I am attempting to retrace steps performed numerous times before and derive Caldeira-Leggett's influence functional found in their paper "Path Integral Approach To Quantum Brownian Motion".

However, I'm getting caught up in what I am sure is a fundamental misunderstanding on my part. I'll pose my question after laying out the mathematical "path" I have followed to get to the current point:

[One of the multiple resources online I've found, and have been following, see pages 16-18 of : http://web.science.uu.nl/itf/Teaching/2006/MxWakker.pdf]

Following the path integral formulism, and obtaining the action for a forced harmonic oscillator from R.P. Feynman and A.R. Hibbs, Quantum Mechanics and Path Integrals,

$ \begin{equation} S^{(k)}_{cl}[x(t),R_{k},R^{\prime}_{k}]=\frac{m\omega_{k}}{2\sin(\omega_{k} T)} \left[ (R_{k}^{2}+R_{k}^{\prime2})\cos(\omega_{k} T)-2R_{k}R^{\prime}_{k} + \\ R_{k}\frac{2}{m \omega_{k}}\int_{t_{a}}^{t_{b}}x(t)\sin(\omega_{k}(t-t_{a}))dt + R^{\prime}_{k}\frac{2}{m \omega_{k}}\int_{t_{a}}^{t_{b}}x(t)\sin(\omega_{k}(t_{b}-t))dt \\ - \frac{2}{m^{2} \omega_{k}^{2}}\int_{t_{a}}^{t_{b}}\int_{t_{a}}^{t}x(t)x(s)\sin(\omega_{k}(t_{b}-t))\sin(\omega_{k}(s-t_{a}))dsdt \right] \end{equation} $ $ \tag{1} $

where $T=t_{b}-t_{a}$, and $x(t)$ is an external force, then the resulting kernel is

$K^{(k)}[x(t),R_{k},R^{\prime}_{k}]= \left(\frac{m\omega_{k}}{2\pi i \sin(\omega_{k} T)}\right)^{\frac{1}{2}} \exp \left[\frac{i}{\hbar} S^{(k)}_{cl}[x(t),R_{k},R^{\prime}_{k}] \right] $ $ \tag{2}$

With this information, the object I am trying to obtain is the influence functional,

$F[x(t),y(t)] \equiv \int \int \int dR^{\prime}dQ^{\prime}dR \rho_{B}(R^{\prime},Q^{\prime},0) \int_{R^{\prime}}^{R} \int_{Q^{\prime}}^{Q} [dQ(t)][dR(t)] \\ \exp \left[\frac{i}{\hbar} (( S^{(k)}_{I}[x(t),R(t)]-S^{(k)}_{I}[y(t),Q(t)]) \\ + ( S^{(k)}_{B}[x(t),R(t)]-S^{(k)}_{B}[y(t),Q(t)])) \right] $ $ \tag{3}$

which can be further simplified using

$ \rho_{B}(R^{\prime},Q^{\prime},0) \equiv \prod_{k} \rho^{(k)}_{B}(R_{}^{\prime},Q_{k}^{\prime},0) \\ \equiv \prod_{k} \left(\frac{m\omega_{k}}{2\pi \hbar \sinh\left(\frac{m\omega_{k}}{k_{b}T}\right)}\right) \exp \left[ \left(\frac{-m\omega_{k}}{2 \hbar \sinh\left(\frac{m\omega_{k}}{k_{b}T}\right)}\right) \left( (R_{k}^{\prime2}+Q_{k}^{\prime2})\cosh\left(\frac{m\omega_{k}}{k_{b}T}\right) - 2 R_{k}^{\prime}Q_{k}^{\prime}\right)\right] $ $ \tag{4}$

which can be substituted into equation (3). A further simplification of (3) is the substitution

$ \int_{R^{\prime}}^{R} \int_{Q^{\prime}}^{Q} [dQ(t)][dR(t)] \\ \exp \left[\frac{i}{\hbar} (( S^{(k)}_{I}[x(t),R(t)]-S^{(k)}_{I}[y(t),Q(t)]) + ( S^{(k)}_{B}[x(t),R(t)]-S^{(k)}_{B}[y(t),Q(t)])) \right] \\ = \prod_{k} K^{(k)}[x(t),R_{k},R^{\prime}_{k}] K^{*(k)}[y(t),R_{k},Q^{\prime}_{k}] $ $\tag{5}$

$K^{(k)}$ is the same as that found in equation (2).

All this substitution will leave a simple, yet tedious, series of Gaussian integrals through $R_{k},R^{\prime}_{k}$ and $Q^{\prime}_{k}$.

My question is this: It appears to me that when you multiply the memory kernel in equation (2) by its conjugate, as is done in equation (5), then the common factor of $R_{k}^{2}$ should cancel, leaving terms of $R_{k}$ in the first order. This would end up being a dirac delta function when being integrated through $R_{k}$. When I did this, first by hand and then with Mathematica, I could not reproduce the answer found in the Caldeira paper. It also goes against the many comments I have read on this derivation, particularly regarding the "multiple tedious gaussian integrals". Is my understanding of calculating the influence functional completely off? Particularly I am wondering if I have misinterpreted equation (5), was simply conjugating the memory kernel (alongside substituting $x(t)$ to $y(t)$), and substituting all the elements I have listed here, naive of me?

Thank you for reading all of that, and any help you can provide!

  • $\begingroup$ I think the upper limit on the 2nd integral on the l.h.s of eq.(5) is $R$, not $Q$. And substituting should work just fine. From what I can tell, even if the terms in $R_k^2$ disappear after multiplying $K^{(k)}$ by its complex conjugate, you are still left with a bunch of Gaussian integrals: integrating the remaining terms in $R_k$ obtains delta functions with arguments of the form $R'_k - Q'_k - f(x, y)$, which means you'd be left with Gaussian integrals over one of $R'_k$ or $Q'_k$. Does your end result include integrating those Gaussians? $\endgroup$ – udrv Mar 26 '17 at 18:21
  • $\begingroup$ Thanks for replying, I really appreciate it! Yes, my end result does include integrating these gaussians. I am left with a prefactor which isn't in the desired result, and a cosectant term multiplied by everything in the exponent. Interestingly, if this cosectant term was not present, I would have an answer very close to that of the result in the paper. $\endgroup$ – Tbone Willsone Mar 27 '17 at 11:41
  • $\begingroup$ If it'll help, I can post/link a screencap of my Mathematica result. $\endgroup$ – Tbone Willsone Mar 27 '17 at 11:46
  • $\begingroup$ Sure, why not. If it's long a link would be better. $\endgroup$ – udrv Mar 27 '17 at 18:26
  • $\begingroup$ Here's a link, I've annotated, hopefully enough so that it's understandable! dropbox.com/s/r15uozx54i9i6y0/CaldeiraLeggettExample.nb?dl=0 $\endgroup$ – Tbone Willsone Mar 28 '17 at 9:52

The integrals you obtained after doing the Gaussians are correct. All you need is an integration trick and a bit of algebra to rearrange them in the final form.

The integration trick: A double integral over a 2-dimensional square domain decomposes as $$ \int_0^t dt' \int_0^t dt" f(t', t") = \int_0^t dt' \int_0^{t'} dt" f(t', t") + \int_0^t dt" \int_0^{t"} dt' f(t', t") $$ But if we interchange the labeling of the integration variables in the 2nd term we get an even simpler form: $$ \int_0^t dt' \int_0^t dt" f(t', t") = \int_0^t dt' \int_0^{t'} dt" f(t', t") + \int_0^t dt' \int_0^{t'} dt" f(t", t') = \\ = \int_0^t dt' \int_0^{t'} dt" \Big( f(t', t") + f(t", t') \Big) $$

For the rest of the answer I will only refer to the integrals in your handwritten notes, leaving aside any irrelevant factors.

Let us apply the above decomposition to your first set of integrals, which reads $$ I_1 =\frac{1}{\sin \omega_k t} \left[ - 2 \int_0^t{dt' \int_0^{t'}{ dt" \Big( x(t')x(t") - y(t')y(t") \Big) \sin\omega_k(t-t') \sin\omega_k t" }} + \\ + \int_0^t{dt'\int_0^t{dt" \Big( x(t') + y(t')\Big) \Big( x(t") - y(t")\Big) \sin\omega_k(t-t') \sin\omega_k t" }} \right] $$ We only want to write the 2nd term, since the 1st is already confined to half the square $[0,t]x[0,t]$: $$ I_1 = \frac{1}{\sin \omega_k t} \left[ - 2 \int_0^t{dt' \int_0^{t'}{ dt" \Big( x(t')x(t") - y(t')y(t") \Big) \sin\omega_k(t-t') \sin\omega_k t" }} + \right.\\ + \int_0^t{dt'\int_0^{t'}{dt" \left[ \Big( x(t') + y(t')\Big) \Big( x(t") - y(t")\Big) \sin\omega_k(t-t') \sin\omega t" + \right.}} \\ + \left.\left.\Big( x(t') - y(t')\Big) \Big( x(t") + y(t")\Big) \sin\omega_k t' \sin\omega_k(t-t") \right]\right] $$ Now we can rewrite everything under the same integrals, and proceed to regroup and simplify the first 2 terms. We obtain eventually $$ I_1 = \frac{1}{\sin \omega_k t} \int_0^t{dt' \int_0^{t'}{ dt" \left[ - \Big( x(t') - y(t') \Big) \Big( x(t") + y(t")\Big) \sin\omega_k(t-t') \sin\omega_k t" + \right.}} $$ $$ + \left. \Big( x(t') - y(t')\Big) \Big( x(t") + y(t")\Big) \sin\omega_k t' \sin\omega_k(t-t") \right] $$ $$ = \frac{1}{\sin \omega_k t} \int_0^t{dt' \int_0^{t'}{ dt" \Big( x(t') - y(t') \Big) \Big( x(t") + y(t")\Big)\cdot \\ \cdot \left[ \sin\omega_k t' \sin\omega_k(t-t") - \sin\omega_k(t-t') \sin\omega_k t" \right]}} $$ But doing the trigonometry in the last square brackets gives $$ \sin\omega_k t' \sin\omega_k(t-t") - \sin\omega_k(t-t') \sin\omega_k t" = $$ $$ = \sin\omega_k t' \sin\omega_k t \cos\omega_k t" - \sin\omega_k t' \cos\omega_k t \sin\omega_k t" - $$ $$ - \sin\omega_k t \cos\omega_k t' \sin\omega_k t" + \cos\omega_k t \sin\omega_k t' \sin\omega_k t" = \sin\omega_k t \sin\omega_k(t'-t") $$ and we finally have $$ I_1 = \int_0^t{dt' \int_0^{t'}{ dt" \Big( x(t') - y(t') \Big) \Big( x(t") + y(t")\Big)\sin\omega_k(t'-t")}} $$ Now for the 2nd set of integrals, which reads $$ I_2 = \frac{1}{\sin^2 \omega_k t} \left[ \int_0^t{dt' \int_0^{t}{ dt" \Big( x(t') - y(t') \Big) \Big( x(t") + y(t")\Big)\sin\omega_k(t-t')\sin\omega_k(t-t') }} + \right. $$ $$ + \int_0^t{dt' \int_0^{t}{ dt" \Big( x(t') - y(t') \Big) \Big( x(t") + y(t")\Big)\sin\omega_kt' \sin\omega_kt" }} + $$ $$ \left.+ 2 \cos\omega_kt \int_0^t{dt' \int_0^{t}{ dt" \Big( x(t') - y(t') \Big) \Big( x(t") + y(t")\Big)\sin\omega_kt'\sin\omega_k(t-t') }} \right] $$ First bring everything under one integral and simplify the trigonometry by expanding the sines as before. I leave the details as an exercise (lots of typing), but the result is $$ I_2 = \frac{1}{\sin^2 \omega_k t} \int_0^t{dt' \int_0^{t}{ dt" \Big( x(t') - y(t') \Big) \Big( x(t") + y(t")\Big) \left[ \sin\omega_k(t-t')\sin\omega_k(t-t') + \right. }}\\ \left. + \sin\omega_kt' \sin\omega_kt" + 2\cos\omega_kt \sin\omega_kt'\sin\omega_k(t-t') \right] = \\ = \frac{1}{\sin^2 \omega_k t} \int_0^t{dt' \int_0^{t}{ dt" \Big( x(t') - y(t') \Big) \Big( x(t") + y(t")\Big) \cdot }}\\ \cdot \left[ \sin\omega_k t \sin\omega_k t' \sin\omega_k t \sin\omega_k t" + \sin\omega_k t \cos\omega_k t' \sin\omega_k t \cos\omega_k t' \right] = $$ So $$ I_2 = \int_0^t{dt' \int_0^{t}{ dt" \Big( x(t') - y(t') \Big) \Big( x(t") + y(t")\Big) \cos\omega_k(t'-t")}} $$ The rest is just a matter of summing up the different contributions so as to write the influence integral neatly.

  • $\begingroup$ @TboneWillsone Welcome. $\endgroup$ – udrv Apr 7 '17 at 18:18

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