Qubits & Oscillators | Zelong's Wiki

Pauli Operators

Use $\sigma_1=\sigma_x$ , $\sigma_2=\sigma_y$ , and $\sigma_3=\sigma_z$ :

\sigma_x= \begin{pmatrix}0&1\\1&0\end{pmatrix},\quad \sigma_y= \begin{pmatrix}0&-i\\i&0\end{pmatrix},\quad \sigma_z= \begin{pmatrix}1&0\\0&-1\end{pmatrix}.

The product, commutator, and anticommutator identities are

\sigma_i\sigma_j=\delta_{ij}I+i\epsilon_{ijk}\sigma_k,\qquad [\sigma_i,\sigma_j]=2i\epsilon_{ijk}\sigma_k,\qquad \{\sigma_i,\sigma_j\}=2\delta_{ij}I .

For a single-qubit pulse about axis $k$ with rotation generator $\hat S_k$ ,

R_{\theta,k}=e^{-i\sigma_k\theta/2}=\cos\frac{\theta}{2}I-i\sin\frac{\theta}{2}\sigma_k, \qquad \hat S_k=\frac{\sigma_k}{2}.

Examples:

R_{\pi,x}=-i\sigma_x,\qquad R_{\pi/2,x}=\frac{\sqrt2}{2}(I-i\sigma_x).

For $j\ne k$ , Pauli rotations transform as

R_{\theta,k}\sigma_j R_{\theta,k}^\dagger = \cos\theta\,\sigma_j+\epsilon_{kji}\sin\theta\,\sigma_i .

For the canonical Hamiltonian

H=\frac{A}{2}p^2+\frac{B}{2}x^2,\qquad [x,p]=i\hbar,

write

x=x_0(a+a^\dagger),\qquad p=\frac{p_0}{i}(a-a^\dagger),

so that

H=\hbar\omega\left(a^\dagger a+\frac12\right),

with

\omega=\sqrt{AB},\qquad x_0=\sqrt{\frac{\hbar}{2}\sqrt{\frac{A}{B}}}, \qquad p_0=\sqrt{\frac{\hbar}{2}\sqrt{\frac{B}{A}}}.

Since $\sigma_k^2=I$ ,

e^{-i\sigma_k\theta/2} =\sum_{n\ {\rm even}}\frac{(-i\theta/2)^n}{n!}I +\sum_{n\ {\rm odd}}\frac{(-i\theta/2)^n}{n!}\sigma_k,

which separates into the sine and cosine expression for $R_{\theta,k}$ .