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1.103: Bloch Sphere

  • Page ID
    158610
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    The eigenfunctions of the Pauli spin matrices

    \[
    \sigma_{\mathrm{Z}} :=\left(\begin{array}{cc}{1} & {0} \\ {0} & {-1}\end{array}\right) \qquad \sigma_{\mathrm{x}} :=\left(\begin{array}{cc}{0} & {1} \\ {1} & {0}\end{array}\right) \qquad \sigma_{\mathrm{y}} :=\left(\begin{array}{cc}{0} & {-\mathrm{i}} \\ {\mathrm{i}} & {0}\end{array}\right)
    \nonumber \]

    are presented mathematically and shown on the Bloch sphere below. The Xu state is highlighted.

    \[
    Z_{u} :=\left(\begin{array}{l}{1} \\ {0}\end{array}\right) \qquad Z_{d} :=\left(\begin{array}{l}{0} \\ {1}\end{array}\right) \\ \mathrm{X}_{\mathrm{u}} :=\frac{1}{\sqrt{2}} \cdot\left(\begin{array}{l}{1} \\ {1}\end{array}\right) \qquad \mathrm{X}_{\mathrm{d}} :=\frac{1}{\sqrt{2}} \cdot\left(\begin{array}{c}{1} \\ {-1}\end{array}\right) \\ \mathrm{Y}_{\mathrm{u}} :=\frac{1}{\sqrt{2}} \cdot\left(\begin{array}{l}{1} \\ {\mathrm{i}}\end{array}\right) \qquad \mathrm{Y}_{\mathrm{d}} :=\frac{1}{\sqrt{2}} \cdot\left(\begin{array}{c}{1} \\ {-\mathrm{i}}\end{array}\right)
    \nonumber \]

    clipboard_e6508caa98f0b6708756d68072c853940.png

    This figure was taken from demonstrations.wolfram.com/QubitsOnThePoincareBlochSphere/ a contribution by Rudolf Muradian.

    The Bloch sphere is prepared in Cartesian coordinates using Mathcad graphics.

    \[
    \text{numpts}:=100 \qquad \mathrm{i} :=0 \ldots \text { numpts } \qquad \mathrm{j} :=0 \ldots \text{numpts} \\ \theta_{\mathrm{i}} :=\frac{\pi \cdot \mathrm{i}}{\text { numpts }} \quad \phi_{\mathrm{j}} :=\frac{2 \cdot \pi \cdot \mathrm{j}}{\text { numpts }} \\ \mathrm{X}_{\mathrm{i}, \mathrm{j}}=\sin \left(\theta_{\mathrm{i}}\right) \cdot \cos \left(\phi_{\mathrm{j}}\right) \qquad \mathrm{Y}_{\mathrm{i}, \mathrm{j}} :=\sin \left(\theta_{\mathrm{i}}\right) \cdot \sin \left(\phi_{\mathrm{j}}\right) \qquad \mathrm{z}_{\mathrm{i}, \mathrm{j}} :=\cos \left(\theta_{\mathrm{i}}\right)
    \nonumber \]

    Next, the coordinates of a quantum qubit are calculated and displayed on the Bloch sphere as a white dot. As the polar and azmuthal angles are changed, you will need to rotate the figure to see where the white dot is on the surface of the Bloch sphere.

    \[
    \theta 1 :=\frac{\pi}{2} \qquad \phi 1 :=0 \\ \Psi(\theta 1, \phi 1) :=\cos \left(\frac{\theta 1}{2}\right) \cdot\left(\begin{array}{c}{1} \\ {0}\end{array}\right)+\exp (\mathrm{i} \cdot \phi 1) \cdot \sin \left(\frac{\theta 1}{2}\right) \cdot\left(\begin{array}{l}{0} \\ {1}\end{array}\right) \quad \Psi(\theta 1, \phi 1)=\left(\begin{array}{l}{0.707} \\ {0.707}\end{array}\right) \\ \mathrm{XX}_{\mathrm{i}, \mathrm{j}}=\sin (\theta 1) \cdot \cos (\phi 1) \qquad \mathrm{YY}_{\mathrm{i}, \mathrm{j}} :=\sin (\theta 1) \cdot \sin (\phi 1) \qquad \mathrm{ZZ}_{\mathrm{i}, \mathrm{j}} :=\cos (\theta 1)
    \nonumber \]

    clipboard_e6bf86e338301900a58e3271571455ad2.png


    This page titled 1.103: Bloch Sphere is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Frank Rioux via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.