# The generator of time translation

1. Jan 3, 2013

### aaaa202

It is not hard to show that for QM what takes the wave function from t->t+Δt is the exponential of the hamiltonian. Yet for some mysterious reason my book decides to note the Hamiltonian as the generator of time translation rather than the exponential of it. What is the reason for this?

2. Jan 3, 2013

### Jazzdude

In the theory of lie groups a generator is defined as the element of the tangential lie algebra that generates the group element. The map from the lie algebra to the lie group is the exponential, so the generator of time evolution is the hamiltonian and the one dimensional subgroup it generates is the time evolution exp(i H t).

You find the same situation for the momentum operator and the spatial translation group it generates. And angular momentum operators generate rotations.

3. Jan 3, 2013

### dextercioby

Books on QM tend to speak from the perspective of group theory, even though they don't/may not contain chapters on Lie groups/algebras and their applications to quantum physics.

4. Jan 4, 2013

### jfy4

The lie algebra has to do with what happens close to the identity. We can take what is given: $e^{-i\hat{H}t}$ is the generator of finite time translations, and see how this portrays the Hamiltonian.
$$e^{-i\hat{H}\epsilon / \hbar}\psi(x,t) = \psi(x,t+\epsilon)$$
now for $\epsilon \approx 0$ we can write
$$\psi(x,t)-\frac{i\hat{H}\epsilon}{\hbar}\psi(x,t) +\mathcal{O}(\epsilon^2) = \psi(x,t)+ \left. \frac{\partial \psi(x,t)}{\partial t}\right|_{t=\epsilon}\epsilon +\mathcal{O}(\epsilon^2)$$
thus
$$\hat{H} = i\hbar \frac{\partial}{\partial t}$$