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.

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

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