Conservation of momentum and translation symmetry

The assertion/assumption

Physics laws should not be changed under translations

Does not imply the statement

Hamiltonian must commute with the generator of translations

If we have a system that has $[H,p]=0$ then it’s a special system where the coordinates of the origin doesn’t matter. It’s a spatially symmetric system. It has translational invariance. For this to happen, we find that $V=text{constant}$.

Consider the following example, we have two infinite boxes. One is empty ($V=0$) and another has a point of interest that enforces $V≠0$. In the first case it doesn’t matter where look at, everything is the same. But that’s not the case with the second one. This is translational invariance. A property of the system.

Now, coming to your assertion that the physical laws should not be changed under translation. It is a restriction on the kinds of (differential) equations we can use to describe nature. It says that whatever law that we have, if we make the transformation $xto x+a$, the law must hold. That is, the structure of the equation must remain invariant under $xto x+a$. Basically, the abstract equation (in this case, the schrodinger equation) should be valid at all points in space.

To summarise, physical laws being valid implies the abstract equation must be valid throughout. Hamiltonian commuting with generator of translation implies that a particular form of the equation remains numerically valid throughout. And they are not saying the same thing.

If you just have the free particle Hamiltonian, with $V = 0$ then you'd expect that the dynamics are the same no matter the position of your system and that is precisely what happens. This corresponds, as you mentioned, to the fact that $left[H,Pright]=0$.

However, when you introduce a potential you break this translational symmetry because now the position of the particle affects its dynamics. For example if you have a quadratic potential at the origin ($V = x^2$) then the dynamics of the system are very different from the ones had you put the quadratic potential very far away from the origin ($V = (x-a)^2, a>>1$).

Now, when we say that the laws of physics are translationally invariant we mean in the sense that if you translate the entire system (which would imply that you'd somehow also translate the potential) then the dynamics you obtain remain the same. This however is not directly related to the commutator $[H,P]$ being zero.

There are two aspects here:

1. If you assuem a fixed (nonconstant) potential $V(x)$, then $H$ and $P$ don't commute. Also, momentum is not conserved, because particles are accelerated by the potential (e.g. are trapped in a potential well and bounce back and forth). Physically, this means that you considfer the potential as external to the system you're investigating, and you ignore the backreaction. The same thing happens in classical mechanics when you e.g. consider masses falling in the gravitational field of the earth and assume earth is fixed.
2. Of course, one assumes that proper physical laws are invariant under translations, momentum is conserved etc. To reconcile this with the firts point, you need to take into account the interaction between your particle and whatever it is that causes the potential (experimental setup, earth, etc.) Then you will find that whenever the particle momentum changes, the change is balanced by the chaneg in the momentum of the "apparatus". It's just that in many situaions, it is a very good approximation, and much simpler, to ignore this effect (consider a ball bouncing on the ground: The change in momentum of the earth is there, but you can completely ignore the change in velocity, so for all practical purposes, the earth is fixed).