Exercise 1: Derivatives of the Dirac delta function

(1)

Flavour Physics Solution 5

FS 21 Prof. G. Isidori Dr. R. Silva Coutinho M. Atzeni, S. Mächler

https://www.physik.uzh.ch/en/teaching/

PHY568/FS2021.html 21.04.2021

Exercise 1: Derivatives of the Dirac delta function

The aim of this exercise sheet is to compute observables in inclusive B → X

c

`ν

`

decays including the leading and next to leading nonperturbative corrections. As a courtesy of the operator product expansion higher order corrections in 1/m

_b

and α

_s

modify the form factors W

_i

which we already computed at the LO. The form factors then are expressed as a double expansion in 1/m

b

and α

s

,

W

_i

=W

_i^(0,0)

+ µ

²_π

2m

²_b

W

_i^π,0

+ µ

²_G

2m

²_b

W

_i^(G,0)

+ ρ

³_D

m

³_b

W

_i^(D,0)

+ ρ

³_LS

m

³_b

W

_i^(LS,0)

+ C

_F

α

_s

π

W

_i^(0,1)

+ µ

²_π

2m

²_b

W

_i^(π,1)

+ µ

²_G

2m

²_b

W

_i^(G,1)

. (1)

If we go to higher orders we nd that the 1/m

_b

power corrections to the form factors T

_i

contain higher powers of (ˆ u = (v − q) ˆ

²

)

⁻¹

, where v is the four velocity of the initial B meson and q is the four momentum carried away by the lepton pair. As the form factors W

i

are obtained as the imaginary parts of the form factors T

_i

using the relation

− 1

π Im 1

(ˆ u + i)

ⁿ

= (−1)

ⁿ⁻¹

(n − 1)!

d

ⁿ⁻¹

dˆ u

ⁿ⁻¹

δ(ˆ u), (2)

we see that they contain derivatives of Dirac delta functions. Here objects with a hat denote dimensionless objects made dimensionless by dividing them by the necessary power of m

_b

, e.g. u ˆ = u/m

²_b

.

(a) As a warmup exercise, compute

Z

∞

−∞

dzf (z)δ

⁰

(z), (3)

where f is a test function and δ

⁰

denotes the derivative of the delta function with respect to its argument.

(b) Compute

Z

∞

−∞

dzf (z)δ

⁰⁰

(z). (4)

(c) Now that you see the pattern, compute Z

∞

−∞

dzf (z)δ

⁽ⁿ⁾

(z), (5)

where δ

⁽ⁿ⁾

denotes the n th derivative of the delta function w.r.t. its argument.

Solution:

(a) We integrate by parts in order to move the derivative away from the delta function and then use the denition of the delta function to perform the remaining integration:

Z

∞

−∞

dzf (z)δ

⁰

(z) = f (z)δ(z)|

^∞_−∞

− Z

∞

−∞

dzf

⁰

(z)δ(z) = −f

⁰

(0) (6)

(2)

(b) In this case we integrate by parts twice:

Z

∞

−∞

dzf (z)δ

⁰⁰

(z) = Z

∞

−∞

dzf

⁰⁰

(z)δ(z) = f

⁰⁰

(0) (7) (c) Now we integrate by parts n times and obtain

Z

∞

−∞

dzf (z) d

ⁿ

dz

ⁿ

δ(z) = (−1)

ⁿ

f

⁽ⁿ⁾

(0). (8) Exercise 2: Decay rate at higher orders in 1/m

b

The inclusive B → X

c

`ν

_`

decay rate for massless leptons is computed by integrating the triple dierential decay rate

d

³

Γ dq

²

dq

⁰

dE

`

= G

²_F

|V

_cb

|

²

24π

³

q

²

W

₁

+

2E

_`

(q

₀

− E

_`

) − q

²

2 W

₂

Θ ˆ q

²

(9) over E

_`

, q

²

, q

₀

. Here it is important to keep the Heaviside function which makes sure that q ˆ

²

is positive.

(a) As the form factors are independent of the energy of the charged lepton, E

_`

, we can perform the integration over E

_`

before plugging in the form factors. Perform this integration. Then change from the variable q

²

= q

²₀

− q

²

to q

²

and from q

0

to the energy of the hadronic nal state ω.

Hint: First consider the decay kinematics to nd the integration boundaries. They are hidden in the solution to exercise sheet 3. You should obtain

d

²

Γ

dq

²

dˆ ω = m

²_b

3 p q ˆ

²

3 (1 − ω) ˆ

²

− q ˆ

²

W

1

+ ˆ q

²

W

2

Θ ˆ q

²

. (10)

(b) In the next step we want to perform the integration over ω ˆ . This integral is computed by making use of the delta functions and the derivatives of delta functions contained in the form factors. The dierential q ˆ

²

distribution is given by

dΓ d ˆ q

²

=

Z

1−

√

ˆ q²

√

ρ+ ˆq²

dˆ ω d

²

Γ

dq

²

dˆ ω . (11)

Use the fact that the double dierential rate can be written as d

²

Γ

dq

²

dˆ ω =

3

X

n=0

(−1)

ⁿ

(n − 1)! f

⁽ⁿ⁾

ω, ˆ q ˆ

²

δ

⁽ⁿ⁾

(ˆ u) (12)

to compute dΓ/d ˆ q

²

.

Hint: It might be useful to change variables from ω ˆ to u ˆ . You should nd dΓ

d ˆ q

²

= 1 2

2

X

n=0

(−1)

ⁿ

(n − 1)!

d

⁽ⁿ⁾

dˆ u

⁽ⁿ⁾

"

f

⁽ⁿ⁾

ω ˆ (ˆ u) , q ˆ

²

p u ˆ + ˆ q

²

+ ρ Θ ˆ q

²

(ˆ u)

#

ˆu=0

. (13)

(c) The dierential distribution dΓ/d ˆ q

²

contains the Heaviside function Θ(ˆ q

²

) and its derivatives w.r.t. u ˆ . We play the same game as above and decompose it as

X ˆ q ˆ

²

=

3

X

n=0

g

⁽ⁿ⁾

q ˆ

²

d

dˆ u

n

Θ

1 + ρ + ˆ u − 2 p

ρ + ˆ q

²

+ ˆ u

u=0ˆ

. (14)

As you can see we need the rst and second derivatives of Θ

1 + ρ + ˆ u − 2 p

ρ + ˆ q

²

+ ˆ u w.r.t.

ˆ

u . Show that d dˆ u Θ

1 + ρ + ˆ u − 2 p

ρ + ˆ q

²

+ ˆ u

ˆu=0

= ρ − 1

2 δ q ˆ

²

− q ˆ

²_max

, (15)

(3)

(d) Now show that d

²

dˆ u

²

Θ

1 + ρ + ˆ u − 2 p

ρ + ˆ q

²

+ ˆ u

u=0ˆ

= 2δ q ˆ

²

− q ˆ

_max²

(1 + ρ)

²

+

p

ρ + ˆ q

²

− 1

2

ρ + ˆ q

²

δ

⁰

1 + ρ − 2 p ρ + ˆ q

²

. (16)

When integrated with a test function h the last term gives Z

^(1−ρ)2

4

0

d ˆ q

²

p

ρ + ˆ q

²

− 1

2

ρ + ˆ q

²

g q ˆ

²

δ

⁰

1 + ρ − 2 p

ρ + ˆ q

²

= − p

ρ + ˆ q

²

d d ˆ q

²





 d ˆ q

²

dx p

ρ + ˆ q

²

− 1

2

ρ + ˆ q

²

g q ˆ

²







ˆ q²= ˆq²_max

(17)

with x = 1 + ρ − 2 p ρ + ˆ q

²

.

(e) The last ingredient we need is the third derivative of the Heaviside function, so show that d

³

dˆ u

³

Θ

1 + ρ + ˆ u − 2 p ˆ

u + ˆ q

²

+ ρ

u=0ˆ

= − 12

(ρ + 1)

⁴

δ q ˆ

²

− q ˆ

_max²

+ 3

p

q ˆ

²

+ ρ − 1 2 ( ˆ q

²

+ ρ)

²

δ

⁰

1 + ρ − 2 p q ˆ

²

+ ρ

+



 p

ˆ

q

²

+ ρ − 1 p q ˆ

²

+ ρ





3

δ

⁰⁰

1 + ρ − 2 p ˆ q

²

+ ρ

. (18) The second and the third term yield

3 2

Z

qˆ²_max 0

d ˆ q

²

p q ˆ

²

+ ρ − 1 ( ˆ q

²

+ ρ)

²

h q ˆ

²

δ

⁰

1 + ρ − 2 p ˆ q

²

+ ρ

= − 3 2

p q ˆ

²

+ ρ d d ˆ q

²

"

d ˆ q

²

dx

p q ˆ

²

+ ρ − 1 ( ˆ q

²

+ ρ)

²

h q ˆ

²

#

ˆ q²= ˆq_max²

. (19)

and

Z

qˆmax

0

d ˆ q

²

p q ˆ

²

+ ρ − 1 p q ˆ

²

+ ρ

!

3

h q ˆ

²

δ

⁰⁰

1 + ρ − 1 p q ˆ

²

+ ρ

=





 1

2 d

d ˆ q

²

+ ˆ q

²

+ ρ d

²

(d ˆ q

²

)

²





 q

q ˆ (x)

²

+ ρ − 1

3

ˆ

q

²

+ ρ h q ˆ

²













qˆ²= ˆq²_max

(20)

respectively when integrated together with a test function h .

(4)

(f) By putting all of the above results together you should now be able to nd Γ

Γ

0

=8 Z

qˆ²_max

0

d ˆ q

²

p

q ˆ

²

g

⁽⁰⁾

q ˆ

²

+

"

ρ − 1 2

p q ˆ

²

g

⁽¹⁾

q ˆ

²

+ 2 p

ˆ

q

²

g

⁽²⁾

q ˆ

²

(1 + ρ)

²

− 12

(ρ + 1)

⁴

g

⁽³⁾

q ˆ

²

#

qˆ²= ˆq_max²

+ p

ρ + ˆ q

²

d d ˆ q

²





 p

ρ + ˆ q

²

− 1

2

p ρ + ˆ q

²

g

⁽²⁾

q ˆ

²

+ 3

2 p q ˆ

²

+ ρ − 1 ( ˆ q

²

+ ρ)

³²

g

⁽³⁾

q ˆ

²







ˆ q²= ˆq_max²

+





 1

2 d

d ˆ q

²

+ ˆ q

²

+ ρ d

²

(d ˆ q

²

)

²





 q

ˆ

q (x)

²

+ ρ − 1

3

ˆ

q

²

+ ρ g

⁽³⁾

q ˆ

²













qˆ²= ˆq²_max

, (21)

with

Γ

0

= G

²_F

|V

_cb

|

²

m

⁵_b

192π

³

. (22)

Solution:

(a) By integrating the triple dierential rate we nd d

²

Γ

dq

²

dˆ ω = Z

E^ˆ^max_`

Eˆ^min_`

d ˆ E

_`

d

³

Γ d ˆ E

_`

dq

²

dˆ ω

= G

²_F

|V

_cb

|

²

m

⁵_b

8π

³

Z

E^ˆ_`^max Eˆ_`^min

ˆ

q

²

W

₁

q ˆ

₀

, q ˆ

²

+

2 ˆ E

_`

q ˆ

₀

W

₂

q ˆ

₀

, q ˆ

²

(23) with E ˆ

_`^(min,max)

=

ˆ q

₀

∓ p

ˆ q

²₀

− q ˆ

²

/2 and change variables to ω ˆ = 1 − q ˆ

₀

we nd d

²

Γ

dq

²

dˆ ω = m

²_b

3 p q ˆ

²

3 (1 − ω) ˆ

²

− q ˆ

²

W

1

+ ˆ q

²

W

2

Θ ˆ q

²

. (24)

(b) We change variables to u ˆ = 2 √ ˆ

ω − q ˆ

²

− ρ and integrate 8π

³

G

_F

|V

_cb

|

²

m

⁵_b

dΓ

d ˆ q

²

= 8π

³

G

_F

|V

_cb

|

²

m

⁵_b

Z

1−

√

ˆ q²

√

ρ+ ˆq²

dˆ ω d

²

Γ d ˆ q

²

dˆ ω

=

Z

1−2

√

ˆ q²−ρ 0

dˆ u 1 2 p

ˆ

u + ˆ q

²

+ ρ h

f

⁽⁰⁾

ω ˆ (ˆ u) , q ˆ

²

δ (ˆ u)

− f

⁽¹⁾

ω ˆ (ˆ u) , q ˆ

²

δ

⁰

(ˆ u) + 1

2 f

⁽²⁾

ω ˆ (ˆ u) , q ˆ

²

δ

⁰⁰

(ˆ u)

i

× Θ ˆ q

²

(ˆ u)

= 1 2

Z

1−2

√

ˆ q²−ρ 0

dˆ u

2

X

n=0

(−1)

ⁿ

(n − 1)!

d

⁽ⁿ⁾

dˆ u

⁽ⁿ⁾

"

f

⁽ⁿ⁾

ω ˆ (ˆ u) , q ˆ

²

p u ˆ + ˆ q

²

+ ρ Θ ˆ q

²

(ˆ u)

# δ (ˆ u)

= 1 2

2

X

n=0

(−1)

ⁿ

(n − 1)!

d

⁽ⁿ⁾

dˆ u

⁽ⁿ⁾

"

f

⁽ⁿ⁾

ω ˆ (ˆ u) , q ˆ

²

p u ˆ + ˆ q

²

+ ρ Θ ˆ q

²

(ˆ u)

#

u=0ˆ

(25)

(5)

(c)

d dˆ u Θ

1 + ρ + ˆ u − 2 p

ρ + ˆ q

²

+ ˆ u

u=0ˆ

= 1 − 1 p ρ + ˆ q

²

! δ

1 + ρ − 2 p ρ + ˆ q

²

= 1 − 1 p ρ + ˆ q

²

! 1 + ρ

2 δ q ˆ

²

−

1 − ρ 2

2

!

= ρ − 1

2 δ q ˆ

²

− q ˆ

²_max

(26) (d)

d

²

dˆ u

²

Θ

1 + ρ + ˆ u − 2 p

ρ + ˆ q

²

+ ˆ u

u=0ˆ

= δ

1 + ρ − 2 p ρ + ˆ q

²

2 (ρ + ˆ q

²

)

³²

+ 1 − 1

p ρ + ˆ q

²

+ ˆ u

! d dˆ u δ

1 + ρ + ˆ u − 2 p

ρ + ˆ q

²

+ ˆ u

u=0ˆ

= 2δ q ˆ

²

− q ˆ

_max²

(1 + ρ)

²

+

p

ρ + ˆ q

²

− 1

2

ρ + ˆ q

²

δ

⁰

1 + ρ − 2 p

ρ + ˆ q

²

(27) (e)

d

³

dˆ u

³

Θ

1 + ρ + ˆ u − 2 p ˆ

u + ˆ q

²

+ ρ

u=0ˆ

= − 12

(ρ + 1)

⁴

δ q ˆ

²

− q ˆ

²_max

+ 3

p

q ˆ

²

+ ρ − 1 2 ( ˆ q

²

+ ρ)

²

δ

⁰

1 + ρ − 2 p q ˆ

²

+ ρ

+



 p

ˆ

q

²

+ ρ − 1 p q ˆ

²

+ ρ





3

δ

⁰⁰

1 + ρ − 2 p ˆ q

²

+ ρ

(28)

(f) The full result is then obtained by combining the above intermediate results.

Exercise 3: Observables Next to the total decay rate in inclusive B decays the rst three moments of the lepton energy and of the hadronic invariant mass are observed. The nth raw lepton energy moment can be dened as

hE

_`ⁿ

i = Z

dq

²

dudE

_`

E

_`ⁿ

d

³

Γ

dq

²

dudE

_`

, (29)

i.e. as integrals of the triple dierential rate weighted with powers of the lepton energy. Analogously for computing the hadronic invariant mass moments, one denes

hω

ⁿ

u

^m

i = Z

dωdudE

_`

ω

ⁿ

u

^m

d

³

Γ

dωdudE

_`

(30)

and uses the relation

m

²_X

= m

²_c

+ u + 2Λω + Λ

²

, (31)

(6)

where Λ = m

_B

− m

_b

, for building central moments of the hadronic invariant mass such as m

²_X

= m

²_c

+ Λ

²

+ 2Λm

_b

h ωi ˆ + m

²_b

hˆ ui (32) D

m

²_X

−

m

²_X

2

E

= m

⁴_b

ˆ u

²

− hˆ ui

²

+ 4Λm

³_b

(hˆ ω ui − hˆ ˆ ωi hˆ ui) + 4Λ

²

m

²_b

ˆ ω

²

− hˆ ωi

²

. (33)

(a) In the Mathematica template the whole computation of exercise 2 is implemented. Adapt the code in such a way that you obtain the rst three lepton energy moments.

(b) One of the building blocks of the hadronic invariant mass moments consists of the moments of u . Can you explain why they are interesting?

Consider the rst moment of u at the LO (no computation needed). What is its value?