Work in progress:
It is now shown that U ~ ( a ) {\displaystyle {\tilde {U}}(a)} is unitary, i.e. U ~ ( a ) † = U ~ ( a ) − 1 {\displaystyle {\tilde {U}}(a)^{\dagger }={\tilde {U}}(a)^{-1}} :
U ~ ( a ) † = e ( i ) † a ℏ ( p ^ ) † = e − i a ℏ p ^ {\displaystyle {\tilde {U}}(a)^{\dagger }=e^{(i)^{\dagger }{\frac {a}{\hbar }}({\hat {p}})^{\dagger }}=e^{-i{\frac {a}{\hbar }}{\hat {p}}}}
⇒ U ~ ( a ) † U ~ ( a ) = e − i a ℏ p ^ e i a ℏ p ^ = [ ∑ m = 0 ∞ ( − i a p ^ ) m m ! ] [ ∑ n = 0 ∞ ( i a p ^ ) n n ! ] = ∑ m = 0 ∞ ∑ n = 0 ∞ ( − i a p ^ ) m ( i a p ^ ) n m ! n ! = ∑ m = 0 ∞ ∑ n = 0 ∞ ( − i a p ^ ) 2 m ( i a p ^ ) n ( 2 m ) ! n ! + ∑ m = 0 ∞ ∑ n = 0 ∞ ( − i a p ^ ) 2 m + 1 ( i a p ^ ) n ( 2 m + 1 ) ! n ! = ∑ m = 0 ∞ ∑ n = 0 ∞ ( i a p ^ ) 2 m ( i a p ^ ) n ( 2 m ) ! n ! − ∑ m = 0 ∞ ∑ n = 0 ∞ ( i a p ^ ) 2 m + 1 ( i a p ^ ) n ( 2 m + 1 ) ! n ! = ∑ n = 0 ∞ ( i a p ^ ) n n ! + ∑ m = 1 ∞ ∑ n = 0 ∞ ( i a p ^ ) 2 m ( i a p ^ ) n ( 2 m ) ! n ! − ∑ n = 0 ∞ ( i a p ^ ) ( n + 1 ) n ! − ∑ m = 1 ∞ ∑ n = 0 ∞ ( i a p ^ ) 2 m + 1 ( i a p ^ ) n ( 2 m + 1 ) ! n ! {\displaystyle {\begin{aligned}\Rightarrow {\tilde {U}}(a)^{\dagger }{\tilde {U}}(a)&=e^{-i{\frac {a}{\hbar }}{\hat {p}}}e^{i{\frac {a}{\hbar }}{\hat {p}}}\\&=\left[\sum _{m=0}^{\infty }{\frac {\left(-ia{\hat {p}}\right)^{m}}{m!}}\right]\left[\sum _{n=0}^{\infty }{\frac {\left(ia{\hat {p}}\right)^{n}}{n!}}\right]\\&=\sum _{m=0}^{\infty }\sum _{n=0}^{\infty }{\frac {\left(-ia{\hat {p}}\right)^{m}\left(ia{\hat {p}}\right)^{n}}{m!\,n!}}\\&=\sum _{m=0}^{\infty }\sum _{n=0}^{\infty }{\frac {\left(-ia{\hat {p}}\right)^{2m}\left(ia{\hat {p}}\right)^{n}}{(2m)!\,n!}}+\sum _{m=0}^{\infty }\sum _{n=0}^{\infty }{\frac {\left(-ia{\hat {p}}\right)^{2m+1}\left(ia{\hat {p}}\right)^{n}}{(2m+1)!\,n!}}\\&=\sum _{m=0}^{\infty }\sum _{n=0}^{\infty }{\frac {\left(ia{\hat {p}}\right)^{2m}\left(ia{\hat {p}}\right)^{n}}{(2m)!\,n!}}-\sum _{m=0}^{\infty }\sum _{n=0}^{\infty }{\frac {\left(ia{\hat {p}}\right)^{2m+1}\left(ia{\hat {p}}\right)^{n}}{(2m+1)!\,n!}}\\&=\sum _{n=0}^{\infty }{\frac {\left(ia{\hat {p}}\right)^{n}}{n!}}+\sum _{m=1}^{\infty }\sum _{n=0}^{\infty }{\frac {\left(ia{\hat {p}}\right)^{2m}\left(ia{\hat {p}}\right)^{n}}{(2m)!\,n!}}-\sum _{n=0}^{\infty }{\frac {\left(ia{\hat {p}}\right)^{(n+1)}}{n!}}-\sum _{m=1}^{\infty }\sum _{n=0}^{\infty }{\frac {\left(ia{\hat {p}}\right)^{2m+1}\left(ia{\hat {p}}\right)^{n}}{(2m+1)!\,n!}}\\\end{aligned}}}
is unitary, i.e. 
:
![{\displaystyle {\begin{aligned}\Rightarrow {\tilde {U}}(a)^{\dagger }{\tilde {U}}(a)&=e^{-i{\frac {a}{\hbar }}{\hat {p}}}e^{i{\frac {a}{\hbar }}{\hat {p}}}\\&=\left[\sum _{m=0}^{\infty }{\frac {\left(-ia{\hat {p}}\right)^{m}}{m!}}\right]\left[\sum _{n=0}^{\infty }{\frac {\left(ia{\hat {p}}\right)^{n}}{n!}}\right]\\&=\sum _{m=0}^{\infty }\sum _{n=0}^{\infty }{\frac {\left(-ia{\hat {p}}\right)^{m}\left(ia{\hat {p}}\right)^{n}}{m!\,n!}}\\&=\sum _{m=0}^{\infty }\sum _{n=0}^{\infty }{\frac {\left(-ia{\hat {p}}\right)^{2m}\left(ia{\hat {p}}\right)^{n}}{(2m)!\,n!}}+\sum _{m=0}^{\infty }\sum _{n=0}^{\infty }{\frac {\left(-ia{\hat {p}}\right)^{2m+1}\left(ia{\hat {p}}\right)^{n}}{(2m+1)!\,n!}}\\&=\sum _{m=0}^{\infty }\sum _{n=0}^{\infty }{\frac {\left(ia{\hat {p}}\right)^{2m}\left(ia{\hat {p}}\right)^{n}}{(2m)!\,n!}}-\sum _{m=0}^{\infty }\sum _{n=0}^{\infty }{\frac {\left(ia{\hat {p}}\right)^{2m+1}\left(ia{\hat {p}}\right)^{n}}{(2m+1)!\,n!}}\\&=\sum _{n=0}^{\infty }{\frac {\left(ia{\hat {p}}\right)^{n}}{n!}}+\sum _{m=1}^{\infty }\sum _{n=0}^{\infty }{\frac {\left(ia{\hat {p}}\right)^{2m}\left(ia{\hat {p}}\right)^{n}}{(2m)!\,n!}}-\sum _{n=0}^{\infty }{\frac {\left(ia{\hat {p}}\right)^{(n+1)}}{n!}}-\sum _{m=1}^{\infty }\sum _{n=0}^{\infty }{\frac {\left(ia{\hat {p}}\right)^{2m+1}\left(ia{\hat {p}}\right)^{n}}{(2m+1)!\,n!}}\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bbda9771eb22dee26235d249782ac7e4a7b35a53)