The generator of spatial translations is momentum. In quantum mechanics this makes a lot of sense to me and so we can write the translation operator like this:

$$\hat{T}(\Delta \textbf{r}) = e^{\frac{i}{\hbar}\Delta \textbf{r} \cdot \hat{p}}.$$ Is there an equivalent expression (writing the translation operator as an exponential using the generator) for classical mechanics?

# Quantum Mechanics – Generator of Translations in Classical Mechanics Explained

classical-mechanicshamiltonian-formalismlie-algebramomentumquantum mechanics

## Best Answer

Yes, absolutely. Copied and pasted from the end of this answer:

That's the tl;dr version. In the remainder of my answer, I will try to explain what each of those things means with as little baggage as possible, and then I will explicitly demonstrate that the momentum observable is the generator of spatial translations.

In Hamiltonian mechanics, the space of possible states of the system is the phase space $\Omega$. A point $x\in \Omega$ can be labeled by a generalized position $q$ and a generalized momentum $p$, so $x=(q,p)$. Note that $q$ and $p$ are

not functions- they are simply coordinates which label a point in phase space.An observable $F$ is a smooth function $F:\Omega\times \mathbb R \rightarrow \mathbb R$ which maps a point in phase space (and possibly the time) to a real number. The two simplest examples of observables are the position observable $$\mathcal Q:\Omega \times \mathbb R \rightarrow \mathbb R$$ $$(q,p,t) \mapsto q$$ and the momentum observable $$\mathcal P:\Omega \times \mathbb R \rightarrow \mathbb R$$ $$(q,p,t) \mapsto p$$

Every smooth function $F$ corresponds to a Hamiltonian vector field $X_F$ given by

$$X_F = \frac{\partial F}{\partial p} \frac{\partial}{\partial q} - \frac{\partial F}{\partial q}\frac{\partial}{\partial p}$$

where I am using the differential geometric understanding of a vector as a directional derivative.

An integral curve $\Gamma:\mathbb R \rightarrow \Omega$ of a vector field $X$ can be obtained by solving the differential equation

$$\frac{d}{dt}\Gamma(t) = X\big|_{\Gamma(t)}$$ where the notation on the right means that the vector field $X$ is evaluated at the phase space point $\Gamma(t)$. Though the context is somewhat sophisticated, this is just the same as the differential equations idea of taking a vector field and finding all of the curves whose tangents are parallel to the vectors at every point.

Given a vector field $X$, we can define a corresponding

flow$\varphi_X: \Omega\times \mathbb R \rightarrow \Omega$ as follows:Start with a point $x_0\in \Omega$. Find the integral curve $\Gamma_{X,x_0}$ of $X$ which starts at $x_0$ (that is, $\Gamma_{X,x_0}(0)=x_0$). Then, we define

$$\varphi_X(x_0,t) = \Gamma_{X,x_0}(t)$$

Intuitively speaking, $\varphi_X$ takes points in the phase space and "pushes" them along the integral curves of $X$.

We are finally equipped to answer your question. Given any smooth function $F$, we can define a Hamiltonian vector field $X_F$, and a corresponding flow $\varphi_{X_F}$. Given some parameter value $t$, $\varphi_{X_F}(\Omega, t)$ constitutes a canonical transformation (i.e. a change of coordinates). We would then call $F$ the generator of the transformation $\varphi_{X_F}$.

Note the similarity to quantum mechanics: Given any self-adjoint operator $F$, we can define a family of unitary operators $\exp[iFt]$. Given some parameter value $t$, $\exp[iFt]$ constitutes a unitary transformation (i.e. a change of basis). We would then call $F$ the generator of the transformation $\exp[iFt]$.

You ask specifically about exponentiation - the set of flows generated by Hamiltonian vector fields is called the Hamiltonian Symplectomorphism Group, whose Lie algebra is given by the Hamiltonian vector fields. In the sense of Lie theory, one obtains the flow by exponentiating the corresponding Hamiltonian vector field.

Let's apply this to the momentum observable. $\mathcal P(q,p,t)=p$, so $$X_{\mathcal P} = \frac{\partial}{\partial q}$$ The integral curves can be found via $$\frac{d}{ds} \Gamma(s) = X_{\mathcal P}\big|_{\Gamma(s)}$$ which, in component form where $\Gamma(s) = (\Gamma^1(s),\Gamma^2(s))$, reads $$\frac{d\Gamma^1}{ds} \frac{\partial}{\partial q} = \frac{\partial}{\partial q} \implies \frac{d\Gamma^1}{ds} = 1$$ $$\frac{d\Gamma^2}{ds} \frac{\partial}{\partial p} = 0 \implies \frac{d\Gamma^2}{ds} = 0$$ which implies that, starting from the initial condition that $\Gamma(0)=(q_0,p_0)$, $$\Gamma^1(s) = q_0 + s$$ $$\Gamma^2(s) = p_0$$ From there, the corresponding flow is $$\varphi_{X_\mathcal P}(q,p,s) = (q+s,p)$$

We therefore see that this flow corresponds to spatial translations, and is generated by the momentum observable $\mathcal P$.