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In my opinion, the existing choice between the canonical (Hamiltonian) and path-integral (Lagrangian) formalisms is a far-reaching consequence of particle-wave dualism in QM. The first emphasizes the spectral aspects, the second can be viewed as a deep generalization of Fermat's principle for rays propagation in optics. Since most of experiments in particle ...

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I believe your question is answered by the LSZ reduction formula. I shall give a brief overview of what I understand of the topic bellow focusing on a scalar field for simplicity. Corrections are welcome. The path integral allows you to compute time-ordered correlation functions: $$\langle \operatorname{T}[\phi(x_1)\cdots \phi(x_n)]\rangle=\int \varphi(x_1)... 0 Yes, you can write down S-matrix elements directly in terms of path integrals. This was figured out by L. Fadeev and is explained in his 1975 Les Houches lecture notes. A review of his work is also in Bailin and Love's gauge field theory textbook. References: Ludwig Faddeev, Introduction to Functional Methods, p. 1-39 in Methods in Field Theory, North ... 1 That's the formula, although not really a closed form'' one:$$\begin{align*} K_{+-}(0,0,x,y)&= \sum_{r=0}^{\lfloor y/2 \rfloor}(-1)^r \binom{(x+y-2)/2}{r}\binom{(y-x-2)/2}{r}\varepsilon^{2r+1}i,\\ K_{++}(0,0,x,y)&= \sum_{r=1}^{\lfloor y/2 \rfloor}(-1)^r \binom{(x+y-2)/2}{r}\binom{(y-x-2)/2}{r-1}\varepsilon^{2r}, \end{align*} $$for each y>|x| ... 1 As far as the single Gaussian x_1 integration goes, OP is doing the correct thing (up to possible typos) in eq. (7). However since the time-increment$$\epsilon~\ll~ \omega^{-1}\tag{i}$$is supposed to be small (in order for Feynman's fudge factor 1/A to be valid), we have under the square root$$ \frac{\sin (\omega \epsilon)}{2\epsilon^2\omega\cos (\...

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The action that goes in the path integral is a functional of the path $x(t)$. Namely, $$S[x(t)]=\int_0^T dt \frac{m}{2}\left(\dot{x}^2-\omega^2 x^2\right)$$ which can be discretized very simply $$S=\frac{m}{2}\sum_{i=1}^N\frac{(x_{i+1}-x_i)^2}{\epsilon}-\epsilon\omega^2 x_i^2$$ The $x$ in this action is not necessarily a classical trajectory, but this is ...

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On one hand, Wikipedia is talking about the underlying basis $(\theta_i)_{i=1,\ldots, n}$ of Grassmann-odd generators for supernumbers. Usually in physics, $n=\infty$. [These basis-elements $\theta_i$ are not needed for practical calculations, and should not be confused with the Grassmann-odd parameters $\theta$ of a superfield $\Phi(x,\theta)$.] On the ...

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Briefly, the action (Lagrangian$^1$ density) in the path integral are functionals (functions), respectively, as opposed to operators. This is a consequence of how the path-integral formalism is derived from the operator formulation (by inserting infinitely many completeness relations). The action & Lagrangian density usually don't depend on Planck's ...

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Right, the exponentially damped Euclidean path integral is mathematically better behaved compared to the oscillatory Minkowski path integral, but it still needs to be regularized, e.g. via zeta function regularization, Pauli-Villars regularization, etc.

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I think this is what you are looking for:Schwinger-Keldysh formalism. Part I: BRST symmetries and superspace As mentioned the Keldysh formalism is the way to go. In this work they aim to provide a more mathematically elegant formulation of such formalism by means of a set of underlying BRST symmetries. It also has a chapter on computation of Out-of-time-...

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Liouville theory possesses conformal invariance on the quantum level, but its classical action isn't conformally invariant (because it contains dimensionful constants such as $\lambda'$).

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The half comes from the fract that the gaussian integral over the bosonic fields gives you the reciprocal of the square root of the Fredholm determinant: In $A$ is an $n$-by-$n$ matrix $$\int d^nx\, e^{ -x^TAx}= \pi^{n/2} [{\rm det}(A)]^{-1/2}.$$

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Chiral Anomaly has already given a good answer. Here is another answer using slightly different words. The $\Theta$-function on the RHS of eq. (12.8) reflects where in momentum space the heavy modes $\hat{\phi}$ live. It is still possible to Fourier transform action terms from position space to momentum space. The $x$-integration becomes part of a Dirac ...

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Since $\det(U) = \det(U^{\dagger})^{-1}$ and the determinant is multiplicative, you get $$\det \left[U \frac{DA^{\mu}}{DA^{\nu}} U^{\dagger}\right] = \det\left[\frac{DA^{\mu}}{DA^{\nu}}\right]$$

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Here are a few hints... The factor $\Theta(k)$ is zero whenever $k$ is in the range where $\hat\phi(k)=0$. This enforces the fact that if $\hat\phi(k)=0$, then the integrand is zero, and therefore the integral must also be zero. The term being questioned comes from a term $\int d^dx\ \phi^2 \hat\phi^2$ in the integrand, which in turn comes from expanding ...

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