QuantumPhysics.dvi

The result is given by

σtot=

∫

4 π

dΩ|f(k′,k)|^2 =

4 π

k^2

∑∞

ℓ=0

(2ℓ+ 1) sin^2 δℓ (12.98)

It is readily checked that the optical theorem is obeyed for this cross section. This may be seen by
computing

Imf(k,k) =

1

2 k

∑∞

ℓ=0

(2ℓ+ 1)

(

1 −cos 2δℓ(k)

)

Pℓ(1) (12.99)

Using now the factPℓ(1) = 1 and that 1−cos 2δℓ(k) = 2 sin^2 δℓ(k), it is easily seen that the relation
of the optical theorem holds.

12.12The example of a hard sphere

Assume we scatter off a spherical body with potential

V(r) =V 0 θ(b−r) V 0 > 0 (12.100)

for some radiusband letV 0 → ∞. This is ahard sphereof radiusb. In this limit, clearly all the
partial wavesRℓ(kr) must vanish atr≤b. Outside the hard sphere, forr > b, the solution is a
linear combination of the spherical Bessel functionsjℓ(kr) andhℓ(kr),

Rℓ(r) =jℓ(kr) +cℓ(k)hℓ(kr) (12.101)

wherecℓ(k) is generally a complex function ofk. Vanishing ofRℓ(b) providescℓ(k),

cℓ(k) =−

jℓ(kb)

hℓ(kb)

(12.102)

From the limit ofhℓ(kr) asr→∞and the result for the phase shift (12.93), we find that

eikr−iπℓ/^2 ×

e^2 iδℓ(k)− 1

2 ik

=−

jℓ(kb)

hℓ(kb)

1

k

eikr−i(ℓ+1)π/^2 (12.103)

we find that

e^2 iδℓ(k)−1 =− 2

jℓ(kb)

hℓ(kb)

(12.104)

Using the relation

jℓ(kb) =

1

2

(hℓ(kb) +h∗ℓ(kb)) (12.105)