Translation invariance in QM In my course we proved that the translational invariance in QM is equivalent to say that $ [\vec{p},H]=0 $.
To show this we started from this constatation:
the system being invariant under translations is equivalent to say that $\forall \psi \,\,\langle\psi\rvert H \lvert\psi\rangle
= \langle \,T_{\vec{a}} \, \psi \,\rvert H \lvert \,T_\vec{a}\,\psi \rangle $.
What I don't understand is why is this an equivalence.
For me if the system is invariant under translation then we have $\forall \psi, \,\, \,\,\langle\psi\rvert H \lvert\psi\rangle
= \langle \,T_{\vec{a}} \, \psi \,\rvert H \lvert \,T_\vec{a}\,\psi \rangle$.
But if we have $\forall \psi, \,\, \langle\psi\rvert H \lvert\psi\rangle
= \langle \,T_{\vec{a}} \, \psi \,\rvert H \lvert \,T_\vec{a}\,\psi \rangle$ we could have no invariance by translation.
Should'nt we prove that : $\forall \phi \forall \psi, \,~\lvert\langle\phi \rvert T_\vec{a}\lvert\psi\rangle\rvert=\lvert\langle\phi \lvert\psi\rangle\rvert$. Indeed if we have this then the the probabilities are conserved under any translation, then all the physics is unchanged under any translation ?
To summarize my two questions :
Why if $\forall \psi, \,\, \langle\psi\rvert H \lvert \psi\rangle=\langle T_\vec{a}\psi\rvert H\lvert T_\vec{a}\psi\rangle $ then the system is invariant by translation ?
Can we say that a system being invariant by translations is equivalent to say: $$\forall \phi \forall \psi, \,~\lvert\langle\phi \rvert T_\vec{a}\lvert\psi\rangle\rvert=\lvert\langle\phi \lvert\psi\rangle\rvert$$
 A: The condition you state,
$$\forall \phi \forall \psi, \,~\lvert\langle\phi \rvert T_\vec{a}\lvert\psi\rangle\rvert=\lvert\langle\phi \lvert\psi\rangle\rvert, \tag 1$$
is not really right, but you can fix it easily by rephrasing it as
$$\forall \phi \forall \psi, \,~\lvert\langle T_\vec{a} \phi \rvert T_\vec{a}\psi \rangle \rvert=\lvert\langle\phi \lvert\psi\rangle\rvert. \tag 2$$
(That is: the inner product is unchanged by translating both $|\phi⟩$ and $|\psi⟩$. Obviously, if you only translate $|\psi⟩$, its inner product with the untranslated $|\phi⟩$ will change.) In this form, it
is a general property of the kinematics of the system: it asks that $T_\vec{a}$ needs to be a unitary operator, which is the very least you wold expect from a transformation that isn't really changing anything.
However, in this form, the condition $(2)$ knows nothing about the hamiltonian, so you could have a completely translation-non-invariant hamiltonian like $H=\frac12 p^2 +\frac12 x^2$ and $(2)$ will still be true. If you want to show that the dynamics are translationally invariant, then you need to refer to $H$ at some point, because $H$ completely determines the dynamics. As it turns out, the correct way to do this is to demand that the matrix elements of $H$ be preserved,
$$\forall\phi \forall \psi \,\,\langle\phi\rvert H \lvert\psi\rangle
= \langle \,T_{\vec{a}} \, \phi \,\rvert H \lvert \,T_\vec{a}\,\psi \rangle , \tag 3$$
which through relatively simple manipulations can be reduced to the condition
$$\forall \psi \,\,\langle\psi\rvert H \lvert\psi\rangle
= \langle \,T_{\vec{a}} \, \psi \,\rvert H \lvert \,T_\vec{a}\,\psi \rangle . \tag 4$$
