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100 views

Free space propagator: reconciling two results

In quantum mechanics, the free space propagator $G(q_f=0,q_i=0;\tau)$ can be easily calculated to be $$\sqrt{\frac{m}{2\pi i \hbar \tau}}$$ by inserting an identity operator. However if we use ...
134 views

Determinant of a propagator

Say I have a path integral $\int D \phi \exp(i S_0)$. $S_0$ is the usual free action $$S_0=\frac{1}{2}\int\phi (-\Box-m^2) \phi=\frac{1}{2}\int \phi G^{-1} \phi,$$ and at the moment I'm not ...
102 views

Regularization of the 1-dimensional Laplacian

Disclaimer: this is a technical question about regularization of functional determinants which comes from a person with (relatively) strong background in QFT, string theory and path integrals, who ...
67 views

What do we take the functional determinant of in computing th effective action in the Background field method?

I have some schematic notes on computing the effective action and I would like someone to help me fill the gaps. We start with \begin{equation*} \int{}\mathcal{D}\phi\,e^{-iS[\phi]} \end{equation*} ...
491 views

Path integral as a functional determinant

In Peskin and Schroeder on pg. 304, the authors call the fermionic path integral: \int {\cal D} \bar{\psi} {\cal D} \psi \exp \left[ i \int \,d^4x \bar{\psi} ( i \gamma_\mu D^\mu - m ...
433 views

Determinant of Dirac operator in flat space?

How would you evaluate $$|iD\!\!\!\!/-m|$$ Where $D_{\mu}=\partial_{\mu}-ieA_{\mu}$. I have an idea of how to do this without the gauge field, because it's essentially \...
332 views

Computing functional determinant for Dirac fermions

In the path integral formulation for quantum field theory, one often encounters functional determinants of operators, for example for a free scalar field $\log \det (\partial^2+m^2)$. For this ...
527 views

What is the status of Witten's and Vafa's argument that the QCD vacuum energy is a minimum for zero $\theta$ angle?

The argument, which I reproduce here from Ramond's `Journies BSM', is originally by Witten and Vafa in ($\it{Phys}$. $\it{Rev}$. $\it{Lett}$. 53, 535(1984)). The argument is that for $\theta = 0$ (...
99 views

functional determinant question

why when we want to evaluate a functional determinant we use the expansion near $t=0$ for the partition function $\sum_{n}e^{-tE_{n}} \sim \sum_{n} a_{n} (t)t^{n}$ instead of just using the ...
116 views

functional determinant evaluaton

given a Hamiltonian and the semiclassical WKB partition function in units $\hbar =2m=1$ $\Theta (t) = \frac{1}{2\pi} \iint dx dp exp(-tp^{2}-tV(x))$ can i use this Theta function to evaluate the ...
Let the Hamiltonian in one dimension be $H+z$, then I would like to evaluate $\det(H+z)$. I have thought that if I know the function $Z(t) = \sum_{n>0}\exp(-tE_{n})$ I can use \sum_{n} (z+E_{n})...
Many mathematical papers concerning the $\zeta$-regularized Determinant of Laplace-type operators refer for motivation to the broad use of such determinants in mathematical physics, especially in ...