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Feynman Path Integral Formulation

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152 5 Semiclassical Gravity<br />

the case d = 0. In this case one has<br />

Q[a] ≃<br />

( a<br />

a 0<br />

) c(1+α)<br />

. (5.52)<br />

It is unclear if this simple model is sophisticated enough to capture the essence<br />

of the more complete calculation, but it does suggest that the behavior at small a<br />

could be strongly influenced by the detailed nature of the short distance cutoff, the<br />

form of the effective action as small distances, and the choice of measure over the h<br />

variables.<br />

5.3 Pair Creation in Constant Electric Fields<br />

It is known that if an atom is subject to a sufficiently strong external uniform electric<br />

field it is possible for the field to create pairs. To be specific, if the atomic potential<br />

is approximated by a function with a minimum −V 0 and going to zero at infinity,<br />

then an external electric field, with potential eEx, will give an ionization probability<br />

for a bound electron<br />

∫ V0 /|eE|<br />

P ≃ exp − dx √ 2m(V 0 −|eE|x) =exp<br />

0<br />

(<br />

− 4 )<br />

√ V 0 2mV0<br />

3 |eE|<br />

, (5.53)<br />

using the WKBformula, with V 0 /|eE| the distance between the two classical turning<br />

points. If one thinks of a negative energy electron as trapped in a potential of depth<br />

V 0 ≈−2m, then one obtains for the probability of pair production in a strong electric<br />

field the semiclassical estimate<br />

)<br />

P ≃ exp<br />

(− 16m2 . (5.54)<br />

3|eE|<br />

Clearly in this last instance one is dealing with a relativistic process involving strong<br />

fields, so a fully relativistic treatment would seem more appropriate.<br />

To this end one considers the external field-dependent <strong>Feynman</strong> amplitude Z[A]<br />

describing the vacuum to vacuum amplitude in the presence of an external electric<br />

field, expressed as a path integral over fermion fields ψ and ¯ψ in a fixed classical<br />

background A,<br />

∫<br />

∫ ]<br />

Z[A] = [dψ][d ¯ψ] exp[<br />

d 4 x ¯ψ(i ̸∂ + e ̸A + m − iε)ψ<br />

= det[i ̸∂ + e ̸A + m − iε] . (5.55)<br />

Using the formal identity detM(A) =exptrlogM(A), the corresponding effective<br />

action S(A) is then given by minus the exponent of the above expression

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