Let's say you have a Hamiltonian with $k$ different terms, \begin{equation} \hat{H}=\sum_{i=1}^k a_k \hat{O}_k \end{equation} with real coefficients $a_k$ and Hermitian operators $\hat{O}_k$.

Now, if there is another Hamiltonian $\hat{H}^{\prime}$ that acts on the same Hilbert space with $k$ operators that have exactly the same commutation/anticommutation relations as $\hat{H}$, is it always the case that there is a unitary transformation $U$ that connects those two? i.e. \begin{equation} H^{\prime}=U H U^{\dagger}. \end{equation}

I was thinking that this should be true even with restricting $U$ to be a Clifford transformation when all $\hat{O}_k$ are (multi-qubit) Pauli operators. However, I found an exception to that, and I got confused.


The "exception" I thought I found turned out to be a mistake. Now I understand that whenever the above assumption is true, there should be a Clifford unitary that corresponds to the transformation. Thanks people!

  • 1
    $\begingroup$ Similarity transformations don't change the spectrum. $\endgroup$
    – jacob1729
    Commented Feb 21, 2022 at 12:17
  • $\begingroup$ Thanks for your comment! I understand that, but I'm kind of asking the reverse, and also am not exactly asking about the spectrum, although I think it's also an equally interesting question. So, when you have two Hamiltonians with exactly the same spectrum, is there always a similarity transformation connecting them? My original question was: given that the commutation relations of all terms are exactly the same, is there always a similarity transformation? $\endgroup$
    – Gitef
    Commented Feb 22, 2022 at 2:45


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