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With the current QED framework, If an electric field is strong enough (say, near a nucleus with $Z > 140$) , pair production will occur spontaneously? Is this a real effect or an artifact before renormalization is carried out?

how can energy be conserved in such scenario?

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This seems similar to my question here: physics.stackexchange.com/questions/7232/… and people claimed Lorentz symmetry means the electric field is not limited. So if QED proposed an effective field limit where the vacuum would start spewing particle pairs to prevent larger fields, then wouldn't that be similar to claiming QED violates Lorentz symmetry? At the very least, it needs to depend on something more than the electric field strength, right? –  John Mar 22 '11 at 2:31
    
@John; I think this case is different from your question because a nucleus (or in Moshe answer, the capacitor) break the lorentz symmetry, so as long as pair production doesn't occur strictly from electromagnetic waves in the vacuum, it's ok (but i wouldn't bet my house on it) –  lurscher Mar 22 '11 at 3:26
    
There is no critical electric field in either case, so it's all consistent. –  user566 Mar 22 '11 at 7:24
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2 Answers 2

up vote 10 down vote accepted

The Schwinger pair production can be understood as follows:

Suppose you have a constant electric field E in some region of space, pointing in the x direction. This is created for example by a large capacitor. Inside this capacitor, the energy of an electron-positron pair separated by some distance $\Delta x$ is $V(\Delta x)= 2m- q E \Delta x$, where $m$ is the mass of the electron (and positron) and $q$ is its charge. The first term is the rest energy of two massive particles, and the second is their potential energy in the presence of the electric field. You can see that for large enough separation the total energy is negative - it becomes energetically favourable to have a pair present instead of an empty capacitor.

This process can be thought of as tunnelling: the configuration of empty space and an electron-positron pair are separated by a barrier $V(\Delta x)$. You therefore have an exponentially small probability to create pairs at the classical turning point $V(\Delta x)=0$, or $\Delta x = \frac{2m} {qE}$. As always in tunnelling process, energy is conserved - it is zero before and after the tunnelling in this case.

Once the particles are created, they accelerate away from each other, and eventually end up neutralizing in part the capacitor, in other words reducing the electric field. Note that there is no critical electric field - pair creation occurs for arbitrarily small electric field, though the probability is exponentially suppressed, roughly as $\exp(- \pi m^2/ q E)$. The derivation of this formula by Schwinger (and corrections to all orders) is a real joy to see, I'd recommend at least for theorists to take a look at the original paper. This may well be the first use of instanton methods in quantum mechanics, though I'm far from being an expert on the history.

Also:

There is no real connection of this to renormalization.

Variants of this calculation are useful in cosmology and QFT in curved spacetime, for example for particle production by time varying backgrounds.

As far as I know, the effect has never been observed due to difficulties creating large enough electric field. I may be wrong on that as well though.

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great answer. +1. I assume this happens even if the electric field is created by a passing electromagnetic wave? –  lurscher Mar 22 '11 at 3:31
    
This applies to electric field which is constant in space and time. I can imagine it still applies for slowly varying fields - waves with very long wavelength or very low frequencies (compared to the other scales in the problem). I don't know if you'd call those a passing wave though –  user566 Mar 22 '11 at 3:40
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I would like to complement rather than to answer: one can see the pair creation as a relaxation mechanism. For example, in a usual capacitor the charges are artificially separated and there is a potential energy of their interaction. In an ideal case of infinite dielectric resistance, the system is stable but in reality there is always a current (leakage) that serves to diminish the system potential energy. Similarly in QED: pair creation serves to neutralize the separated charges. Of course, the energy is conserved. In reality it is transformed into heat due to resistance of the dielectric, for example.

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I see what you are saying, but it was difficult to follow at first. You don't clearly indicate that you're talking about QED in the medium ("vacuum") between the two capacitor plates. And if that is what you're talking about then I don't see you saying anything different from what @Moshe said. –  user346 Mar 22 '11 at 14:09
    
I would like to underline that a charged capacitor is a metastable system, a system out of its minimum energy. –  Vladimir Kalitvianski Mar 22 '11 at 14:21
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