It is best to consider this problem mathematically first. We setx= (E−Ei)/(2 ̄h), and thus
seek to obtain the following largetbehavior (tx≫1 forxfixed)
lim
t→∞sin^2 (tx)
tx^2(13.42)
Clearly, whenx 6 = 0, this quantity tends to zero ast→∞. So, it can have support only atx= 0.
In fact, atx= 0, its value is justt, so we suspect that the limit is of the formδ(x). To prove this,
integrate the quantity against a smooth functionf(x),
lim
t→∞∫∞−∞dxf(x)
sin^2 (tx)
tx^2
= lim
t→∞∫∞−∞dyf(y/t)
sin^2 y
y^2= f(0)∫∞−∞dy
sin^2 y
y^2=πf(0) (13.43)As a result, we conclude that
tlim→∞sin^2 (tx)
tx^2
=πδ(x) (13.44)The object of interest may now be handled as follows,
4
(E−Ei)^2
sin^2(E−Ei)t
2 ̄h
∼t×4
(E−Ei)^2 t
sin^2(E−Ei)t
2 ̄h(13.45)
ast→∞. The second factor now has a limit, as computed above, and we have
4
t(E−Ei)^2sin^2
(E−Ei)t
2 ̄h∼t×
2 π
̄hδ(E−Ei) (13.46)Putting all together, we obtain still for larget,
dP(E,t)
dE
∼t×2 π
̄h
ρ(E)|VE,Ei|^2 δ(E−Ei) (13.47)It is customary to express this result in terms of thetransition rateper unit time, by taking the
time derivative on both sides,
ωi→E=
d^2 P(E,t)
dE dt=
2 π
̄hρ(E)|VE,Ei|^2 δ(E−Ei) (13.48)This formula is referred to asFermi’s Golden Rule.
The Fermi Golden Rule formula must be interpreted with some care. For finitet, as we are of
course always assuming, it is not strictly true that only a state with energyE=Eiwill contribute.
This is a simple reflection of the energy time uncertainty relation,
∆t∆E≥ ̄h (13.49)Thus, for finite time, we should really integrate overEin a range aroundEigiven by ∆E∼ ̄h/t,
wheretis large. The density of states may be assumed to be a smooth function ofE, but the
matrix elements may or may not vary smoothly. Thus, one defines an average matrix element
squared|VE,Ei|^2 in terms of which we have,
ωi→E=2 π
̄h
ρ(E)|VE,Ei|^2 (13.50)