Tag Info

New answers tagged

1

It does vanish and it should, because in a non-broken symmetry phase the current operator cannot create a single particle state. You should have summed over all intermediate states, including the vacuum and all multi particle states, whereas you seem to have only inserted a single particle state.


2

You might end up slapping yourself in the forehead because it's probably simpler than what you were thinking. $$\left(\frac{4\pi}{\Delta}\right)^{\epsilon/2} = \exp\left[\frac{\epsilon}{2}\log \left(\frac{4\pi}{\Delta}\right)\right] = 1 + \frac{\epsilon}{2}\log \left(\frac{4\pi}{\Delta}\right) +\mathcal{O}(\epsilon^2)$$ whence the result follows.


2

As you said regularization is necessary to even begin making sense of the diagrams which appear in perturbation theory. The word "perturbation" already contains a hint towards the answer you may be looking for. If you are trying to make sense of the interacting theory $d\nu$ by perturbation, this implies that there is somebody being perturbed, namely a free ...


2

Singularity in Force Laws If force laws were fundamental to nature, this would be a serious problem. Imagine, for example, the gravitational energy between photons. They are Bosons and can hence occupy the same quantum state; crucially, more than one of them can be and stay in the same position where the gravitational force (they have energy and hence, ...


1

... r=0 is trivially excluded (for macroscopic objects at least) because they have well defined excluded volumes and cannot occupy the same space at the same time, hence one may argue that the divergence at r=0 case is a mathematical artifact Radius of elementary particles can be 0 if they are point particles (electrons are so far best thought of as ...


3

As noted already, within classical physics, singularities such as $1/r^2$ signal a break down of the theory. If we are really interested in what is happening at the point of the singularity, we should use quantum physics. You can think of $1/r^2$ as the asymptotic scaling form of the quantum theory for large $r$. The actual singularity is not physical. On ...


2

I can't speak to singularities in the sense of general relativity, but your example of $1/r^2$ laws in classical physics is actually solved most of the time by internal structure. One of my physics professors used to always say that nature solves infinities with internal structure. For example, for a charged sphere of uniform charge density, the electric ...


3

I can only offer a partial answer to the first portion of your question. one comes across functions that diverge at a given point, typical examples would be the Coulomb or the gravitational forces being ∝ 1/r², clearly they diverge at r=0... Gravitational force isn't actually proportional to 1/r². Take a look at the plot of gravitational potential on ...


0

The usual way to compute such a path integral is by writing the fields (in your case: the paths) as "classical configuration" (the straight line) plus "quantum fluctuations". So if you write your paths as $\gamma(\tau) = a + \tau b + \hat\gamma(\tau)$ (with $\tau\in[0,1]$ a parameter describing the path), then $\hat\gamma$ will be the "quantum fluctuation" ...



Top 50 recent answers are included