0
$\begingroup$

I came across an expression which I don't understand for the density matrix $\rho$ given by the path integral method (Fradkin, p.760) -

$$ \left< \phi(x) \left| \rho\right| \phi\left(x'\right) \right>=\frac{1}{Z}\intop \mathcal{D}\phi\,\prod_{x}\delta\left( \phi\left(x,\,\tau=0\right) - \phi\left(x\right) \right)\prod_{x}\delta\left( \phi\left(x,\,\tau=\beta\right) - \phi'\left(x\right) \right)e^{-S_E\left[\phi\right]} $$

My questions are:

  1. Intuitively I understand the deltas "choose" to integrate only over the field configurations $\phi(x)$ and $\phi(x)$, but since this is a product of delta, wouldn't it vanish anyway?
  2. The exponent depended on a single argument, does this imply $\left< \phi(x) \left| \rho\right| \phi\left(x'\right) \right>=0$ when $\phi\left(x\right)\neq\phi'\left(x\right)$? It makes since for the density matrix to be diagonal, but it does not have to be so.
  3. How is the partition function achieved from this expression? One would have to set $\phi\left(x\right)=\phi'\left(x\right)$ and trace over but I can't see how it turns into $$Z=\intop \mathcal{D}\phi e^{-S_E\left[\phi\right]}$$
$\endgroup$
  • 1
    $\begingroup$ 1) no. the deltas give two constraints the field $\phi(x,\tau)$ \emph{at different times.} Its initial value is given by the field on the ket, and it's final value by the field on the bra. These conditions are not incompatible, hence the deltas don't vanish. 2) I don't get how you imply that from the exponent... are you sure your notation is correct? 3)you are almost there. You put $\phi=\phi'$ and integrate over $\phi$, meaning that in the functional integral the deltas only enforce the periodicity of the field but not it's inital/final values. $\endgroup$ – tbt Apr 20 at 9:10
  • $\begingroup$ @tbt 1. The integral measure $\mathcal{D}$ is over fields $\phi$ at what times? Ob the right product is it possible to write $\prod_{x}\delta(\phi'(x,\tau=\beta)-\phi'(x))$? Shouldn't a density matrix have elements at equal times? or since imaginary time is periodic everything is ok? 2. the exponent is $e^{-S_E[\phi]}$ and not (of course) $e^{-S_E[\phi,\phi']}$ but what does the whole integral mean then? I would appreciate a reference for a full derivation of this expression ,I couldn't find anything. $\endgroup$ – proton Apr 21 at 14:29
  • $\begingroup$ Sorry for the late reply. The matrix element is evaluated at the same "time", but in doing the manipulations that bring you to that expression of the partition function one has to introduce the time interval $[0,\beta]$. For bosonic field the two extremes are identified, so there's no discrepancy in the appearance of both $\phi(\tau=0)$ and $\phi(\tau=\beta)$ . You can find a complete derivation of the formula $Z=\int D\phi \ldots$ in the first chapters of the book by Negele&Orland with all the details. $\endgroup$ – tbt Apr 26 at 20:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.