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I am reading this article (arXiv:1505.01908 ) in which author is calculating linear response of a perturbation. The perturbation Hamiltonian is $H$ (Eq. 2 of article) given as $$ H = \frac{JS}{a^3}\int d^3r \bigg( [\frac{\partial}{\partial t}b^\dagger(r,t) ]\nabla b(r,t) + \nabla b^\dagger(r,t) [\frac{\partial}{\partial t} b(r,t) ] \bigg) \mathbf{A} (t) $$ here $b(r,t)$ are field operators and $\mathbf{A}$ is a vector field. The author define Fourier transform of field operators as $$ b(r,t)=\sqrt{\frac{a^3}{V}}\int \frac{d\omega}{2\pi} \sum_q e^{i(rq-\omega t)} \; b_{q,\omega} $$ he assumes that $\Omega$ is an infinitesimal external angular frequency, and then he gets final result in Fourier representation as (Eq 42):

$$ H =-\frac{2JS}{V}\int \frac{d\omega}{2\pi} \int \frac{d\Omega}{2\pi} \sum_q e^{i\Omega t} q\omega A(-\Omega) b^\dagger_{q,\omega+\frac{\Omega}{2}} b_{q,\omega-\frac{\Omega}{2}} $$

I am trying to understand how exactly he reaches to this result. My attempt is given below:

My Attempt

After putting above definition of Fourier representation in $H$, we get $$ H = -\frac{ JS}{V} \int\frac{d\omega}{2\pi} \int \frac{d\omega'}{2\pi}\sum_{q,q'} \int d^3r \bigg( \omega q' + q \omega' \bigg) e^{ir(q'-q)} e^{it(\omega -\omega')} b^\dagger_{q,\omega} b_{q',\omega'} \mathbf{A} (t) $$ Use $\int d^3r e^{irk}=\delta_{k,0}$, and as the field $\mathbf{A(t)}$ also depend upon time, it should also be represented in Fourier space, I defined it as $\mathbf{A(t)}=\int \frac{d\Omega}{2\pi}e^{-i\Omega t}A(\Omega)$,

$$ H = -\frac{ JS}{V} \int\frac{d\omega}{2\pi} \int \frac{d\omega'}{2\pi} \sum_{q} q\big( \omega + \omega' \big) e^{it(\omega -\omega')} b^\dagger_{q,\omega} b_{q,\omega'} \int\frac{d\omega''}{2\pi}e^{-i\omega'' t}A(\omega'') \\ H = -\frac{ JS}{V} \int\frac{d\omega}{2\pi} \int \frac{d\omega'}{2\pi}\int\frac{d\Omega}{2\pi}\sum_{q} q\big( \omega + \omega' \big) e^{it(\omega -\omega'-\Omega)} A(\Omega) b^\dagger_{q,\omega} b_{q,\omega'} $$ This looks almost like the one given by author of that article. But it has an extra integration over frequency and I don't know how do we get $\Omega/2$ terms with operators $b_{q,\omega}$. Any comment or suggestion will be highly appreciated

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from my understanding of the article and of the calculation you are doing above, the difference is that there is an integration difference between you definition of the hamiltonian and his own

$\tilde{\mathcal{H}}=\int dt H$

This would in your last computation line induce a factor Dirac delta $2\pi\delta(\omega-\omega'-\Omega)$and simplify your last expression.

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