In phase estimation algorithms, we have $U|\psi\rangle = e^{2\pi i\theta}|\psi\rangle$, where $|\psi\rangle$ is an eigenvector and $ e^{2\pi i\theta}$ is the corresponding eigenvalue. Since $U$ represents the time-evolution of the quantum system, suppose the original Hamiltonian has an eigenvalue $E_1$, then after certain time-evolution, the value becomes $e^{-iE_1t}$, I wonder to simulate the time-evolution in the algorithm, how can we determine the time $t$?

Should we let $-iE_1t = 2\pi i\theta$ to solve for $t$? Also, given that $\theta\in[0,1]$, to make phase estimation less confusing, can we scale the Hamiltonian being simulated to make its eigenvalues within $[0,2\pi]$?


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    $\begingroup$ See Appendix F in "Elucidating Reaction Mechanisms on Quantum Computers", 2016. arxiv.org/abs/1605.03590 $\endgroup$
    – Ron Cohen
    Feb 28, 2022 at 9:10

1 Answer 1


In general, you're doing phase estimation because you don't know the eigenvalue $E_1$. So, while you might ideally like to choose a specific value of $\theta$ and rearrange for $t$, because you don't know $E_1$, you cannot.

Instead, you need to try and determine some basic properties of your Hamiltonian $H$. For example, if you can bound all the energies $E$ to be between $0\leq E\leq E_\max$, then you can choose $tE_{\max}=2\pi$ which will guarantee that whatever the corresponding value of $\theta$ is, it certainly falls in the range 0 to $2\pi$.

Note that if your lower bound is non-zero, you can always change $H\rightarrow H-E_{\min}I$ and it'll work just the same.

Yes, you could certainly choose to rescale $H$ to incorporate the factor of $2\pi$ if you wanted to.

  • $\begingroup$ Thanks so much for the answer! If we set $tE_{max}=2\pi$, are we assuming we know the maximum energy value? If so, can I let $t = 2\pi/E_{max}$ to run the time-evolution? $\endgroup$
    – IGY
    Feb 28, 2022 at 16:07
  • $\begingroup$ Yes, exactly. The challenge is to find out $E_{\max}$ or at least a reasonable upper bound on it. This can often be inferred from some properties of your Hamiltonian. $\endgroup$
    – DaftWullie
    Mar 1, 2022 at 7:41

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