# Energy conservation in a vacuum bubble

I'm a bit confused by vacuum "fluctuations" (I know there is nothing fluctuating since vacuum is lorentz invariant) and their interpretation/representation by Feynman diagrams.

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...)

$D(x-y) D(x-y) = \int d^4k \int d^4k' \frac{e^{-ik(x-y)} e^{-ik'(x-y)}}{(k^2-m^2)(k'^2-m^2)} =\int d^4k \int d^4k' \frac{e^{-i(k+k')(x-y)} }{(k^2-m^2)(k'^2-m^2)}$

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:
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.
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)

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mfb
Mentor
Is this then also true for the zero-component, i.e. the energy (so that the energy of one branch is negative)?
Right.
How is such a negative-energy state interpreted?
Like all virtual particles, as a model with limited physical reality.
Am I right that this is simply nonsense, because what is violated is the mass-shell condition, not energy conservation?
I think so.

• 1 person
DarMM
Gold Member
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?
It isn't interpreted. Propagators in Feynman diagrams are just terms which appear in the expansion of
correlation functions, they do not correspond to any state in the Hilbert Space.

• 1 person
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...)

$D(x-y) D(x-y) = \int d^4k \int d^4k' \frac{e^{-ik(x-y)} e^{-ik'(x-y)}}{(k^2-m^2)(k'^2-m^2)} =\int d^4k \int d^4k' \frac{e^{-i(k+k')(x-y)} }{(k^2-m^2)(k'^2-m^2)}$

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.
this diagram mainly occur in the middle of a feynman diagram,where foregoing line will not have k=0.(if it is simple vacuum diagram,it is not counted).This will lead you to only one integration over momenta i.e. only one d4k will appear because of energy momentum conservation at the vertex.In fact the integral diverges because energy momentum conservation does not fix k.It can be very high or very low.However there is observable effect of vacuum polarization(in qed) which is accounted in calculating lamb shift where it gives a contribution of -27 megacycles and it is necessary to include it for a correct estimation.

if it is simple vacuum diagram,it is not counted
Yes, I do understand that isolated vacuum bubbles simply divide out when I calculate the amplitude for a process (like scattering etc.).
However, in principle the vacuum-to-vacuum-transition amplitude contains such diagrams, does it not?

And since you mention vacuum polarisation, here is a follow-up question on the vacuum state:
In the Schrödinger picture, the state is a wave functional, and the vacuum is a superposition of SHO ground states, one for each k-value (i.e., the amplitude for each k-value is Gaussian, centered at zero). (And in a single-particle-state with a particle at a certain k-value, the amplitude for that k-value is given by the first excited state of the QHO etc.).
How would this state look like in the presence of a charge? If I consider, for example, a simple charged spin-0 quantum field, in the vacuum state, the probability to find a positive or negative particle would be the same everywhere. If a (say negative) test charge is present, would the amplitude in space for the field component corresponding to the positive charge be larger than that for the negative charge close to my test charge?

tom.stoer
Am I right that this is simply nonsense, because what is violated is the mass-shell condition, not energy conservation?
Yes.

Internal lines are off-shell, i.e. pμpμ ≠ m2 but 4-momentum is conserved at each vertex.

• 1 person

Yes, I do understand that isolated vacuum bubbles simply divide out when I calculate the amplitude for a process (like scattering etc.).
However, in principle the vacuum-to-vacuum-transition amplitude contains such diagrams, does it not?
what do you mean by divide out.They come with an infinity,cancel and give no contribution.So you can even in principle left these term,there is no need for them.

tom.stoer
I think he is talking about disconnected diagrams

I think he is talking about disconnected diagrams
But the vacuum to vacuum amplitude is given by only the sum over connected diagram.

tom.stoer
But the vacuum to vacuum amplitude is given by only the sum over connected diagram.
Only b/c you chose the correct PI not generating disconnected diagrams, i.e. ##W[J] = -i \ln Z[J]##

Only b/c you chose the correct PI not generating disconnected diagrams
yes,you can say in principle they exist but not counted.

@tomstoer
Yes, that's what I meant - isn't
Z(J) = Z(J=0) exp(iW(J))
and usually we only look at the second part because we compare to the vacuum?

However, I'm trying to understand the vacuum state in the different formulations of QFT, for that I would like to find an answer to the above questions.