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Below is the cost Hamiltonian for an unweighted max-cut problem, I don't understand what the point of the half coefficient is. Why couldn't we omit it?

$C_\alpha = \frac{1}{2}\left(1-\sigma_{z}^j\sigma_{z}^k\right),$

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  • $\begingroup$ I imagine it's because it scales the range of $C_{\alpha}$ to between 0 and 1. $\endgroup$
    – DaftWullie
    Feb 5 at 13:54
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You are correct that a constant factor in the cost Hamiltonian isn't relevant. This factor is included so that the eigenvalues of the cost Hamiltonian can be interpreted in terms of the number of cuts in the graph. In this simple case of a two-node graph, the eigenvalues are 0 or 1 when the factor of 1/2 is included, which correspond to 0 or 1 cuts.

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  • $\begingroup$ Wait so it isn't always the half coefficient? $\endgroup$ Feb 5 at 15:20
  • $\begingroup$ How are the eigenvalues 0 or 1 when the coefficient of 1/2 is included? What does that mean? $\endgroup$ Feb 5 at 15:21

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