# Relationship between those two “exponentials”

Let $G$ be a Lie group and $L(G)$ it's Lie algebra. We know that every left-invariant vector field $X$ in $G$ is complete, and so one can consider the integral curve defined for all $t\in \mathbb{R}$ by $t\mapsto \exp(tX)$. If $M$ is a smooth manifold on which $G$ acts on the left, then if $a\in M$ we can consider the curve in $M$

$$\gamma_a(t) = \exp(tX)\cdot a$$

which "moves the point $a$ an amount $t$" using the infinitesimal transformation induced by $X$ on $M$.

On the other hand, in Quantum Mechanics, if $\Psi$ is the wavefunction of a particle in one-dimension, for example, we have some interesting things. First of all, let $\varepsilon\in \mathbb{R}$ and suppose $\Psi$ has Taylor series in both variables, then

$$\Psi(x+\varepsilon,t)=\sum_{n=0}^\infty \dfrac{1}{n!}\dfrac{\partial^n\Psi}{\partial x^n}(x,t)\varepsilon^n = \sum_{n=0}^\infty \dfrac{1}{n!}\left(\varepsilon\dfrac{\partial}{\partial x}\right)^n\Psi(x,t),$$

using then the fact that the momentum operator is $\hat{p}=-i\hbar \dfrac{\partial}{\partial x}$ together with the definition of the exponential of an operator we have then

$$\Psi(x+\varepsilon, t)=\exp\left[\dfrac{i\varepsilon }{\hbar}\hat{p}\right]\Psi(x,t).$$

In the same way we can show that

$$\Psi(x,t+\varepsilon)=\exp\left[-\dfrac{i\varepsilon}{\hbar}\hat{H}\right]\Psi(x,t).$$

Those two equations are often used to say that $\hat{p}$ is the generator of spatial translations and $\hat{H}$ is the generator of time translations, in the same way that the vector field $X$ on the Lie group is the generator of the diffeomorphisms it induces on the manifold.

These two situations seems to be very similar. In both cases we have something generating one transformation, where the actual transformation is performed using some notion of exponential (the first is the exponential map on the Lie group, the second is the exponential of the operator).

So I was wondering, is there a relationship between those things? Or it is just a coincidence that they seem to be so closely related?

• There is definitely a connection, but I'll leave the explanation to someone who knows more than I do. As an aside, great job figuring it out on your own! I remember when I first discovered that $\sum (\varepsilon\ d/dx)^n / n! = \exp(\varepsilon\ d/dx)$ is the reason that momentum is the generator of translations; suddenly everything made sense. – Javier Jun 3 '15 at 22:51
• These are just manifestations of Stone's theorem. – ACuriousMind Jun 4 '15 at 13:23

Yes, there is a relation between both things. Let $$g(t)=\exp(tX), \qquad g(0)=e$$ Be a curve on $G$, such that $\gamma_a(t)=g(t)\cdot a$ Then, by deffinition, for every smooth function $f:G\to\mathbb{R}$ this curve satisfies the differential equation $$\frac{d}{dt}(f\circ g)(t)=X(f\circ g)(t)$$ or directly $$\frac{d}{dt}g(t)=X(g(t))$$
In an analogous way, if $\xi:\mathbb{R}\to\mathcal{H}=L^2(\mathbb{R})$ is a curve on the Hilbert space that maps $s\mapsto \xi(s)=\psi(x-s)$ that is a solution to the differential equation $$i\hbar\frac{d}{d s}\xi(s)=\hat{p}\xi(s),$$ and if $\gamma:\mathbb{R}\to\mathcal{H}=L^2(\mathbb{R})$ is a curve on the Hilbert space that maps $t\mapsto \gamma(t)=\psi(x,t)$ that is a solution to the differential equation $$i\hbar\frac{d}{d t}\gamma(t)=\hat{H}\gamma(t),$$ since $\hat{p}$ and $\hat{H}$ are self-adjoint operators, by the Spectral theorem, these equations have solutions of the form $$\xi(s)=\exp\left(-\frac{i}{\hbar}\hat{p} s\right)\xi(0)=\exp\left(-\frac{i}{\hbar}\hat{p} s\right)\psi(x)=\psi(x-s),\\ \gamma(t)=\exp\left(-\frac{i}{\hbar}\hat{H} t\right)\xi(0)=\exp\left(-\frac{i}{\hbar}\hat{H} t\right)\psi(x,0)=\psi(x,t)$$ Hence the operators are said to be the generators of spatial and time translation, respectively. Note that while your "proof" is instructive, it lacks rigor, because you can not define the exponential of an unbounded operator such as $\hat{p}$ or $\hat{H}$ through a Taylor series for the whole Hilbert space $\mathcal{H}$, but you can with the Spectral Theorem.
• Here is one such condition: If $G$ admits a bi-invariant metric (for example, if $G$ is compact, connected), then the exponential map with respect to this metric will be the same as the Lie group exponential map. – Sean Pohorence Jun 27 '16 at 15:56