Published Sep 28, 2026
Last updated Sep 28, 2026
Qubit ¶ In quantum computing, the fundamental unit of information is the quantum bit or qubit . A qubit can exist in a superposition of states, represented as:
∣ ψ ⟩ = α ∣ 0 ⟩ + β ∣ 1 ⟩ ∈ C 2 , \ket{\psi} = \alpha\ket{0} + \beta\ket{1} \in \mathbb{C}^2, ∣ ψ ⟩ = α ∣ 0 ⟩ + β ∣ 1 ⟩ ∈ C 2 , where ∣ 0 ⟩ = [ 1 0 ] \ket{0} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} ∣ 0 ⟩ = [ 1 0 ] ∣ 1 ⟩ = [ 0 1 ] \ket{1} = \begin{bmatrix} 0 \\ 1 \end{bmatrix} ∣ 1 ⟩ = [ 0 1 ] α , β ∈ C 2 \alpha, \beta \in \mathbb{C}^2 α , β ∈ C 2 ∣ 0 ⟩ , ∣ 1 ⟩ \ket{0}, \ket{1} ∣ 0 ⟩ , ∣ 1 ⟩ 0 , 1 0, 1 0 , 1
∣ ψ ⟩ = [ α β ] . \ket{\psi} = \begin{bmatrix}
\alpha\\
\beta
\end{bmatrix}. ∣ ψ ⟩ = [ α β ] . Quantum state normalization ¶ For an arbitrary quantum state, it has to satisfy the normalization condition, the sum of the squares of their magnitudes equals one. Mathematically speaking,
∣ α ∣ 2 + ∣ β ∣ 2 = 1 |\alpha|^2 + |\beta|^2 = 1 ∣ α ∣ 2 + ∣ β ∣ 2 = 1 .
Here are examples for a valid quantum state of qubits:
∣ ψ 1 ⟩ = ∣ 0 ⟩ + 0 ⋅ ∣ 1 ⟩ = ∣ 0 ⟩ , ( α = 1 , β = 0 ) \ket{\psi_1} = \ket{0} + 0\cdot\ket{1} = \ket{0}, \quad (\alpha = 1, \beta = 0) ∣ ψ 1 ⟩ = ∣ 0 ⟩ + 0 ⋅ ∣ 1 ⟩ = ∣ 0 ⟩ , ( α = 1 , β = 0 )
∣ ψ 2 ⟩ = 0 ⋅ ∣ 0 ⟩ + ∣ 1 ⟩ = ∣ 1 ⟩ , ( α = 0 , β = 1 ) \ket{\psi_2} = 0\cdot\ket{0} +\ket{1} = \ket{1}, \quad (\alpha = 0, \beta = 1) ∣ ψ 2 ⟩ = 0 ⋅ ∣ 0 ⟩ + ∣ 1 ⟩ = ∣ 1 ⟩ , ( α = 0 , β = 1 )
∣ ψ 3 ⟩ = 1 2 ( ∣ 0 ⟩ + ∣ 1 ⟩ ) , ( α = β = 1 2 ) \ket{\psi_3} = \frac{1}{\sqrt{2}}(\ket{0} + \ket{1}), \quad (\alpha = \beta = \frac{1}{\sqrt{2}}) ∣ ψ 3 ⟩ = 2 1 ( ∣ 0 ⟩ + ∣ 1 ⟩ ) , ( α = β = 2 1 )
∣ ψ 4 ⟩ = 1 2 ( ∣ 0 ⟩ + i ∣ 1 ⟩ ) , ( α = 1 2 , β = i 1 2 ) \ket{\psi_4} = \frac{1}{\sqrt{2}}(\ket{0} + i\ket{1}), \quad (\alpha = \frac{1}{\sqrt{2}}, \beta = i\frac{1}{\sqrt{2}}) ∣ ψ 4 ⟩ = 2 1 ( ∣ 0 ⟩ + i ∣ 1 ⟩ ) , ( α = 2 1 , β = i 2 1 )
You can check if they satisfied the normalization condition. I think they do.
Geometric representation ¶ So how do we visualize a qubit? Starting with the complex space of α , β \alpha, \beta α , β
α = r α e i ϕ α , r α ∈ R ≥ 0 , ϕ α ∈ [ 0 , 2 π ) β = r β e i ϕ β , r β ∈ R ≥ 0 , ϕ β ∈ [ 0 , 2 π ) . \begin{aligned}
\alpha & = r_\alpha e^{i\phi_\alpha}, \quad r_\alpha \in \mathbb{R}_{\geq 0}, \phi_\alpha \in [0, 2\pi)\\
\beta & = r_\beta e^{i\phi_\beta}, \quad r_\beta \in \mathbb{R}_{\geq 0}, \phi_\beta \in [0, 2\pi).
\end{aligned} α β = r α e i ϕ α , r α ∈ R ≥ 0 , ϕ α ∈ [ 0 , 2 π ) = r β e i ϕ β , r β ∈ R ≥ 0 , ϕ β ∈ [ 0 , 2 π ) . And the magnitude of α \alpha α β \beta β
∣ α ∣ = r α ∣ β ∣ = r β . \begin{aligned}
|\alpha| & = r_\alpha \\
|\beta| & = r_\beta.
\end{aligned} ∣ α ∣ ∣ β ∣ = r α = r β . Plug them into the normalization condition:
∣ α ∣ 2 + ∣ β ∣ 2 = 1 ⇔ ∣ r α ∣ 2 + ∣ r β ∣ 2 = 1 ⇔ r β = 1 − r α 2 . \begin{aligned}
& |\alpha|^2 + |\beta|^2 = 1 \quad
\Leftrightarrow & \left| r_\alpha \right|^2 + \left|r_\beta \right|^2 = 1 \quad
\Leftrightarrow & r_\beta = \sqrt{1 - r_\alpha^2}.
\end{aligned} ∣ α ∣ 2 + ∣ β ∣ 2 = 1 ⇔ ∣ r α ∣ 2 + ∣ r β ∣ 2 = 1 ⇔ r β = 1 − r α 2 . Now let plug them back to the (1)
∣ ψ ⟩ = r α e i ϕ α ∣ 0 ⟩ + 1 − r α 2 e i ϕ β ∣ 1 ⟩ \ket{\psi} = r_\alpha e^{i\phi_\alpha}\ket{0} + \sqrt{1 - r_\alpha^2} e^{i\phi_\beta}\ket{1} ∣ ψ ⟩ = r α e i ϕ α ∣ 0 ⟩ + 1 − r α 2 e i ϕ β ∣ 1 ⟩ Now we can factor out the common term of e i ϕ α e^{i\phi_\alpha} e i ϕ α e i ϕ β e^{i\phi_\beta} e i ϕ β
∣ ψ ⟩ = r α e i ϕ α ∣ 0 ⟩ + 1 − r α 2 e i ϕ α e i ( ϕ β − ϕ α ) ∣ 1 ⟩ = e i ϕ α ( r α ∣ 0 ⟩ + 1 − r α 2 e i ( ϕ β − ϕ α ) ∣ 1 ⟩ ) . \begin{aligned}
\ket{\psi} &= r_\alpha e^{i\phi_\alpha}\ket{0} + \sqrt{1 - r_\alpha^2} e^{i\phi_\alpha}e^{i(\phi_\beta-\phi_\alpha)}\ket{1} \\
&= e^{i\phi_\alpha}\left(r_\alpha \ket{0} + \sqrt{1 - r_\alpha^2} e^{i(\phi_\beta-\phi_\alpha)} \ket{1} \right).
\end{aligned} ∣ ψ ⟩ = r α e i ϕ α ∣ 0 ⟩ + 1 − r α 2 e i ϕ α e i ( ϕ β − ϕ α ) ∣ 1 ⟩ = e i ϕ α ( r α ∣ 0 ⟩ + 1 − r α 2 e i ( ϕ β − ϕ α ) ∣ 1 ⟩ ) . Finally, since r α 2 + r β 2 = 1 r_\alpha^2 + r_\beta^2 = 1 r α 2 + r β 2 = 1 r α , r β r_\alpha, r_\beta r α , r β
r α = cos θ 2 , r β = sin θ 2 . r_\alpha = \cos{\frac{\theta}{2}}, \quad r_\beta = \sin{\frac{\theta}{2}}. r α = cos 2 θ , r β = sin 2 θ . And let ϕ α = γ , ϕ β − ϕ α = φ \phi_\alpha = \gamma, \phi_\beta-\phi_\alpha = \varphi ϕ α = γ , ϕ β − ϕ α = φ ψ \psi ψ
∣ ψ ⟩ = e i γ ( cos θ 2 ∣ 0 ⟩ + e i φ sin θ 2 ∣ 1 ⟩ ) . \ket{\psi} = e^{i\gamma}\left(\cos{\frac{\theta}{2}}\ket{0} + e^{i\varphi}\sin{\frac{\theta}{2}}\ket{1}\right). ∣ ψ ⟩ = e iγ ( cos 2 θ ∣ 0 ⟩ + e i φ sin 2 θ ∣ 1 ⟩ ) . From this form, we have three parameters that completely specify the qubit state: γ \gamma γ θ \theta θ φ \varphi φ γ \gamma γ
∣ ψ ⟩ = cos θ 2 ∣ 0 ⟩ + e i φ sin θ 2 ∣ 1 ⟩ . \ket{\psi} = \cos{\frac{\theta}{2}}\ket{0} + e^{i\varphi}\sin{\frac{\theta}{2}}\ket{1}. ∣ ψ ⟩ = cos 2 θ ∣ 0 ⟩ + e i φ sin 2 θ ∣ 1 ⟩ . Therefore, a qubit would be represented by a point on a unit three-dimensional sphere, known as the Bloch sphere. And it is often called the Bloch sphere representation of a qubit.
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