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.
  • 1
    $\begingroup$ 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. $\endgroup$ Commented Nov 29, 2012 at 18:32
  • $\begingroup$ Since your Hamiltonian is periodic in time this link might be helpful : en.wikipedia.org/wiki/Floquet_theory $\endgroup$
    – jjcale
    Commented Nov 29, 2012 at 19:51
  • $\begingroup$ 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. $\endgroup$
    – Oiale
    Commented Nov 29, 2012 at 20:00
  • $\begingroup$ Try reading about the Magnus, Fer and Wilcox expansions: sciencedirect.com/science/article/pii/S0370157308004092 repositori.uji.es/xmlui/bitstream/handle/10234/32076/… $\endgroup$
    – Benjamin
    Commented Apr 4, 2015 at 1:07
  • $\begingroup$ A situation a lot like this is solved in these lecture notes that might be useful. $\endgroup$ Commented Jul 2, 2015 at 12:53

1 Answer 1


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_1 > t_2\\ H(t_2) H(t_1) & t_1 < t_2 \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.

  • 6
    $\begingroup$ I think you wanted to write this for time ordering: $ H(t_2) H(t_1) t_2 > t_1 $ and $ H(t_1)H(t_2) t_2 < t_1 $. If I am not wrong. $\endgroup$
    – L.K.
    Commented Mar 4, 2017 at 19:24

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