- #1

- 229

- 11

In a normal Feynman diagram, you have energy conservation at each vertex, so overall energy conservation is ensured as well (no such thing as "borrowing energy by HUP" or such).

How does this work for a vacuum bubble, though? If I look at a single Feynman diagram of a simple bubble, the bubble starts at some point, the two lines separate and then join again, right?

Since k=0 before the bubble exists, each component of k of one branch should be the opposite of that of the other branch: k=-k'.

Is this then also true for the zero-component, i.e. the energy (so that the energy of one branch is negative)? How is such a negative-energy state interpreted?

If I try to calculate this formally (I'm very bad at these things, though), I would think that I can write a simple vacuum bubble as D(x-y)D(x-y) (where D is the propagator), because I have two lines propagating from x to y. Inserting a free propagator, this would mean (disregarding the i-epsilon-terms for laziness...)

[itex]D(x-y) D(x-y) = \int d^4k \int d^4k'

\frac{e^{-ik(x-y)} e^{-ik'(x-y)}}{(k^2-m^2)(k'^2-m^2)}

=\int d^4k \int d^4k'

\frac{e^{-i(k+k')(x-y)} }{(k^2-m^2)(k'^2-m^2)}

[/itex]

Probably I can use some fourier transform equation for δ to use this to show that k=-k' - is this right? That would indeed show that the k_0-component is negative on one branch.

And, final question, the wikipedia enttry on vacuum fluctuations http://en.wikipedia.org/wiki/Quantum_fluctuation says:

Am I right that this is simply nonsense, because what is violated is the mass-shell condition, not energy conservation? (I know there are some threads here discussing this, but I'm not sure I understand them completely and there seem to be somewhat conflicting views around, so I'm looking for clarification)That means that conservation of energy can appear to be violated, but only for small times. This allows the creation of particle-antiparticle pairs of virtual particles.