Prove that there is a unitary operator U ~ ( a ) {\displaystyle {\tilde {U}}(a)} , which is a function of p ^ = ℏ i d d x {\displaystyle {\hat {p}}={\frac {\hbar }{i}}{\frac {d}{dx}}} , such that for some wavefunction ψ ( x ) {\displaystyle \psi (x)} , U ~ ( a ) ψ ( x ) = ψ ( x + a ) {\displaystyle {\tilde {U}}(a)\psi (x)=\psi (x+a)} .
ψ ( x + a ) = ∑ n = 0 ∞ ψ ( n ) ( x ) | x = 0 ⋅ ( x + a ) n n ! = ∑ n = 0 ∞ ( ψ ( n ) ( x ) | x = 0 n ! ⋅ ∑ m = 0 n ( n m ) x n − m a m ) = ∑ n = 0 ∞ ( ψ ( n ) ( x ) | x = 0 n ! ⋅ ∑ m = 0 n n ( n − 1 ) ⋯ ( n − ( m − 1 ) ) m ! x n − m a m ) = ∑ n = 0 ∞ ( ψ ( n ) ( x ) | x = 0 n ! ⋅ ∑ m = 0 n a m m ! d m d x m x n ) = ∑ n = 0 ∞ [ ψ ( n ) ( x ) | x = 0 n ! ⋅ ( ∑ m = 0 n a m m ! d m d x m x n + ∑ m = n + 1 ∞ 0 ) ] = ∑ n = 0 ∞ [ ψ ( n ) ( x ) | x = 0 n ! ⋅ ( ∑ m = 0 n a m m ! d m d x m x n + ∑ m = n + 1 ∞ d m d x m x n ) ] = ∑ n = 0 ∞ ( ψ ( n ) ( x ) | x = 0 n ! ⋅ ∑ m = 0 ∞ a m m ! d m d x m x n ) = ∑ n = 0 ∞ ∑ m = 0 ∞ ψ ( n ) ( x ) | x = 0 n ! a m m ! d m d x m x n = [ ∑ m = 0 ∞ a m m ! d m d x m ] [ ∑ n = 0 ∞ ψ ( n ) ( x ) | x = 0 n ! x n ] = exp ( a ⋅ d m d x m ) ψ ( x ) = exp ( a ⋅ i ℏ ⋅ ℏ i d m d x m ) ψ ( x ) = e i a ℏ p ^ ψ ( x ) {\displaystyle {\begin{aligned}\psi (x+a)&=\sum _{n=0}^{\infty }\left.\psi ^{(n)}(x)\right\vert _{x=0}\cdot {\frac {(x+a)^{n}}{n!}}\\&=\sum _{n=0}^{\infty }\left({\frac {\left.\psi ^{(n)}(x)\right\vert _{x=0}}{n!}}\cdot \sum _{m=0}^{n}{\binom {n}{m}}x^{n-m}a^{m}\right)\\&=\sum _{n=0}^{\infty }\left({\frac {\left.\psi ^{(n)}(x)\right\vert _{x=0}}{n!}}\cdot \sum _{m=0}^{n}{\frac {n\,(n-1)\cdots (n-(m-1))}{m!}}x^{n-m}a^{m}\right)\\&=\sum _{n=0}^{\infty }\left({\frac {\left.\psi ^{(n)}(x)\right\vert _{x=0}}{n!}}\cdot \sum _{m=0}^{n}{\frac {a^{m}}{m!}}{\frac {d^{m}}{dx^{m}}}x^{n}\right)\\&=\sum _{n=0}^{\infty }\left[{\frac {\left.\psi ^{(n)}(x)\right\vert _{x=0}}{n!}}\cdot \left(\sum _{m=0}^{n}{\frac {a^{m}}{m!}}{\frac {d^{m}}{dx^{m}}}x^{n}+\sum _{m=n+1}^{\infty }0\right)\right]\\&=\sum _{n=0}^{\infty }\left[{\frac {\left.\psi ^{(n)}(x)\right\vert _{x=0}}{n!}}\cdot \left(\sum _{m=0}^{n}{\frac {a^{m}}{m!}}{\frac {d^{m}}{dx^{m}}}x^{n}+\sum _{m=n+1}^{\infty }{\frac {d^{m}}{dx^{m}}}x^{n}\right)\right]\\&=\sum _{n=0}^{\infty }\left({\frac {\left.\psi ^{(n)}(x)\right\vert _{x=0}}{n!}}\cdot \sum _{m=0}^{\infty }{\frac {a^{m}}{m!}}{\frac {d^{m}}{dx^{m}}}x^{n}\right)\\&=\sum _{n=0}^{\infty }\sum _{m=0}^{\infty }{\frac {\left.\psi ^{(n)}(x)\right\vert _{x=0}}{n!}}{\frac {a^{m}}{m!}}{\frac {d^{m}}{dx^{m}}}x^{n}\\&=\left[\sum _{m=0}^{\infty }{\frac {a^{m}}{m!}}{\frac {d^{m}}{dx^{m}}}\right]\left[\sum _{n=0}^{\infty }{\frac {\left.\psi ^{(n)}(x)\right\vert _{x=0}}{n!}}x^{n}\right]\\&=\exp \left(a\cdot {\frac {d^{m}}{dx^{m}}}\right)\psi (x)\\&=\exp \left(a\cdot {\frac {i}{\hbar }}\cdot {\frac {\hbar }{i}}{\frac {d^{m}}{dx^{m}}}\right)\psi (x)\\&=e^{i{\frac {a}{\hbar }}{\hat {p}}}\psi (x)\end{aligned}}}
So,
U ~ ( a ) = e i a ℏ p ^ {\displaystyle {\tilde {U}}(a)=e^{i{\frac {a}{\hbar }}{\hat {p}}}}
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 ^ = e i ( − a ) ℏ p ^ = U ~ ( − a ) {\displaystyle {\tilde {U}}(a)^{\dagger }=e^{(i)^{\dagger }{\frac {a}{\hbar }}({\hat {p}})^{\dagger }}=e^{-i{\frac {a}{\hbar }}{\hat {p}}}=e^{i{\frac {(-a)}{\hbar }}{\hat {p}}}={\tilde {U}}(-a)}
⇒ U ~ ( a ) † U ~ ( a ) ψ ( x ) = U ~ ( − a ) U ~ ( a ) ψ ( x ) = U ~ ( − a ) ψ ( x + a ) = ψ ( x + a − a ) = 1 ψ ( x ) {\displaystyle {\begin{aligned}\Rightarrow {\tilde {U}}(a)^{\dagger }{\tilde {U}}(a)\psi (x)&={\tilde {U}}(-a){\tilde {U}}(a)\psi (x)\\&={\tilde {U}}(-a)\psi (x+a)\\&=\psi (x+a-a)\\&=\mathbf {1} \psi (x)\\\end{aligned}}}
⇒ U ~ ( a ) † U ~ ( a ) = 1 ⇒ U ~ ( a ) † U ~ ( a ) U ~ ( a ) − 1 = 1 U ~ ( a ) − 1 ∴ U ~ ( a ) † = U ~ ( a ) − 1 {\displaystyle {\begin{aligned}&\Rightarrow {\tilde {U}}(a)^{\dagger }{\tilde {U}}(a)=\mathbf {1} \\&\Rightarrow {\tilde {U}}(a)^{\dagger }{\tilde {U}}(a){\tilde {U}}(a)^{-1}=\mathbf {1} {\tilde {U}}(a)^{-1}\\&\therefore {\tilde {U}}(a)^{\dagger }={\tilde {U}}(a)^{-1}\\\end{aligned}}}
, which is a function of 
, such that for some wavefunction 
, 
.
![{\displaystyle {\begin{aligned}\psi (x+a)&=\sum _{n=0}^{\infty }\left.\psi ^{(n)}(x)\right\vert _{x=0}\cdot {\frac {(x+a)^{n}}{n!}}\\&=\sum _{n=0}^{\infty }\left({\frac {\left.\psi ^{(n)}(x)\right\vert _{x=0}}{n!}}\cdot \sum _{m=0}^{n}{\binom {n}{m}}x^{n-m}a^{m}\right)\\&=\sum _{n=0}^{\infty }\left({\frac {\left.\psi ^{(n)}(x)\right\vert _{x=0}}{n!}}\cdot \sum _{m=0}^{n}{\frac {n\,(n-1)\cdots (n-(m-1))}{m!}}x^{n-m}a^{m}\right)\\&=\sum _{n=0}^{\infty }\left({\frac {\left.\psi ^{(n)}(x)\right\vert _{x=0}}{n!}}\cdot \sum _{m=0}^{n}{\frac {a^{m}}{m!}}{\frac {d^{m}}{dx^{m}}}x^{n}\right)\\&=\sum _{n=0}^{\infty }\left[{\frac {\left.\psi ^{(n)}(x)\right\vert _{x=0}}{n!}}\cdot \left(\sum _{m=0}^{n}{\frac {a^{m}}{m!}}{\frac {d^{m}}{dx^{m}}}x^{n}+\sum _{m=n+1}^{\infty }0\right)\right]\\&=\sum _{n=0}^{\infty }\left[{\frac {\left.\psi ^{(n)}(x)\right\vert _{x=0}}{n!}}\cdot \left(\sum _{m=0}^{n}{\frac {a^{m}}{m!}}{\frac {d^{m}}{dx^{m}}}x^{n}+\sum _{m=n+1}^{\infty }{\frac {d^{m}}{dx^{m}}}x^{n}\right)\right]\\&=\sum _{n=0}^{\infty }\left({\frac {\left.\psi ^{(n)}(x)\right\vert _{x=0}}{n!}}\cdot \sum _{m=0}^{\infty }{\frac {a^{m}}{m!}}{\frac {d^{m}}{dx^{m}}}x^{n}\right)\\&=\sum _{n=0}^{\infty }\sum _{m=0}^{\infty }{\frac {\left.\psi ^{(n)}(x)\right\vert _{x=0}}{n!}}{\frac {a^{m}}{m!}}{\frac {d^{m}}{dx^{m}}}x^{n}\\&=\left[\sum _{m=0}^{\infty }{\frac {a^{m}}{m!}}{\frac {d^{m}}{dx^{m}}}\right]\left[\sum _{n=0}^{\infty }{\frac {\left.\psi ^{(n)}(x)\right\vert _{x=0}}{n!}}x^{n}\right]\\&=\exp \left(a\cdot {\frac {d^{m}}{dx^{m}}}\right)\psi (x)\\&=\exp \left(a\cdot {\frac {i}{\hbar }}\cdot {\frac {\hbar }{i}}{\frac {d^{m}}{dx^{m}}}\right)\psi (x)\\&=e^{i{\frac {a}{\hbar }}{\hat {p}}}\psi (x)\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/226a4eb9c3449d267fc89271e9376c2104004da4)
is unitary, i.e. 
:
