# Fourier transform of a vector

For a function $$f(t)$$ the Fourier transform $$\mathcal{F}[f(t)] = \tilde{f}(\omega) = \int_{-\infty}^{\infty} e^{-i \omega t} f(t)$$ of its derivation i.e., $$\mathcal{F}\left[\frac{d}{dt} f(t) \right] = -i \omega \tilde{f}(\omega).$$

Now consider a state vector denoted by $$\begin{pmatrix} \textbf{x} \\ \textbf{p} \end{pmatrix} = \begin{pmatrix} x_1\\ x_2 \\ p_1\\ p_2 \end{pmatrix}$$ where $$x_i$$ and $$p_i$$ are the position and momentum coordinates. It is argued in this reference (equation S29) that the symplectic matrix $$\Omega = \begin{pmatrix} \textbf{0} & \textbf{I} \\ - \textbf{I} & \textbf{0} \end{pmatrix}$$ plays the role of the imaginary unit i=$$\sqrt{-1}$$. Therefore, can one define the Fourier transform of the derivative (analogous to the above equation) as:

$$\mathcal{F} \left[\frac{d}{dt}\begin{pmatrix} \textbf{x} \\ \textbf{p} \end{pmatrix} \right] = - \Omega \omega \begin{pmatrix} \tilde{\textbf{x}} \\ \tilde{\textbf{p}} \end{pmatrix}.$$

Does this make sense? If yes, can it be proved?

I don't have access to the reference you provide, but it is perfectly possible to associate a well-defined Fourier transform of functions over a phase space $$M$$.

Let $$M$$ be a $$2n$$-dimensional manifold. It can be rendered symplectic with the introduction of a canonical form

$$\omega = \sum_{i=1}^n \text{d}p_i \wedge \text{d}x_i \quad ,$$

which for cartesian $$\mathbb{R}^{2n}$$ has precisely the matrix form of your $$\Omega$$. By splitting $$\mathbb{R}^{2n} = \mathbb{R}^n \oplus \mathbb{R}^n$$, with positions inhabiting the first component and momenta the second, it is easy to see that, in this case,

$$\omega(u,v) = u \cdot \Omega v = u_x \cdot u_p - u_p \cdot u_x \quad ,$$

with $$u_x$$ and $$v_x$$ in $$\mathbb{R}^n \oplus 0$$ and $$u_p$$ and $$v_p$$ in $$0 \oplus \mathbb{R}^n$$.

Let $$f:\mathbb{R}^{2n} \longrightarrow \mathbb{R}$$ be a continuous function and $$\eta \in \mathbb{R}^{2n}$$. The symplectic Fourier transform of $$f$$ is defined as

$$\widetilde{f}(\xi) = \frac{1}{(2\pi)^n} \int_{\mathbb{R}^{2n}} \text{d}\eta \, f(\eta) \exp \left( i \, \eta \cdot \Omega \xi \right) \quad .$$

One can show this operation fulfills everything that is expected from a Fourier transform on phase space. For details see Symplectic Geometry and Quantum Mechanics, by Maurice de Gosson, section 6.2.1.

Edit: Consider the case of one degree of freedom. We then have

\begin{align} \widetilde{\nabla_\xi f(\xi_x,\xi_p)} &= \frac{1}{(2\pi)^n} \int_{\mathbb{R}^{2}} \text{d}\eta_x \, \text{d}\eta_p \nabla_\xi f(\eta_x,\eta_p) \exp \left[ i \left( \eta_x \, \xi_p - \eta_p \, \xi_x \right) \right] . \end{align}

The result you want follows from integration by parts, where one needs to consider that $$f$$ belongs to some adequate Schwartz space to cancel the boundary term. This is exactly the same result of the usual Fourier transform, but applied to position and momentum simultaneously. The need for the symplectic matrix $$\Omega$$ is: (1) classically, the different signs in Hamilton's equations; (2) quantum mechanically, due to the generators of translations in position and momentum representations having phaes with different signs.

• Thanks, @QuantumBrick. Does it lead to the result for the derivative that I wrote? I don't see that! Jul 14 at 21:00
• Consider $\xi$ as a function of $t$ and take the derivative. Then a $i\,\Omega \dot{xi(t)}$ will come down the exponential and you will end up with $i \, \Omega \dot{xi(t)} \widetilde{f}(\xi)$. I think this gives you what you want. Let me know if you want me to include this in the answer. Jul 14 at 21:07
• Could you be a little more explicit and include this in your answer? Jul 15 at 5:49
• @Zubin I added an explicit expression in 2D, from which it must be obvious that what you are looking for is there. The symplectic Fourier transform is just a way of treating position and momentum Fourier transforms simultaneously, so all the properties of the individual components apply in the symplectic case. You can actually even define the transform without the $\Omega$, at the cost terrible sign manipulation. Jul 15 at 9:41

I don't have access to the reference you provide, so I don't know the version of the Fourier transform that you are using, but will try using this version of the Fourier transform. $$\mathcal{F} = \int_{-\infty}^{\infty} f(t) e^{-\Omega t w} dt$$

The proof follows:

Let $$u = \begin{pmatrix} \mathbf{x} \\ \mathbf{p} \end{pmatrix}$$, so $$u' = \frac{\partial}{\partial t}\begin{pmatrix} \mathbf{x} \\ \mathbf{p} \end{pmatrix}$$

$$u = \frac{1}{2\pi} \int_{-\infty}^{\infty} [(\mathcal{F}u)(w)] e^{\Omega w t} dw$$ $$u' = \frac{\partial}{\partial t} \frac{1}{2\pi} \int_{-\infty}^{\infty} [(\mathcal{F}u)(w)] e^{\Omega w x} dw$$ $$u' = \frac{1}{2\pi} \int_{-\infty}^{\infty} [(\mathcal{F}u)(w)] \frac{\partial}{\partial t} e^{\Omega w x} dw$$ $$u' = \frac{1}{2\pi} \int_{-\infty}^{\infty} [(\mathcal{F}u)(w)] \Omega w e^{\Omega w x} dw$$ Therefore $$(\mathcal{F}u') = \frac{\Omega w}{2\pi} (\mathcal{F}u)$$