# How to show that scalar fields are translation invariant?

Classical scalar fields governed by Klein-Gordon equation, $$\left(-\frac{\partial^2}{\partial t^2}+\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}+m^2\right)\phi(\vec{x},t)=0,$$ can be solved by taking its Fourier transform, $$\phi(\vec{x},t)=\int\frac{d^3k}{\left(2\pi\right)^{3/2}}\phi(\vec{k},t)e^{i\vec{k}\cdot\vec{x}},$$ and imposing the on-shell condition $$k^2+m^2=0$$. Now physically this this theory should be translation invariant. Under a transformation $$\vec{x}\rightarrow\vec{x}+\vec{a}$$ the theory should be invariant. Now how to show this mathematically working in the Fourier space? I guess this is related to the shift property of Fourier transform but I am unable to see how to show it formally.

The same problem can be considered in quantum field theory. This time the Fourier expansion changes to something like, $$\phi(\vec{x},t)=\int\frac{d^3k}{(2\pi)^{3/2}\sqrt{2E_k}}\left(c_{\vec{k}}e^{-i\vec{k}\cdot\vec{x}}+c_{\vec{k}}^\dagger e^{i\vec{k}\cdot\vec{x}}\right).$$ The one has to show that the vacuum state which is annihilated by the annihilation operator is invariant under the transformation $$e^{i\vec{P}\cdot\vec{a}}$$, where, $$P$$ is momentum operator or the generator of translation.

Recall that a theory being invariant under some symmetry does not imply that a solution will be invariant under those symmetries.

The first part of the question is about classical field theory. A solution for $$\phi$$ will break translational invariance if the initial conditions you give to the equations break it.

The second part of the question is about a different thing: it is the invariance of the theory vacuum. Typically it is the case that the symmetries you have in the theory are also symmetries of the vacuum, and it is definitely the case for the free field. But this is not necessarily true in general due to the phenomenon of spontaneous symmetry breaking.

Now physically this theory should be translation invariant. Under a transformation $$\vec{x}\to\vec{x}+\vec{a}$$ the theory should be invariant. Now how to show this mathematically working in the Fourier space?
If $$\phi(\vec{x},t)$$ is a solution of the Klein-Gordon equation, then $$\phi'(\vec{x},t)=\phi(\vec{x}+\vec{a},t) \tag{1}$$ is a solution as well.
Let's denote the Fourier-transformed fields by a $$\tilde{}$$.
Then the Fourier-transform of equation (1) is $$\tilde{\phi}'(\vec{k},t)=e^{-i\vec{k}\vec{a}}\ \tilde{\phi}(\vec{k},t). \tag{2}$$ So a translation in $$\vec{x}$$-space becomes a phase-transformation in $$\vec{k}$$-space.
Now, the Fourier-transform of the Klein-Gordon equation is $$\left(-\frac{\partial^2}{\partial t^2}-k_x^2-k_y^2-k_z^2+m^2\right)\phi'(\vec{k},t)=0. \tag{3}$$
If $$\tilde{\phi}(\vec{k},t)$$ is a solution of the Fourier-transformed Klein-Gordon eequation (3), then $$\tilde{\phi}'(\vec{k},t)=e^{-i\vec{k}\vec{a}}\ \tilde{\phi}(\vec{k},t)$$ is a solution as well.