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Quantum Computation and Quantum Information

  1. Mar 1, 2006 #1
    Hey all, I'm a beginner of the Quantum Computation and Quantum Information. For a long time, I feel very confuse about the question bellow. Could you do me a favor and show me the proof? Many thanks!
    Show that any qubit can be expressed as
    for real numbers γ,θ and Φ. The phase factor exp(iγ) has no observational effect and can be dropped.
    And then show that there is a one to one correspondence between qubits
    and the points on the unit sphere in R(3) called the Bloch sphere, with and as the spherical coordinates of a point of the sphere.
  2. jcsd
  3. Mar 2, 2006 #2
    Start by expressing any normalized state vector as [tex]|\psi \rangle = a |0\rangle + b|1\rangle [/tex], where normalization entails [tex] a^2 + b^2 =1 [/tex]. [tex] a,b [/tex] are complex numbers, which can always be expressed as [tex] a = r_a e^{i\gamma_a} [/tex], [tex] b = r_b e^{i\gamma_b} [/tex]. This means that [tex] a^2 + b^2 =r_a^2+r_b^2=1 [/tex], and since [tex] r_a, r_b [/tex] are positive we can express them as [tex] r_a = \cos(\theta /2), r_b= \sin(\theta/2) [/tex] where [tex] 0 \leq \theta \leq \pi [/tex]. Consequently, we can write [tex]|\psi \rangle = \cos(\theta /2) e^{i\gamma_a} |0\rangle + \sin(\theta/2) e^{i\gamma_b} |1\rangle = e^{i\gamma}(\cos(\theta /2)|0\rangle + \sin(\theta/2) e^{i\Phi} |1\rangle )[/tex], where [tex] \gamma = \gamma_a, \Phi = \gamma_b-\gamma_a [/tex].
  4. Mar 2, 2006 #3
    Thank you, Davidk!
  5. Mar 3, 2006 #4


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    how is it that you can say r_a = cos (theta/2) when r_a is a radius with a real length and cos(theta) is a dimensionless quantity?
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