# How to perform Fourier transform of this Hamiltonian?

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

$$\tilde{\mathcal{H}}=\int dt H$$
This would in your last computation line induce a factor Dirac delta $$2\pi\delta(\omega-\omega'-\Omega)$$and simplify your last expression.