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Can we always find a set of coefficients ${k_i}$ (where not every $k_i = 0$) for a given Hamiltonian $H = \sum k_i H_i$, such that the unitary operator becomes the identity operation: $e^{-iH} = e^{i\alpha}I$, where $\alpha$ is a real phase? The $H_i$'s are given arbitrary Hermitian matrices that may or may not commute with each other.

For instance, when $H = k_1\sigma_x$, any $k_1 = n \frac{\pi}{2}, n = 1,2,3,\dotsc$ would make $e^{-iH} \propto I$. Similarly, for $H = k_1\sigma_x + k_2 \sigma_z$, one can find suitable $k_1$ and $k_2$ since $(\sigma_x + \sigma_z)^2 \propto I$. The question is whether this can be generalized to any arbitrary Hamiltonian, where $H_i$ are arbitrary Hermitian matrices. My conjecture is that this is always possible, but I'm not sure how to rigorously prove it.

Furthermore, as an extension of the question, does the property still hold when $H = H_0 + \sum k_i H_i$, where $H_0$ now has a fixed coefficient of $1$? Unlike my original question stated above, I suspect that the answer for this question is no.

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    $\begingroup$ I’m not sure if it matters, but do you put any bounds on $i$ and also the number of qubits that each term acts on? For example are you local, with each $H_i$ only acting on a subset of the $n$ qubits? $\endgroup$ Aug 2, 2023 at 21:55
  • $\begingroup$ @MarkSpinelli I'm not assuming any locality on $H_i$, i.e., $H_i$ can be arbitrary Hermitian (and perhaps traceless, if needed) operators. For instance, $H(t) = k_1\sigma_Z^{(1)}\sigma_Z^{(10)} + k_2 \sigma_X^{(4)}\sigma_X^{(6)}\sigma_X^{(9)}$ where $^{(i)}$ indicates the index of qubit that the operator is acting on. $\endgroup$
    – Hailey Han
    Aug 2, 2023 at 22:29

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I don't believe that this is always possible. For instance, what if my set of $\{H_i\}$s comprise a single term that I can construct to be arbitrarily awkward? The key feature will be gaps between eigenvalues. If I pick two gaps $\Delta_i$ that are irrational multiples of each other, it will be impossible to find a time $t$ such that $\Delta_i t\equiv 0\text{ mod }2\pi$ for both $i$ simultaneously. For example, why not try: $$ H=0|0\rangle\langle 0|+|1\rangle\langle 1|+\sqrt{2}|2\rangle\langle 2|. $$ The time evolution is now $$ e^{-iHt}=|0\rangle\langle 0|+e^{it}|1\rangle\langle 1|+e^{i\sqrt{2}t}|2\rangle\langle 2|. $$ Thus, at and time $t=2\pi n$, the $|1\rangle$ term has the appropriate phase, but the $|2\rangle$ term cannot (although it can be arbitrarily close).

I'm sure one could now construct more complex examples with multiple terms. Analysis is easier if you let them all commute.

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    $\begingroup$ Thanks for the wonderful answer. I now agree with you that this isn't always possible. But as you pointed out, I believe it can be arbitrarily close to $I$. I have been thinking about it yesterday, and from the compactness of unitary group it immediately follows that there always exist a positive integer $k$ such that $\vert\vert U^k - I \vert\vert < \epsilon$ for any $\epsilon >0$. $\endgroup$
    – Hailey Han
    Aug 3, 2023 at 19:20
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    $\begingroup$ This is (essentially) the Quantum Recurrence Theorem, journals.aps.org/pr/abstract/10.1103/PhysRev.107.337 (admittedly, they talk about it in the context of a specific state recurring rather than a whole unitary). The main concern is how long it might take (often, we're talking age of the universe scales!) $\endgroup$
    – DaftWullie
    Aug 4, 2023 at 6:54

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