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 .


1 Answer 1


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 $A$.

  • $\begingroup$ hmm... snarky but thanks. $\endgroup$
    – George
    Jan 31 at 14:40
  • 4
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.