Fourier transform of the Hamiltonian of harmonic oscillator from $k$-space to $x$-space

Question

Conventionally, I have the sum of an infinite number of harmonic oscillators with the free Hamiltonian $H_{0}$ to be $$ \begin{equation} H_{0} = \sum_{k}\hbar\omega_{k}\left(a_{k}^{\dagger}a_{k} + \frac{1}{2}\right) \tag{1} \end{equation} $$ where $a_{k}$ represents the annihilation operator in the $k$-space. I wish to obtain a field theoretic representation of $a_{k}$ and transforming it to a field operator, namely, $a(x)$. In particular my transformation should lead to $$ H_{0} = \int dx\left\{a^{\dagger}(x)(-i\hbar v\frac{\partial}{\partial x})a(x)\right\}\,\,\,\,\,\,\,(2) $$ where $v$ is the phase velocity of the the oscillator. Here are my initial steps: I have the definition of the FT of $a_{k}$ to be $$ a_{k} = \frac{1}{\sqrt{L}}\int dx e^{-ikx}a(x)\,\,\,\,\,\,\,(3) \\ a_{k}^{\dagger} = \frac{1}{\sqrt{L}}\int dx e^{-ikx}a^{\dagger}(x)\,\,\,\,\,\,\,(4) $$ where $L$ is the normalization length (1 dimensional). I know that going from summation in $k$ to an integral is given as $$ \sum_{k} \longrightarrow \frac{L}{2\pi}\int dk\,\,\,\,\,\,\,(5) $$ Using Eqs. (3)-(5) and substituting into Eq. (1), I have $$ H_{0} = \frac{L}{2\pi}\int dk\,\hbar\omega_{k}\left(\frac{1}{\sqrt{L}}\int dx\,e^{-ikx}a^{\dagger}(x)\right)\left(\frac{1}{\sqrt{L}}\int dx^{\prime}\,e^{-ikx^{\prime}}a^{\dagger}(x^{\prime})\right) \\ = \frac{\hbar}{2\pi}\int dk\,\omega_{k}\int\int dx\,dx^{\prime}\left(e^{-ik(x+x^{\prime})}a^{\dagger}(x)a(x^{\prime})\right)\,\,\,\,\,\,\,(6) $$ At this point, I know that I could use the definition of dirac delta. Since $$ \delta(x+x^{\prime}) = \frac{1}{2\pi}\int dk\,e^{ik(x+x^{\prime})} \,\,\,\,\,\,\,(7) $$ but what of $\omega_{k}$? If I blindly substitute my definition for the dirac delta, I have $$ H_{0}=\hbar\int\int\omega_{k}a^{\dagger}(x)a(x^{\prime})\delta(-x-x^{\prime})dx\,dx^{\prime} \\ = \hbar\omega_{k}\int dx\,a^{\dagger}(x)a(-x) \,\,\,\,\,\,\,(8) $$ which I absolutely do not believe, since $\omega_k$ should have been absorbed into the $dk$ integral. How should I proceed in general?

Edit: Looking at Eq.(2), it contains $-i\hbar\partial_{x}$ which suggests that the momentum operator is at play somehow. I know that $\omega_{k} = vk$, so if $k$ is an operator, I would be able to act on $a^{\dagger}(x)a(x^{\prime})$ somehow to give me the form of Eq.(2), but $\omega_k$ is scalar in this case so I'm not sure how to proceed.

Edit 2: As suggested by @Heidar, I should start with the IFT of Eq.(3). As such I have that $$ a(x) = \frac{1}{\sqrt{L}}\int dk\,e^{ikx}a_{k} $$ Taking $\partial_{x}$ $$ \partial_{x}a(x) = \frac{i}{\sqrt{L}}\int dk\,ke^{ikx}a_{k} \\ -i\partial_{x}a(x) = \frac{1}{\sqrt{L}}\int dk\,ke^{ikx}a_{k} $$ at this point, I'm not sure of how i should get rid of the $k$ so that I have something like $-i\partial_{x}a(x) = ka_{k}$. Any help is appreciated.

Edit: Looking at this problem again after several months. I noticed that there was a mistake in @Heidar 's answer. In particular, the equality $$ [\partial_x a^\dagger(x)]^\dagger=\partial_{x}a(-x) $$ is not correct. If $$ a(x)\propto \int dk \exp(ikx)a_k \\ a^\dagger(x)\propto \int dk \exp(ikx)a_k^\dagger $$ then $$ \partial_x a(x)\propto \int dk \exp(ikx)ik a_k \\ \partial_x a^\dagger(x)\propto \int dk \exp(ikx)ik a_k^\dagger $$ But $$ [\partial_x a^\dagger(x)]^{\dagger}\propto \int dk \exp(ikx)(-ika_k) $$ and $$ \partial_x a(-x)\propto \int dk \exp(-ikx)ik a_k $$ and the right hand side of the last two equations are not equivalent. How should one correct for this? Or am I missing something here?

Answer 1

I'll start by changing the phases: $$ a_{k} = \frac{1}{\sqrt{L}}\int dx e^{ikx}a(x)\,\,\,\,\,\,\,(3) \\ a_{k}^{\dagger} = \frac{1}{\sqrt{L}}\int dx e^{-ikx}a^{\dagger}(x)\,\,\,\,\,\,\,(4). $$ We can continue to find $$ H_{0} = \frac{L}{2\pi}\int dk\,\hbar\omega_{k}\left(\frac{1}{\sqrt{L}}\int dx\,e^{-ikx}a^{\dagger}(x)\right)\left(\frac{1}{\sqrt{L}}\int dx^{\prime}\,e^{ikx^{\prime}}a(x^{\prime})\right) \\ = \frac{\hbar}{2\pi}\int dk\,\omega_{k}\int\int dx\,dx^{\prime}\left(e^{-ik(x-x^{\prime})}a^{\dagger}(x)a(x^{\prime})\right)\,\,\,\,\,\,\,(6). $$ We then use your property $\omega_{k} = vk$ to write $$H_{0} = \frac{\hbar}{2\pi}\int dk\,\int\int dx\,dx^{\prime} \left(\left[iv\partial_xe^{-ik(x-x^{\prime})}\right]a^{\dagger}(x)a(x^{\prime})\right).$$ Then we pull out the derivative and rearrange the order of the integrals to first integrate over $k$, which as you point out looks like a delta function $$ \delta(x-x^{\prime}) = \frac{1}{2\pi}\int dk\,e^{ik(x-x^{\prime})} \,\,\,\,\,\,\,(7). $$ So $$H_{0} = \hbar\int\int dx\,dx^{\prime} \left(\left[iv\partial_x \int dk\,e^{-ik(x-x^{\prime})}\right]a^{\dagger}(x)a(x^{\prime})\right)\\ = \hbar\int\int dx\,dx^{\prime} \left(\left[iv\partial_x \delta(x-x^{\prime})\right]a^{\dagger}(x)a(x^{\prime})\right).$$ We integrate by parts or use properties of derivatives of delta functions to do the integral: $$H_{0} = - \hbar\int\int dx\,dx^{\prime} \left(\left[iv \delta(x-x^{\prime})\right]\partial_x a^{\dagger}(x)a(x^{\prime})\right).$$ Then we just integrate over $x^\prime$ using the delta function to finally get $$H_{0} = \hbar\int dx \left(-i\hbar v\partial_x a^{\dagger}(x)\right)a(x).$$

Now of course your equation wanted the derivative to be on the second term, so we can integrate by parts again $$H_{0} = - \hbar\int dx a^{\dagger}(x)\left(-i\hbar v\partial_x a(x)\right).$$ And that's it! The difference in sign might be an error of mine that you can track down or may be related to how we defined our phases.

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Or we can use OP's original phase conventions: $$ a_{k} = \frac{1}{\sqrt{L}}\int dx e^{-ikx}a(x)\,\,\,\,\,\,\,(3) \\ a_{k}^{\dagger} = \frac{1}{\sqrt{L}}\int dx e^{-ikx}a^{\dagger}(x)\,\,\,\,\,\,\,(4). $$ We can continue to find $$ H_{0} = \frac{L}{2\pi}\int dk\,\hbar\omega_{k}\left(\frac{1}{\sqrt{L}}\int dx\,e^{-ikx}a^{\dagger}(x)\right)\left(\frac{1}{\sqrt{L}}\int dx^{\prime}\,e^{-ikx^{\prime}}a(x^{\prime})\right) \\ = \frac{\hbar}{2\pi}\int dk\,\omega_{k}\int\int dx\,dx^{\prime}\left(e^{-ik(x+x^{\prime})}a^{\dagger}(x)a(x^{\prime})\right)\,\,\,\,\,\,\,(6). $$ We then use your property $\omega_{k} = vk$ to write $$H_{0} = \frac{\hbar}{2\pi}\int dk\,\int\int dx\,dx^{\prime} \left(\left[iv\partial_xe^{-ik(x+x^{\prime})}\right]a^{\dagger}(x)a(x^{\prime})\right).$$ Then we pull out the derivative and rearrange the order of the integrals to first integrate over $k$, which as you point out looks like a delta function (the sign is irrelevant because the delta function is even in its arguments) $$ \delta(x+x^{\prime}) = \frac{1}{2\pi}\int dk\,e^{-ik(x+x^{\prime})} \,\,\,\,\,\,\,(7). $$ So $$H_{0} = \hbar\int\int dx\,dx^{\prime} \left(\left[iv\partial_x \int dk\,e^{-ik(x+x^{\prime})}\right]a^{\dagger}(x)a(x^{\prime})\right)\\ = \hbar\int\int dx\,dx^{\prime} \left(\left[iv\partial_x \delta(x+x^{\prime})\right]a^{\dagger}(x)a(x^{\prime})\right).$$ We integrate by parts or use properties of derivatives of delta functions to do the integral: $$H_{0} = - \hbar\int\int dx\,dx^{\prime} \left(\left[iv \delta(x+x^{\prime})\right]\partial_x a^{\dagger}(x)a(x^{\prime})\right).$$ Then we just integrate over $x^\prime$ using the delta function to finally get $$H_{0} = \hbar\int dx \left(-i\hbar v\partial_x a^{\dagger}(x)\right)a(-x).$$

Now since the Hamiltonian is Hermitian, we can equally say$$H_{0} = \left[\hbar\int dx \left(-i\hbar v\partial_x a^{\dagger}(x)\right)a(-x)\right]^\dagger = \hbar\int dx \left[a(-x)\right]^\dagger\left[\left(-i\hbar v\partial_x a^{\dagger}(x)\right)\right]^\dagger.$$ In your Fourier transform convention, we have that $a^\dagger(x)\equiv \left[a(-x)\right]^\dagger$. We also have $[\partial_x a^\dagger(x)]^\dagger=\partial_{x}a(-x)$, as may be verified by taking the inverse Fourier transforms $a(x)\propto \int dk \exp(ikx)a_k$ and $a^\dagger(x)\propto \int dk \exp(ikx)a_k^\dagger$, calculating $\partial_x a(x)\propto \int dk \exp(ikx)ik a_k$ and $\partial_x a^\dagger(x)\propto \int dk \exp(ikx)ik a_k^\dagger$, and inspecting the Hermitian conjugate. These together yield $$H_{0} = \hbar\int dx a^\dagger(-x)\left(+i\hbar v\partial_x a(-x)\right).$$ Then we simply use $\partial_x=-\partial_{-x}$ to find the desired $$H_{0} = \hbar\int dx a^\dagger(-x)\left(-i\hbar v\partial_{-x} a(-x)\right)= \hbar\int dx a^\dagger(x)\left(-i\hbar v\partial_{x} a(x)\right),$$ where the integration variable can be changed like this because we are really doing a definite integral over all space so that $\int_{-\infty}^\infty dx=-\int_\infty^{-\infty} dx=\int_{-\infty}^\infty d(-x)$. Of course there will be quicker ways of achieving this goal but one has to be careful with Fourier transform tricks arising from different conventions.