# Does quantum pair creation violate conservation of energy?

1. Mar 11, 2007

### CosmologyHobbyist

Hi! A question from a beginner in quantum physics... If the quantum field randomly produces an electron-positron pair, and the pair annhialate, a gamma ray photon is produced. The net result of this process is gamma ray photons coming out of the quantum field at random. This means the quantum field is constantly producing energy out, with no energy inputs. I have never read that the quantum field randomly absorbs photons to provide energy for pair creation. This seems to me to be in violation of conservation of energy. Does the quantum field constantly pour energy into the universe, or is there a balancing side to this equation?

2. Mar 11, 2007

### lalbatros

Remember the uncertainty principle.

3. Mar 11, 2007

### Ratzinger

To make these 'vitual particles' in the vaccum real (longer lasting than time- uncertainty permits), you have to put in energy.

4. Mar 11, 2007

### vanesch

Staff Emeritus
This is a question that comes back regularly. The answer is: depends on exactly what one understands by "virtual particles". In a commonly used version of it, there is, at each interaction, perfect conservation of energy and momentum. Only, the "virtual particles" are not exactly like their real counter parts. One sometimes says that they are "off-shell", and they can carry imaginary mass for instance. It is only when one insists on them to ressemble a bit more to their real counterparts than they formally are, that one seems to run into a problem of conservation of energy.

Just let us give an example. Imagine a photon, with energy-momentum (k,0,0,k) (in units where c = 1). Now, imagine that this photon "splits" into a virtual electron-positron pair.
Well, the electron will have energy-momentum (k-e1, l1, l2, k+l3) and the positron will have energy-momentum (e1,-l1,-l2,-l3).
Normally, for a real electron, we have the on-shell condition that
(k-e1)^2 = m^2 + l1^2 +l2^2 +(k+l3)^2, but in this case, this will not hold. What will hold is that the sum of the 4-momentum of the virtual positron and electron will equal the 4-momentum (k,0,0,k), so there will be perfect conservation of energy and momentum!
But the particles will not have the right "mass" (can even have imaginary mass).

5. Mar 12, 2007

### CosmologyHobbyist

Thanks guys! That clears things up. I think I better locate an easy read on the basics!

6. Mar 12, 2007

### quetzalcoatl9

can you please explain this a little more?

7. Mar 17, 2007

### PhillipKP

From what I understand.

Nothing is real until it is "measured". So if the virtual particles are created and annihilate each other prior to any type of measurement (interaction with other particles). Then there is no net change in energy and thus no violation of the conversation of energy law.

Someone correct me if I am wrong.

8. Mar 18, 2007

### Mentz114

The electromagnetic field, under the right conditions, can produce actual, as opposed to virtual, electron/positron pairs. Energy is perfectly conserved in this reaction.

9. Mar 19, 2007

### lalbatros

The time-energy uncertainty relation reads: DE.Dt > hbar .
Energy conservation can be violated during short times.

10. Mar 19, 2007

### masudr

Isn't it more like

$$\Delta E \Delta A}{\left | \frac{\mbox{d}t}{\mbox{d}\langle A \rangle}\right |} \ge \frac{\hbar}{2}$$

Where $\Delta A = \sqrt{\langle A^2 \rangle - \langle A \rangle^2}$ where $\langle A \rangle = \langle \psi | A | \psi \rangle$

Last edited: Mar 19, 2007
11. Mar 19, 2007

### FunkyDwarf

as far as i know convservation of energy is never violated, its simply that the particles dont have energies you would normally predict and as such are virtual and exist for a short amount of time

12. Mar 19, 2007

### masudr

Off-shell (a.k.a. "virtual") particles are ways to keep track of the integrals when doing perturbations in QFT. The way the machinery works means that these off-shell particles can have any energy/momentum. NB: these are only convenient ways to keep track of the integral sums.

We know that they are never detected: Whether or not they actually exist is up for debate. In the strictest Popperian approach, they clearly do not exist on a par with ordinary particles.

13. Mar 19, 2007

### Demystifier

Real photons are never produced from the true vacuum. Instead, there must be a source (e.g. a black hole or a charged source of an electric field) that will decrease its energy when real photons get created. Thus, energy is always conserved in real physical processes.

Next, virtual particles do not exist. They are not even a part of the exact theory. They are nothing but a pictorial name for some mathematical terms appearing in certain approximative methods of solving otherwise difficult and abstract quantum equations.

Finally, energy of a closed system is always conserved, even during a short time. Only the energies of certain subsystems may be uncertain, but this is always compensated by energies of other subsystems, such that the total energy is conserved exactly.

14. Mar 22, 2007

### reilly

If the field, what ever that is, emits photons it absorbs them as well, as pretty is much is guaranteed by the basics of QED. this means in practice photon and pair currents are dancing together in the field = universe?, and conserving all that's supposed to be onserved.
Regards,
Reilly Atkinson

15. Mar 26, 2007

### 6Stang7

Think of the interaction between two electrons, which occures when one electron emits a virtual photon that is then absorbed by the other electron. To understand this idea of a virtual photon, you must understand that an electron can't emit a real photon and just remain an electron, because this would violate the conservation of energy and momentum. This is easy to see if you think of the initial electron at rest. The final state, after emitting a real photon, would consist of an electron and a photon moving off back-to-back, and that configuration necessarily has more energy than the initial at-rest electron.
With the uncertainty principle, this is easy to resolve. The uncertianty realtion, dEdt>h-bar, allows for deviations from energy conservation, but only for time dt determined by the degree of energy uncertianty. During the time period dt the photon is said to be virtual, and as long as the photon is reabsored by the other electron quickly enough, there is no measurable violation of conservation of energy.

16. Mar 29, 2007

### Sunset

The photon decays into a virtual electron (mass m) and a virtual positron (mass M) for a certain time t.

(k-a+ E1)²= m² + 1² + 2² + (k+3)²
(a+ E2)²= M² + (-1)² + (-2)² + (-3)²

E1 and E2 are the uncertainties in energy
t1 und t2 are the uncertainties in time

The process only is allowed if t1 > h/E1 and t2> h/E2
that means if for example t1>t2, the two particles must anihilate after at least t=t2 into a photon again.