
#1
Apr308, 11:36 PM

P: 213

Do vacuum fluctuations produce only particle  antiparticle of photons Or it would also produce pairs of electronspositron and so on...if it also produce various pairs of particleanti particle other than photons then what it is probability of the production of each such different pair?




#2
Apr408, 06:06 AM

P: 981

It produces all of them.




#3
Apr408, 10:52 PM

P: 213

sometimes it is producing photon pair and sometimes it is producing electron pair so what determines this ? 



#4
Apr508, 04:50 AM

P: 1,142

Vacuum fluctuations[tex]{E} = \begin{matrix} \frac{1}{2} \end{matrix} \hbar \omega \ [/tex] [tex] \left[ \frac{\mathbf{p}^2}{2m} + V(\mathbf{r}) \right] \psi(t)\rang = i \hbar \frac{\partial}{\partial t} \psi(t)\rang,[/tex] Taking into account every point in space according to field theory and then renormalising appropriately. But since I know nothing particularly profound about field theory, your guess is as good as mine. Probably something like this. In other words I don't know, but I am keen to know why. I'd imagine the probabilities are related to this Hamiltonian in some way. [tex] \left[\phi(\mathbf{r}) , \phi(\mathbf{r'}) \right] = 0 \quad,\quad \left[\phi^\dagger(\mathbf{r}) , \phi^\dagger(\mathbf{r'}) \right] = 0 \quad,\quad \left[\phi(\mathbf{r}) , \phi^\dagger(\mathbf{r'}) \right] = \delta^3(\mathbf{r}  \mathbf{r'}) [/tex] And the expectation value of field theory. [tex]\left\langle F\right\rangle=\frac{\int \mathcal{D}\phi F[\phi]e^{i\mathcal{S}[\phi]}}{\int\mathcal{D}\phi e^{i\mathcal{S}[\phi]}}[/tex] Trouble is even after reading an article on it I have no idea exactly what this means, precisely? Except that this and the equation below and above are used to model expectation energies from the vacuum or anywhere else. I bet there's some really simple equation that takes all this and converts it into [itex]E_e=\int \mathcal{D}\phi[/itex] So in short? I don't know, but I'm sure someone does somewhere... 



#5
Apr508, 04:57 AM

P: 213





#6
Apr508, 05:05 AM

P: 1,142





#7
Apr508, 06:09 AM

P: 981

As you say, this is something which is quite nontrivial. It would help you understood how nonquantum statistical mechanics treats fluctuations  the quantum field theory is very similar, just boosted up a few dimensions.
A more heuristic argument would simply note that there must be a dependence of the probability on the mass of the created pair. Now that here "mass" means the total relativistic energy. A guess would be that [tex]p \propto e^{m}[/tex], which is not too far off the answer (I think...) 



#8
Apr508, 05:16 PM

P: 410

I don't think the probability of the production of each pair is going to be a well defined answer. Infinitely many such pairs are created at almost all energy levels everywhere... For instance, asking the question about massless particles (e.g. photons) will give you what is called an infrared divergence.
But these particles are intermediate states which we can't observe, so it's more useful to think of them as terms in a series expansion of a function known as the partition function (alternatively, the time evolution operator). They do have an effect though, for example, in vacuum polarization, zitterbewegung, etc. 



#9
Apr508, 05:46 PM

P: 863




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