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So I have this question: Given an operator, find some Hamiltonian implementing this operator/gate. I have realized that this is a swap gate and I know the matrix for it. I also know that $U = \text{exp}(-iHt)$-- in this case, $U = G$. I also know that $\text{exp}(iA) = I + A + \frac{A^2}{2}! + \frac{A^3}{3}! + \cdots$ I also know that a $CNOT=∣0⟩⟨0∣⊗I+∣1⟩⟨1∣⊗X$ and that $G$ operator is composed

My questions/confusions are:

I know $H$ will be some combination of tensor product of pauli matrices--is there a direct way to do this/ calculate it or is this just a case by case basis--observing things like that $XX$ swaps the states $∣01⟩$ and $∣10⟩$ and $ZZ$ leaves $∣00⟩$ and $∣11⟩$ unchanged but adds a negative phase to $∣01⟩$ and $∣10⟩$... etc.? I guess related to this is, given some matrix--is there a specific process to follow to translate it from a matrix representation to a tensor-product of pauli matrices representation? I need help with this question.[question in image: Consider operator G acting on Hilbert space x Hilbert_space as follows: $∣\phi⟩ ∣\psi⟩ \to ∣\psi⟩ ∣\phi⟩$ for all $\psi$, $\phi$ in $H$. Find some Hamiltonian implementing the $G$ gate if $G$ is linear and unitary. 1 .

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There are many ways that you could go about doing this. (You might also want to take a look at what you"know" about exponentiating a matrix, because you've exponentiated $A$ not $iA$.)

One method, whose usage is a bit limited, but works very well in this case, is to think about any special cases of matrix exponentiation you might know about. There's a common identity for $e^{iAt}$ where $A^2=I$ which you might be able to reverse engineer to find something useful...

More generally, you might find the exponential easier to think about in terms of the spectral decomposition: $$ e^{iA}=\sum_ne^{i\lambda_n}|\lambda_n\rangle\langle\lambda_n| $$ where $A=\sum_n\lambda_n|\lambda_n\rangle\langle\lambda_n|$. So, given the unitary, you can figure out some properties of the eigenvectors and eigenvalues of $A$.

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  • $\begingroup$ hmm... snarky but thanks. $\endgroup$
    – George
    Jan 31 at 14:40
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    $\begingroup$ not intended to be snarky, but trying to walk the line of not saying too much given that this is probably (or at least could be) a homework problem, so I shouldn't just be giving you the answers. $\endgroup$
    – DaftWullie
    Jan 31 at 15:44

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