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Please let me make questions after showing what I am studying.
We first consider two particles (they may be either leptons or photons) with initial (i.e. before collision) four momentum ##p_i = (E_i, \mathbf p_i)##, ##i=1,2##. These two collide and produce ##N## final particles with momentum ##p'_f = (E'_f, \mathbf p'_f)##, ##f=1,...,N##. In this post we assume that initial and final particles are in definite polarization states.
I learned in the previous chapter of the book that the S-matrix elements can be written as
$$\langle f| S |i \rangle = \delta_{fi} + \Big[ (2\pi)^4 \delta^{(4)} (P_f-P_i) \Pi_e \Big(\frac{m}{VE}\Big)^{1/2}\Pi_e \Big(\frac{1}{2V \omega}\Big)^{1/2} \Big] \mathscr{M} \ \ \ \ (1)$$
Where the e attached to the products stands for external fermions and photons.
In our case ##(1)## becomes
$$S_{fi} = \delta_{fi} + (2 \pi)^4 \delta^{(4)} (\sum p'_f - \sum p_i) \Pi_i \Big(\frac{1}{2V E_i}\Big)^{1/2} \Pi_f \Big(\frac{1}{2V E'_f}\Big)^{1/2}\Pi_l (2m_l)^{1/2}\mathscr{M} \ \ \ \ (2)$$
Where ##l## runs over all external leptons involved in the collision.
Eq. ##(2)## applies to an infinite time interval (i.e ##T \rightarrow \infty##) and an infinite volume (i.e ##V \rightarrow \infty##).
For finite intervals we'd get the exact same as Eq. ##(2)## but ##(2 \pi)^4 \delta^{(4)} (\sum p'_f - \sum p_i)## being replaced by ##\delta_{TV} (\sum p'_f - \sum p_i)##, as
$$(2 \pi)^4 \delta^{(4)} (\sum p'_f - \sum p_i) = lim_{T \rightarrow \infty, \ V \rightarrow \infty} \delta_{TV} (\sum p'_f - \sum p_i) \ \ \ \ (3)$$
In deriving the cross section we better take ##V## and ##T## to be finite. In such a case the transition probability per unit time becomes
$$w=|S_{fi}|^2/T \ \ \ \ (4)$$
We notice that ##|S_{fi}|^2## means we have to deal with ##\Big( \delta_{TV} (\sum p'_f - \sum p_i) \Big)^2##, which equals to the following
$$\Big( \delta_{TV} (\sum p'_f - \sum p_i) \Big)^2 = TV (2\pi)^4 \delta^{(4)} (\sum p'_f - \sum p_i) \ \ \ \ (5)$$
With errors which tend to zero as ##T \rightarrow \infty## and ##V \rightarrow \infty##. Once we know this ##(4)## becomes
$$w=V(2\pi)^4 \delta^{(4)} (\sum p'_f - \sum p_i) \Pi_i \Big(\frac{1}{2V E_i}\Big)^{1/2} \Pi_f \Big(\frac{1}{2V E'_f}\Big)^{1/2} \Pi_l (2m_l)^{1/2} |\mathscr{M}|^2 \ \ \ \ (6)$$
Eq. ##(6)## is the transition rate to one definite final state.
If we wish to obtain the transition rate to a group of final states with momentum in the interval ##(\mathbf p'_f, \mathbf p'_f +d\mathbf p'_f)##, ##f=1,...,N##, we have to multiply ##w## by the number of these states, which is
$$\Pi_f \Big(\frac{V d^3 \mathbf p'_f}{(2\pi)^3}\Big)$$
The differential cross section is the transition rate into this group of final states for one scattering center and unit incident flux (which is also known as luminosity; I've read that particle physicists use more the latter term). With our choice of normalization for the states, the volume V contains one scattering center and the incident flux is ##\frac{v_{rel}}{V}##, where ##v_{rel}## is the relative velocity of the colliding particles.
At this point we have everything to set up the equation for the differential cross section
$$d \sigma = w \frac{V}{v_{rel}} \Pi_f \Big(\frac{V d^3 \mathbf p'_f}{(2\pi)^3}\Big) \ \ \ \ (7)$$
Plugging ##(6)## into ##(7)## we get
$$d \sigma = (2\pi)^4 \delta^{(4)} (\sum p'_f - \sum p_i) \frac{1}{4 E_1 E_2 v_{rel}} \Pi_l (2m_l) \Pi_f \Big(\frac{d^3 \mathbf p'_f}{(2\pi)^3 2 E'_f}\Big) |\mathscr{M}|^2$$
My questions are:
1) I do not see why ##(2)## corresponds to the limit of an infinite time interval and an infinite volume.
2) I do not understand why we have to replace ##(2 \pi)^4 \delta^{(4)} (\sum p'_f - \sum p_i)## by ##\delta_{TV} (\sum p'_f - \sum p_i)##. Besides, I do not understand equation ##(3)##.
3) I do not see how to get equation ##(5)##.
4) I first studied differential cross-section from the nice book by Griffiths: Introduction to Quantum Mechanics; (second edition; chapter 11). Then I learned that the larger the infinitesimal patch of cross-sectional area ##d\sigma## is the bigger ##d \Omega## will be, which means we can establish the following proportionality factor ##D(\theta)##
$$d \sigma=D(\theta)d \Omega \ \ \ \ (8)$$
Thus (8) and (7) must be equal to each other but I do not see it.Any help is appreciated.
Thanks
We first consider two particles (they may be either leptons or photons) with initial (i.e. before collision) four momentum ##p_i = (E_i, \mathbf p_i)##, ##i=1,2##. These two collide and produce ##N## final particles with momentum ##p'_f = (E'_f, \mathbf p'_f)##, ##f=1,...,N##. In this post we assume that initial and final particles are in definite polarization states.
I learned in the previous chapter of the book that the S-matrix elements can be written as
$$\langle f| S |i \rangle = \delta_{fi} + \Big[ (2\pi)^4 \delta^{(4)} (P_f-P_i) \Pi_e \Big(\frac{m}{VE}\Big)^{1/2}\Pi_e \Big(\frac{1}{2V \omega}\Big)^{1/2} \Big] \mathscr{M} \ \ \ \ (1)$$
Where the e attached to the products stands for external fermions and photons.
In our case ##(1)## becomes
$$S_{fi} = \delta_{fi} + (2 \pi)^4 \delta^{(4)} (\sum p'_f - \sum p_i) \Pi_i \Big(\frac{1}{2V E_i}\Big)^{1/2} \Pi_f \Big(\frac{1}{2V E'_f}\Big)^{1/2}\Pi_l (2m_l)^{1/2}\mathscr{M} \ \ \ \ (2)$$
Where ##l## runs over all external leptons involved in the collision.
Eq. ##(2)## applies to an infinite time interval (i.e ##T \rightarrow \infty##) and an infinite volume (i.e ##V \rightarrow \infty##).
For finite intervals we'd get the exact same as Eq. ##(2)## but ##(2 \pi)^4 \delta^{(4)} (\sum p'_f - \sum p_i)## being replaced by ##\delta_{TV} (\sum p'_f - \sum p_i)##, as
$$(2 \pi)^4 \delta^{(4)} (\sum p'_f - \sum p_i) = lim_{T \rightarrow \infty, \ V \rightarrow \infty} \delta_{TV} (\sum p'_f - \sum p_i) \ \ \ \ (3)$$
In deriving the cross section we better take ##V## and ##T## to be finite. In such a case the transition probability per unit time becomes
$$w=|S_{fi}|^2/T \ \ \ \ (4)$$
We notice that ##|S_{fi}|^2## means we have to deal with ##\Big( \delta_{TV} (\sum p'_f - \sum p_i) \Big)^2##, which equals to the following
$$\Big( \delta_{TV} (\sum p'_f - \sum p_i) \Big)^2 = TV (2\pi)^4 \delta^{(4)} (\sum p'_f - \sum p_i) \ \ \ \ (5)$$
With errors which tend to zero as ##T \rightarrow \infty## and ##V \rightarrow \infty##. Once we know this ##(4)## becomes
$$w=V(2\pi)^4 \delta^{(4)} (\sum p'_f - \sum p_i) \Pi_i \Big(\frac{1}{2V E_i}\Big)^{1/2} \Pi_f \Big(\frac{1}{2V E'_f}\Big)^{1/2} \Pi_l (2m_l)^{1/2} |\mathscr{M}|^2 \ \ \ \ (6)$$
Eq. ##(6)## is the transition rate to one definite final state.
If we wish to obtain the transition rate to a group of final states with momentum in the interval ##(\mathbf p'_f, \mathbf p'_f +d\mathbf p'_f)##, ##f=1,...,N##, we have to multiply ##w## by the number of these states, which is
$$\Pi_f \Big(\frac{V d^3 \mathbf p'_f}{(2\pi)^3}\Big)$$
The differential cross section is the transition rate into this group of final states for one scattering center and unit incident flux (which is also known as luminosity; I've read that particle physicists use more the latter term). With our choice of normalization for the states, the volume V contains one scattering center and the incident flux is ##\frac{v_{rel}}{V}##, where ##v_{rel}## is the relative velocity of the colliding particles.
At this point we have everything to set up the equation for the differential cross section
$$d \sigma = w \frac{V}{v_{rel}} \Pi_f \Big(\frac{V d^3 \mathbf p'_f}{(2\pi)^3}\Big) \ \ \ \ (7)$$
Plugging ##(6)## into ##(7)## we get
$$d \sigma = (2\pi)^4 \delta^{(4)} (\sum p'_f - \sum p_i) \frac{1}{4 E_1 E_2 v_{rel}} \Pi_l (2m_l) \Pi_f \Big(\frac{d^3 \mathbf p'_f}{(2\pi)^3 2 E'_f}\Big) |\mathscr{M}|^2$$
My questions are:
1) I do not see why ##(2)## corresponds to the limit of an infinite time interval and an infinite volume.
2) I do not understand why we have to replace ##(2 \pi)^4 \delta^{(4)} (\sum p'_f - \sum p_i)## by ##\delta_{TV} (\sum p'_f - \sum p_i)##. Besides, I do not understand equation ##(3)##.
3) I do not see how to get equation ##(5)##.
4) I first studied differential cross-section from the nice book by Griffiths: Introduction to Quantum Mechanics; (second edition; chapter 11). Then I learned that the larger the infinitesimal patch of cross-sectional area ##d\sigma## is the bigger ##d \Omega## will be, which means we can establish the following proportionality factor ##D(\theta)##
$$d \sigma=D(\theta)d \Omega \ \ \ \ (8)$$
Thus (8) and (7) must be equal to each other but I do not see it.Any help is appreciated.
Thanks