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- Thread starter spidey
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- #2

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It produces all of them.

- #3

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It produces all of them.

then what is the probability of those different virtual pairs to be created?

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

- #4

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then what is the probability of those different virtual pairs to be created?

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

I'm sure someone can explain how QM predicts the particles that should exist in a given metric of space in the vacuum. It's enough to make you believe in the ether again

[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...

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- #5

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I'm sure someone can explain how QM predicts the particles that should exist in a given metric of space in the vacuum. It's enough to make you believe in the ether again

[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]

ok..i dont understand anything from this...

- #6

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ok..i dont understand anything from this...

Me neither. It was my way of highlighting how complicated the maths is when talking about vacuum fluctuations, if you'll pardon the Greek. I think the Hamiltonian (total energy in a system) determines the probability of any particle pair being created, ie the energy involved, predicts the likelihood of the wave function at any given point in space, which is where field theory comes in, apparently it states that the expanse of this result is applicable at every point in space, at least naively, which is where renormalisation comes in, we need to have something that doesn't involve infinities and that is practicable. This is where I got confused though, I don't actually know how precisely the model determines what we expect from the vacuum. All I know is that it's something to do with the maths squiggles above. Hopefully someone will be along presently to clear up the maths? Or to tell me to shut up. One of the two. I've found that the best way of getting attention, if you really are dying to know something, is to be wrong about something or at least misguided, being the helpful and friendly place PF is I'm sure there's someone somewhere who can answer the question.

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- #7

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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

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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

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Yes, as genneth said, it produces all of them. Photons unlike mass particles are their own antiparticles. What this means is that you can take two photons 180 degrees out of phase and the result is no photon. We can cancel sound waves essentially the same way. These particles are the result of the Heisenberg Uncertainty Principle (HUP), i.e., random fluctuations in space. A good illustration of virtual photons is the Casimir effect. When two metal plates are placed very close together, a few micrometers. Longer wavelengths of virtual photons cannot exist between the plates, only wavelengths that can fit in those few micrometers. This means the vacuum energy outside the plates is higher so the plates will attract. This is the force often referred to as zero point energy with lots of wacky attempts to utilize it.

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