Quantum Dynamics

Time Evolution Operator

Let $\left| \psi(t) \right>$ be the state of a system at time $t$. The time evolution of $\left| \psi(t) \right>$ is governed by the Schrodinger equation $$i\hbar\frac{d}{dt}\left|\psi(t)\right> = \hat{E}\left|\psi(t)\right>,$$ where $\hat{E}$, the Hamiltonian, is the energy operator of the system. Let $U(t)$ be a time evolution operator such that $$\left|\psi(t)\right> = U(t) \left|\psi(0)\right>.$$ It is known that $U(t)$ is an unitary operator because $\left|\psi(t)\right>$ and $\left|\psi(0)\right>$ are normalized. At $t = 0$, $$\left|\psi(0)\right> = U(0) \left|\psi(0)\right>.$$ So, $$U(0) = I$$ By Schrodinger equation, $$\begin{align} i\hbar\frac{d\left|\psi(t)\right>}{dt} &= \hat{E}\left|\psi(t)\right> \\ i\hbar\dot{U(t)}\left|\psi(0)\right> &= \hat{E}U(t)\left|\psi(0)\right>. \end{align}$$ So, $$i\hbar\frac{dU(t)}{dt} = \hat{E}U(t).$$

Time-independent Hamiltonian

If the Hamiltonian is time-independent, then $$U(t) = e^{-i\hat{E}t/\hbar} = \sum^\infty_{n=0}\frac{1}{n!}\left(-i\frac{\hat{E}t}{\hbar}\right)^n$$ is a solution to $$i\hbar\frac{dU(t)}{dt} = \hat{E}U(t).$$ If we express the Hamiltonian in terms of its eigenvalue and eigenvectors, $$\hat{E}=\sum_n\lambda_n\left| n\right>\left< n \right|,$$ then the kth power of the Hamiltonian is $$\hat{E}^k = \sum_n\lambda^k_n \left| n \right> \left< n \right|.$$ Substitute back into the time evolution operator, we have $$U(t) = e^{-i\hat{E}t/\hbar} = \sum_n e^{-i\lambda_nt/\hbar}\left| n \right>\left< n \right|.$$ The general solution of the state vector will be $$\begin{align} \left|\Psi(t)\right> &= U(t)\left|\Psi(0)\right> \\ &= U(t)\left(\sum_n c_n\left| n \right>\right) \\ &= \sum_m e^{-i\lambda_mt/\hbar}\left| m \right>\left< m \right|\left(\sum_n c_n\left| n \right>\right) \\ &= \sum_n c_n e^{-i\lambda_n t/\hbar}\left| n \right>, \end{align}$$ which is the same result as what we got using separation of variable in previous post.

Time Dependent Hamiltonian

The solution of the time evolution operator in previous no longer holds for time-dependent Hamiltonian. We might transform the Hamiltonian to a time-independent one $$i\hbar\frac{d}{dt}\left|\psi(t)\right> = \hat{E}(t)\left|\psi(t)\right>.$$ Let $\left|\psi(t)\right>=T(t)\left|\phi(t)\right>$, where $TT^\dagger = T^\dagger T = I$. Substitute into the Schrodinger equation, $$\begin{align} i\hbar\left[T(t)\frac{d\left|\phi(t)\right>}{dt} + \frac{dT(t)}{dt}\left|\phi(t)\right>\right] &= \hat{E}(t)T(t)\left|\phi(t)\right>\\ i\hbar T^\dagger(t) \left[T(t)\frac{d\left|\phi(t)\right>}{dt} + \frac{dT(t)}{dt}\left|\phi(t)\right>\right] &= T^\dagger(t)\hat{E}(t)T(t)\left|\phi(t)\right> \\ i\hbar\frac{d\left|\phi(t)\right>}{dt} + i\hbar T^\dagger(t)\frac{dT(t)}{dt}\left|\phi(t)\right> &= T^\dagger(t)\hat{E}(t)T(t)\left|\phi(t)\right> \\ i\hbar\frac{d\left|\phi(t)\right>}{dt} &= \left[T^\dagger(t)\hat{E}(t)T(t)-i\hbar T^\dagger(t)\frac{dT(t)}{dt}\right]\left|\phi(t)\right>. \end{align}$$ Define the transformed Hamiltonian as $$\hat{E}' = \left[T^\dagger(t)\hat{E}(t)T(t)-i\hbar T^\dagger(t)\frac{dT(t)}{dt}\right].$$ If the transformed Hamiltonian is time-independent, then we can use back the time-independent method to find the time evolution operator. However, such a transformation does not always exists.

Heisenberg's Picture

Since $\left|\psi(t)\right> = U(t) \left|\psi(0)\right>$, the expectation value of an operator $M$ is $$\left< \psi \right| M \left| \psi(t)\right> = \left<\psi(0) \right| U^\dagger(t)MU(t) \left|\psi(0)\right>.$$ Define the operator $$M(t) = U^\dagger(t)MU(t),$$ in which $M(0)=M$, which $M$ is the original Schrodinger operator. Then, the time evolution of the operator $M$ is $$\begin{align} \frac{d}{dt}M(t) &= \frac{d}{dt}\left[U^\dagger(t)MU(t)\right]\\ &= \dot{U}^\dagger(t)MU(t) + U^\dagger(t)M\dot{U}(t) \\ &= \frac{i}{\hbar}U^\dagger(t)\hat{E}MU(t) - \frac{i}{\hbar}U^\dagger(t)M\hat{E}U(t) \\ &= \frac{i}{\hbar}U^\dagger(t)(\hat{E}M - M\hat{E})U(t)\\ &= \frac{i}{\hbar}[\hat{E}(t), M(t)]. \end{align}$$

Example

Find $x(t)$ of a free particle in terms of $x(0)$ and $p(0)$.

Solution: The Hamiltonian is $$\hat{E} = \frac{p^2}{2m}.$$ Then, $$\frac{dx(t)}{dt} = \frac{i}{\hbar}\left[\frac{p^2}{2m},x(t)\right] = \frac{p(t)}{m}.$$ By $[H,p]=0$, $p(t)=p(0)$, so integrating the above equation, we have $$x(t) = x(0) + \frac{p(0)}{m}t$$