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Suppose we have a parameterized time-dependent Hamiltonian $$H(t; \theta) = H_0 + f(t; \theta)V_0$$, is there any way to obtain the derivative of time-evolution operator $\frac{d}{d\theta} U(t, t_0; \theta)$ analytically (in terms of $H_0$, $V_0$ and $f(t; \theta)$)?

Ultimately my goal is to find the derivative

$$\frac{d}{d\theta} \langle \phi | U^\dagger(t, t_0; \theta) A U(t, t_0; \theta) | \phi \rangle$$.

Here is my attempt:

Let $t_0 = 0$ for now. The time-evolution operator in the interation picture can be written in Dyson series:

$$U_I(t, 0; \theta) = \sum_{n=0}^\infty U_n(t, 0;\theta)$$

$$U_n(t, 0; \theta) = (-i)^n \int_0^t dt_1 \cdots \int_0^{t_{n-1}} dt_n V_I(t_1; \theta) \cdots V_I(t_n; \theta) $$

$$V_I(t; \theta) = f(t; \theta) e^{iH_0t} V_0 e^{-iH_0t}$$

The derivative to each term $U_n(t, 0;\theta)$ is

$$\frac{d}{d\theta} U_n(t, 0; \theta) = (-i)^n \int_0^t dt_1 \cdots \int_0^{t_{n-1}} dt_n \frac{d}{d\theta} \left( V_I(t_1; \theta) \cdots V_I(t_n; \theta) \right)$$

Simplifying the problem even more, assume time and parameter dependency are separable: $f(t; \theta) = g(\theta) h(t)$. The derivative above becomes

$$\frac{d}{d\theta} U_n(t, 0; \theta) = n \frac{g'(\theta)}{g(\theta)}U_n(t, 0; \theta)$$

So, the derivative to the time-evolution operator is

$$\frac{d}{d\theta} U_I(t, 0; \theta) = \frac{g'(\theta)}{g(\theta)} \sum_{n=0}^\infty n U_n(t, 0;\theta)$$

Here is where I am stuck. I have no idea how to handle the summation $\sum_{n=0}^\infty n U_n(t, 0;\theta)$.

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  • $\begingroup$ I suppose $f(t;\theta)$ is meant to be a c-number function commuting with $H_0$ and $V_0$? The answer to your question depends on what you mean be an "analytic solution". As a first try you could write down the (trivial) solution of the (admittedly not very interesting) special case where $H_0$ and $V_0$ commute. If you like the form of this simple solution, you may still proceed to the general case... $\endgroup$
    – Hyperon
    Dec 26, 2022 at 7:22
  • $\begingroup$ @Hyperon Well if $H_0$ and $V_0$ commute, then $V_I(t;\theta)=f(t;\theta)V_0$ and each term of the Dyson series can be calculated directly as $U_n(t, 0; \theta)=\frac{1}{n!}(-iV_0\int_0^t d\tau f(\tau; \theta))^n$, so the time-evolution operator has a neat form: $U(t, 0; \theta) = e^{-iV_0\int_0^t d\tau f(\tau; \theta)}$, and the derivative wrt $\theta$ follows pretty obviously. I like the form of this solution (very much), but attacking the problem from this angle doesn't yield the desired result as shown in my attempt above. $\endgroup$
    – haoyu
    Dec 26, 2022 at 9:47

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