Here's a supplement to ACuriousMind's excellent answer, which:
- Goes into more depth about the meaning of "walking in the direction specified by [an] operator."
- Connects this to the more familiar operation of exponentiating a complex number.
- Gives another take on the physical meaning of exponentiation.
Sorry it's so long! If you don't feel like reading the whole thing, you might want to skip to the last section, which doesn't depend too much on the earlier ones.
Let's start out looking at the Lie group $U(1)$, the group of rotations of the complex plane. Geometrically, $U(1)$ looks like a circle. In fact, it's natural to identify $U(1)$ with the unit circle in $\mathbb{C}$, because if $a$ is a complex number with $|a| = 1$, the transformation "multiply by $a$" is a rotation about the origin—the rotation that sends $1$ to $a$. The identity $\mathbf{1} \in U(1)$ corresponds to the point $1$ on the unit circle, because "multiply by $1$" is rotation by zero radians.
As ACuriousMind said, the Lie algebra of a Lie group can be defined as the tangent space at the identity—the space of possible velocities of paths through the identity. If you think about $U(1)$ as the unit circle in $\mathbb{C}$, you can see this tangent space very explicitly: it's the line tangent to the unit circle at $1$, which can be naturally identified with the "imaginary line" $i\mathbb{R}$.
Now, here's a cool fact about Lie groups: if you pick a velocity at the identity element, it naturally extends to a velocity field defined everywhere in the group! Here's how.
If you're walking along a Lie group $G$, and you find yourself getting lonely, you can imagine you have a twin walking a path which is just like yours, but "translated" by an element $g$ of $G$: when you're at $h$, your twin is at $gh$. Whenever you walk through the identity, your twin walks through $g$, so your velocity at the identity corresponds to a twin velocity at $g$. The connection between your motion and your twin's motion gives a correspondence between the tangent spaces $T_{\mathbf{1}}G$ and $T_gG$.
If you're still lonely, you can imagine you have a twin for every element of $G$, all walking with you in lockstep. When you're at the identity, your $g$ twin is at $g$, your $g'$ twin is at $g'$, and so on—you have a twin at every point in $G$. Thus, your velocity at the identity corresponds to a velocity at every point in $G$. Through your twins, your velocity at the identity is extended to a velocity field defined everywhere in the group.
A velocity field that comes from a bunch of people walking in lockstep like this is called a left-invariant vector field. Using this vocabulary, we can restate the cool fact from before by saying that...
In a Lie group, picking a velocity at the identity is the same as picking a left-invariant vector field.
For a concrete example, let's see how a velocity at the identity of $U(1)$ extends to a velocity field defined everywhere on $U(1)$:
As you might expect, the left-invariant vector fields on $U(1)$ are precisely the rotation-symmetric vector fields on the unit circle.
Now, let's say you have an element $v$ of the Lie algebra $\mathfrak{u}(1)$, which we just saw is the same as having a left-invariant vector field on $U(1)$. If you're standing at the identity in $U(1)$, and you're bored, you can entertain yourself by trying to walk so that your velocity always matches the velocity field $v$. It turns out there's always exactly one way to do this, so if you do it for time $t$, you'll always end up at the same place. This group element is called $\exp_t v$. You can think of $\exp_t$ as a function that turns Lie algebra elements into Lie group elements; it's called the Lie group exponential. Writing an element $a$ of $U(1)$ in exponential form means finding a left-invariant vector field $v \in \mathfrak{u}(1)$ that will take you to $a$ in one unit of time, so $a = \exp_1 v$.
If you think about $U(1)$ as the unit circle in $\mathbb{C}$, there's a very concrete way to compute $\exp_t v$. We saw earlier that $\mathfrak{u}(1)$ can be naturally identified with $i\mathbb{R}$, so we can think of $v$ as an imaginary number. You can turn this number into a velocity field defined over the whole complex plane by telling people at $z \in \mathbb{C}$ to move with velocity $vz$:
If you're a fancy-pants scientist, "telling people at $z$ to move with velocity $vz$" sounds too undignified, so instead you write this sentence as the differential equation
$$\frac{dz}{dt} = vz.$$
In this equation, $v$ is a constant imaginary number, and $z$ is a time-dependent complex number. If you know a little calculus, you can verify that $z(t) = e^{vt}$ satisfies the equation, with $z(0) = 1$. Hence, if you start at the identity of $U(1)$ and follow the velocity field $v$ for time $t$, you'll end up at $e^{vt}$. That means
$$\exp_t v = e^{vt}.$$
We've discovered that, if you look at things in the right way, the Lie group exponential for $U(1)$ matches the familiar exponential of complex numbers!
As your textbook says, every unitary operator has an exponential form: in other words, any element $g$ of the Lie group $U(n)$ can be written as $\exp_1 v$ for some left-invariant vector field $v \in \mathfrak{u}(n)$. What's the physical meaning of this?
In quantum mechanics, unitary operators represent symmetries, so one physical consequence of the fact above is that...
In quantum mechanics, any symmetry can be made continuous.
Once you express a symmetry as an exponential $\exp_1 v$, you see it comes embedded in a whole family of symmetries $\exp_t v$. When $t = 1$, you get your original symmetry back, when $t = 0$, you get the identity, and when $t \in (0, 1)$ you get something in between.
Here's an example of this fact in action. One of the most common kinds of symmetry in physics—so common that we often don't even think of it as a symmetry—is time evolution. Let's say you have a physical system that evolves in discrete time steps. The symmetry that moves you one step forward in time will be represented by a unitary operator $U$. If you write $U$ in exponential form as $\exp_1 v$, you can make the time evolution continuous by declaring that $\exp_t v$ moves you forward $t$ units of time, for any real number $t$.
Of course, there's a catch: there are usually many different ways to write a given unitary operator in exponential form. For example, the operator $\frac{1}{2} + i\frac{\sqrt{3}}{2}$ in $U(1)$ can be expressed as $\exp_1 i\frac{\pi}{3}$, but it can also be expressed as $\exp_1 i\left(\frac{\pi}{3} + 2\pi\right)$. So, if you want to turn a discrete symmetry into a continuous one, there's usually no single "right way" to do it.
