1
$\begingroup$

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 $$ \hat{U_{\tau}}(\hat{H})|\psi(t) \rangle=\left[ \lim\limits_{N \to \infty}\left(\hat{I} + \frac{i}{\hbar}\frac{\tau}{N}\hat{H}\right)^{N}\right]|\psi(t) \rangle = e^{\tfrac{i}{\hbar}\tau \hat{H}} | \psi(t) \rangle = \psi(t-\tau) \rangle. \tag{04} $$

$\endgroup$

closed as off-topic by ACuriousMind, user36790, CuriousOne, Gert, honeste_vivere Jun 7 '16 at 13:00

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – ACuriousMind, Community, CuriousOne, Gert, honeste_vivere
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 2
    $\begingroup$ 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} $\endgroup$ – Frobenius Jun 5 '16 at 18:56
  • 2
    $\begingroup$ ..... 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} $\endgroup$ – Frobenius Jun 5 '16 at 18:57