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4

There is a mistake, we have $$\text{Tr}(\sigma_{i_1}\sigma_{j_1}\otimes\sigma_{i_2}\sigma_{j_2}\otimes\sigma_{i_3}\sigma_{j_3}) = \delta_{i_1,j_1}\delta_{i_2,j_2}\delta_{i_3,j_3}\text{Tr}(I) = 8\delta_{i_1,j_1}\delta_{i_2,j_2}\delta_{i_3,j_3}.$$ So that $\sum_{i_1,i_2,i_3}|u_{i_1,i_2,i_3}|^2=1$. The same is true for every $n$. The other equalities that we ...


1

If you set $P_y(t)=0$ for a particular length of time, then you just get an $X$ rotation. Similarly, if you set $P_x(t)=0$ for a certain length of time, you just get a $Y$ rotation. So, you do a sequence that looks something like \begin{align*} P_X(t)=\left\{\begin{array}{cc} J & 0\leq t\leq \frac{d}{2J} \\ 0 & 0<t-\frac{d}{2J}\leq \frac{c}{2J} \\ ...


5

What the author wrote is completely correct, they did not make a mistake. The subgroup of Cliffords fixing $X_n$ and $Z_n$ is indeed isomorphic to $C_{n-1}$ as a group, this is simply because this subgroup acts by assumption as $$ U (\sigma_1 \otimes \dots \otimes \sigma_n) U^\dagger = \tilde U (\sigma_1\otimes\dots\otimes\sigma_{n-1})\tilde U^\dagger \...


2

A Clifford $C_n$, defined by how it maps each of $X_i$ and $Z_i$ for $1 \leq i \leq n$, via the functions $g_i(\sigma_i)$ where $$\sigma_i = \{\pm I_i, \pm X_i, \pm Y_i, \pm Z_i\},$$ can be seen as the operation $g_1(\sigma_1) \cdot g_2(\sigma_2) \cdots \cdot g_n(\sigma_n)$ that acts on any arbitrary Pauli, $$P = \sigma_1 \cdot \sigma_2 \cdots \cdot \sigma_n....


2

I think the confusion arises from the fact we usually measure the energy of the spins in a magnetic field. In that case, the energy is given by a Hamiltonian $$ H=-\vec{\mu}\vec{B} $$ where $\vec{\mu} = -1/2g\mu_B\vec{\sigma}\approx \mu_B\vec{\sigma}$ and $\mu_B$ is a positive constant called Bohr magneton. When B is only in the $z$-direction, it becomes $$ ...


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