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

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?

• Your mistake is that you cannot take $\omega_k$ out of the $k$ integral. The integral thus won't give a delta function. It's easier to see how things work by starting the other way around. Write $a(x)$ in terms of $a_k$ (inverse of the Fourier transform you've written). Then compute $\partial_xa(x)$. You should be able to see that by Fourier transform you have roughly $-i \hbar\partial_xa(x) \leftrightarrow ka_k$. Feb 14, 2022 at 18:57
• Thanks for pointing this out. I think the correct way would be integration by parts acting on the $\omega_{k}\,e^{ik(x+x^{\prime})}$ term. Even then, I still have $\omega_k$ out of the integral. Can you elaborate more on your last point? I don't see how $k$ can be an operator since it can only come from $\omega_k$ which is scalar Feb 14, 2022 at 20:09
• Your definitions of $a_k$ and $a_k^\dagger$ are not Hermitian conjugates of each other - the phase needs to be fixed Feb 14, 2022 at 20:22
• @QuantumMechanic thanks for your comment. By default, $a_{k}$ and $a_{k}^{\dagger}$ are not Hermitian conjugates of one another. They are the usual annihilation and raising operators which are not Hermitian. Feb 14, 2022 at 20:26
• @kowalski since when? They should be conjugates of each other, not each be Hermitian themselves Feb 14, 2022 at 20:27

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.

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.

• Thanks for your answer. However this was the issue that I pointed out in my main post, namely I don't know how you can justify letting the scalar $k$ to be an operator $k$, since $\omega_{k}=vk$ and $\omega_k$ is scalar Feb 14, 2022 at 20:54
• @kowalski I explicitly used the fact that $\partial_x \exp(-ikx)=-ik\exp(-ikx)$. This is the whole point of using different representations, where one operator looks trivial when acting on an eigenstate. (Recall intro QM where the position representation has the operator $\hat{x}$ get replaced by the "scalar" $x$.) The relation I quoted always holds, regardless of context (it's not a quantum relation, it's a functional relation). Equivalently... why claim $\omega_k$ is a scalar? You could get it from acting with the Hamiltonian operator on an energy eigenstate Feb 14, 2022 at 20:59
• Yes, I agree that $k$ doesn't have to be an operator. I was focused on the fact that the $-i\hbar\partial_{x}$ must necessarily come from an operator but you can get $\partial_{x}$ simply through algebraic manipulation. Feb 14, 2022 at 22:28
• I however, am still skeptical about the definition of the phase that you use. It seems to me that my definition is the conventional way of defining FT in operators. Here's another link whereby the OP asked a similar question math.stackexchange.com/questions/1786924/… although it lacks a satisfactionary answer. Feb 14, 2022 at 22:30
• @kowalski no problem - this was still an exercise you could have done using the answer that I originally wrote! Regardless I've updated with an answer using your Fourier transform convention, so that you may compare the two Feb 15, 2022 at 15:15