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The time translation is given by a finite unitary transformation $$\hat{U_{\tau}}(\hat{H}) = e^{\big(\frac{i}{\hbar}\tau \bar{H}\big)}.$$ Where $$\hat{U_{\tau}}(\hat{H})|\psi(t) \rangle = |\psi(t-\tau)\rangle).$$ We can show that the hamiltonian is invariant under the finite unitary translation given by $$\hat{U}(t,t_0) = e^{-\frac{i(t-t_0)\hat{H}}{\hbar}}$$ so we have $\hat{H} = e^{\frac{-i(t-t_0)\hat{H}}{\hbar}} \hat{H} e^{\frac{i(t-t_0)\hat{H}}{\hbar}} = \hat{H}$. Aslo we have that $[\hat{H}, \hat{H}] = 0$, thus $$\frac{d}{dt} \langle \hat{H} \rangle = \frac{1}{i \hbar} \langle [ \hat{H}, \hat{H} ] \rangle + \langle \frac{\partial \hat{H}}{\partial t} \rangle = 0.$$

So $\hat{H}$ is conserved.

How does all of this imply that if $\psi(t)$ satisfies the time dependent Schrodinger equation then so does the time displaced $\psi(t- \tau)?$

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Since $[\hat{H},\hat{H}]=0$, you also have $[\hat{H},\hat{U}(\tau)]=0$. So, if we consider Schrodinger's equation, we have (neglecting factors of $\hbar$ for simplicity)

$$ i\frac{d}{dt}|\psi(t)\rangle = \hat{H}|\psi(t)\rangle $$ Multiplying both sides by $\hat{U}(\tau)$ gives $$ i\hat{U}(\tau)\frac{d}{dt}|\psi(t)\rangle = \hat{U} (\tau)\hat{H}|\psi(t)\rangle $$

Now, $\frac{d}{dt}$ commutes with $\hat{U}(\tau)$ since $\hat{U}(\tau)$ does not depend on $t$, and we already said $\hat{U}(\tau)$ commutes with $\hat{H}$. So, we then get

$$ i\frac{d}{dt}\hat{U}(\tau)|\psi(t)\rangle = \hat{H}\hat{U}(\tau)|\psi(t)\rangle $$

or

$$ i\frac{d}{dt}|\psi(t-\tau)\rangle = \hat{H}|\psi(t-\tau)\rangle $$

So the time-displaced kets also satisfy Schrodinger's equation.

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