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: $$\forall \phi \forall \psi, \,~\lvert\langle\phi \rvert T_\vec{a}\lvert\psi\rangle\rvert=\lvert\langle\phi \lvert\psi\rangle\rvert$$

  • $\begingroup$ How is $\langle \phi \vert T_a \vert \psi\rangle$ "a translation"? A translation acts on each vector as $\lvert \psi \rangle \mapsto T_a \lvert \psi \rangle$. $\endgroup$ – ACuriousMind Nov 7 '16 at 13:22

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 $$\forall \psi \,\,\langle\psi\rvert H \lvert\psi\rangle = \langle \,T_{\vec{a}} \, \psi \,\rvert H \lvert \,T_\vec{a}\,\psi \rangle . \tag 4$$


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