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1

First of all, let us compute the probability of success of this algorithm. If you are given the state $|\psi\rangle\langle\psi|\otimes|\psi\rangle\langle\psi|$, the SWAP test will return the state $|0\rangle$ with probability $1$, which is the probability of success of the algorithm in this case. Let us now consider the second case. The initial state is: $$\...


2

I will suppose that you meant: $H_{\rm int}=g(X_1X_2+Y_1Y_2+Z_1Z_2)$ (it is unclear as you did not define your notation). If you want to calculate: $$U(t)=\exp[-i g t (X_1X_2+Y_1Y_2+Z_1Z_2)],$$ in that case note that $[X_1X_2,Y_1Y_2]=[Y_1Y_2,Z_1Z_2]=[X_1X_2,Z_1Z_2]=0$, so we can separate the exponential as $$U(t)=\exp[-i g t X_1X_2]\exp[-i g t Y_1Y_2]\exp[-i ...


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....


1

The spaces spanned by the different ladder operators commute, so the problem is reduced to showing the desired property for a Hamiltonian of the form $H=a^\dagger a$ for some ladder operator $a$ (this can be $a_k$ for any $k$). You then remember that $a^\dagger a$ is diagonal in the number basis, and thus $f(a^\dagger a)$ is also diagonal in the same basis, ...


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