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In many undergraduate texts on quantum mechanics (I'm using Modern Quantum Mechanics 2nd Edition by Sakurai as reference here), the treatment of angular momentum goes something along the lines of:

We start with some rotation matrcices $R_x, R_y, R_z$ and then apply some power series expansion to the sines and cosines in these rotation matricies and discard terms of order $O(\epsilon^3)$ to get the expressions for infinitesimal rotations.

Then we use the "infinitesimal forms" of the corresponding unitary operator:

$$ U_\epsilon = 1 - i G \epsilon $$ where $G$ is our corresponding operator on the hilbert space. We then impose the corresponding commutation relations we got from rotation matricies $R_x, R_y, R_z$ previously.

Afterwards, the claim is that for finite rotations, the unitary operator becomes an exponential of the form $U = e^{-iG\phi}$

Here's what I don't understand:

1) Why are we using "infinitesimal" rotations as part of this derivation and where does the formula for the unitary of an infinitesimal operator come from exactly?

2) Why does the corresponding unitary for a finite rotation take the form of an exponential? ($ U = e^{-iG\phi}$). For (2), I know that the operators for observables form a Lie group (Galilean group) characterized by their commutation relations, and that the unitaries are elements of the Lie algebra which is the tangent space of the Lie group.

In my mind this means that the these unitaries have something to do with how our observables change. I also vaguely recall that the exponential map relates the Lie group and Lie algebra in some way but I'm not sure how exactly this ties everything together.

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    $\begingroup$ Interestingly I would classify Modern QM by sakurai as a graduate level textbook $\endgroup$ – Triatticus Nov 24 '18 at 3:12
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The use of infinitesimal forms is a convenience, not a necessity. Stone's theorem explains why we can get away with it. The last part of this answer mentions two reasons why we might want to get away with it.

First, here's one way to think about the relationship between a generator $G$ and the corresponding one-parameter unitary group $U(x)=\exp(iGx)$. If we're given a one-parameter group of (mutually commuting) unitary operators $U(x)$, we can define $G$ by $$ \frac{d}{dx}U(x)=iGU(x). \tag{1} $$ Conversely, given $G$, we can define $U(x)$ by equation (1) together with the condition $$ U(0)=1, \tag{2} $$ where $1$ denotes the identity operator. Here's a simple example: Suppose that we are given $$ G = -i\left(\begin{matrix} 0&-1\\ 1&0\end{matrix}\right), \tag{3} $$ and we want to determine $U(x)=\exp(iGx)$. Equations (1)-(2) are satisfied by $$ G = \left(\begin{matrix} \cos x &-\sin x\\ \sin x&\cos x\end{matrix}\right). \tag{3} $$ Since (1) is a first-order differential equation, its solution is uniquely specified by the initial condition (2), so (3) is the only solution to (1)-(2).

We write the solution as an exponential because equations (1)-(2) imply $$ U(x)U(y)=U(x+y), \tag{4} $$ which is the characteristic property of exponentials. For infinitesimal $x$, equations (1)-(2) imply $U(x)=1+iGx+O(x^2)$. This again is consistent with referring to $U(x)$ as the "exponential" of $iGx$.

Here are two reasons why we might want to work with a Lie algebra (aka the generators of the Lie group) instead of working with the Lie group itself:

  • Although unitary representations of the rotation group SO(3) are sufficient for describing how observables transform, observables are often constructed using other quantities that transform instead according to a unitary representation of the covering group of SO(3), which is SU(2). Here's a related post, which shows how this idea extends from the rotation group to the Lorentz group: https://physics.stackexchange.com/a/437221/206691. One of the nice things about working with the generators of rotations instead of the rotations themselves is that the generators of SO(3) satisfy the same algebra (commutation relations) as the generators of the covering group SU(2). By working with the generators, we can do some parts of some calculations without specifying which of these two groups is involved. More generally, by working with representations of a Lie algebra, we are automatically including all representations of the (simply-connected) covering group. This is expressed more carefully in section 8.1 of Fulton and Harris (1991), Representation Theory: A First Course.

  • Even when dealing with linear translations (instead of rotations), working with the infinitesimal forms is often convenient. For translations in space, the infinitesimal forms are the (total) momentum operators; and for translations in time, the infinitesimal form is the Hamiltonian (total energy) operator. Writing down an explicit expression for the Hamiltonian is easier than writing down an explicit expression for the unitary operator that implements a finite translation in time. In other words, writing down the Schrödinger equation is easier than writing down the most general solution of the Schrödinger equation... and that's an understatement.

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    $\begingroup$ I would add a third item to your list of reasons to work with the Lie algebra instead of the Lie group: It lets you generate conserved quantities from unitary symmetries. A (continuous) unitary symmetry of the Hamiltonian implies that there is some Hermitian operator that also commutes with the Hamiltonian, namely the derivative of the unitary symmetry. This immediately gives a conserved observable. Conversely, given a conserved observable, you can exponentiate it to generate a unitary symmetry. $\endgroup$ – Jahan Claes Nov 24 '18 at 18:22

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