Question

Derive an expression for momentum operator and energy operator

Answer 1

The translation operator is denoted T(ε), where ε represents the length of the translation. It satisfies the following identity:

{\displaystyle T(\varepsilon )|\psi \rangle =\int dxT(\varepsilon )|x\rangle \langle x|\psi \rangle } T(\varepsilon )|\psi \rangle =\int dxT(\varepsilon )|x\rangle \langle x|\psi \rangle

that becomes

{\displaystyle \int dx|x+\varepsilon \rangle \langle x|\psi \rangle =\int dx|x\rangle \langle x-\varepsilon |\psi \rangle =\int dx|x\rangle \psi (x-\varepsilon )} \int dx|x+\varepsilon \rangle \langle x|\psi \rangle =\int dx|x\rangle \langle x-\varepsilon |\psi \rangle =\int dx|x\rangle \psi (x-\varepsilon )

Assuming the function ψ to be analytic (i.e. differentiable in some domain of the complex plane), one may expand in a Taylor series about x:

{\displaystyle \psi (x-\varepsilon )=\psi (x)-\varepsilon {\frac {d\psi }{dx}}} \psi (x-\varepsilon )=\psi (x)-\varepsilon {\frac {d\psi }{dx}}

so for infinitesimal values of ε:

{\displaystyle T(\varepsilon )=1-\varepsilon {d \over dx}=1-{i \over \hbar }\varepsilon \left(-i\hbar {d \over dx}\right)} T(\varepsilon )=1-\varepsilon {d \over dx}=1-{i \over \hbar }\varepsilon \left(-i\hbar {d \over dx}\right)

As it is known from classical mechanics, the momentum is the generator of translation, so the relation between translation and momentum operators is:

{\displaystyle T(\varepsilon )=1-{i \over \hbar }\varepsilon {\hat {p}}} T(\varepsilon )=1-{i \over \hbar }\varepsilon {\hat {p}}

thus

{\displaystyle {\hat {p}}=-i\hbar {d \over dx}.} {\hat {p}}=-i\hbar {d \over dx}.