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A N-particles system is described by the following Hamiltonian $$H=H_0+H'(t)$$ where $H_0$ is the unperturbated Hamiltonian and $H'(t)$ is the perturbation $$H'(f)=-A\cdot\mathcal{F}(t)$$ written as the coupling of a "characteristic lenght" related to the system ($A$) and a "generalized external force" ($\mathcal{F}(t)$): infact, the product of the two has to be energy-dimensional.

The phase-space distribution function of this N-particles system is written as follows: $$\rho(t)=\rho_0(t)+\Delta\rho(t)$$ where the first contribution is the one related to the unperturbated system and the second one is the one due to the perturbation (first-order only, being interested in LINEAR response theory!).

Now, defining the Liouville Operator as $$\mathcal{L}\equiv i\{H,\cdot\}$$ it is possible to to write the equation of motion of $\rho(t)$ as $$\frac{\partial\rho(t)}{\partial t}=-i\mathcal{L_0}\rho(t)-\{A,\rho(t)\}\mathcal{F}(t)$$ ($\mathcal{L_0}\equiv i\{H_0,\cdot\}$) and - substituing the expression $\rho(t)=\rho_0(t)+\Delta\rho(t)$ in it, simplifying and neglecting higher-order terms - one can derive the following equation: $$\frac{\partial\Delta\rho(t)}{\partial t}=-i\mathcal{L_0}\Delta\rho(t)-\{A,\rho_0(t)\}\mathcal{F}(t)$$

My book says that integrating the last differential equation (having set the origin at $t=-\infty$) one can obtain the solution $$\Delta\rho(t)=-\int_{-\infty}^{t} e^{-i(t-s)\mathcal{L_0}} \{A,\rho_0(s)\}\mathcal{F}(s) ds$$ I think I understand the physical meaning of the last equation but here comes my question: how can I derive it myself? I don't seem to be able to do the math, because I don't know how to work with operators (the Liouville one) in this type of problems.

Hope someone will help me, thank you in advance for your time!

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    $\begingroup$ Just as a hint--think of the equation you wrote down for $\Delta \rho$ as a normal ODE, and just think of $-i\mathcal{L}_0$ as a constant, and $\{A,\rho_0(t)\} \mathcal{F}(t)$ as a known function of time, so your ODE would be something like $\dot{\Delta \rho}=C\Delta \rho + F(t)$. Would you know how to solve this ODE? $\endgroup$ – Andrew Jul 8 '15 at 22:14
  • $\begingroup$ Thank you Andrew! Yes, I'd know how to solve that and that's what I tried but then I get things like $e^{-\int_{-\infty}^{t} -i\mathcal{L_0} ds}$ and I don't know how to handle them... $\endgroup$ – lucia de finetti Jul 8 '15 at 22:23
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    $\begingroup$ OK good! So I would say you do know how to derive that equation and your main question is how to interpret the exponential of a differential operator (which is a pretty valid thing to wonder about!). As written that expression is pretty formal. To actually use that equation, probably you either want to expand the exponential as a Taylor series, or you want to expand $\{A,\rho \} \mathcal{F}$ in terms of eigenfunctions of $\mathcal{L}_0$, since if $\mathcal{L}_0 f = \lambda f$, then $e^{i \mathcal{L}_0} f = e^{i \lambda}f$ (at least at a physics level of rigor.) $\endgroup$ – Andrew Jul 8 '15 at 22:28
  • $\begingroup$ Thank you! Still I don't know how to solve the integral in the exponential I wrote ($e^{-\int_{-\infty}^{t} -i\mathcal{L_0} ds}$), exponential that figures in the solution of the DE...how should I proceed? $\endgroup$ – lucia de finetti Jul 9 '15 at 6:50
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    $\begingroup$ I don't think you should think of $e^{i \int\mathcal{L}_0}$ as an integral per se. You should think of it as a differential operator that acts on $\{A,\rho\} \mathcal{F}$. I think probably the point is that you need to expand $\{ A,\rho \} \mathcal{F}$ in terms of eigenfunctions of $\mathcal{L}_0$, then the whole thing $\int e^{i\mathcal{L}_0}\{ A,\rho \} \mathcal{F}$ should turn into a sum of ordinary integrals. For example, if $\mathcal{L}_0$ is the laplacian $\nabla^2$, the eigenfunctions are $e^{ikx}$, in which case you want to take the Fourier transform of $\{A,\rho\}\mathcal{F}$. $\endgroup$ – Andrew Jul 9 '15 at 12:25
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We'll give some details based on Andrew's comments.

We are trying to solve $$ \frac{ \partial \Delta \rho(t) }{ \partial t } = -i\mathcal L_0 \Delta \rho(t) -\{A, \rho_0(t) \} \, \mathcal F(t). \qquad (1) $$

Homogeneous solution

To develop some familiarity with the Liouville operator, let us first consider the special case of $A = 0$. Then we have a homogeneous equation: $$ \frac{ \partial \Delta \rho(t) }{ \partial t } = -i\mathcal L_0 \Delta \rho(t) \qquad (2) $$ Note that the Liouville operator $\mathcal L_0$ does not explicitly depend on $t$ because the unperturbed Hamiltonian $H_0(q, p)$ is not time-dependent. That is, for any function $f(q, p)$, we have $$ \begin{align} -i\mathcal L_0 f(q, p) &= \{H_0(q, p), f(q, p)\} \\ &= \sum_i \frac{ \partial H_0} { \partial q_i } \frac{ \partial f} { \partial p_i } - \frac{ \partial H_0} { \partial p_i } \frac{ \partial f} { \partial q_i }, \end{align} $$ which is a time-independent expression. In short, $\mathcal L_0$, albeit its operator nature, is a constant of time. This means that we can integrate $(2)$ as if it were a regular ODE: $$ \begin{aligned} \Delta \rho(t) = e^{-i t\mathcal L_0} g(q, p), \end{aligned} \qquad (3) $$ where $g(q, p)$ is some time-independent function. To be safe, we can plug this into the $(2)$ to directly verify that this is indeed a solution: $$ \begin{aligned} \mbox{l.h.s.}= \frac{\partial \Delta \rho(t) }{\partial t} = i \mathcal L_0 \,e^{-i t\mathcal L_0} g(q, p) = i \mathcal L_0 \, \Delta \rho(t) = \mbox{r.h.s.}. \end{aligned} $$

Special solution

We can similar borrow the technique for regular ODEs to solve $(1)$. Denote $$ B(t) \equiv -\{A, \rho_0(t)\} F(t), $$ then $(1)$ is equivalent to $$ \frac{ \partial \Delta \rho(t) }{ \partial t } = -i\mathcal L_0 \Delta \rho(t) +B(t). \qquad (4) $$ To solve $(4)$, we shall now treat $g(q, p)$ in the homogeneous solution to be time dependent, $g(q, p, t)$, i.e., we postulate a trial solution $$ \begin{aligned} \Delta \rho(t) = e^{-i t\mathcal L_0} g(q, p, t). \end{aligned} \qquad (5) $$ Then $$ -i\mathcal L_0 e^{-i t\mathcal L_0} g(q, p, t) + e^{-i t\mathcal L_0} \frac{\partial}{\partial t} g(q, p, t) = -i\mathcal L_0 e^{-i t\mathcal L_0} g(q, p, t) +B(t), $$ or $$ \frac{\partial}{\partial t} g(q, p, t) = e^{i t\mathcal L_0} B(t). $$ The right hand side is just a function, so $$ g(q, p, t) = \int_{-\infty}^t e^{i s\mathcal L_0} B(s) \, ds, $$ assuming $B(-\infty) = 0$.

Using this in $(5)$, we get $$ \begin{aligned} \Delta \rho(t) &= e^{-i t\mathcal L_0} \int_{-\infty}^t e^{i s\mathcal L_0} B(s) \, ds \\ &= \int_{-\infty}^t e^{-i (t - s) \mathcal L_0} B(s) \, ds \\ &= -\int_{-\infty}^t e^{-i (t - s) \mathcal L_0} \{A, \rho_0(s)\} \mathcal F(s) \, ds. \end{aligned} $$ This is the desired result. To ensure nothing went wrong, we can always explicitly plug this solution in $(1)$.

In summary, our solution is a simple copy of the general solution of a regular ODE. The key is to remember that $\mathcal L_0(q, p)$, unlike $\mathcal L(q, p, t)$, does not explicitly depend on time.

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