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I've been reading a number of papers about approximate algorithms and they mention that when the standard Ising Hamiltonian of the form

$$ H_{c}=\sum_{} c_{i} Z_{i} +\sum_{} J_{i j} Z_{i} Z_{j}, $$

has integer eigenvalues, the value of $$ F_{p}(\boldsymbol{\gamma}, {\boldsymbol{\beta}})= \left\langle\psi_{p}(\boldsymbol{\gamma}, {\boldsymbol{\beta}})\left|H_{C}\right| \psi_{p}(\boldsymbol{\gamma}, {\boldsymbol{\beta}})\right\rangle, $$

where, $$ |\psi_{p}(\boldsymbol{\gamma}, \boldsymbol{\beta})\rangle=U_{B}\left(\beta_{p}\right) U_{C}\left(\gamma_{p}\right) \cdots U_{B}\left(\beta_{1}\right) U_{C}\left(\gamma_{1}\right)|+\rangle^{\otimes N} $$

is periodic, but the value when $H_{c}$ is not integer eigenvalued isn't: why is this ?


Cross-posted on physics.SE

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    $\begingroup$ could you link the paper that these equations come from? $\endgroup$
    – rjh324
    Apr 25 at 22:08
  • $\begingroup$ What are $U_B$ and $U_C$? $\endgroup$ Apr 27 at 16:37
  • $\begingroup$ To confirm: when eigenvalues are non-integer rational numbers then the claim is that $F_p$ is not periodic? $\endgroup$ Apr 27 at 16:40
  • $\begingroup$ Please refrain from vandalising your own post. If for whatever reason you need to hide some information, like references to papers, you can just delete the question altogether $\endgroup$
    – glS
    May 11 at 16:30
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If you diagonalize $ H_c = \sum \lambda_i |u_i \rangle \langle u_i| $ and the eigenvalues $\lambda_i$ are integers, then: $$ e^{- i (\gamma + 2\pi) H_c} = \sum e^{- i (\gamma + 2\pi) \lambda_i} |u_i \rangle \langle u_i| = \sum e^{- i \gamma \lambda_i} |u_i \rangle \langle u_i| = e^{- i \gamma H_c} $$

since $ e^{- i 2\pi k} = 1, \forall k \text{ integer}$.

This means $ |\psi_p(\gamma, \beta) \rangle = |\psi_p(\gamma + 2\pi, \beta) \rangle \implies F_p(\gamma, \beta) = (\gamma + 2\pi, \beta)$.

A similar argument holds for $ \beta $ if the mixer Hamiltonian $H_B$ has integer eigenvalues.

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  • $\begingroup$ What about the other direction? Also, the claim appears to be that if $\lambda_i$ are non-integer rational numbers then $F_p$ is not periodic, but this proof suggests otherwise (TBH, my bet is your proof is correct and the claim is incorrect, but the question isn't clear enough to say with certainty). $\endgroup$ Apr 27 at 16:40
  • $\begingroup$ @AdamZalcman I don't think you can make this claim. If you take $ a H_c $ where $ a $ non-integer and $ H_c $ has integer eigenvalues, you have non-integer eigenvalues and a periodic $ F_p $ $\endgroup$
    – tsgeorgios
    Apr 27 at 17:29
  • $\begingroup$ I agree, yet AFAICT the question implies that claim. $\endgroup$ Apr 27 at 17:35

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