# How Virtual Photons form?

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1. May 21, 2015

### Prashan Shan

1).Virtual Photons forms in pairs like particles and anti particles? or
2).by borrowing energy from future? or
3) in both ways?

2. May 21, 2015

### phinds

Virtual photons are a mathematical convenience. They do not have actual existence.

3. May 21, 2015

### Prashan Shan

but i saw on a website that "Physicists create light out of nothing" and it mentioned that it is due to Casimir effect.

4. May 21, 2015

### phinds

"saw it on a website" is not a valid citation on this forum. I'm not an expert on this but it has been discussed over and over and over here on this forum and it is my understanding that, as I said, virtual photons are just a mathematical convenience to help explain certain phenomena and do not actually exist.

For example, you will almost always hear Hawking Radiation described as the effect of one photon of a virtual photon pair falling into a black hole and the other one escaping, and thus causing a reduction in the mass/energy of the black hole. BUT ... Hawking himself has said that this was just as close as he could come in English to describing a process that really can only be described with math, and that the "virtual particle" explanation is nothing more than an analogy

5. May 21, 2015

### ShayanJ

As phinds pointed out, virtual particles don't exist. They're just mathematical tools. But why you read about them in pop science(or even sometimes scientific) texts? That's because although it is wrong, its the best translation of the math of QFT to english. Which means you shouldn't trust any such translation.

6. May 21, 2015

### ChrisVer

Virtual particles are exactly mathematical entities, so they don't exist.
However given the fluctuations of the vacuum, you can create things "out of nothing" (except for the energy you are giving) just by letting the fields excite. In particular an virtual particle is an off-shell particle, however given enough energy the particle can become on-shell.
The vacuum energy is infinite, however the infinite+some more, can give you something real ... (the strangeness of translating QFT in english)

7. May 21, 2015

### Prashan Shan

ok let it be, but why it cannot exist?

8. May 21, 2015

### ChrisVer

Because so far no off-shell particle has been observed and special relativity works out fine. If you wanted to observe them, you would have to give enough energy to turn them real.

9. May 21, 2015

### ChrisVer

As for the casimir effect it exists because the "energy" of the vacuum you can have between the plates is infinite , but it's still less infinite than the energy of the vacuum you can have outside the plates. And this difference is calculable.

10. May 21, 2015

### phinds

You are somewhat asking the wrong question. Why SHOULD it exist? Unicorns do not exist. Do you ask why they cannot exist? Lots of things don't exist. Do you worry about all of them?

11. May 21, 2015

### OCR

Lol...

12. May 22, 2015

### haushofer

If they do, they would already exist at the classical level, because treelevel diagrams contain "virtual particles".

Conceptually, we like particles, so we like to translate mathematics into objects we think we understand, even though this gives them unphysical properties like not obeying Einstein's energy relations. It is meant as a metaphor, and often this metaphor gives an intuitive explanation of what is going on. See e.g. vacuum polarization by "virtual particles". But that's how they should be understood: as metaphor, resulting from mathematical bookkeeping tools.

13. May 22, 2015

### Andrekosmos

We approximate using Feynman diagrams and the concept of virtual particles in this case is part of perturbation theory. Virtual particles have no other role other than in Feynman diagram calculations.

I think the vacuum fluctuations of quantum field theory will be of further relevance.

14. May 24, 2015

### A_s_a_d

If it has no physical reality, would you please explain how Casimir force originates between two plates in an empty vacuum?

15. May 24, 2015

### ChrisVer

The vacuum in QFT has infinite energy (which you can drop out since you only care about energy differencies in physics - to put it simply)...
The thing with the Casimir effect is that there is a countable energy difference between the vacuum between the plates and the vacuum outside the plates (put again simply, taking the difference of the infinity of the one region and the infinity of the other, you get a fixed number). If you wish further explanation, I could illustrate the simplest scenario proof.

16. May 24, 2015

### A_s_a_d

It would be very much helpful. :)

17. May 24, 2015

### PWiz

I'm sorry, but I can't follow you. How can the difference between two "infinities" result in a finite number? After all, there is no number known as "infinity" - it is singular in its kind, and the difference should come out to 0. (You can't have different kinds of infinities)

18. May 24, 2015

### rolotomassi

Virtual particles can violate conservation laws and normal physics. For instance virtual photons which are used to describe EM force between electrons, can have negative mass and energy, which is of course non physical so cannot exist. They can also violate energy conservation. But these virtual photons only exist within the uncertainty principle and so the apparent violations can never be measured.

I know in Feynman diagrams you can have scenarios in which an electron absorbs a virtual photon its yet to emit. I guess it wrong to think about them as ' physically forming,' rather a trick to get more accurate predictions to measurements, as other people have said.

19. May 24, 2015

### ChrisVer

The simplest scenario concerns an 1 dimensional problem... so you have two conductor plates at a distance $L$.
The Hamiltonian of the photons inbetween the plates can be expressed as:
$H = \sum_r \sum_k ( a_{kr}^\dagger a_{kr} + \frac{1}{2} \omega_k )$

Where $r$ counts the handedness of the photons (right/left handed) and $k$ is the wavenumber. $a_{kr}^{(\dagger)}$ are the ladder operators annihilating(creating) photon states with wavenumber $k$ and polarization $r$ , and their combination above is the Number operator measuring the number of photons in a given state. I am using the notation that $\hbar=c=1$. Now the vacuum is defined as the state $|\Omega>$ that contains no photons.
So:
$H |\Omega> = \sum_{r,k}\frac{1}{2} \omega_k |\Omega>$ and so the energy is:
$E = \sum_{r,k} \frac{1}{2}\omega_k = \sum_k \omega_k$

Now $\omega_k =k = \frac{\pi n}{L}$ in-between the plates. And so:

$E_{vac}^{with} = \frac{\pi}{L} \sum_{n=1}^\infty n \rightarrow E_{vac}^{with} = \frac{\pi \hbar c}{L} \sum_{n=1}^\infty n$

This is obviously infinite...

Now let's go to the same space but without the cavity of the conductors... There the vacuum energy looks exactly similar, but the infinite sum has to be replaced by an integral (the modes are not discrete since there's no cavity):

$E_{vac}^{w/o} = \frac{ \pi \hbar c}{L} \int_0^\infty n~ dn$

Which again is infinite.

So there appears a difference between the vacuum energy with the cavity and when there is not cavity... That difference is:

$\Delta E = E_{vac}^{with} -E_{vac}^{w/o} = \frac{\hbar c \pi}{L} \Big[ \sum_{n=1}^\infty n - \int_0^\infty \nu~ d\nu \Big]$

This difference can be calculated by the Euler-Maclaurin summation formula and gives:

$\Delta E= - \frac{\hbar c \pi}{12 L}$

And that's the Casimir effect in fact. This difference between the vacuum energy with and without the cavity leads to the appearence of an attractive force between the plates when you put them close together ($F= \partial \Delta E / \partial L= \frac{\hbar c \pi}{12L^2}$ ). It's a quantum effect because of the quantum vacuum.

Last edited: May 24, 2015
20. May 24, 2015

### ChrisVer

For the Euler -Maclaurin summation formula you use:

$\sum_{m=0}^\infty f(m) = (\int_0^\infty dx f(x)) + \frac{1}{2} f(0) - \frac{1}{12} f^{\prime}(0) +\frac{1}{720} f^{\prime \prime \prime } (0) - \frac{1}{30240} f^{\prime \prime \prime \prime \prime} (0) +...$

$\sum_{n=1}^\infty n = \sum_{n=0}^\infty n = (\int_0^\infty dx ~x )- \frac{1}{12}$

Since $f(n)=n$ , $f(0)=0~,~f^\prime(0) = 1$ and higher derivatives are zero.