1000 Solved Problems in Modern Physics

(Romina) #1

588 10 Particle Physics – II


10.91 Let aK^0 beam be formed through a strong interaction likeπ−+p→K^0 +Λ.


Neither


∣K^0 〉nor



∣K^0


is an eigen state of|cp〉. However, linear combinations
can be formed.

|Ks〉=

1


2

(∣

∣K^0 〉+



∣K^0

〉)

cp|Ks〉=

1


2

(∣∣

∣K^0


+


∣K^0 〉

)

=|Ks〉

cp=+ 1

|KL〉=

1


2

(∣

∣K^0





∣K^0

〉)

cp|KL〉=

1


2

(∣


∣K^0




∣K^0

〉)

=−|KL〉

cp=− 1

whileK^0 andK^0 are distinguished by their mode of production,Ksand
KL are distinguished by the mode of decay. Typical decays areKs →
π^0 π^0 ,π+π−,KL→π+π−π^0 ,πμν.
Att=0, the wave function of the system will have the form

ψ(0)=


∣K^0 〉=√^1

2

(|Ks〉+|KL〉)

As time developsKsandKLamplitudes decay with their characteristic life-
times. The intensity ofKsorKLcomponents can be obtained by squaring
the appropriate coefficient inΨ(t). The amplitudes therefore contain a factor
e−iEt/which describes the time dependence of an energy eigen function in
quantum mechanics.
In the rest frame of theK^0 we can write the factore−iEt/ase−imc

(^2) t/
,
where m is the mass. The complete wavefunction for the system can therefore
be written as
ψ(t)=


1


2

[

|Ks〉e
−t

( 1
2 τs+
imsc^2


)
+|KL〉e

−t

(
21 τ
L+

imLc^2


)]

=

1


2

e−imsc

(^2) t/[
|Ks〉e−
2 tτ
s+|KL〉eiΔmc
(^2) t/]
whereΔm=ms−mLand we have neglected the factore−t/^2 τLwhich varies
slowly (τL∼= 70 τs). Reexpressing|K 1 〉and|K 2 〉in terms of


∣∣

K^0


and

∣∣

∣K^0

Free download pdf