Uniform dynamics

This is a time-indepndent Hamiltonian H(t)=HH(t)=H.

We use an ODE for UU which we derived earlier.

iℏdUdt=HU  ⟺  dUdt=−iHℏUi\hbar\frac{dU}{dt} = HU \iff \frac{dU}{dt} = -\frac{iH}{\hbar}U

This is a differential

dU^U^(t)=−iH^ℏdt\frac{d\hat{U}}{\hat{U}(t)}=-\frac{i\hat{H}}{\hbar}dt ⇒∫1U^(t)dU^=−iH^ℏ∫dt\Rightarrow\quad\int\frac{1}{\hat{U}(t)}d\hat{U}=-\frac{i\hat{H}}{\hbar}\int dt ⇒ln⁡∣U^(t)∣=−iH^ℏt+C\Rightarrow\quad\ln|\hat{U}(t )|=-\frac{i\hat{H}}{\hbar}t+C ⇒∣U^(t)∣=eCe−iℏHt/ℏ\Rightarrow\quad |\hat{U}(t)|=e^Ce^{-i\hbar{H}t/\hbar}

at t=0t=0, U^(0)=eC\hat{U}(0)=e^C thus

U^(t)=U(0)e−iH^t/ℏ\hat{U}(t) = U(0)e^{-i\hat{H}t/\hbar}

Recall matrix exponential Hence true.

Let's see what U(t,t0)U(t,t_0) is.

Note U0U_0 is some Integration constant which we haven’t defined yet. We find U0U_0

I=U(t0,t0)=e−iHt0/ℏU0  ⟹  U0=eiHt0/ℏI = U(t_0,t_0) = e^{-iHt_0/\hbar}U_0 \implies U_0 = e^{iHt_0/\hbar}   ⟹  U(t,t0)=e−iHt/ℏeiHt0/ℏ\implies U(t,t_0) = e^{-iHt/\hbar}e^{iHt_0/\hbar}

via Baker, Campbell, and Hausdorff (1897-1906), since [H,H]=0[H,H]=0 (trivially as any operator commutes with itself)

U(t,t0)=e−iHt/ℏeiHt0/ℏU(t,t_0) = e^{-iHt/\hbar}e^{iHt_0/\hbar} ⇒U(t,t0)=e−iH(t−t0)/ℏ\Rightarrow\boxed{U(t,t_0) = e^{-iH(t-t_0)/\hbar}}

Hence this solves SE

iℏddtU(t,t0)=HU(t,t0)i\hbar \frac{d}{dt}U(t,t_0) = HU(t,t_0)

Proof