Qubits, Bloch Sphere, Pauli Algebra, and One-Qubit Gates

Multiplying the entries of the matrices in [F1] gives the nine row-by-column products $X^2=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)=I$, $XY=\left(\begin{array}{cc}i& 0\\ 0& -i\end{array}\right)=iZ$, $XZ=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)=-iY$, $YX=\left(\begin{array}{cc}-i& 0\\ 0& i\end{array}\right)=-iZ$, $Y^2=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)=I$, $YZ=\left(\begin{array}{cc}0& i\\ i& 0\end{array}\right)=iX$, $ZX=\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)=iY$, $ZY=\left(\begin{array}{cc}0& -i\\ -i& 0\end{array}\right)=-iX$, and $Z^2=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)=I$.

By [F1] and [F2], $X^{\ast }=X$, $Y^{\ast }=Y$, and $Z^{\ast }=Z$; the diagonal entries of each product $X^{\ast }X,Y^{\ast }Y,Z^{\ast }Z$ are $1$ and its off-diagonal entries are $0$, so each product is $I$. By [F4] each is a unitary operator, in the sense of [F5].

The six off-diagonal products in step 1.1 have opposite signs when their order is reversed, so the matrices anticommute pairwise and the table is exactly the $\delta ,ϵ$ product formula in the Statement. The same matrices have $\mathrm{tr}(X)=\mathrm{tr}(Y)=\mathrm{tr}(Z)=0$ and $\mathrm{tr}(I)=2$ by [F3]. Taking traces in step 1.1 gives $\mathrm{tr}({\sigma }_j{\sigma }_k)=2$ when $j=k$ and $0$ otherwise.

By [F1] and [F2], $X^{\ast }=X$, $Y^{\ast }=Y$, and $Z^{\ast }=Z$. Multiplying the displayed matrices gives $X^{\ast }X=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)=I$, $Y^{\ast }Y=\left(\begin{array}{cc}0& -i\\ i& 0\end{array}\right)\left(\begin{array}{cc}0& -i\\ i& 0\end{array}\right)=I$, and $Z^{\ast }Z=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)=I$. By [F4] each fixed Pauli matrix is unitary in the sense of [F3].

By [L1] and [F2], the displayed matrix for $H$ is self-adjoint, and direct multiplication gives $H^{\ast }H=H^2=\frac{1}{2}\left(\begin{array}{cc}1& 1\\ 1& -1\end{array}\right)\left(\begin{array}{cc}1& 1\\ 1& -1\end{array}\right)=I$. Thus $H$ is unitary by [F4] and [F3].

From [L1], $S^{\ast }=\mathrm{diag}(1,-i)$, so $S^{\ast }S=\mathrm{diag}(1,1)=I$. Also $T^{\ast }T=\mathrm{diag}(1,\mathrm{\mid }{e}^{i\pi \mathrm{/}4}{\mathrm{\mid }}^2)=I$ by [F5]. Hence $S,T$ are unitary by [F4] and [F3].

For $\mathbf{n}$ and $\theta$ as in the Given clause, $B=-i\theta (\mathbf{n}\cdot \mathit{\sigma })\mathrm{/}2$ is a linear endomorphism of $Q_2$. By [F6], $Q_2$ is Banach and $B$ is bounded, so [F7] makes the series defining $\exp (B)$ converge absolutely in operator norm. Thus $R_{\mathbf{n}}(\theta )$ is well-defined as a matrix.

For arbitrary real $\mathbf{r}$, the matrices in [F6] give $\rho (\mathbf{r})=\frac{1}{2}\left(\begin{array}{cc}1+r_z& r_x-i{r}_y\\ r_x+i{r}_y& 1-r_z\end{array}\right)$. Its diagonal entries are real, its lower-left entry is the conjugate of the upper-right, and its trace is $1$, so it is Hermitian trace one.

Write the arbitrary Hermitian trace-one matrix $\rho =\left(\begin{array}{cc}a& z\\ \bar{z}& 1-a\end{array}\right)$ with $a\in \mathbb{R}$: [F2] makes the computational basis orthonormal, and [F3]–[F4] give the conjugate-transpose matrix criterion for ${\rho }^{\ast }=\rho$, so its diagonal entries are real and its lower-left entry is the conjugate of the upper-right; trace one fixes the second diagonal entry. Put $r_x=2\mathrm{Re}z$, $r_y=-2\mathrm{Im}z$, and $r_z=2a-1$, which are real. The matrices in [F6] and formula in [F8] give $\rho (\mathbf{r})=\frac{1}{2}\left(\begin{array}{cc}1+r_z& r_x-i{r}_y\\ r_x+i{r}_y& 1-r_z\end{array}\right)=\left(\begin{array}{cc}a& z\\ \bar{z}& 1-a\end{array}\right)=\rho$. This proves existence.

For the entries $a,z$ of step 1.2, direct multiplication using [F6] and summing diagonal entries by [F9] gives $\mathrm{tr}(\rho X)=z+\bar{z}=2\mathrm{Re}z=r_x$, $\mathrm{tr}(\rho Y)=i(z-\bar{z})=-2\mathrm{Im}z=r_y$, and $\mathrm{tr}(\rho Z)=a-(1-a)=2a-1=r_z$. If $\rho =\rho (\mathbf{s})$ for another real vector $\mathbf{s}$, [F7] applied to [F8] gives $\mathrm{tr}(\rho {\sigma }_j)=s_j$ for each $j$, so $\mathbf{s}=\mathbf{r}$ and uniqueness follows. When $\rho$ is a density operator, [F5] makes it Hermitian and trace one, and [F8] identifies these same real trace coordinates with the map $\mathbf{r}(\rho )$.

By [F3], ${\mathrm{\Sigma }}^2={\sum }_j{n}_j^2{\sigma }_j^2+{\sum }_{j<k}{n}_j{n}_k({\sigma }_j{\sigma }_k+{\sigma }_k{\sigma }_j)=({\sum }_j{n}_j^2)I=I$. Also ${\mathrm{\Sigma }}^{\ast }=\mathrm{\Sigma }$, since the $n_j$ are real and every ${\sigma }_j$ is Hermitian.

Let $B=-i\theta \mathrm{\Sigma }\mathrm{/}2$. By [F1], $Q_2$ is finite-dimensional; [F9] makes it Banach and [F10] makes $B$ bounded, so [F6] makes ${\sum }_{k\ge 0}{B}^k\mathrm{/}k!$ absolutely convergent. From [step 1.1], induction gives $B^{2m}=(-1{)}^m(\theta \mathrm{/}2{)}^{2m}I$ and $B^{2m+1}=-i(-1{)}^m(\theta \mathrm{/}2{)}^{2m+1}\mathrm{\Sigma }$. Grouping its even and odd terms is valid by absolute convergence; [F4] identifies the two scalar series as $\cos (\theta \mathrm{/}2)$ and $\sin (\theta \mathrm{/}2)$. Thus $R_{\mathbf{n}}(\theta )=\cos (\theta \mathrm{/}2)I-i\sin (\theta \mathrm{/}2)\mathrm{\Sigma }$. When $\theta =0$, this is $I$.

Since ${\mathrm{\Sigma }}^{\ast }=\mathrm{\Sigma }$, the adjoint of the displayed formula is $\cos (\theta \mathrm{/}2)I+i\sin (\theta \mathrm{/}2)\mathrm{\Sigma }$. Multiplying this by $R_{\mathbf{n}}(\theta )$ and using ${\mathrm{\Sigma }}^2=I$ from step 1.1 gives $R_{\mathbf{n}}(\theta {)}^{\ast }{R}_{\mathbf{n}}(\theta )=({\cos }^2(\theta \mathrm{/}2)+{\sin }^2(\theta \mathrm{/}2))I=I$ by [F5]. Hence [F7] makes $R_{\mathbf{n}}(\theta )$ unitary in the sense of [F8].

Using [F1] and [F2], direct multiplication gives $HXH=\frac{1}{2}\left(\begin{array}{cc}1& 1\\ 1& -1\end{array}\right)\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)\left(\begin{array}{cc}1& 1\\ 1& -1\end{array}\right)=Z$, $HZH=\frac{1}{2}\left(\begin{array}{cc}1& 1\\ 1& -1\end{array}\right)\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)\left(\begin{array}{cc}1& 1\\ 1& -1\end{array}\right)=X$, and $HYH=\frac{1}{2}\left(\begin{array}{cc}1& 1\\ 1& -1\end{array}\right)\left(\begin{array}{cc}0& -i\\ i& 0\end{array}\right)\left(\begin{array}{cc}1& 1\\ 1& -1\end{array}\right)=-Y$. The products agree with the Pauli multiplication table in [F3], including the negative sign.

The first column of $H$ in [F1] gives $H\mathrm{\mid }0⟩=(\mathrm{\mid }0⟩+\mathrm{\mid }1⟩)\mathrm{/}\sqrt{2}=\mathrm{\mid }+⟩$, and its second gives $H\mathrm{\mid }1⟩=(\mathrm{\mid }0⟩-\mathrm{\mid }1⟩)\mathrm{/}\sqrt{2}=\mathrm{\mid }-⟩$. By [F2], $X\mathrm{\mid }+⟩=\mathrm{\mid }+⟩$ and $X\mathrm{\mid }-⟩=-\mathrm{\mid }-⟩$. Expanding with [F4] gives $⟨+,+{⟩}_{\mathrm{r}\mathrm{o}\mathrm{o}\mathrm{t}}=\frac{1}{2}(1+0+0+1)=1$, $⟨-,-{⟩}_{\mathrm{r}\mathrm{o}\mathrm{o}\mathrm{t}}=\frac{1}{2}(1-0-0+1)=1$, and $⟨+,-{⟩}_{\mathrm{r}\mathrm{o}\mathrm{o}\mathrm{t}}=\frac{1}{2}(1-0+0-1)=0$. Thus these vectors form the $X$-eigenbasis.

Put $t=\mathrm{\parallel }\mathbf{r}{\mathrm{\parallel }}_2$ and $\mathrm{\Sigma }=\mathbf{r}\cdot \mathit{\sigma }$. By [F3], ${\mathrm{\Sigma }}^{\ast }=\mathrm{\Sigma }$ and ${\mathrm{\Sigma }}^2={\sum }_j{r}_j^2{\sigma }_j^2+{\sum }_{j<k}{r}_j{r}_k({\sigma }_j{\sigma }_k+{\sigma }_k{\sigma }_j)=t^{2}I$. Using the displayed matrices in [F4], $\mathrm{\Sigma }=\left(\begin{array}{cc}{r}_z& r_x-i{r}_y\\ r_x+i{r}_y& -r_z\end{array}\right)$, so $\mathrm{tr}(\mathrm{\Sigma })=0$ and $\det (\mathrm{\Sigma })=-t^2$. Thus its characteristic polynomial is ${\lambda }^2-t^2$. If $t>0$, its two eigenvalues are $t$ and $-t$; if $t=0$, $\mathrm{\Sigma }=0$. Therefore $\rho (\mathbf{r})=\frac{1}{2}(I+\mathrm{\Sigma })$ has eigenvalues $(1+t)\mathrm{/}2$ and $(1-t)\mathrm{/}2$, both $1\mathrm{/}2$ at the zero vector.

Since [F2] gives trace one, [F5]–[F7] show that $\rho (\mathbf{r})$ is a density operator exactly when both eigenvalues are nonnegative. The eigenvalue $(1+t)\mathrm{/}2$ is always positive; $(1-t)\mathrm{/}2\ge 0$ exactly when $t\le 1$. At $t=1$ the zero eigenvalue is allowed by non-negativity; at $t=0$ both eigenvalues are $1\mathrm{/}2$. Hence the condition is precisely $\mathrm{\parallel }\mathbf{r}{\mathrm{\parallel }}_2\le 1$.

Let $\mathrm{\Sigma }=\mathbf{r}\cdot \mathit{\sigma }$. From [F3] and [F4], $\mathrm{tr}({\rho }^2)=\frac{1}{4}\mathrm{tr}(I+2\mathrm{\Sigma }+{\mathrm{\Sigma }}^2)=\frac{1}{4}(2+2\mathrm{\parallel }\mathbf{r}{\mathrm{\parallel }}_2^2)=\frac{1}{2}(1+\mathrm{\parallel }\mathbf{r}{\mathrm{\parallel }}_2^2)$, because ${\mathrm{\Sigma }}^2=\mathrm{\parallel }\mathbf{r}{\mathrm{\parallel }}_2^{2}I$ by the Pauli anticommutation relation in [F4].

Let ${\lambda }_{+},{\lambda }_{-}$ be the eigenvalues from [F5]. They are nonnegative and sum to one by [F1]. Then $\mathrm{tr}({\rho }^2)={\lambda }_{+}^2+{\lambda }_{-}^2=1$ exactly when $2{\lambda }_{+}{\lambda }_{-}=0$, equivalently the eigenvalue pair is $(1,0)$. By [F2], $\rho$ has an orthonormal eigenbasis; in this case it is the orthogonal projection onto its one-dimensional $1$-eigenspace, so surjectivity in [F6] makes it the projector of a pure-state ray. Conversely, if $\rho$ is pure, [F7] gives a unit representative $u$ with $\rho =P_u$, and [F6] makes $P_u$ an orthogonal projection, so ${\rho }^2=\rho$. By [F1], $\mathrm{tr}(\rho )=1$, hence $\mathrm{tr}({\rho }^2)=1$. Thus purity is equivalent to $\mathrm{tr}({\rho }^2)=1$; step 1.1 makes this equivalent to $\mathrm{\parallel }\mathbf{r}{\mathrm{\parallel }}_2=1$.