When I studied QM I'm only working with time independent Hamiltonians. In this case the unitary evolution operator has the form $$\hat{U}=e^{-\frac{i}{\hbar}Ht}$$ that follows from this equation
$$
i\hbar\frac{d}{dt}\hat{U}=H\hat{U}.
$$
And in this case, Hamiltonian in Heisenberg picture ($H_{H}$) is just the same as the Hamiltonian in Schrödinger picture ($H_{S}$), i.e. it commutes with $\hat{U}$.
Now I have a Hamiltonian that depends explicitly on time. Specifically,
$$H_{S}=\frac{\hat{p}^2}{2m}+\frac{1}{2}m\omega \hat{q}^2-F_0 \sin(\omega_0t)\hat{q}.$$
And in my problem I need to calculate $H_H$ (Hamiltonian in Heisenberg picture).
I've found that differential equation for $\hat{U}$ (I've mentioned about it above.) has generally solution in the form (with $U(0)=1$)
$$U(t)=1+\xi\int\limits_0^t H(t')\,dt'+ \xi^2\int\limits_0^t H(t')\,dt'\int\limits_0^t' H(t'')\,dt''+\xi^3\int\limits_0^t H(t')\,dt'\int\limits_0^t' H(t'')\,dt''\int\limits_0^t'' H(t''')\,dt'''+…$$
So my questions are:
- Is there other ways to calculate $\hat{U}$, could give a link or tell me about them?
- If you know form of the solution for my case, please tell me.
- If you know any articles or papers articles on this topice, please link them to me, too.
Best Answer
Yes, you are on the right track. The series you have there is called Dyson's series.
First note that the $n$'th term looks like $$ U_n = \left(-\frac{i}{\hbar}\right)^n\int_0^t dt_1 \cdots\int_0^{t_{n-1}} dt_{n} H(t_1)\cdots H(t_n) $$
The order of the Hamiltonians is important, since we work with operators. Each term in the series possess a nice symmetry, allowing us to write:
\begin{align} U_n = \left(-\frac{i}{\hbar}\right)^n \int_0^t dt_1 \cdots\int_0^{t_{n-1}} dt_{n}\ H(t_1)\cdots H(t_n) = \frac{\left(-\frac{i}{\hbar}\right)^n}{n!}\int_0^t dt_1 \cdots\int_0^t dt_{n} \mathcal{T}\left[H(t_1)\cdots H(t_n)\right] \end{align}
Two things happened: first, we "overcount" by making the upper limits equal to $t$ on all the integrals. This is compensated by the factor of $\frac{1}{n!}$. You'll need to convince yourself why this factor is needed ;)
Second, by this change of integration area we mess up the ordering of the Hamiltonians in the process. This is where the time-ordering symbol $\mathcal{T}$ comes in. Basically, this operator ensures that the Hamiltonians are always ordered in the correct way. For instance for $n=2$ it operates as
\begin{align} \mathcal{T}[H(t_1) H(t_2)] = \begin{cases} H(t_1) H(t_2) & t_2 > t_1\\ H(t_2) H(t_1) & t_2 < t_1 \end{cases} \end{align}
Putting everything together we have
$$ U(t, t') = 1 + \sum_{n=1}^\infty \frac{\left(-\frac{i}{\hbar}\right)^n}{n!} \int_{t'}^t dt_1 \cdots\int_{t'}^t dt_n \mathcal{T}[H(t_1)\cdots H(t_n)] $$ Frequently, this is denoted symbolically as
$$ U(t, t') = \mathcal{T}\exp\left(-\frac{i}{\hbar} \int_{t'}^t H(t_1) dt_1\right) $$ This notation is understood as representing the power series.