# Derivation involving finite unitary transformation [closed]

Hi I just want to confirm a short derivation involving a particular finite unitary transformation which is important in QM. My working is as follows:

Given the finite unitary transformation defined by $$\hat{U_{\tau}}(\hat{H}) = \lim\limits_{N \to \infty}\left(\hat{I} + \frac{i}{\hbar}\frac{\tau}{N}\hat{H}\right)^{N} = e^{\tfrac{i}{\hbar}\tau \hat{H}}. \tag{01}$$ Using the Schrodinger equation in the first equation and the Taylor expansion in the second we get: $$\left(\hat{I} + \frac{i}{\hbar}\frac{\tau}{N}\hat{H}\right)|\psi(t) \rangle = |\psi(t) \rangle - \frac{\tau}{N}\frac{\partial |\psi(t) \rangle}{\partial t} \approx \bigg| \psi \left(t-\frac{\tau}{N}\right) \Big\rangle. \tag{02}$$ Hence we have $$\lim\limits_{N \to \infty}\left(\hat{I} + \frac{i}{\hbar}\frac{\tau}{N}\hat{H}\right)^{N}|\psi(t) \rangle = | \psi(t-\tau) \rangle. \tag{03}$$ Therefore it follows that $$\hat{U_{\tau}}(\hat{H})|\psi(t) \rangle=\left[ \lim\limits_{N \to \infty}\left(\hat{I} + \frac{i}{\hbar}\frac{\tau}{N}\hat{H}\right)^{N}\right]|\psi(t) \rangle = e^{\tfrac{i}{\hbar}\tau \hat{H}} | \psi(t) \rangle = \psi(t-\tau) \rangle. \tag{04}$$

## closed as off-topic by ACuriousMind♦, user36790, CuriousOne, Gert, honeste_vivereJun 7 '16 at 13:00

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• It seems to be OK. As the momentum operator is the generator of displacements in space ..... \begin{align} & \mathbf{p} =-i\hbar \boldsymbol{\nabla}\\ & U_{\mathbf{x}}\left(\boldsymbol{\rho}\right)\psi\left(\mathbf{x}\right) = e^{-i \tfrac{\rho \boldsymbol{\cdot} \mathbf{p}}{\hbar}}\psi\left(\mathbf{x}\right)=\psi\left(\mathbf{x}-\boldsymbol{\rho}\right) \tag{01} \end{align} – Frobenius Jun 5 '16 at 18:56
• ..... so the Hamiltonian is the generator of displacements in time \begin{align} & H =i\hbar \dfrac{\partial}{\partial t}\\ & U_{t}\left(\tau \right)\psi\left( t \right) = e^{i \tfrac{\tau\cdot H}{\hbar}}\psi\left (t \right)=\psi \left(t-\tau\right) \tag{02} \end{align} – Frobenius Jun 5 '16 at 18:57