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 $\psi(t- \tau)?$


1 Answer 1


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

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


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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.