The following is from Ref. 1.

Given the (Euclidean) action for a particle ($q$) coupled to a bath of harmonic oscillators $q_\alpha$. Goal is to find an effective action for the particle, e.g integrate out the bath degrees of freedom.

$$ S_{particle} = \int_0^\beta d\tau'\left(\frac{m}{2}(\partial_{\tau'}q)^2+V(q) \right)$$ $$ S_{bath} = \sum_\alpha\frac{m_\alpha}{2}\int_0^\beta d\tau'\left((\partial_{\tau'}q_{\alpha})^2+\omega_\alpha^2q_{\alpha}^2 \right)$$ and coupling $$ S_c = \sum_\alpha\int_0^\beta d\tau'\left( f_\alpha[q]q_\alpha+\frac{f_\alpha[q]^2}{2m_\alpha\omega_\alpha^2}\right) $$ $f_\alpha$ are some generic functions and the $q$ and $q_\alpha$ obey periodic boundary conditions. My first instinct is switching to frequency space: $$ S_{bath} = \sum_{\alpha,n} \frac{m_\alpha}{2}q_{\alpha,n}\underbrace{(\omega_n^2+\omega_\alpha^2)}_{A_{n,-n}}q_{\alpha,-n} $$

My problem is a) The matrix $A$ is not diagonal and b) the coupling term still reads $$ S_c = \sum_{\alpha,n} \int_0^\beta d\tau' f_\alpha[q]q_{\alpha,n}\exp(i\omega_n\tau') $$ and I don't know how to get rid of the integration.

Does somebody have a hint on how to perform the gaussian integration?

The effective action should read $$ S_{eff} = S_{particle}[q] + \frac{\beta}{2}\sum_{\alpha,n}\frac{\omega_n^2f_\alpha[q(\omega_n)]f_\alpha[q(-\omega_n)]}{m_\alpha\omega_\alpha^2(\omega_\alpha^2+\omega_n^2)} $$


  1. A. Altland and B. Simons, Condensed Matter Field Theory, Second edition (2010), p.130.
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    $\begingroup$ The interacting part of $S_c$ could be written something like: $S_c \sim \Sigma_{\alpha, n} (f_\alpha[q(-\omega_n)]q_{\alpha,n}+f_\alpha[q(\omega_n)]q_{\alpha,-n})$. $\endgroup$
    – Trimok
    Sep 13, 2013 at 17:20
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    $\begingroup$ And the other term of $S_c$ could be written something like $\sim \Sigma_{\alpha, n} \frac{f_\alpha[q(-\omega_n)]f_\alpha[q(\omega_n)]}{2 m_\alpha \omega_\alpha ^2}$ $\endgroup$
    – Trimok
    Sep 13, 2013 at 17:31
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    $\begingroup$ The global matrix is block-diagonal in the set ($q_{\alpha,n},q_{\alpha,-n}$), and the restricted matrix to this set is proportionnal to $\begin{pmatrix}0&1\\1&0\end{pmatrix}$, so easily invertible. $\endgroup$
    – Trimok
    Sep 13, 2013 at 17:37
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    $\begingroup$ @Trimok That's the answer to the question! Post it up so I can upvote :) $\endgroup$ Sep 14, 2013 at 0:01
  • $\begingroup$ @MarkMitchison : Not quite, Mark... I shamely have a problem with the $\beta$ factor, in the last expression, which had to be there because of dimensional analysis $(f \sim m\omega^2 q)$. But I don't see exactly where the $\beta$ factor comes, while I suspect the answer would be easy.... $\endgroup$
    – Trimok
    Sep 14, 2013 at 8:17

1 Answer 1


First, a notational simplification. I write the particle coordinate as $x = q$. The bath coordinates $q_{\alpha}$ are uncoupled, so the functional integration over the bath coordinates splits into a product of independent integrals. Therefore, it suffices to consider just a single bath coordinate denoted $y = q_{\alpha} $, with frequency $\nu = \omega_{\alpha} $ and mass $m = m_{\alpha} $. At the end we just add together the contributions from each coordinate in the effective action.

Going to frequency space, we expand the coordinate $y$ in Fourier components as $$ y(\tau) = \sum\limits_k e^{i \omega_k \tau} y_k, \qquad \omega_k = \frac{2\pi k}{\beta}, \quad k \in \mathbb{Z}$$ so that the inverse transformation is $$ y_k = \frac{1}{\beta} \int\limits_0^{\beta} e^{-i\omega_k \tau} y(\tau). $$ I denote Fourier components of the coupling function as $$ f_k = \frac{1}{\beta} \int\limits_0^{\beta} e^{-i\omega_k \tau} f(x(\tau)), $$ which in Altland & Simons is written $f(x(\omega_k))$ by a slight abuse of notation. Note also that $\omega_{-k} = -\omega_k$. The frequency-space action then reads $$ S_{bath} = \frac{m\beta}{2} \sum\limits_k y_k (\omega_k^2 + \nu^2) y_{-k},$$ $$S_C = \beta\sum\limits_k \left ( f_{-k} y_k + \frac{ f_k f_{-k}}{2 m \nu^2} \right).$$ The second term on the RHS above is called the counter-term, it does not depend on $y$ so we ignore it for the moment and focus on performing the integration.

The integral is of the Gaussian form$^{\ast}$ $$ \int \prod\limits_i \mathrm{d} y_i\, \exp \left[-\frac{1}{2} (y_j A_{jk} y_k + J_k y_k) \right] \propto \exp\left(\frac{1}{2} J_j A^{-1}_{jk} J_k\right), $$ where summation over repeated indices is implied, and we defined $$ A_{jk} = m\beta (\omega_j^2 + \nu^2)\delta_{j,-k}$$ $$ J_j = \beta f_{-j}, $$ and $A^{-1}_{jk}$ are matrix elements of the inverse of $A$. As Trimok pointed out, the matrix $A_{jk}$ is block diagonal, therefore so is its inverse. Each block of $A_{jk}$ is proportional to the Pauli matrix $\sigma^x$ in the subspace spanned by $\{y_k,y_{-k}\}$. Since $\sigma^x$ is self-inverse, each block of $A_{jk}^{-1}$ is also proportional to $\sigma^x$. Given these considerations we can basically write down the inverse by inspection: $$ A_{jk}^{-1} = \frac{1}{m\beta} \frac{1}{\omega_j^2 + \nu^2} \delta_{j,-k}.$$ Adding the counterterm back in, we find the effective action (remember the minus sign in the Euclidean functional measure $Dx \,e^{-S[x]}$) \begin{align} S_{eff} & = S_{particle}[x] + \frac{\beta}{2m}\sum\limits_k f_k f_{-k} \left(\frac{1}{\nu^2} - \frac{1}{\omega_k^2 + \nu^2} \right) \nonumber \\ & = S_{particle}[x] + \frac{\beta}{2m}\sum\limits_k\frac{\omega_k^2\, f_k f_{-k}}{\nu^2(\omega_k^2 + \nu^2)}. \end{align}

$^{\ast}$ I have ignored the fluctuation determinant $\det(A)^{-1/2}$ that also results from the integration over bath coordinates. This factor is independent of $x$, and thus does not affect any observables for the particle only.

  • $\begingroup$ Thanks Mark, very helpful so far. But it still eludes me, how the $f(x(\omega_k))$ come about. For example in the interaction: $ \sum_k\int d\tau f(x(\tau))y_k\exp(i\omega_k\tau) = \sum_k f(x(\omega_k))y_k $? I'm a bit thick right now, sorry. As to the $\beta$, it depends on which convention you use for the fourier-transformation. In the original text a footnote states, that the the symmetric convention is used($\sqrt{\beta}^{-1}$). In that case, you wouldn't have any $\beta$ at all. $\endgroup$
    – Nephente
    Sep 14, 2013 at 13:43
  • $\begingroup$ Ok, that makes sense. I'll try to do a careful dimensional analysis. Shouldn't be too hard. Thanks! Also @Trimok. $\endgroup$
    – Nephente
    Sep 14, 2013 at 14:33
  • $\begingroup$ I have to correct myself... They actually use the convention where $q(\tau) = \sum_k q_k\exp(i\omega_k\tau)$ with $\beta^{-1}$ in the reverse transformation. Then one ends up with the prefactor stated in the book. But your derivation is of course equally valid, since upon switching back to real(or in that case imaginary) time, it doesn't matter anyhow. So I'd say you're both correct and there's no typo! $\endgroup$
    – Nephente
    Sep 14, 2013 at 15:49
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    $\begingroup$ @MarkMitchison : +1 : I think it is correct, that is : all is coherent with your definition of $y_k$ in function of $y(\tau)$ (of course, if the book has an other definition, maybe it is coherent with itself...). I would add only one minor point : there should be a term $\sim log (det A)$ that could be recasting in some constant (or renormalized), I think, so it does not matter, in fact. $\endgroup$
    – Trimok
    Sep 14, 2013 at 15:50
  • $\begingroup$ Answer has been edited to agree with Altland & Simons Fourier transform conventions, and clarify my notation. $\endgroup$ Sep 17, 2013 at 18:23

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