Phy5645:Problem 4.1 Solution

Problem

Prove that there is a unitary operator ${\tilde {U}}(a)$, which is a function of ${\hat {p}}={\frac {\hbar }{i}}{\frac {d}{dx}}$, such that for some wavefunction $\psi (x)$, ${\tilde {U}}(a)\psi (x)=\psi (x+a)$.

Solution

${\begin{aligned}\psi (x+a)&=\sum _{n=0}^{\infty }\left.\psi ^{(n)}(x)\right\vert _{x=0}\cdot {\frac {(x+a)^{n}}{n!}}\\&=\sum _{n=0}^{\infty }\left({\frac {\left.\psi ^{(n)}(x)\right\vert _{x=0}}{n!}}\cdot \sum _{m=0}^{n}{\binom {n}{m}}x^{n-m}a^{m}\right)\\&=\sum _{n=0}^{\infty }\left({\frac {\left.\psi ^{(n)}(x)\right\vert _{x=0}}{n!}}\cdot \sum _{m=0}^{n}{\frac {n\,(n-1)\cdots (n-(m-1))}{m!}}x^{n-m}a^{m}\right)\\&=\sum _{n=0}^{\infty }\left({\frac {\left.\psi ^{(n)}(x)\right\vert _{x=0}}{n!}}\cdot \sum _{m=0}^{n}{\frac {a^{m}}{m!}}{\frac {d^{m}}{dx^{m}}}x^{n}\right)\\&=\sum _{n=0}^{\infty }\left[{\frac {\left.\psi ^{(n)}(x)\right\vert _{x=0}}{n!}}\cdot \left(\sum _{m=0}^{n}{\frac {a^{m}}{m!}}{\frac {d^{m}}{dx^{m}}}x^{n}+\sum _{m=n+1}^{\infty }0\right)\right]\\&=\sum _{n=0}^{\infty }\left[{\frac {\left.\psi ^{(n)}(x)\right\vert _{x=0}}{n!}}\cdot \left(\sum _{m=0}^{n}{\frac {a^{m}}{m!}}{\frac {d^{m}}{dx^{m}}}x^{n}+\sum _{m=n+1}^{\infty }{\frac {d^{m}}{dx^{m}}}x^{n}\right)\right]\\&=\sum _{n=0}^{\infty }\left({\frac {\left.\psi ^{(n)}(x)\right\vert _{x=0}}{n!}}\cdot \sum _{m=0}^{\infty }{\frac {a^{m}}{m!}}{\frac {d^{m}}{dx^{m}}}x^{n}\right)\\&=\sum _{n=0}^{\infty }\sum _{m=0}^{\infty }{\frac {\left.\psi ^{(n)}(x)\right\vert _{x=0}}{n!}}{\frac {a^{m}}{m!}}{\frac {d^{m}}{dx^{m}}}x^{n}\\&=\left[\sum _{m=0}^{\infty }{\frac {a^{m}}{m!}}{\frac {d^{m}}{dx^{m}}}\right]\left[\sum _{n=0}^{\infty }{\frac {\left.\psi ^{(n)}(x)\right\vert _{x=0}}{n!}}x^{n}\right]\\&=\exp \left(a\cdot {\frac {d^{m}}{dx^{m}}}\right)\psi (x)\\&=\exp \left(a\cdot {\frac {i}{\hbar }}\cdot {\frac {\hbar }{i}}{\frac {d^{m}}{dx^{m}}}\right)\psi (x)\\&=e^{i{\frac {a}{\hbar }}{\hat {p}}}\psi (x)\end{aligned}}$

So,

${\tilde {U}}(a)=e^{i{\frac {a}{\hbar }}{\hat {p}}}$

It is now shown that ${\tilde {U}}(a)$ is unitary, i.e. ${\tilde {U}}(a)^{\dagger }={\tilde {U}}(a)^{-1}$:

${\tilde {U}}(a)^{\dagger }=e^{(i)^{\dagger }{\frac {a}{\hbar }}({\hat {p}})^{\dagger }}=e^{-i{\frac {a}{\hbar }}{\hat {p}}}=e^{i{\frac {(-a)}{\hbar }}{\hat {p}}}={\tilde {U}}(-a)$

${\begin{aligned}\Rightarrow {\tilde {U}}(a)^{\dagger }{\tilde {U}}(a)\psi (x)&={\tilde {U}}(-a){\tilde {U}}(a)\psi (x)\\&={\tilde {U}}(-a)\psi (x+a)\\&=\psi (x+a-a)\\&=\mathbf {1} \psi (x)\\\end{aligned}}$

${\begin{aligned}&\Rightarrow {\tilde {U}}(a)^{\dagger }{\tilde {U}}(a)=\mathbf {1} \\&\Rightarrow {\tilde {U}}(a)^{\dagger }{\tilde {U}}(a){\tilde {U}}(a)^{-1}=\mathbf {1} {\tilde {U}}(a)^{-1}\\&\therefore {\tilde {U}}(a)^{\dagger }={\tilde {U}}(a)^{-1}\\\end{aligned}}$