Does energy conservation imply time invariance?

Question

This is similar to this question: Is the converse of Noether's first theorem true: Every conservation law has a symmetry?. However, the answer given there is very technical and general. I am only interested in the specific case of energy conservation (mostly because dark energy seems to break energy conservation / time invariance).

Answer 1

It is actually the other way around. Time translation symmetry refers to Energy conservation. We define the Hamiltonian as $$\normalsize {H} = \Large{\Sigma_i}\normalsize{p_i\overset{.}{q_i} - L} $$ This says that the Hamiltonian in other words the energy is conserved when the Lagrangian has no explicit time dependance. i.e. $$\frac{dH}{dt}=\frac{\partial L}{\partial t}$$

This means as long as the laws of motion are time translation invariant, the energy of the system in consideration is conserved.

The converse is true as well. As you can see the equations say that

$$\frac{dH}{dt}=\frac{\partial L}{\partial t}$$

Which means that if energy is conserved it means that the Lagrangian has no explicit time dependance. Now even though it might seem in certain systems that the energy is not conserved, we must remember that the system is not necessarily isolated, so when we see that the energy is not conserved it just means that the flow of energy is from the surroundings

Answer 2

You may possibly want this explained with regards to Noether's theorem or something similar, but the answer is yes if we're dealing with quantum mechanics. In quantum mechanics the time-translation operator is given as

\begin{equation} \hat{T}(t) = \exp{(-i\hat{H}t/\hbar)} \end{equation}

such that $T(t_{0})|\psi(t)\rangle = |\psi(t + t_{0})\rangle$.

In the Heisenberg picture we know

\begin{equation} \frac{\text{d}\hat{A}(t)}{\text{d}t} = \frac{i}{\hbar}[\hat{H}, \hat{A}] \end{equation}

Energy conservation imples $\frac{\text{d}\hat{H}}{\text{d}t} = 0$ which is easily shown by the above equation:

\begin{equation} \frac{\text{d}\hat{H}}{\text{d}t} = [\hat{H}, \hat{H}] = 0 \end{equation}

We can rephrase this in terms of the time-translation operator as

\begin{equation} \begin{split} [e^{i\hat{H}t/\hbar}, \hat{H}] &= \\ [\hat{T}(t), \hat{H}] &= 0 \end{split} \end{equation}

We can start with this final equation and work through the above derivation backwards to show that time-translational symmetry does indeed imply energy conservation.