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PHY211 - KTI Exercise Sheet 5

Herbstsemester 2017 Prof. N. Serra

P. Owen, D. Lancierini

http://www.physik.uzh.ch/de/lehre/PHY211/HS2017.

html

Issued: 24.11.2017 Due: 29.11.2017 10:15

Exercise 1: QED Vertex

Explain why the QED vertex shown in figure can not be a real process

e

e

γ

(Hint: use conservation of 4-momentum)

Exercise 2: Weak Decays

The Ξ

is a strange baryon whose quark content is Ξ

(ssd). After drawing the Feynman diagrams for the following weak decays, determine which is more likely:

a) Ξ

(ssd) → Λ

0

(usd) + π

(d¯ u) b) Ξ

(ssd) → n(udd) + π

(d¯ u)

The D

0

(c¯ u) is a charmed meson. After drawing the Feynman diagrams for the following weak decays, determine which is more likely:

a) D

0

(c¯ u) → K

(s¯ u) + π

+

(u d) ¯ b) D

0

(c¯ u)π

(d¯ u) + π

+

(u d) ¯

c) D

0

(c¯ u) → K

+

su) + π

ud)

(2)

Exercise 3: Isospin conservation and strong decays Consider the following strong processes.

a) π

+

+ pπ

+

+ p elastic scattering b) π

+ pπ

+ p elastic scattering c) π

+ pπ

0

+ n charge exchange

The cross section is proportional to the square of the matrix element connecting initial and final states.

Using Dirac notation of h | and | i for the in and out states, one can write

σ ∝ hψ

f

|O|ψ

i

i

2

= M

2

Where O is an isospin operator equalling O

1/2

if it operates on initial and final states of isospin I = 1/2 and O

3/2

for states of I = 3/2. By conservation of isospin, there is no operator connecting initial and final states of different isospin. Let

M

1/2

= hψ

f

|O

1/2

i

i M

3/2

= hψ

f

|O

3/2

i

i

The strong process a) involves the transition between two states of I = 3/2 and I

z

= +3/2, therefore

σ

a

= K |M

3/2

|

2

Where K is some constant. By using the decomposition law of angular momentum deduce the cross- section ratios

σ

a

: σ

b

: σ

c

In terms of M

1/2

and M

3/2

.

In the same spirit of the previous calculation, determine the ratios of the cross sections of the strong processes

a) Σ

∗0

(uds) → Σ

+

(uus) + π

(d¯ u) b) Σ

∗0

(uds) → Σ

0

(dus) + π

0

(d d) ¯

c) Σ

∗0

(uds) → Σ

(dds) + π

(u d) ¯

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