# How flexible can a unitary transformation be?

Let's say you have a Hamiltonian with $$k$$ different terms, $$$$\hat{H}=\sum_{i=1}^k a_k \hat{O}_k$$$$ with real coefficients $$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. $$$$H^{\prime}=U H U^{\dagger}.$$$$

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.

(Edit)

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!

• Similarity transformations don't change the spectrum. Commented Feb 21, 2022 at 12:17
• 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? Commented Feb 22, 2022 at 2:45