Let's consider the unitary group $\hat{S_{\tau}^†}$ such that :$$\hat{S^†_{\tau}}|\psi(t)\rangle=|\psi(t-\tau)\rangle$$ Since we know that: $$\hat{U}(t,t_0)|\psi(t_0)\rangle=|\psi(t)\rangle$$ Where ${U}(t,t_0)$ is the time evolution operator, we can conclude that: $$\hat{S^†_{\tau}}=U(t-\tau,t).$$ Now we also know that $\hat{U}$ solves: $$iℏ\partial_t\hat{U}=\hat{H}\hat{U}$$ So that if $\hat{H}$ does not depend explicitly on time we have: $$U(t,t_0)=e^{-\frac{i}{ℏ}(t-t_0)\hat{H}} (*)$$ Hence: $$\hat{S^†_{\tau}}=U(t-\tau,t)=e^{\frac{i}{ℏ}\tau\hat{H}}$$ $$\hat{S_{\tau}}=e^{-\frac{i}{ℏ}\tau\hat{H}}$$ And we can surely say that $\hat{H}$ is the generator of time translations. Now what happens when $\hat{H}$ has an explicit dependence on time ? Equation $(*)$ no longer holds and I am not confident in saying that $\hat{H}$ is still the generator of the group $\hat{S_{\tau}}$. The book I'm reading doesn't distinguish between these two cases. Is it because it only considers infinitesimal translations?
1 Answers
Consider the solution of: $$i\hbar{\partial\over \partial t}\mathcal U(t,t_0)=H\mathcal U(t,t_0).$$ If the Hamiltonian is time dependent and if it commutes with itself at different times, $$[H(t),H(t^\prime)]=0,\;t\neq t^\prime.$$ Then the time development is given by: $$\mathcal U(t,t_0)=e^{-{i\over\hbar}\int_{t_0}^t H(t^\prime)\;dt^\prime}.$$ If the Hamiltonian does not commute with itself at different times: $$[H(t),H(t^\prime)]\neq 0,$$ then the time development operator is given as: $$\mathcal U(t,t_0)=1+\sum\limits_{n=1}\limits^{\infty}\bigg({-i\over\hbar}\bigg)^n\int_{t_0}^t\;dt_1\int_{t_0}^{t_1}\;dt_2\cdots \int_{t_0}^{t_{n-1}}\;dt_n\;H(t_1)H(t_2)\cdots H(t_n). $$ This last result is known as a Dyson series and is important in quantum field theory.
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