# Integral form of solution of Dyson series and differentiation of the exponential form

The solution to time dependent hamiltonian equation is: $$\frac{\partial}{\partial t}U(t) = -\frac{i}{\hbar}H(t)U(t)$$

The immediate integral form solution is

1. $U(t) = I - \frac{i}{\hbar}\int_{0}^{t}H(t')U(t')dt'$

having used $U(0) = I$. I have some trouble deriving this solution. Why the time-ordering is absent in the RHS integral, but have to be present in the formal solution:

$$\mathcal{T}\left\{e^{-\frac{i}{\hbar}\int_{0}^{t}H(t')dt'}\right\} = \sum_{n=0}^{\infty}\frac{1}{n!}\left(-\frac{i}{\hbar}\right)^{n}\mathcal{T} \left\{\left(\int_{0}^{t}H(t')dt'\right)^n\right\}$$

1. What is the first and second derivative with respect to time of $\mathcal{T}\left\{e^{-\frac{i}{\hbar}\int_{0}^{t}H(t')dt'}\right\}$, (which is essentially $e^{-\frac{i}{\hbar}\mathcal{T}\int_{0}^{t}H(t')dt'}$)? At first glance, they are:

1st derivative: $\mathcal{T} \left(\int_{0}^{t}H(t')dt'\right)e^{-\frac{i}{\hbar}\mathcal{T}\int_{0}^{t}H(t')dt}$

2nd derivative: $\mathcal{T} \left(\int_{0}^{t}H(t')dt'\right) \mathcal{T} \left(\int_{0}^{t}H(t')dt'\right)e^{-\frac{i}{\hbar}\mathcal{T}\int_{0}^{t}H(t')dt}$.

However, I doubt that. I think in the second derivative, the 2nd derivative should have an additional time-ordering:

$$\mathcal{T}(\mathcal{T} \left(\int_{0}^{t}H(t')dt'\right) \mathcal{T} \left(\int_{0}^{t}H(t'')dt''\right))e^{-\frac{i}{\hbar}\mathcal{T}\int_{0}^{t}H(t')dt}$$. Is my thinking right?

P.S. My argument comes from that $$I - \frac{i}{\hbar} \int_{0}^{t} dt' H(t') + \left(-\frac{i}{\hbar}\right)^2 \frac{1}{2} \mathcal{T}\left(\int_{0}^{t} dt' H(t')\right)^2+ \left(-\frac{i}{\hbar}\right)^3 \mathcal{T}\left(\frac{1}{3!}\left(\int_{0}^{t} dt' H(t')\right)^3\right) ...$$ is the same as $$I - \frac{i}{\hbar}\int_{0}^{t}dt' H(t') + \left(-\frac{i}{\hbar}\right)^2 \int_{0}^{t}dt' \int_{0}^{t'}dt'' H(t') H(t'')\\ + \left(-\frac{i}{\hbar}\right)^3 \int_{0}^{t}dt' \int_{0}^{t'}dt'' \int_{0}^{t''}dt''' H(t') H(t'') H(t''') + ...$$

• – Qmechanic Jul 21 '16 at 12:29

## 1 Answer

The evolution operator generated by a time-dependent Hamiltonian has two parameters, not just one (i.e. the initial and final time).

Let's denote such evolution by $(U(t,s))_{(t,s)\in\mathbb{R}^2}$. Formally, we can write $U(t,s)=e^{-i\int_s^t H(\tau)d\tau}$ (or with time-ordering if you want), but this has to be intended as follows.

$(U(t,s))_{(t,s)\in\mathbb{R}^2}$ is the time-dependent two-parameter family of evolutions for $(H(t))_{t\in\mathbb{R}}$ if it satisfies the following properties $\forall t,s,r\in\mathbb{R}$:

• $U(t,s)$ is unitary;
• $U(t,t)=I$;
• $U(t,s)U(s,r)=U(t,r)$ [$\Rightarrow U(t,s)^*=U(t,s)^{-1}=U(s,t)$];
• $i\partial_t U(t,s)=H(t)U(t,s)$, $i\partial_s U(t,s)=-U(t,s)H(s)$, in a suitable sense.

The existence and uniqueness of $U(t,s)$ depends from the regularity properties of $t\mapsto H(t)$; from the latter also depends in which sense the last point is satisfied. The easiest example is when the map is continuous in the operator norm, and then the Dyson series is rigorously defined and provides the unique $U(t,s)$.

Concerning OP's points:

1. Indeed if $U(t,s)$ is the evolution, $U(t,s)=I-i\int_s^t H(\tau)U(\tau,s)d\tau$ holds (in a suitable dense domain).

2. The formal expression with time-ordering has always to be intended as the evolution that satisfies the properties above (and that becomes a rigorous expression only for a norm-continuous $t\mapsto H(t)$). Therefore the derivatives are given as expected: the first derivative is "the Schrödinger equation", the second derivative is formally $\partial_t^2 U(t,s)= -H(t)^2U(t,s) - \dot{H}(t)U(t,s)$.

• Yes sorry I have updated my initial condition. Is my 2nd derivative expression correct? – diff Jul 21 '16 at 11:01
• @diff It does not look correct; in the derivative you should have no integrals outside the exponential. – yuggib Jul 21 '16 at 13:58