At every cutoff, the free-field generator applied to the Gaussian penalty is exactly minus the cost strength times the cutoff Wick square; consequently each cutoff operator equals the discount term minus the generator applied to the test function minus the penalty, plus half the gradient energy, minus the running cost, up to the cutoff sum of the difference between the free-field variances and the counterterm.
In the setting of The Wick-Square Problem on the Torus: Standing Notation, let be the Gaussian penalty, the free-field variances and the cutoff Wick square; mode derivatives are those of Derivatives of a Function on the Sobolev Space of Order -1 along the Fourier Modes §derivatives, and and are the free-field generator and gradient energy at cutoff . Then the following hold.
1. (Mode derivatives of the penalty) is twice differentiable along the modes, and for every and
2. (The corrector identity) For every and ,
3. (Penalised form of the cutoff operators) Let and , let and be the cutoff operators with counterterm and the Wick-ordered cutoff operators for , and let be twice differentiable along the modes. Then is twice differentiable along the modes, and for every and
in particular
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