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Flux-flux correlation function under Feynman's path integral

  1. Mar 25, 2017 #1
    This is a chemically inspired problem, but the path is fully quantum mechanics and a bunch of integrals.

    How does one calculate fully quantum mechanical rate ($\kappa$) in the golden-rule approximation for two linear potential energy surfaces?

    Attempt:

    Miller (83) proposes $$\kappa=\int{Tr[\exp{(-\beta\hat{H})}\hat{F}\exp{(-i\hat{H}t/\hbar)}\hat{F}\exp{(i\hat{H}t/\hbar)}]}dt$$

    Where integrand is simply the flux-flux correlation function: $C_{ff}(t)$. Which can be calculated under Feynman's path integral formalism. My attempt (which is in vain) at calculating $C_{ff}(t)$ is as follows:

    $$C_{ff}(t)=Tr[\exp{(-\beta\hat{H})}\hat{F}\exp{(-i\hat{H}t/\hbar)}\hat{F}\exp{(i\hat{H}t/\hbar)}]$$

    $$Tr[\exp{(-\beta\frac{\hat{H}}{2})}\hat{F}\exp{(-\beta\frac{\hat{H}}{2})}\exp{(-i\hat{H}t/\hbar)}\hat{F}\exp{(i\hat{H}t/\hbar)}]$$

    By cyclicly permuting the operators we reach at:

    $$Tr[\exp{(i\hat{H}t/\hbar)}\exp{(-\beta\frac{\hat{H}}{2})}\hat{F}\exp{(-\beta\frac{\hat{H}}{2})}\exp{(-i\hat{H}t/\hbar)}\hat{F}]$$

    The boltzmann operator and quantum mechanical propagator can be combined as follows:

    $$Tr[\exp{\hat{H}(\frac{it}{\hbar}-\frac{\beta}{2})}\hat{F}\exp{\hat{H}(\frac{-it}{\hbar}-\frac{\beta}{2})}\hat{F}]$$

    In the golden-rule (non-adiabatic) case, we have two electronic states 0 and 1. So F is simply a projection operator. Hence one can obtain:

    $$Tr[\exp{\hat{H_0}(\frac{it}{\hbar}-\frac{\beta}{2})}\exp{\hat{H_1}(\frac{-it}{\hbar}-\frac{\beta}{2})}]$$

    This basically is kernel corresponding to two potential energy surfaces $V_0$ and $V_1$. For trajectory starting at $x_a$ and ending at $x_b$, we have

    $$C_{ff}(t)=\int{\int{K_0(x_a,x_b,\frac{it}{\hbar}-\frac{\beta}{2})K_1(x_b,x_a,\frac{-it}{\hbar}-\frac{\beta}{2})}}dx_adx_b$$

    For a linear potential energy surfaces (PES), where my PES looks as follows:

    $$V_0=k_0 x$$

    $$V_1=k_1 x$$

    My kernels are:

    $$K_0=\sqrt{\frac{m}{2\pi t_0}}\exp{(-S_0)}$$

    $$K_1=\sqrt{\frac{m}{2\pi t_1}}\exp{(-S_1)}$$

    $S's$ correspond to action which is:

    $$S_n(x_a,x_b,t_n)=\frac{m(x_a-x_b)^2}{2 t_n}-\frac{(x_a+x_b)k_nt_n}{2}-\frac{k_n^2t_n^3}{24m}$$

    The problem is the integral for flux flux correlation function doesn't seem to be converging with the imaginary argument for $t$'s. I am trying to integrate w.r.t $x_a$, $x_b$ and $t$ from -Inf to +Inf. My final answer for rate should look something like this:

    $$\exp{\frac{k_0^2k_1^2\hbar^2\beta^3}{24m(k_0-k_1)^2}}$$

    Is it a gaussian integral with respect to $x_a$ and $x_b$? One has to be careful because there is also an imaginary parts in the exponent. How does one reach the final answer for rate with those integrals? Really confused! Any help is appreciated.
     
  2. jcsd
  3. Mar 26, 2017 #2

    DrDu

    User Avatar
    Science Advisor

    So ##\hat{F}=\frac{1}{2} \{p,\delta(x)\}##?
     
  4. Mar 26, 2017 #3
    Nope! F is just flux operator from reactant to product ##(|0><1|-|1><0|)##. This is for my case (or non-adiabatic case). However, in adiabatic case it is ##(p \delta(x-s)+\delta(x-s) p)##. But I am mainly interested in non-adiabatic case, where there are two electronic states. Basically, in adiabatic case, there is one potential energy surface, and for non-adiabatic case there are two potential energy surfaces.
     
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