# Unexpected symmetry of wave equations in momentum representation

In the $$x$$-representation, the translational invariance implies that $$\mathcal{D}[\psi(\vec{x},t)]=0\quad \Longrightarrow\quad \mathcal{D}[\operatorname{e}^{i\vec{a}\hat{\vec{P}}}\psi(\vec{x},t)]=0$$ where $$\hat{\vec{P}} = -i\nabla$$ and $$\mathcal{D}$$ is the differential operator corresponding to some translationally-invariant equation. Let's explore how this works in the momentum representation. For definiteness, let's use the KG equation. After the Fourier transform, it reads as $$\left[\partial^2_t+(\vec{p}^2+m^2)\right]\psi(\vec{p},t)=0$$ Since in this representation $$\hat{\vec{P}} = \vec{p}\cdot$$, we have: $$\operatorname{e}^{i\vec{a}\hat{\vec{P}}}\psi(\vec{p},t) = \operatorname{e}^{i\vec{a}\vec{p}}\psi(\vec{p},t)$$ Consequently, $$\left[\partial^2_t+(\vec{p}^2+m^2)\right]\left(\operatorname{e}^{i\vec{a}\vec{p}}\psi(\vec{p},t)\right) =\operatorname{e}^{i\vec{a}\vec{p}}\left[\partial^2_t+(\vec{p}^2+m^2)\right]\psi(\vec{p},t) =0$$ So, if $$\psi(\vec{p},t)$$ is a solution, then $$\operatorname{e}^{i\vec{a}\vec{p}}\psi(\vec{p},t)$$ is a solution as well.

However, by looking at the last equation, we conclude that we can actually multiply $$\psi(\vec{p},t)$$ by an arbitrary function of $$\vec{p}$$, and it still will be a solution: $$\left[\partial^2_t+(\vec{p}^2+m^2)\right]\left(f(\vec{p})\psi(\vec{p},t)\right) =f(\vec{p})\left[\partial^2_t+(\vec{p}^2+m^2)\right]\psi(\vec{p},t) =0$$

Here comes the question: if $$f(\vec{p})\psi(\vec{p},t)$$ is always a solution, why it is only the operator of multiplying by $$\vec{p}$$ who generates symmetries? Naively, I would expect some infinite-dimensional Lie algebra to emerge.

I would like to have an answer not relying on Fourier-transforming back to the position space but rather at the level of the equation in the momentum space.

• You seem to have rediscovered that a superposition of plane waves, say, is still a solution of the wave equation. – user178876 Oct 2 '18 at 17:00
• Unfortunately, I don't see how this discovery provides an immediate answer to my question. – mavzolej Oct 2 '18 at 17:22
• Consider $\phi(x)=\int\!\mathrm{d}^4p\,a(p)\,\mathrm{e}^{-\mathrm{i} p\cdot x}+\text{cc}$, where $p\cdot x=Et-\vec p\cdot \vec x$. – user178876 Oct 2 '18 at 17:25
• The vanishing commutator ${\left[\partial^2_t+(\vec{p}^2+m^2),\,f(\vec{p})\right]}$. – mavzolej Oct 2 '18 at 17:55
• Indeed, you do get an infinite-dimensional algebra. You've rediscovered BMS supertranslations. Congrats! – AccidentalFourierTransform Oct 2 '18 at 18:57