I'll make a few comments about these rules, and leave further elaboration to a quantum mechanics course. First, recall that Hermitian operators have real eigenvalues. This means that we can only measure real values! This is physically sensible, and in fact, is the motivation for making observables Hermitian in the first place.

Secondly, rule 3 and rule 5 tell totally different stories about how the system evolves. Rule 5, often called the "collapse of the wavefunction", rudely interrupts the Schrödinger evolution and resets the system to an eigenvector of the observable. This seems very strange, and very ugly, and it is.

Let's talk more about the spectrum of values we can measure for an observable $\hat{A}$. From rule 4, these are just the eigenvalues of $\hat{A}$. (By the way, this is where the name of the spectral theorem comes from.) To begin with, suppose $V$ is finite-dimensional. We can use the second version of the spectral theorem to deduce that there exist self-adjoint Hermitian operators $P_i$ and scalars $\lambda_i$ for $i=1, \ldots, k$, such that:

1. $\lambda_i \neq \lambda_j$ if $i\neq j$;
2. $P_i^2 = P_i$;
3. $\sum_{i=1}^k P_i = 1_V$; and
4. $\sum_{i=1}^k \lambda_i P_i = 1$.

In the finite-dimensional case, let's check this reduces to the familiar spectral theorem. Set
\[
E(\lambda) \equiv \sum_{\lambda_i < \lambda} P_i
\]and interpret the integrals in properties 3 and 4 using
\[
\int_{-\infty}^\infty f(\lambda)\, dE(\lambda) \equiv \sum_{i}f(\lambda_i) P_i.
\]I'll leave it as an exercise to check that this has properties 1-4. In the infinite-dimensional case, we would make the sum an integral, so heuristically we have something like
\[
E(\lambda) = \int_{-\infty}^\lambda P(\lambda') \, d\lambda'.
\]I won't tell you how to integrate an infinite-dimensional operator (this requires functional analysis), but hopefully it doesn't seem too far removed from the finite-dimensional case. The moral is that even in infinite dimensions, we can "diagonalise" a Hermitian operator (write it as a sum of projection operators) and interpret the projection operators in quantum mechanical terms.

Quantising classical systems

So far I haven't discussed the link between the classical and the quantum description of a system. It turns out you need such a link in order to figure out the operators $\hat{A}$ corresponding to observables; we can't deduce them from the rules alone. The process of going from a classical to a quantum description of the same system is called quantisation. Again, I'll leave most of the details to a quantum mechanics course. However, since the evolution of the system is governed by the energy operator $\hat{H}$ (also called the Hamiltonian) let's see how we calculate it, in principle.

In the classical setup, there is an energy operator $H$ which takes the state of the system and spits out its energy (a real number). For the single particle system described above, it's just the sum of kinetic energy (associated with motion) and potential energy (to do with moving around in the force field $\mathbf{F}(\mathbf{x})$). If the force is conservative, i.e. satisfies
\[
\mathbf{F}(\mathbf{x}) = - \nabla V(\mathbf{x})
\]for some scalar potential $V(\mathbf{x})$, and the particle has mass $m$, then the energy operator is
\[
H(\mathbf{x}, \mathbf{p}) = \frac{|\mathbf{p}|^2}{2m} + V(\mathbf{x}).
\]The energy operator in the quantum description is related very simply to this: it is just
\[
\hat{H} = \frac{\hat{\mathbf{p}}^2}{2m} + V(\hat{\mathbf{x}}),
\]where $\hat{\mathbf{x}}$ and $\hat{\mathbf{p}}$ are the operators associated to the position and momentum observables. In other words, just replace dynamical variables with their operators! So the usual form of Schrödinger's equation,
\[
i\hbar \frac{d}{dt}\Psi = -\frac{\hbar^2}{2m}\nabla^2\Psi + V\Psi
\]just comes from replacing $\mathbf{p}$ with its quantum counterpart $\hat{\mathbf{p}} = -i\hbar \nabla$. (It's not at all obvious that momentum should become a differential operator, but it turns out to be right. Consult your local quantum mechanics course for details on how to find $\hat{\mathbf{x}}$ and $\hat{\mathbf{p}}$.)

Classical symmetries

Finally, it's time for group theory to make an entrance. Recall that we can describe a classical particle in 3D using position $\mathbf{x}$ and momentum $\mathbf{p}$. We can smush them together into a vector $v \equiv (\mathbf{x}, \mathbf{p}) \in \mathbb{R}^3\times\mathbb{R}^3$. The combined position-momentum space $M \equiv \mathbb{R}^3\times\mathbb{R}^3$ is called phase space. A symmetry of the system will be an invertible transformation of phase space $T: M \to M$ that leaves it "invariant", that is, unchanged with respect to an equivalence relation of our choice. We choose "invariant" to mean "the energy of the system isn't changed by $T$, whatever state it's in".
Formally, for all $v \in M$,
\[
H(Tv) = H(v).
\]The collection of all the symmetries forms a group $\mathcal{G}$, as you can check. [Technical comment: The evolution of the system can be related to the derivatives of the energy function by Hamilton's equations. This guarantees that the physics will be the same under a symmetry transformation, given the way we've defined it.]

Some examples

Nature seems to favour groups which are simple enough for us to figure out, but mathematically non-trivial. A few physically important examples:

• rotations of $\mathbb{R}^n$, also known as the orthogonal group $\mathrm{O}(n)$, which arise in systems with rotational invariance;
• rotations of $\mathbb{C}^n$, also known as the unitary group $\mathrm{U}(n)$, which often come from conservation of probability in quantum mechanics;
• translations, spatial rotations, and Newtonian "boosts" of $\mathbb{R}^4$, also called the Galilean group, which connect (Newtonian) physics in different inertial frames;
• rotations and relativistic "boosts" of $\mathbb{R}^4$, also called the Lorentz group $\mathrm{SO}(1,3)$, which connect inertial frames in relativistic mechanics;
• the Poincaré group, which just adds translations to the Lorentz group.

Note that "translations and rotations" includes pure translations, pure rotations, and combinations of the two; the same remark applies to boosts.

Quantum symmetries and representations

Now, suppose we want to quantise a classical system with symmetry group $G$. Symmetries are not always preserved when we quantise, but if they are, what should they look like? Well, before $G$ acted on the phase space $M$. Now it should act on the Hilbert space of the quantum theory, $G \hookrightarrow V$. Since $V$ is a vector space, and we want symmetries to preserve the vector space structure, they should act as linear transformations.

Thus, each group element $T \in G$ should be assigned a matrix, $\rho(T) \in \mathrm{GL}(V)$, where $\mathrm{GL}(V)$ denotes the invertible linear operators on $V$. To ensure that these matrices define a group action, we require the matrix assignment function $\rho: G \to \mathrm{GL}(V)$ to be a homomorphism:
\[
\rho(T\cdot S) = \rho(T) \cdot\rho(S).
\]Since "matrix assignment homomorphism" is a bit of a mouthful, $\rho$ is instead called a representation of $G$. In fact, we've already seen representations in the guise of matrix groups, e.g., the general linear group $\mathrm{GL}_n(\mathbb{F})$, the orthogonal group $\mathrm{O}(n)$, and the unitary group $\mathrm{U}(n)$. Finally, instead of leaving the energy invariant, the closest condition in quantum mechanics is that $\rho(G)$ not interfere with the measurement of energy. More formally, each $\rho(g)$ should commute with the energy operator $\hat{H}$: $$ [\hat{H}, \rho(g)] = \hat{H} \rho(g) - \rho(g)\hat{H} = 0. $$ Note that a group $G$ may act on $V$ without commting with $\hat{H}$, but it is not then a symmetry of the system.

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