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Let $N\geq1$ be an integer and let $H:[0,1]^2\to\mathbb C^{N\times N}$ be a pointwise hermitean matrix valued function. For $y\in[0,1]$ and $0\leq a\leq b\leq 1$, let $U_y(b,a)$ be the time evolution operator (solution operator) to the Schrödinger equation $i\psi'(x)=H(x,y)\psi(x)$. More precisely, the operator is defined so that for any $\phi\in\mathbb C^N$ and $a\in[0,1]$ the solution of the initial value problem $$ \begin{cases} i\psi'(x)=H(x,y)\psi(x)\\ \psi(a)=\phi \end{cases} $$ is $\psi(x)=U_y(x,a)\phi$. (Alternatively, we could define $U_y(b,a)$ to be the solution of $U_y(a,a)=I$, $\partial_aU_y(a,b)=-iH(a,y)U_y(a,b)$ and $\partial_bU_y(a,b)=iU_y(a,b)H(b,y)$.)

I'm interested in differentiating this time evolution operator with respect to the parameter $y$. A sketchy calculation and physical intuition suggest that $$ \partial_yU_y(1,0) = -i\int_0^1U_y(1,x)\partial_yH(x,y)U_y(x,0)dx. $$ What regularity of $H$ do I need to assume for this to be true? Is there a reference for this result? I feel that giving a proof instead of a reference would be a distraction from what I want (readers of what I'm writing) to focus on, unless there is a particularly short and neat proof.

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  • $\begingroup$ That formula looks awfully familiar. If I remember where I've seen it before I'll post a reference. $\endgroup$ Commented Apr 29, 2015 at 10:19
  • $\begingroup$ In the case that $H$ is time-independent (so independent of $x$), I think your formula is the same as Theorem 2.19 in section IX.2.6 in Kato's Perturbation theory. $\endgroup$ Commented Apr 29, 2015 at 12:42
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    $\begingroup$ You can just differentiate the ODE solved by $U$ wrt $y$ and then apply variation of constants. That this is correct is discussed in any advanced ODE text (for example, Coddington-Levinson). $\endgroup$ Commented Apr 29, 2015 at 14:13

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Come to think of it, though, since you are in the ODE world, isn't what you wrote a direct consequence of variation of parameters?

I write it in a slightly easier (for me) notations.

Proposition: Let $S(t_1,t_2)$ be the solution operator for $\phi'(t) = A(t) \phi(t)$ with initial data prescribed at $t_2$, and let $T(t_1,t_2)$ be the solution operator for $\phi'(t) = B(t) \phi(t)$ with initial data prescribed at $t_2$, we have $$ S(t_1,t_2)x - T(t_1,t_2)x = \int_{t_2}^{t_1} T(t_2, s) [A(s) - B(s)] S(s,t_1) ~\mathrm{d}s .$$

Proof:

$$ \begin{align} (S(t,0) x - T(t,0)x)' &= A(t)S(t,0) x - B(t) T(t,0)x \\ &= [A(t)- B(t)]S(t,0) x + B(t) (S(t,0)x - T(t,0)x) \end{align}$$ so by Duhamel's principle we have $$ S(t,0) x - T(t,0)x = \int_0^t T(t,s) [A(s) - B(s)] S(s,0) x ~\mathrm{d}s $$


Now if $A(t) = B(t) + D(t) y + O(y^2)$, you immediately have for each fixed $t$

$$ S(t,0) x - T(t,0) x = y \int_0^t T(t,s) D(s) T(s,0) \mathrm{d}s + O(y^2) $$

which should imply the result you want.

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  • $\begingroup$ Thanks! I knew the integral identity for $S-T$, but I used it for something very different and it somehow didn't occur to me that it would be useful here. Silly me. $\endgroup$ Commented Apr 29, 2015 at 17:30

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