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\input{./header}
\input{./shortcuts}

\graphicspath{{images/}{image_install/}}


\newcommand{\sheetnr}{1}
\newcommand{\issued}{25.03.2017}
\newcommand{\dueUni}{31.03.2017}
%\newcommand{\version}{1}   % if you need to release a corrected version, uncomment and increase this counter

\newcommand{\dd}[1][]{\text{d}^{#1}}
\newcommand{\Mod}{{\rm mod\,}}
\DeclareMathOperator{\tr}{Tr}

\showsolutions

\startsheet

\setcounter{exercise}{1}
\exercise[3.5]{Conservation laws}


Under which interactions are the following interactions possible (please not that some of the processes) are not possible at all): 

\begin{itemize}
\item $\pi^- p \to \pi^0 n$
\item $\pi^0 \to \gamma \gamma \gamma$
\item $\pi^0 \to \gamma \gamma$
\item $\pi^+ \to \mu^+ \nu_{\mu}$
\item $\pi^+ \to \mu^+ \bar{\nu}_{\mu}$
\item $p \bar{p} \to \Lambda_0 \Lambda_0$
\item $p \bar{p} \to \gamma$
\end{itemize}

\setcounter{exercise}{2}
\exercise[2.0]{Forbade decays}


Which conservation laws forbade the following decays:

\begin{itemize}
\item $n \to p e^-$
\item $n \to \pi^+ e^-$
\item $n \to p \pi^-$
\item $n \to p \gamma^-$
\end{itemize}


\setcounter{exercise}{3}
\exercise[2.0]{Suppress decays}


Which conservation laws forbade or suppress the following processes:

\begin{itemize}
\item $p n \to p \Lambda^0$
\item $K^+ \to \pi^- \pi^+ \pi^- \pi^+  \pi^+ \pi^0 $
\item $\Lambda_0 \to K^0 \pi^0$
\item $K \to \pi \gamma$
\item $K^- \to \pi^0 e^-$
\item $K^+ \to \pi^+ \pi^- \pi^0$
\end{itemize}



\setcounter{exercise}{4}
\exercise[16]{Pion decay}


Please calculate the matrix element of the dacay $\pi^- \to \mu \nu_{\mu}$. The form factor for the pion has the form of $F^{\mu} = p_{\mu} f_{\pi}$, where $f_{\pi}$ is so called pion decay constant and is calculated on lattice to be $f_{\pi}=130~\rm MeV$. Using the matrix element calculate the $\Gamma$. Calculate also the $\Gamma$ for the $\pi^- \to e \nu_{e}$. Why is the electron mode different then the muon one?

Please read about the ''Casimir trick'' in the D. Griffiths handbook (Sec. 7.7).


\setcounter{exercise}{5}
\exercise[16]{Muon decay}

Calculate the matrix element for the dacay of the muon: $\mu^- \to e^- \nu_{\mu} \bar{\nu_e}$. Using ''Golden Rule'' calculate the calculate the $\Gamma$ and the lifetime of the muon.

\setcounter{exercise}{6}
\exercise[12]{Muon decay simulation}

Please simulate the muon decay from exercise 5 using ROOT. Please assume for the moment flat phase space (aka matrix element =1). The example can be found:

\url{https://root.cern.ch/root/html/tutorials/physics/PhaseSpace.C.html}

having simulate this decay please calculate the electron energy in the muon central of mass and draw it for your simulated events. Simulate at least 100.000 events.



\clearpage
\iffalse
\begin{solution}
\begin{subtasks}

\task The answers are:
\begin{center}
\begin{tabular}{| l | l | l |}
\hline
Decimal:					& Hexadecimal:					&	Binary: 				\\ \hline
$12$							& C												& 1100						\\ \hline
$53$							& 35											& 110101				\\ \hline
$123$						& 7B											& 1111011				\\ \hline
$431$						& 1AF										& 110101111			\\ \hline
\end{tabular}
\end{center}

A simple algorithm to convert an integer $N$ from decimal to hexadecimal is the following:
\begin{enumerate}
\item[1.] set $i=1$.
\item[2.] Compute $h_i = N_{\Mod 16}$, i.e.~the remainder of the integer (Euclidean) division $N\div16$.
\item[3.] Use the hexadecimal “dictionary" to convert $h_i$:

\begin{tabular}{| l | l | l | l | l | l | l | l | l | l | l | l | l | l | l | l | l |}
\hline
$h$:	& 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 \\
\hline
hex: & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & A & B & C & D & E & F	\\
\hline
\end{tabular}
\item[4.] Replace $N \rightarrow (N - h_i)/16$, set $i=i+1$ and go back to step 2 until $N=0$.
\item[5.] The number in hexadecimal representation is: $h_n h_{n-1} \cdots h_2 h_1$
\end{enumerate}
This algorithm is actually valid for every base change.

Let us apply the algorithm above. 

$N=12$:
\begin{align*}
& i=1; 		&		\\
& h_1 = 12_{\Mod 16} = 12 \rightarrow \boxed{C};		&	\\
& N \rightarrow (N - 12)/16 =0; \\
&{\rm end:~} 12_{10} = \boxed{C_{16}}
\end{align*}
%
$N=53$:
\begin{align*}
& i=1																			&&	i=2	\\
& h_1 = 53_{\Mod 16} = 5 \rightarrow \boxed{5};		&	& h_2 = 3_{\Mod 16} = 3 \rightarrow \boxed{3} \\
& N \rightarrow (N - 5)/16 =3; 													&&	 N \rightarrow (N - 3)/16 =0; \\
&&&{\rm end:~} 53_{10} = \boxed{35_{16}}
\end{align*}
%
$N=123$:
\begin{align*}
& i=1 																				&&	i=2	\\
& h_1 = 123_{\Mod 16} = 11 \rightarrow \boxed{B};		&	& h_2 = 7_{\Mod 16} = 7 \rightarrow \boxed{7} \\
& N \rightarrow (N - 11)/16 =7; 													&&	 N \rightarrow (N - 7)/16 =0; \\
&&&{\rm end:~} 123_{10} = \boxed{7B_{16}}
\end{align*}
%
$N=431$:
\begin{align*}
& i=1 																				&&	i=2	  					&&i=3\\
& h_1 = 431_{\Mod 16} = 15 \rightarrow \boxed{F};		
&& h_2 = 26_{\Mod 16} = 10 \rightarrow \boxed{A} 
&& h_3 = 1_{\Mod 16} = 1 \rightarrow \boxed{1} 
\\
& N \rightarrow (N - 11)/16 =26; 		&&	 N \rightarrow (N - 10)/16 =1;  &&	 N \rightarrow (N - 1)/16 =0; \\
&&&&&{\rm end:~} 431_{10} = \boxed{1AF_{16}}
\end{align*}
%

The former algorithm can be applied also for the conversion in binary. One just need to change $\Mod 16 \rightarrow \Mod 2$ in step 2 and to divide by $2$ (not by $16$) in step 4.

$N=12$:
\begin{align*}
& i=1 		&&		i=2  && i=3 && i=4 \\
& h_1 = 12_{\Mod 2} = \boxed{0} 		&& h_2 = 6_{\Mod 2} = \boxed{0}	
&& h_3 = 3_{\Mod 2} = \boxed{1} && h_4 = 1_{\Mod 2} = \boxed{1} \\
& N \rightarrow (N - 0)/2 =6;  && N \rightarrow (N - 0)/2 =3; 
&& N \rightarrow (N - 1)/2 =1;  && N \rightarrow (N - 1)/2 =0; \\
&&&&&&&{\rm end:~} 12_{10} = \boxed{1100_{2}}
\end{align*}
%
$N=53$:
\begin{align*}
& i=1 		&&		i=2  && i=3 && i=4 \\
& h_1  = 53_{\Mod 2} = \boxed{1} 		&& h_2 = 26_{\Mod 2} = \boxed{0}	
&& h_3 = 13_{\Mod 2} = \boxed{1} && h_4 = 6_{\Mod 2} = \boxed{0} \\
& N \rightarrow  (N - 1)/2 =26;  && N \rightarrow (N - 0)/2 =13; 
&& N \rightarrow (N - 1)/2 =6;  && N \rightarrow (N - 0)/2 =3; \\
%
\\
%
& i=5 		&&		i=6  \\
& h_5  = 3_{\Mod 2} = \boxed{1} 		&& h_6 = 1_{\Mod 2} = \boxed{1}	\\
& N \rightarrow  (N - 1)/2 =1;  && N \rightarrow (N - 1)/2 =0;  \\
&&&{\rm end:~} 53_{10} = \boxed{110101_{2}}
\end{align*}
%
$N=123$:
\begin{align*}
& i=1; 		&&		i=2  && i=3 && i=4 \\
& h_1  = 123_{\Mod 2} = \boxed{1} 		&& h_2 = 61_{\Mod 2} = \boxed{1}	
&& h_3 = 30_{\Mod 2} = \boxed{0} && h_4 = 15_{\Mod 2} = \boxed{1} \\
& N \rightarrow  (N - 1)/2 =61;  && N \rightarrow (N - 1)/2 =30; 
&& N \rightarrow (N)/2 =15;  && N \rightarrow (N - 1)/2 =7; \\
%
\\
%
& i=5 		&&		i=6  && i=7\\
& h_5  = 7_{\Mod 2} = \boxed{1} 		&& h_6 = 3_{\Mod 2} = \boxed{1} && h_7 = 1_{\Mod 2} = \boxed{1}	\\
& N \rightarrow  (N - 1)/2 =3; && N \rightarrow  (N-1)/2 =1;  && N \rightarrow (N - 1)/2 =0;  \\
&&&{\rm end:~} 53_{10} = \boxed{1111011_{2}}
\end{align*}
%
$N=431$:
\begin{align*}
& i=1 		&&		i=2  && i=3 && i=4,5,6,7,8,9 \\
& h_1  = 431_{\Mod 2} = \boxed{1} 		&& h_2 = 215_{\Mod 2} = \boxed{1}	
&& h_3 = 107_{\Mod 2} = \boxed{1} && \dots ({\rm see~}N=53) \\
& N \rightarrow  (N - 1)/2 =215;  && N \rightarrow (N - 1)/2 = 107; 
&& N \rightarrow (N )/2 =53;  && \dots \\
\\
&	{\rm end:~} 431_{10} = \boxed{110101111_{2}}
\end{align*}
%

\task One needs to understand that the decimal representation of an integer is just a short-cut notation. With $124_{10}$ one really means:
\begin{equation*}
124_{10} = 1 \cdot 10^2 + 2 \cdot 10^1 + 4 \cdot 10^0 ~.
\end{equation*}
%
This is valid in \emph{every} basis. Thus is easy to find:
%
\begin{align*}
10011_{2} &= (1 \cdot 2^4 + 0 \cdot 2^3 + 0 \cdot 2^2 + 1 \cdot 2^1 + 1 \cdot 2^0)_{10}
						= (16+2+1)_{10} = 19_{10}
\\
1101_{2} &=(1 \cdot 2^3 + 1 \cdot 2^2 + 0 \cdot 2^1 + 1 \cdot 2^0)_{10}
					= (8+4+1)_{10} = 13_{10}
\\
A2_{16} &=(A \cdot 16^1 + 2 \cdot 16^0)_{10}
					= (10 \cdot 16 + 2)_{10} = 162_{10}
\\
1AD_{16} &=(1 \cdot 16^2 + A \cdot 16^1 + D \cdot 16^0)_{10}
					= (1 \cdot 256 + 10 \cdot 16 + 13)_{10} = 429_{10}
\end{align*}
%


\task Before converting a value into a floating number, one should convert it into a binary expression. We already know how to convert $-431_{10}$. To convert a non-integer number $N$ into binary, one can use an extension of the algorithm used in exercise (1a).
\begin{enumerate}
\item[1.] i=0.
\item[2.] Take $\lfloor N \rfloor$ (i.e.~integer part of $N$) and convert it into a binary expression 
(call it $h_i$).
\item[3.] Substitute $N \rightarrow 2(N - \lfloor N \rfloor)$.
\item[4.] Set $i=i+1$ and go back to step 2 until $N=0$ or you have reached the desired precision.
\item[5.] Your number is $h_0. h_1 h_2 \dots h_n$
\end{enumerate}
%

Let us convert $0.3125$ with the mentioned algorithm:
%
\begin{align*}
& i=0 		&&		i=1  && i=2 \\
& h_0  = \lfloor 0.3125 \rfloor = \boxed{0} 		&& h_1 =  \lfloor 0.625 \rfloor = \boxed{0}
&& h_2 =  \lfloor 1.25 \rfloor = \boxed{1}  \\
& N \rightarrow  2(N - \lfloor N \rfloor) = 0.625;  && N \rightarrow 2(N - \lfloor N) \rfloor) = 1.25; 
&& N \rightarrow 2(N - \lfloor N) \rfloor) = 0.5;    \\
\\
& i=3 		&&		i=4  \\
& h_3  = \lfloor 0.5 \rfloor = \boxed{0} 		&& h_1 =  \lfloor 1 \rfloor = \boxed{1}\\
& N \rightarrow  2(N - \lfloor N \rfloor) = 1;  && N \rightarrow 2(N - \lfloor N) \rfloor = 0; 
  \\
&&&	{\rm end:~} 0.3125_{10} = \boxed{0.0101_{2}}
\end{align*}
%
Thus $0.3125_{10} = 0.0101_{2} = (1.01 \cdot 2^{-2})_2$. Thus:
%
\begin{itemize}
\item[-] The sign is $+ \rightarrow 0$
\item[-] The exponent is $-2$. Exponent+bias: $-2+15=13=01101_2$.
\item[-] The mantissa is $1.010$. The first digit ($1.$) is not stored, and we store $0100000000$.
\end{itemize}
%
Thus we have:
%
\begin{equation*}
0.3125_{10} \rightarrow 	\underbrace{0}_{\rm sign}
														\underbrace{01101}_{\rm exp+bias}
														\underbrace{0100000000}_{\rm mantissa}
							\rightarrow	\underbrace{00110101}_{\rm byte~1} 
													~ \underbrace{00000000}_{\rm byte~2}
							\rightarrow	\underbrace{35~~00}_{\rm hexadecimal}
\end{equation*}
%

For $-431_{10}$ the exercise is similar. We have already computed $431_{10}=110101111 = 1.10101111 \cdot 2^{8}$. Thus:
%
\begin{itemize}
\item[-] The sign is $- \rightarrow 1$
\item[-] The exponent is $8$. Exponent+bias: $8+15=23=10111_2$.
\item[-] The mantissa is $1.10101111$. The first digit ($1.$) is not stored, and we store $1010111100$.
\end{itemize}
%
Thus we have:
%
\begin{equation*}
-431_{10} \rightarrow 	\underbrace{1}_{\rm sign}
														\underbrace{10111}_{\rm exp+bias}
														\underbrace{1010111100}_{\rm mantissa}
							\rightarrow	\underbrace{11011110}_{\rm byte~1} 
													~ \underbrace{10111100}_{\rm byte~2}
							\rightarrow	\underbrace{\rm DE~~BC}_{\rm hexadecimal}
\end{equation*}
%


\task We have already found the hexadecimal bytes representation for our floating numbers: $0.3125_{10}=35~00$ and $-431_{10}={\rm DE~BC}$.
In big endian, these bytes would be sequentially stored in the memory “from left to right", while in little endian the sequential storage would go in the other direction. Explicitly:
%
\begin{align*}
0.3125_{10}  \rightarrow ~& {\rm big~endian:~}
					\underbrace{\boxed{35}}_{\rm slot~i}~\underbrace{\boxed{00}}_{\rm slot~i+1}
&
-431_{10}  \rightarrow ~& {\rm big~endian:~}
					\underbrace{\boxed{\rm DE}}_{\rm slot~i}~\underbrace{\boxed{\rm BC}}_{\rm slot~i+1}
\\
					\rightarrow ~& {\rm little~endian:~}
					\underbrace{\boxed{00}}_{\rm slot~i}~\underbrace{\boxed{35}}_{\rm slot~i+1}
&
					\rightarrow ~& {\rm little~endian:~}
					\underbrace{\boxed{\rm BC}}_{\rm slot~i}~\underbrace{\boxed{\rm DE}}_{\rm slot~i+1}
\end{align*}
%


\task  
The number $2050$ expressed in binary is $100000000010$.

In {\it binary16} its representation is:
%
\begin{equation}
	\underbrace{0}_{\rm sign}
	\underbrace{11010}_{\rm exp+bias}
	\underbrace{0000000001}_{\rm mantissa}
	~~\Rightarrow~~
	+ \;
	2^{26 - 15} \times
	1.0000000001
	=
	1.0000000001 \times 2^{11}
\end{equation}
%
From this representation is already clear that the {\it binary16} precision doesn't allow to store numbers like $2049$ or $2051$ at all, since this would require a 12 digit precision.

Thus, both in INPUT A and in INPUT B, the code is requiring the machine a precision it cannot reach. As it should be clear from how the number $2050$ is stored, the machine has reached it maximal precision and it cannot store the number $2051$ whatsoever. This means that at each step in which the number $2051$ is met, the machine truncate it consistently with its maximum working precision, returning $2050$.

INPUT C is well written and works fine. Differently from INPUT B, this time the machine works \emph{at first} with numbers of order 1, which are manipulated without difficulties. For example, the {\it binary16} representation of $5$ would be:
%
\begin{equation}
	\underbrace{0}_{\rm sign}
	\underbrace{10001}_{\rm exp+bias}
	\underbrace{0100000000}_{\rm mantissa -1}
	~~\Rightarrow~~
	+ \;
	2^{17 - 15} \times
	1.0100000000
	=
	1.01 \times 2^2 = 101_2
\end{equation}
%
where indeed $101_2=5_{10}$. After having dealt with all order-1 numbers, it finally sums them to 2050, getting 2060 which is a perfectly storable number.

In conclusion, one should be aware of the bad consequences of requiring the machine to work with more significant digits than its maximum.

\end{subtasks}
\end{solution}



\fi


\end{document}
