$\require{cancel} \newcommand{\Ket}[1]{\left|{#1}\right\rangle} \newcommand{\Bra}[1]{\left\langle{#1}\right|} \newcommand{\Braket}[1]{\left\langle{#1}\right\rangle} \newcommand{\Rsr}[1]{\frac{1}{\sqrt{#1}}} \newcommand{\RSR}[1]{1/\sqrt{#1}} \newcommand{\Verti}{\rvert} \newcommand{\HAT}[1]{\hat{\,#1~}} \DeclareMathOperator{\Tr}{Tr}$

Kous Delta Operator

• 3 Aug 2018 - First draft with theorem proven after failed attempt without $\delta_\psi(\theta)$.
• 2 Jan 2019 - Starting to analyse the behaviour of $\delta_\psi(\theta)$.

This article is to prove that the Bloch rotation Operator about $\Ket\psi$ is $\boxed{~R_\psi(\theta) =\Braket{X}_\psi R_x(\theta)+\Braket{Y}_\psi R_y(\theta)+\Braket{Z}_\psi R_z(\theta)+\delta_\psi(\theta)~},$ where $\Braket{Q}_\psi$ is the expectation value of measurement on Pauli axis $Q$, and $\delta_\psi(\theta)$ is the Kous Delta Operator. $\boxed{\delta_\psi(\theta):=I\cos\frac{\theta}{2}\left(1-\Braket{X}_\psi-\Braket{Y}_\psi-\Braket{Z}_\psi\right) .}$

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$\delta_\psi(\theta)=0$ when $\Ket\psi$ is $X, Y$ or $Z$, or when $\theta=\pi.$

Let us define $B_\psi:=R_\psi(\pi)=\Braket{X}_\psi R_x(\pi)+\Braket{Y}_\psi R_y(\pi)+\Braket{Z}_\psi R_z(\pi)=\Braket{X}_\psi X+\Braket{Y}_\psi Y+\Braket{Z}_\psi Z.~~~~ B_\psi^2=I,$ as $R_\psi(2\pi)=I.$

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The givens:

$R_\psi(\theta) =e^{-i\frac{\theta}{2}B_\psi} =I\cos\frac{\theta}{2}-iB_\psi\sin\frac{\theta}{2} .~~~~ \left(\text{If }A^2=I,~~e^{i\alpha A}=\cos(\alpha)~I+i\sin(\alpha)~A .\right)$

Ref: "Rotation: Block Sphere" and http://www.vcpc.univie.ac.at/~ian/hotlist/qc/talks/bloch-sphere-rotations.pdf, in which $B$ is represented by $\HAT n\cdot\vec{\,\sigma~}.$

$R_x(\theta) =e^{-i\frac{\theta}{2}X} =I\cos\frac{\theta}{2}-iX\sin\frac{\theta}{2} ,$

$R_y(\theta) =e^{-i\frac{\theta}{2}Y} =I\cos\frac{\theta}{2}-iY\sin\frac{\theta}{2} ,$

$R_z(\theta) =e^{-i\frac{\theta}{2}Z} =I\cos\frac{\theta}{2}-iZ\sin\frac{\theta}{2} .$

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$\mathrm{RHS} =\left(\Braket{X}_\psi I\cos\frac{\theta}{2}-i\Braket{X}_\psi X\sin\frac{\theta}{2}\right) +\left(\Braket{Y}_\psi I\cos\frac{\theta}{2}-i\Braket{Y}_\psi Y\sin\frac{\theta}{2}\right) +\left(\Braket{Z}_\psi I\cos\frac{\theta}{2}-i\Braket{Z}_\psi Z\sin\frac{\theta}{2}\right) +\delta_\psi(\theta) .$

$=\Big(\big(\Braket{X}_\psi+\Braket{Y}_\psi+\Braket{Z}_\psi\big)~I\cos\frac{\theta}{2}+\delta_\psi(\theta)\Big) -i\Big(\Braket{X}_\psi X+\Braket{Y}_\psi Y+\Braket{Z}_\psi Z\Big)\sin\frac{\theta}{2} =I\cos\frac{\theta}{2}-iB_\psi\sin\frac{\theta}{2}=\mathrm{LHS~~~~QED}.$

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KEPT - The following might be useful.

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$R_\psi(\theta)=\Braket{X}_\psi R_x(\theta)+\Braket{Y}_\psi R_y(\theta)+\Braket{Z}_\psi R_z(\theta) +\left(1-\Braket{X}_\psi-\Braket{Y}_\psi-\Braket{Z}_\psi\right)~I\cos\frac{\theta}{2} .$

$\delta=1-\Braket{X}_\psi-\Braket{Y}_\psi-\Braket{Z}_\psi =\Braket{X}_\psi^2+\Braket{Y}_\psi^2+\Braket{Z}_\psi^2-\Braket{X}_\psi-\Braket{Y}_\psi-\Braket{Z}_\psi =-\Braket{X}_\psi(1-\Braket{X}_\psi)-\Braket{Y}_\psi(1-\Braket{Y}_\psi)-\Braket{Z}_\psi(1-\Braket{Z}_\psi) .$

$\delta=1-\sin\theta\cos\phi-\sin\theta\sin\phi-\cos\theta =1-{1\over2}(\sin(\theta+\phi)+\sin(\theta-\phi))+{1\over2}(\cos(\theta+\phi)-\cos(\theta-\phi))-\cos\theta ={1\over2}(\sin^2(\theta+\phi)-\sin(\theta+\phi)+\sin^2(\theta-\phi)-\sin(\theta-\phi)+\cos^2(\theta+\phi)+\cos(\theta+\phi)+\cos^2(\theta-\phi)-\cos(\theta-\phi))-\cos\theta .$

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$B_X=X,~B_Y=Y,~B_Z=Z.$

Let $\Ket\psi =\alpha\Ket0+\beta\Ket1 =\cos\theta/2\Ket0+e^{i\phi}\sin\theta/2\Ket1 .$ We will have

$\Braket{X}_\psi$ $=\Braket{\psi\lvert X\rvert\psi} =(\alpha^*\Bra0+\beta^*\Bra1)~(\Ket0\Bra1+\Ket1\Bra0)~(\alpha\Ket0+\beta\Ket1) =\alpha^*\beta+\alpha\beta^* \\ =\cos\theta/2\cdot e^{i\phi}\sin\theta/2+\cos\theta/2\cdot e^{-i\phi}\sin\theta/2 =\sin(\theta)\cosh(i\phi) =\sin(\theta)\cos(\phi) ,$

$\Braket{Y}_\psi$ $=\Braket{\psi\lvert Y\rvert\psi} =(\alpha^*\Bra0+\beta^*\Bra1)~(-i\Ket0\Bra1+i\Ket1\Bra0)~(\alpha\Ket0+\beta\Ket1) =-i\alpha^*\beta+i\alpha\beta^* \\ =-i\cos\theta/2\cdot e^{i\phi}\sin\theta/2+i\cos\theta/2\cdot e^{-i\phi}\sin\theta/2 =-i\sin(\theta)\sinh(i\phi) =\sin(\theta)\sin(\phi) ,~~\text{and}$

$\Braket{Z}_\psi$ $=\Braket{\psi\lvert Z\rvert\psi} =(\alpha^*\Bra0+\beta^*\Bra1)~(\Ket0\Bra0-\Ket1\Bra1)~(\alpha\Ket0+\beta\Ket1) =\alpha^*\alpha-\beta^*\beta \\ =\cos^2\theta/2-\sin^2\theta/2 =\cos\theta .$

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Let $\HAT\psi=\left(\Braket{X}_\psi,\Braket{Y}_\psi,\Braket{Z}_\psi\right).~~~\HAT\psi\cdot\HAT\psi^T=1$

?? $R_\psi(\theta) =I\cos\frac{\theta}{2}-iB_\psi\sin\frac{\theta}{2} =\HAT\psi\cdot\HAT\psi^T~I\cos\frac{\theta}{2} -i~\HAT\psi\cdot(X,Y,Z)^T\sin\frac{\theta}{2} =\HAT\psi\cdot\left(\left(\Braket{X}_\psi,\Braket{Y}_\psi,\Braket{Z}_\psi\right)^TI\cos\frac{\theta}{2}-i(X,Y,Z)^T\cos\frac{\theta}{2}\right) .$