 3 2 ∣ 0 ⟩ + e i π / 3 2 ∣ 1 ⟩ |\psi'\rangle = \frac{e^{i\pi/3}}{\sqrt{2}}|0\rangle + \frac{e^{i\pi/3}}{\sqrt{2}}|1\rangle ∣ ψ ′ ⟩ = 2 e iπ /3 ∣0 ⟩ + 2 e iπ /3 ∣1 ⟩
P ψ ′ ( 0 ) = ∣ ⟨ 0 ∣ ψ ′ ⟩ ∣ 2 = ∣ e i π / 3 2 ∣ 2 = ∣ e i π / 3 ∣ 2 ⋅ 1 2 = 1 ⋅ 1 2 = 1 2 P_{\psi'}(0) = \left|\langle 0|\psi'\rangle\right|^2 = \left|\frac{e^{i\pi/3}}{\sqrt{2}}\right|^2 = |e^{i\pi/3}|^2 \cdot \frac{1}{2} = 1 \cdot \frac{1}{2} = \frac{1}{2} P ψ ′ ( 0 ) = ∣ ⟨ 0∣ ψ ′ ⟩ ∣ 2 = ∣ ∣ 2 e iπ /3 ∣ ∣ 2 = ∣ e iπ /3 ∣ 2 ⋅ 2 1 = 1 ⋅ 2 1 = 2 1
P ψ ′ ( 1 ) = ∣ ⟨ 1 ∣ ψ ′ ⟩ ∣ 2 = ∣ e i π / 3 2 ∣ 2 = 1 2 P_{\psi'}(1) = \left|\langle 1|\psi'\rangle\right|^2 = \left|\frac{e^{i\pi/3}}{\sqrt{2}}\right|^2 = \frac{1}{2} P ψ ′ ( 1 ) = ∣ ⟨ 1∣ ψ ′ ⟩ ∣ 2 = ∣ ∣ 2 e iπ /3 ∣ ∣ 2 = 2 1 Conclusion : the measurement probabilities are exactly the same! The global phase factor e i π / 3 e^{i\pi/3} e iπ /3
(b) Measurement in the x x x
First expand ∣ ψ ⟩ |\psi\rangle ∣ ψ ⟩ x x x
∣ ψ ⟩ = 1 2 ∣ 0 ⟩ + 1 2 ∣ 1 ⟩ = ∣ + ⟩ |\psi\rangle = \frac{1}{\sqrt{2}}|0\rangle + \frac{1}{\sqrt{2}}|1\rangle = |+\rangle ∣ ψ ⟩ = 2 1 ∣0 ⟩ + 2 1 ∣1 ⟩ = ∣ + ⟩ Therefore:
P ψ ( + ) = ∣ ⟨ + ∣ ψ ⟩ ∣ 2 = ∣ 1 ∣ 2 = 1 P_\psi(+) = |\langle +|\psi\rangle|^2 = |1|^2 = 1 P ψ ( + ) = ∣ ⟨ + ∣ ψ ⟩ ∣ 2 = ∣1 ∣ 2 = 1
P ψ ( − ) = ∣ ⟨ − ∣ ψ ⟩ ∣ 2 = ∣ 0 ∣ 2 = 0 P_\psi(-) = |\langle -|\psi\rangle|^2 = |0|^2 = 0 P ψ ( − ) = ∣ ⟨ − ∣ ψ ⟩ ∣ 2 = ∣0 ∣ 2 = 0 For ∣ ψ ′ ⟩ |\psi'\rangle ∣ ψ ′ ⟩
∣ ψ ′ ⟩ = e i π / 3 ( 1 2 ∣ 0 ⟩ + 1 2 ∣ 1 ⟩ ) = e i π / 3 ∣ + ⟩ |\psi'\rangle = e^{i\pi/3}\left(\frac{1}{\sqrt{2}}|0\rangle + \frac{1}{\sqrt{2}}|1\rangle\right) = e^{i\pi/3}|+\rangle ∣ ψ ′ ⟩ = e iπ /3 ( 2 1 ∣0 ⟩ + 2 1 ∣1 ⟩ ) = e iπ /3 ∣ + ⟩ Therefore:
P ψ ′ ( + ) = ∣ ⟨ + ∣ ψ ′ ⟩ ∣ 2 = ∣ e i π / 3 ∣ 2 = 1 P_{\psi'}(+) = |\langle +|\psi'\rangle|^2 = |e^{i\pi/3}|^2 = 1 P ψ ′ ( + ) = ∣ ⟨ + ∣ ψ ′ ⟩ ∣ 2 = ∣ e iπ /3 ∣ 2 = 1
P ψ ′ ( − ) = ∣ ⟨ − ∣ ψ ′ ⟩ ∣ 2 = 0 P_{\psi'}(-) = |\langle -|\psi'\rangle|^2 = 0 P ψ ′ ( − ) = ∣ ⟨ − ∣ ψ ′ ⟩ ∣ 2 = 0 Conclusion : the measurement probabilities are the same in every basis.
Part 2: Expectation Values
(c) The expectation value of σ ^ z \hat{\sigma}_z σ ^ z
For ∣ ψ ⟩ |\psi\rangle ∣ ψ ⟩
⟨ ψ ∣ σ ^ z ∣ ψ ⟩ = 1 2 ( 1 1 ) ( 1 0 0 − 1 ) ( 1 1 ) \langle \psi|\hat{\sigma}_z|\psi\rangle = \frac{1}{2}\begin{pmatrix} 1 & 1 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\begin{pmatrix} 1 \\ 1 \end{pmatrix} ⟨ ψ ∣ σ ^ z ∣ ψ ⟩ = 2 1 ( 1 1 ) ( 1 0 0 − 1 ) ( 1 1 )
= 1 2 ( 1 1 ) ( 1 − 1 ) = 1 2 ( 1 − 1 ) = 0 = \frac{1}{2}\begin{pmatrix} 1 & 1 \end{pmatrix}\begin{pmatrix} 1 \\ -1 \end{pmatrix} = \frac{1}{2}(1 - 1) = 0 = 2 1 ( 1 1 ) ( 1 − 1 ) = 2 1 ( 1 − 1 ) = 0 For ∣ ψ ′ ⟩ = e i π / 3 ∣ ψ ⟩ |\psi'\rangle = e^{i\pi/3}|\psi\rangle ∣ ψ ′ ⟩ = e iπ /3 ∣ ψ ⟩
⟨ ψ ′ ∣ σ ^ z ∣ ψ ′ ⟩ = ⟨ ψ ∣ e − i π / 3 σ ^ z e i π / 3 ∣ ψ ⟩ = e − i π / 3 e i π / 3 ⟨ ψ ∣ σ ^ z ∣ ψ ⟩ = 0 \langle \psi'|\hat{\sigma}_z|\psi'\rangle = \langle \psi|e^{-i\pi/3}\hat{\sigma}_z e^{i\pi/3}|\psi\rangle = e^{-i\pi/3}e^{i\pi/3}\langle \psi|\hat{\sigma}_z|\psi\rangle = 0 ⟨ ψ ′ ∣ σ ^ z ∣ ψ ′ ⟩ = ⟨ ψ ∣ e − iπ /3 σ ^ z e iπ /3 ∣ ψ ⟩ = e − iπ /3 e iπ /3 ⟨ ψ ∣ σ ^ z ∣ ψ ⟩ = 0 Conclusion : the expectation values are the same.
(d) The expectation value of σ ^ x \hat{\sigma}_x σ ^ x
For ∣ ψ ⟩ |\psi\rangle ∣ ψ ⟩
⟨ ψ ∣ σ ^ x ∣ ψ ⟩ = 1 2 ( 1 1 ) ( 0 1 1 0 ) ( 1 1 ) \langle \psi|\hat{\sigma}_x|\psi\rangle = \frac{1}{2}\begin{pmatrix} 1 & 1 \end{pmatrix}\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 1 \\ 1 \end{pmatrix} ⟨ ψ ∣ σ ^ x ∣ ψ ⟩ = 2 1 ( 1 1 ) ( 0 1 1 0 ) ( 1 1 )
= 1 2 ( 1 1 ) ( 1 1 ) = 1 2 ( 1 + 1 ) = 1 = \frac{1}{2}\begin{pmatrix} 1 & 1 \end{pmatrix}\begin{pmatrix} 1 \\ 1 \end{pmatrix} = \frac{1}{2}(1 + 1) = 1 = 2 1 ( 1 1 ) ( 1 1 ) = 2 1 ( 1 + 1 ) = 1 For ∣ ψ ′ ⟩ |\psi'\rangle ∣ ψ ′ ⟩
⟨ ψ ′ ∣ σ ^ x ∣ ψ ′ ⟩ = 1 \langle \psi'|\hat{\sigma}_x|\psi'\rangle = 1 ⟨ ψ ′ ∣ σ ^ x ∣ ψ ′ ⟩ = 1 (by the same reasoning)
Part 3: Density Matrices
(e) Computing the density matrices
For ∣ ψ ⟩ = 1 2 ( 1 1 ) |\psi\rangle = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 1 \end{pmatrix} ∣ ψ ⟩ = 2 1 ( 1 1 )
ρ = ∣ ψ ⟩ ⟨ ψ ∣ = 1 2 ( 1 1 ) ( 1 1 ) = 1 2 ( 1 1 1 1 ) \rho = |\psi\rangle\langle\psi| = \frac{1}{2}\begin{pmatrix} 1 \\ 1 \end{pmatrix}\begin{pmatrix} 1 & 1 \end{pmatrix} = \frac{1}{2}\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} ρ = ∣ ψ ⟩ ⟨ ψ ∣ = 2 1 ( 1 1 ) ( 1 1 ) = 2 1 ( 1 1 1 1 ) For ∣ ψ ′ ⟩ = e i π / 3 2 ( 1 1 ) |\psi'\rangle = \frac{e^{i\pi/3}}{\sqrt{2}}\begin{pmatrix} 1 \\ 1 \end{pmatrix} ∣ ψ ′ ⟩ = 2 e iπ /3 ( 1 1 )
ρ ′ = ∣ ψ ′ ⟩ ⟨ ψ ′ ∣ = e i π / 3 2 ( 1 1 ) ⋅ e − i π / 3 2 ( 1 1 ) \rho' = |\psi'\rangle\langle\psi'| = \frac{e^{i\pi/3}}{\sqrt{2}}\begin{pmatrix} 1 \\ 1 \end{pmatrix} \cdot \frac{e^{-i\pi/3}}{\sqrt{2}}\begin{pmatrix} 1 & 1 \end{pmatrix} ρ ′ = ∣ ψ ′ ⟩ ⟨ ψ ′ ∣ = 2 e iπ /3 ( 1 1 ) ⋅ 2 e − iπ /3 ( 1 1 )
= e i π / 3 e − i π / 3 2 ( 1 1 1 1 ) = 1 2 ( 1 1 1 1 ) = ρ = \frac{e^{i\pi/3}e^{-i\pi/3}}{2}\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} = \frac{1}{2}\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} = \rho = 2 e iπ /3 e − iπ /3 ( 1 1 1 1 ) = 2 1 ( 1 1 1 1 ) = ρ Check : ρ = ρ ′ \rho = \rho' ρ = ρ ′
(f) Explaining the physical equivalence
The density matrix ρ \rho ρ
If two states have the same density matrix (ρ = ρ ′ \rho = \rho' ρ = ρ ′ every possible experiment and are therefore completely equivalent physically.
The global phase factor cancels automatically when the density matrix is formed:
ρ ′ = ( e i γ ∣ ψ ⟩ ) ( e − i γ ⟨ ψ ∣ ) = e i γ e − i γ ∣ ψ ⟩ ⟨ ψ ∣ = ∣ ψ ⟩ ⟨ ψ ∣ = ρ \rho' = (e^{i\gamma}|\psi\rangle)(e^{-i\gamma}\langle\psi|) = e^{i\gamma}e^{-i\gamma}|\psi\rangle\langle\psi| = |\psi\rangle\langle\psi| = \rho ρ ′ = ( e iγ ∣ ψ ⟩) ( e − iγ ⟨ ψ ∣ ) = e iγ e − iγ ∣ ψ ⟩ ⟨ ψ ∣ = ∣ ψ ⟩ ⟨ ψ ∣ = ρ Part 4: Observability of a Relative Phase
(g) Measurement in the computational basis
For ∣ ϕ ⟩ = 1 2 ∣ 0 ⟩ + e i π / 3 2 ∣ 1 ⟩ |\phi\rangle = \frac{1}{\sqrt{2}}|0\rangle + \frac{e^{i\pi/3}}{\sqrt{2}}|1\rangle ∣ ϕ ⟩ = 2 1 ∣0 ⟩ + 2 e iπ /3 ∣1 ⟩
P ϕ ( 0 ) = ∣ 1 2 ∣ 2 = 1 2 P_\phi(0) = \left|\frac{1}{\sqrt{2}}\right|^2 = \frac{1}{2} P ϕ ( 0 ) = ∣ ∣ 2 1 ∣ ∣ 2 = 2 1
P ϕ ( 1 ) = ∣ e i π / 3 2 ∣ 2 = 1 2 P_\phi(1) = \left|\frac{e^{i\pi/3}}{\sqrt{2}}\right|^2 = \frac{1}{2} P ϕ ( 1 ) = ∣ ∣ 2 e iπ /3 ∣ ∣ 2 = 2 1 Observation : the measurement probabilities are the same as those of ∣ ψ ⟩ |\psi\rangle ∣ ψ ⟩
(h) Measurement in the x x x
Expand ∣ ϕ ⟩ |\phi\rangle ∣ ϕ ⟩ x x x
∣ ϕ ⟩ = 1 2 ∣ 0 ⟩ + e i π / 3 2 ∣ 1 ⟩ |\phi\rangle = \frac{1}{\sqrt{2}}|0\rangle + \frac{e^{i\pi/3}}{\sqrt{2}}|1\rangle ∣ ϕ ⟩ = 2 1 ∣0 ⟩ + 2 e iπ /3 ∣1 ⟩ Using ∣ 0 ⟩ = 1 2 ( ∣ + ⟩ + ∣ − ⟩ ) |0\rangle = \frac{1}{\sqrt{2}}(|+\rangle + |-\rangle) ∣0 ⟩ = 2 1 ( ∣ + ⟩ + ∣ − ⟩) ∣ 1 ⟩ = 1 2 ( ∣ + ⟩ − ∣ − ⟩ ) |1\rangle = \frac{1}{\sqrt{2}}(|+\rangle - |-\rangle) ∣1 ⟩ = 2 1 ( ∣ + ⟩ − ∣ − ⟩)
∣ ϕ ⟩ = 1 2 ( ∣ + ⟩ + ∣ − ⟩ ) + e i π / 3 2 ( ∣ + ⟩ − ∣ − ⟩ ) |\phi\rangle = \frac{1}{2}(|+\rangle + |-\rangle) + \frac{e^{i\pi/3}}{2}(|+\rangle - |-\rangle) ∣ ϕ ⟩ = 2 1 ( ∣ + ⟩ + ∣ − ⟩) + 2 e iπ /3 ( ∣ + ⟩ − ∣ − ⟩)
= 1 + e i π / 3 2 ∣ + ⟩ + 1 − e i π / 3 2 ∣ − ⟩ = \frac{1 + e^{i\pi/3}}{2}|+\rangle + \frac{1 - e^{i\pi/3}}{2}|-\rangle = 2 1 + e iπ /3 ∣ + ⟩ + 2 1 − e iπ /3 ∣ − ⟩ Compute 1 + e i π / 3 = 1 + cos ( π / 3 ) + i sin ( π / 3 ) = 1 + 1 / 2 + i 3 / 2 = 3 + i 3 2 1 + e^{i\pi/3} = 1 + \cos(\pi/3) + i\sin(\pi/3) = 1 + 1/2 + i\sqrt{3}/2 = \frac{3 + i\sqrt{3}}{2} 1 + e iπ /3 = 1 + cos ( π /3 ) + i sin ( π /3 ) = 1 + 1/2 + i 3 /2 = 2 3 + i 3
∣ 1 + e i π / 3 ∣ 2 = ∣ 3 + i 3 2 ∣ 2 = 9 + 3 4 = 3 |1 + e^{i\pi/3}|^2 = \left|\frac{3 + i\sqrt{3}}{2}\right|^2 = \frac{9 + 3}{4} = 3 ∣1 + e iπ /3 ∣ 2 = ∣ ∣ 2 3 + i 3 ∣ ∣ 2 = 4 9 + 3 = 3 Therefore:
P ϕ ( + ) = ∣ 1 + e i π / 3 ∣ 2 4 = 3 4 P_\phi(+) = \frac{|1 + e^{i\pi/3}|^2}{4} = \frac{3}{4} P ϕ ( + ) = 4 ∣1 + e iπ /3 ∣ 2 = 4 3
P ϕ ( − ) = ∣ 1 − e i π / 3 ∣ 2 4 = 1 4 P_\phi(-) = \frac{|1 - e^{i\pi/3}|^2}{4} = \frac{1}{4} P ϕ ( − ) = 4 ∣1 − e iπ /3 ∣ 2 = 4 1 Observation : these differ from those of ∣ ψ ⟩ |\psi\rangle ∣ ψ ⟩ P ψ ( + ) = 1 P_\psi(+) = 1 P ψ ( + ) = 1 P ϕ ( + ) = 3 / 4 P_\phi(+) = 3/4 P ϕ ( + ) = 3/4
Key conclusion : a relative phase can be observed by choosing a suitable measurement basis.
(i) Comparing the density matrices
ρ ϕ = ∣ ϕ ⟩ ⟨ ϕ ∣ = 1 2 ( 1 e i π / 3 ) ( 1 e − i π / 3 ) \rho_\phi = |\phi\rangle\langle\phi| = \frac{1}{2}\begin{pmatrix} 1 \\ e^{i\pi/3} \end{pmatrix}\begin{pmatrix} 1 & e^{-i\pi/3} \end{pmatrix} ρ ϕ = ∣ ϕ ⟩ ⟨ ϕ ∣ = 2 1 ( 1 e iπ /3 ) ( 1 e − iπ /3 ) = 1 2 ( 1 e − i π / 3 e i π / 3 1 ) = \frac{1}{2}\begin{pmatrix} 1 & e^{-i\pi/3} \\ e^{i\pi/3} & 1 \end{pmatrix} = 2 1 ( 1 e iπ /3 e − iπ /3 1 ) whereas the earlier one is:
ρ = 1 2 ( 1 1 1 1 ) \rho = \frac{1}{2}\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} ρ = 2 1 ( 1 1 1 1 ) Key differences :
The off-diagonal entries of ρ \rho ρ ρ 01 = ρ 10 = 1 / 2 \rho_{01} = \rho_{10} = 1/2 ρ 01 = ρ 10 = 1/2
The off-diagonal entries of ρ ϕ \rho_\phi ρ ϕ ( ρ ϕ ) 01 = e − i π / 3 / 2 (\rho_\phi)_{01} = e^{-i\pi/3}/2 ( ρ ϕ ) 01 = e − iπ /3 /2 ( ρ ϕ ) 10 = e i π / 3 / 2 (\rho_\phi)_{10} = e^{i\pi/3}/2 ( ρ ϕ ) 10 = e iπ /3 /2
The off-diagonal entries carry the relative phase information ; this is the mathematical signature of quantum coherence!
(j) Summary of the comparison
Property Global phase e i γ ∣ ψ ⟩ e^{i\gamma}|\psi\rangle e iγ ∣ ψ ⟩ Relative phase α ∣ 0 ⟩ + e i ϕ β ∣ 1 ⟩ \alpha|0\rangle + e^{i\phi}\beta|1\rangle α ∣0 ⟩ + e i ϕ β ∣1 ⟩ Physically observable No Yes Affects measurement probabilities No (in every basis) Yes (in some bases) Density matrix Unchanged Off-diagonal entries change Bloch sphere Same point Different points Interference effects No effect Decisive effect
Core distinction :
Global phase : all coefficients rotate together, and their relative relationships do not change
Relative phase : the phase difference between coefficients, which determines the interference pattern
Part 5: The Bloch Sphere Representation
(k) Parametrization
The standard Bloch sphere form: ∣ ψ ( θ , ϕ ) ⟩ = cos θ 2 ∣ 0 ⟩ + e i ϕ sin θ 2 ∣ 1 ⟩ |\psi(\theta,\phi)\rangle = \cos\frac{\theta}{2}|0\rangle + e^{i\phi}\sin\frac{\theta}{2}|1\rangle ∣ ψ ( θ , ϕ )⟩ = cos 2 θ ∣0 ⟩ + e i ϕ sin 2 θ ∣1 ⟩
For ∣ ψ ⟩ |\psi\rangle ∣ ψ ⟩ :
∣ ψ ⟩ = 1 2 ∣ 0 ⟩ + 1 2 ∣ 1 ⟩ = cos π / 2 2 ∣ 0 ⟩ + e i ⋅ 0 sin π / 2 2 ∣ 1 ⟩ |\psi\rangle = \frac{1}{\sqrt{2}}|0\rangle + \frac{1}{\sqrt{2}}|1\rangle = \cos\frac{\pi/2}{2}|0\rangle + e^{i \cdot 0}\sin\frac{\pi/2}{2}|1\rangle ∣ ψ ⟩ = 2 1 ∣0 ⟩ + 2 1 ∣1 ⟩ = cos 2 π /2 ∣0 ⟩ + e i ⋅ 0 sin 2 π /2 ∣1 ⟩ Hence θ = π / 2 \theta = \pi/2 θ = π /2 ϕ = 0 \phi = 0 ϕ = 0 x x x
For ∣ ψ ′ ⟩ |\psi'\rangle ∣ ψ ′ ⟩ :
∣ ψ ′ ⟩ = e i π / 3 ( 1 2 ∣ 0 ⟩ + 1 2 ∣ 1 ⟩ ) |\psi'\rangle = e^{i\pi/3}\left(\frac{1}{\sqrt{2}}|0\rangle + \frac{1}{\sqrt{2}}|1\rangle\right) ∣ ψ ′ ⟩ = e iπ /3 ( 2 1 ∣0 ⟩ + 2 1 ∣1 ⟩ ) By global phase invariance, factor out e i π / 3 e^{i\pi/3} e iπ /3
∣ ψ ′ ⟩ ≡ 1 2 ∣ 0 ⟩ + 1 2 ∣ 1 ⟩ |\psi'\rangle \equiv \frac{1}{\sqrt{2}}|0\rangle + \frac{1}{\sqrt{2}}|1\rangle ∣ ψ ′ ⟩ ≡ 2 1 ∣0 ⟩ + 2 1 ∣1 ⟩ Hence θ = π / 2 \theta = \pi/2 θ = π /2 ϕ = 0 \phi = 0 ϕ = 0 the same as ∣ ψ ⟩ |\psi\rangle ∣ ψ ⟩ ).
For ∣ ϕ ⟩ |\phi\rangle ∣ ϕ ⟩ :
∣ ϕ ⟩ = 1 2 ∣ 0 ⟩ + e i π / 3 2 ∣ 1 ⟩ = cos π / 2 2 ∣ 0 ⟩ + e i π / 3 sin π / 2 2 ∣ 1 ⟩ |\phi\rangle = \frac{1}{\sqrt{2}}|0\rangle + \frac{e^{i\pi/3}}{\sqrt{2}}|1\rangle = \cos\frac{\pi/2}{2}|0\rangle + e^{i\pi/3}\sin\frac{\pi/2}{2}|1\rangle ∣ ϕ ⟩ = 2 1 ∣0 ⟩ + 2 e iπ /3 ∣1 ⟩ = cos 2 π /2 ∣0 ⟩ + e iπ /3 sin 2 π /2 ∣1 ⟩ Hence θ = π / 2 \theta = \pi/2 θ = π /2 ϕ = π / 3 \phi = \pi/3 ϕ = π /3
(l) Explanation on the Bloch sphere
∣ ψ ⟩ |\psi\rangle ∣ ψ ⟩ ∣ ψ ′ ⟩ |\psi'\rangle ∣ ψ ′ ⟩ the same point ( π / 2 , 0 ) (\pi/2, 0) ( π /2 , 0 )
∣ ϕ ⟩ |\phi\rangle ∣ ϕ ⟩ a different point ( π / 2 , π / 3 ) (\pi/2, \pi/3) ( π /2 , π /3 )
Geometric intuition :
Each point on the Bloch sphere represents a physically inequivalent quantum state
Global phase invariance means that a phase “rotation” around the unit circle, ∣ ψ ⟩ ↦ e i γ ∣ ψ ⟩ |\psi\rangle \mapsto e^{i\gamma}|\psi\rangle ∣ ψ ⟩ ↦ e iγ ∣ ψ ⟩
Only the position on the sphere (θ , ϕ \theta, \phi θ , ϕ
This is exactly why quotienting C 2 \mathbb{C}^2 C 2 S 2 S^2 S 2
Global phase :
∣ ψ ′ ⟩ = e i γ ∣ ψ ⟩ |\psi'\rangle = e^{i\gamma}|\psi\rangle ∣ ψ ′ ⟩ = e iγ ∣ ψ ⟩ All coefficients are multiplied by the same e i γ e^{i\gamma} e iγ
Completely unobservable physically
The density matrix is unchanged: ρ ′ = ρ \rho' = \rho ρ ′ = ρ
The same point on the Bloch sphere
Relative phase :
∣ ψ ⟩ = α ∣ 0 ⟩ + e i ϕ β ∣ 1 ⟩ |\psi\rangle = \alpha|0\rangle + e^{i\phi}\beta|1\rangle ∣ ψ ⟩ = α ∣0 ⟩ + e i ϕ β ∣1 ⟩ The phase difference ϕ \phi ϕ
Physically observable (in a suitable basis)
Affects the off-diagonal entries of the density matrix
Different points on the Bloch sphere
Why is a global phase unobservable?
All physical quantities are sesquilinear forms: ⟨ ψ ∣ A ^ ∣ ψ ⟩ \langle \psi|\hat{A}|\psi\rangle ⟨ ψ ∣ A ^ ∣ ψ ⟩ ∣ ⟨ n ∣ ψ ⟩ ∣ 2 |\langle n|\psi\rangle|^2 ∣ ⟨ n ∣ ψ ⟩ ∣ 2
The global phase cancels between the ket and the bra
Only the relative phase survives in the computation
Mathematical essence :
The space of quantum states is really the projective Hilbert space P H \mathbb{P}\mathcal{H} P H
The equivalence relation: ∣ ψ ⟩ ∼ e i γ ∣ ψ ⟩ |\psi\rangle \sim e^{i\gamma}|\psi\rangle ∣ ψ ⟩ ∼ e iγ ∣ ψ ⟩ γ ∈ R \gamma \in \mathbb{R} γ ∈ R
The Bloch sphere is a geometric realization of this projective space
This idea is essential for understanding quantum interference, the Berry phase, phase gates in quantum computing, and more.
