Time translation invariance of Hamiltonian

I am learning about the time translation invariance of the Hamiltonian. I read that

the time translation invariance is already manifest in the fact that our Hamiltonian is chosen an instantaneous function of time—we have assumed that the dynamics only depends on position and velocity at the current time, rather than the full history of the particle’s trajectory.

I do not quite understand this. The Hamiltonian for a one-dimensional point particle can be written as $$H(x,t)=\frac{[p(x,t)]^2}{2m}+V(x,t)$$ where $$p(x,t)$$ and $$V(x,t)$$ are the momentum and potential repectively. If the time translation is $$t \rightarrow t+ t_0$$, the Hamiltonian will become $$H(x,t+t_0)=\frac{[p(x,t+t_0)]^2}{2m}+V(x,t+t_0).$$ Why can one say that $$H(x,t)=H(x,t+t_0)$$?

• The underlying confusion here seems to be about what the Hamiltonian is actually a function of. See ŧhis answer of mine, which among other things explains why the Hamiltonian is not a function of $x$ and $t$. Commented Feb 2, 2019 at 19:27
• There's nothing too deep here. If a function (of time) is invariant under time translation then it is constant. In other words the Hamiltonian does not depends explicitly on time.
– lcv
Commented Feb 2, 2019 at 19:45

Here is the setting. You have a system that is described by some generalized coordinates $$q_i$$ and some generalized momenta $$p_i$$ (in the Lagrangian formalism, you'd use $$\dot{q_i}$$ instead). Hamilton's equations tell you how these $$q_i$$ and $$p_i$$ change with time.

$$\dot{q_i} = \frac{\partial H}{\partial p_i}, \dot{p_i} = -\frac{\partial H}{\partial q_i}$$

The Hamiltonian, $$H = H(q_i, p_i, t)$$, is a function of the $$q_i$$, the $$p_i$$ and possibly time $$t$$. The question about time invariance of the Hamiltonian is simply asking if there is any explicit dependence of $$H$$ on $$t$$. Consider

$$H = H(p,q) = \frac{p^2}{2m} + V(q)$$

This has no explicit dependence on time. You could easily imagine a situation where you vary the potential with time so that $$V(q)$$ becomes $$V(q, t)$$. Now, the time dependence is explicit and the Hamiltonian is not time translational invariant.

If a function (of time) is invariant under time translation then it is constant. In other words the Hamiltonian does not depends explicitly on time.

The (admittedly a bit sibylline) paragraph that you cite wants to stress the fact that, from the get go, the Hamiltonian is assumed to be time local i.e., to be a function of time only at time $$t$$. You can have (master) equations where the generator depends on the whole history of times from the beginning to the present time. For example something of this form

$$\frac{d\rho(t)}{dt} = \int_0^t dt' K(t,t') \rho(t') \, ,$$

[$$\rho$$ could be the vector $$(q,p)$$]. However, we assume that the Hamiltonian is simply of the form $$H(q,p,t)$$. In this case time translation invariance simply means that $$H$$ does not depends explicitly on $$t$$, i.e., effectively $$H=H(q,p)$$.

The potential V(x,t), if its time dependent it will make the Hamiltonian as a whole time dependent. But dependence inside of momentum p doesnt count. For example in an oscillator, $$H=p^2+x^2$$, and p and x are time dependent but because the potential, $$V=x^2$$ is not explicitly time dependent, therefore the total energy is conserved. There is a theorem which says:

$${dH \over dt} = {\partial H \over \partial t}$$

meaning that the total change in energy is all times equal to the change in energy due to the explicit time dependence in the potential.

I understand that you want to explicitly calculate $$dH = H(x,t) - H(x,t+t_0) = 0$$

It would be easier to show the invariance of $$H$$ under spatial translations $$x_0$$, expressing explicitly $$H$$ as $$1/2mv^2 + U(x)$$

He clearly shows how to calculate $$dL$$. By following the example, you can probably do it yourself for $$H$$.