# Evolution operator for time-dependent Hamiltonian

When i studyed QM I'm only working with non time-dependent Hamiltonians. In this case 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 that Hamiltonian in Schrödinger picture ($H_{S}$), i.e. it commutes with $\hat{U}$. Now i have Hamiltonian that depens explicitly on time. Specifically, $$H_{S}=\frac{\hat{p}^2}{2m}+\frac{1}{2}m\omega \hat{q}^2-F_0sin(\omega_0t)\hat{q}$$.

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.
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Can you explain what exactly you want to do? Generally this sort of problem is more tractable in the Interaction Picture generated by $H_0 = \frac{p^2}{2m} + \frac{1}{2} m \omega^2 q^2$, since you already know the eigenstates of $H_0$. Then you can calculate the evolution of observables or states using the Dyson series, which is essentially what you have calculated. Your perturbation is time-periodic, so you might also find Floquet theory useful. –  Mark Mitchison Nov 29 '12 at 18:32
Since your Hamiltonian is periodic in time this link might be helpful : en.wikipedia.org/wiki/Floquet_theory –  jjcale Nov 29 '12 at 19:51
I want to calculate $E=\langle\hat{H}\rangle$ at time $T$ i.e. $\langle \psi(t)|\hat{H}| \psi(t) \rangle$ or in Heisenberg picture $\langle \psi |\hat{H}(t)| \psi \rangle$. For this thing I need evolution operator to calculate $\hat{H(t)}$, as I think. –  Oiale Nov 29 '12 at 20:00

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 $$U_n = (-\frac{i}{\hbar})^n\int_0^t dt_1 \cdots\int_0^{t_{n-1}} dt_{n} H(t_1)\cdots H(t_n)$$

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 =(-\frac{i}{\hbar})^n \int_0^t dt_1 \cdots\int_0^{t_{n-1}} dt_{n}\ H(t_1)\cdots H(t_n) = \frac{(-\frac{i}{\hbar})^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 thatthe 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

$$U(t,t') = 1 + \sum_{n=1}^\infty \frac{(-\frac{i}{\hbar})^n}{n!} \int_{t'}^t dt_1 \cdots\int_{t'}^t dt_n \mathcal{T}[H(t_1)\cdots H(t_n)]$$ 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.

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good job, very enlightening, 1up –  Dylan Sabulsky Nov 30 '12 at 2:48