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I am trying to build a sort of time propagator matrix. I have a matrix $H$ that is 500x500 and i am trying to evaluate $\exp(-H t)$. I need to find this in a symbolic way because i will need to do $\frac{d}{dt}\langle i|\exp(-H t)|i_o \rangle $ for given vectors $|i \rangle$ and $|i_0 \rangle$. I have done it using the Schur and Jordan decomposition, but i can't do it. My pc either freezes, abort the evaluation or it says it can't perform the evaluation.

I know that the problem is this symbolic $t$ variable since if a give a numerical value to it, Mathematica solves it in a few seconds. But how can i do it for any $t$? Am I asking to much of it? My pc is has **Intel® Core™ i7-8550U CPU @ 1.80GHz × 8 ** and i am using Mathematica 11.0.1 under Ubuntu. Can you guys help me?

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    $\begingroup$ Do you really need it symbolically for any t? Can you not just evaluate it for many close-together t and take the derivative numerically? $\endgroup$
    – evanb
    Commented Aug 21, 2020 at 19:52
  • $\begingroup$ Are the matrix $H$ and the vector $i_0$ numerical? If so, why not use NDSolveValue? $\endgroup$
    – Carl Woll
    Commented Aug 21, 2020 at 20:17
  • $\begingroup$ You should consider the limitations that play into these sorts of large symbolic computations. Depending upon how many free parameters you have, you might consider a purely numerical procedure using the suggestions that others have offered. That said, I suggest you investigate if there are any symmetries in your matrix that you can exploit. Why are you trying to evaluate this married exponential? Is H a function of t? If it is indeed a time propagator problem, perhaps you can use the Magnus Expansion? $\endgroup$
    – CA Trevillian
    Commented Aug 21, 2020 at 23:52
  • $\begingroup$ My matrix $H$ is independent of t. All it's components are real number in the range [0,1] . I need it for every t because i have to evaluate $\int_0^{\infty} dt t \frac{d}{dt} <i| \exp(-Ht)|i_o>$. Due to the form of those vectors, $|i> = {0,0,...1,...0} $ with the $1$ on the $i-th$ component, $<i| \exp(-Ht)|j>$ is the component $ij$-th component of the $\exp(-Ht)$. Is there a way to evaluate just that component and not the hole matrix? $\endgroup$
    – Hugo Andrade
    Commented Aug 22, 2020 at 2:35

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Erm. Isn't $\frac{\mathrm{d}}{\mathrm{d} t} \exp(- t H) = -H \exp(- t H) = - \exp(- t H)H$? The latter two can be evaluated numerically....

Addendum:

This is more and more becoming clear that this is an XY-question. From the comments it appears that you seek to compute $$ \int_0^\infty t \, \frac{\mathrm{d}}{\mathrm{d} t} \langle u | \exp(- t H) | v\rangle \, \mathrm{d} t $$ Provided that $H$ is invertible, integration by parts leads to $$\begin{aligned} & \int_0^T t \, \frac{\mathrm{d}}{\mathrm{d} t} \langle u | \exp(- t H) | v\rangle \, \mathrm{d} t \\ &= \Big[ t \, \langle u | \exp(- t H) | v\rangle \Big]_{t=0}^{t=T} - \int_0^T \langle u | \exp(- t H) | v\rangle \, \mathrm{d} t \\ &= \Big[ t \, \langle u | \exp(- t H) | v\rangle \Big]_{t=0}^{t=T} + \Big[ \langle u | H^{-1} \exp(- t H) | v\rangle \Big]_{t=0}^{t=T} \\ &= T \, \langle u | \exp(- T H) | v\rangle - 0 + \langle u | H^{-1} \exp(- T H) | v\rangle - \langle u | H^{-1} | v\rangle. \end{aligned} $$ Provided that the real parts of the eigenvalues of $H$ are contained in $[\varepsilon,\infty)$ for some $\varepsilon > 0$, $\langle u | H^{-1} \exp(- T H) | v\rangle$ converges rapidly to $0$ for $T \to \infty$, so we can apply the limit to obtain $$ \int_0^\infty t \, \frac{\mathrm{d}}{\mathrm{d} t} \langle u | \exp(- t H) | v\rangle \, \mathrm{d} t = - \langle u | H^{-1} | v\rangle . $$

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  • $\begingroup$ My matrix $H$ is independent of t. All it's components are real number in the range [0,1] . I need it for every t because i have to evaluate $\int_0^{\infty} dt t \frac{d}{dt} <i| \exp(-Ht)|i_o>$. Due to the form of those vectors, $|i> = {0,0,...1,...0} $ with the $1$ on the $i-th$ component, $<i| \exp(-Ht)|j>$ is the component $ij$-th component of the $\exp(-Ht)$. Is there a way to evaluate just that component and not the hole matrix? $\endgroup$
    – Hugo Andrade
    Commented Aug 22, 2020 at 2:39
  • $\begingroup$ I am sorry. i write it in a cumbersome way. Let me rewrite it, $\int_0^{\infty}t*\frac{d}{dt}<u|exp(-t*H)|v> dt$ $\endgroup$
    – Hugo Andrade
    Commented Aug 22, 2020 at 11:26
  • $\begingroup$ I see. I edited the post. ($dtt$ really looked like a typo.) $\endgroup$
    – Henrik Schumacher
    Commented Aug 22, 2020 at 13:47
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    $\begingroup$ It is probably worth mentioning that the formula you derive is the partial case of the Laplace transform of matrix exponential $\int_0^\infty e^{-ts}e^{tH}\,dt=(sI-H)^{-1}$. Good connection to Green's functions and resolvents. $\endgroup$
    – yarchik
    Commented Aug 22, 2020 at 14:08
  • $\begingroup$ I think your integration by parts output shouldn’t have d/dt inside the integral $\endgroup$
    – Carl Woll
    Commented Aug 22, 2020 at 17:25

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