Why do we care about the spectrum of the Hamiltonian?

Aside of the physical interpretation, one always would like to have the eigenvalues and eigenvectors of a given operator. Think e.g. in time evolution: an arbitrary vector state can be written in the eigenbasis of your operator in question, in this case $H$. The time evolution simply amounts to

\begin{equation} \begin{split} |\Psi(t)\rangle&=U(t)|\Psi(t=0)\rangle\\&=U(t)\sum_{n=1}^N c_n |E_n\rangle \\ &= \sum_{n=1}^N c_n \exp(-itH/\hbar)|E_n\rangle\\ &=\sum_{n=1}^N c_n \exp(-itE_n/\hbar)|E_n\rangle. \end{split} \end{equation}

Simple, time evolution amounts to multiply a real phase, which is this case are simply the energies $E_n$, in each of the eigenvectors of the aforementioned decomposition.

I can give you two reasons for studying the eigenvalue problem for the Hamiltonian operator.

  1. In Quantum Mechanics states are unit vectors in a Hilbert space and obserables are hermitian operators in this space. The postulates state that given one observable ${\cal O}$ the possible values it may attain are the elements of its spectrum $\sigma({\cal O})$. Now for simplicity assume the spectrum is discrete and non-degenerate, so that for each $\lambda_i\in \sigma({\cal O})$ there is one unique state $|i\rangle$ such that $${\cal O}|i\rangle=\lambda_i|i\rangle.$$ Then the postulates also tell you that the probability distribution for the measurement of ${\cal O}$ given the state of the system is $|\psi\rangle$ is $$p_i = |\langle i|\psi\rangle|^2$$ So given any observable if you want to find what are its possible values, find what are the probabilities for each value in a given state, and find out the expected value for that obserable, you need to solve its eigenvalue problem. The allowed values are the elements of the spectrum and the probabilities are encoded in projections on the associated eigenstates. The Hamiltonian is one observable you may be interested in and so you could follow this procedure with it to understand the allowed energies of the system and the corresponding probabilities and mean values in some state.

  2. There is, however, a second reason, which is simply to understand the dynamics of the system. Put simply we are often really interested in the dynamics of the system: how it evolves in time according to some interaction. It turns out that this can be answered by the Schrödinger equation: the Hamiltonian is the generator of time translations, so that if $|\psi(t)\rangle$ is the state of the system at time $t$ it obeys $$i\hbar \dfrac{d}{dt}|\psi(t)\rangle = H|\psi(t)\rangle,$$ where we prescribe some initial condition $|\psi_0\rangle$ at some $t_0$. Now one simple strategy to deal with this is to define one operator which we call the time evolution operator $U(t,t_0)$. It is defined so that $$|\psi(t)\rangle = U(t,t_0)|\psi_0\rangle.$$ Now the interesting part is that it obeys a simple equation. In fact employing the Schrödinger equation for $|\psi(t)\rangle$ we find $$i\hbar \dfrac{d}{dt}U(t,t_0)|\psi_0\rangle = HU(t,t_0)|\psi_0\rangle,$$ and since this should work for any initial condition we must have $$i\hbar \dfrac{d}{dt}U(t,t_0)=HU(t,t_0).$$ If $H$ is time-independent then this equation can be solved by the exponential. In fact we have $$U(t,t_0)=e^{-\frac{i}{\hbar}H(t-t_0)}.$$ Now it turns out that functions of hermitian operators are best understood in their bases of eingenstates. In fact, for a general hermitian operator ${\cal O}$ with basis $|i\rangle$ as in the previous item, if $f(x)$ is some real function one defines $f({\cal O})$ to be the hermitian operator given by $$f({\cal O})|i\rangle = f(\lambda_i)|i\rangle.$$ This defines $f({\cal O})$ completely because it specifies how it acts in one particular basis. In that case $U(t,t_0)$ acts very simply in the basis of $H$: $$U(t,t_0)|i\rangle = e^{-\frac{i}{\hbar}E_i(t-t_0)}|i\rangle.$$ But now this fully solves the dynamical problem. Indeed, let $|\psi_0\rangle$ be your initial state. We want to act with $U(t,t_0)$ upon it. So expand it in the basis of $H$. This gives you $$|\psi_0\rangle = \sum_i c_i(0)|i\rangle.$$ Now act with $U(t,t_0)$ upon it to obtain $$U(t,t_0)|\psi_0\rangle = \sum_i c_i(0) e^{-\frac{i}{\hbar}E_i(t-t_0)}|i\rangle.$$ But now notice that this gives you the expansion coefficients $c_i(t)$ of $|\psi(t)\rangle$ in one basis. So effectively you know the state at time $t$.

So in summary I have given two reasons why one may care about the eigenvalue problem for the Hamiltonian: it tells you about the values of one particular observable, the energy of the system, and it also tells you how to define the time evolution operator so that you can evolve your states in time.

This is more of a linear algebra question than a quantum mechanics question.

What this looks like pre-quantum mechanics

So an operator is a generalization of a matrix and a wavefunction is a vector with a continuous, rather than discrete, index. The inner product trades out a summation sign for an integral sign, but other than that you have a simple $\vec u \cdot \vec v = \sum_n u_n^* ~v_n$ being traded out for the slightly more complicated $\langle u|v\rangle = \int_{\mathbb R}\mathrm dx ~u^*(x)~v(x)$ in QM. A lot of QM is just linear algebra in funny hats.

So let’s take the discrete case. Suppose you have, say, a Markov matrix describing a flow of classical probability among three states, $$\begin{bmatrix}p_1(t_n)\\p_2(t_n)\\p_3(t_n)\end{bmatrix} = \begin{bmatrix}0.5&0.2&0.1\\ 0.5&0.3&0.4\\ 0&0.5&0.5\end{bmatrix} \begin{bmatrix}p_1(t_{n-1})\\p_2(t_{n-1})\\p_3(t_{n-1})\end{bmatrix}$$So maybe I start with $[1; 0; 0]$ and then I find myself the next timestep in the probability distribution $[0.5; 0.5; 0]$ and then the next timestep I find myself in the distribution $[0.35; 0.40; 0.25]$, and so on. We could also form a continuous-time version of this system; if that matrix is $\mathbf M$ then we could consider $\mathrm d\mathbf v/\mathrm dt = (\mathbf M - I) \mathbf v$ for example.

Now one thing that might be really helpful is to know that this particular Markov matrix $\mathbf M$ has eigenvector $\mathbf v_1 = [3/13, 5/13, 5/13]$ with eigenvalue $1$ and some other eigenvalues $\lambda_\pm = (3 \pm \sqrt{29})/20,$ so that essentially there is a geometric/exponential decay of whatever probability vector I feed it, down to that stationary distribution $\mathbf v_1.$ So knowing the eigenvalues and eigenvectors really helps to analyze the system! It tells me that if I just wait a few timesteps, the system will more or less relax to its “fixed point” $\mathbf v_1$ and indeed it suggests that even if $\mathbf M$ gets large I can probably analyze $\mathbf M^n$ for some large $n$ and get some simpler physics once all the irrelevant info decays out.

How QM complicates the story

Unfortunately, quantum mechanics does not have stationary distributions at first glance. Quantum mechanics is information-preserving. You will much later model these sorts of physics in your solid-state classes, but only with complicated arguments involving infinite reservoirs of particles, and the “density matrix” or “state matrix” formalism allowing you to create “impure” states which have lost their quantum “coherence.” So it is not quite this simple in quantum mechanics.

But still there is some value. Why did the eigenvalues and eigenvectors help above? It helps because it diagonalizes the evolution operator. And that is huge.

The problem with vectors is that vectors have a magnitude and a direction. That direction is awfully complicated in general! Then I have to think about a lot of different components of my vector along different axes and they’re generally all evolving in coupled equations. But an eigenvalue says, “along this axis $\mathbf v$, the action of this operator is purely a scaling, $\mathbf M~\mathbf v = \lambda \mathbf v$: it does not rotate that axis into any other axis.” And therefore I can solve the problem along that axis directly. If you let me diagonalize, you give me an $n$-dimensional problem and $n$ linearly independent axes along which you tell me what $\mathbf M$ does. That is really helpful, that is almost always sufficient to fully tell me what happens to every state!

“Almost always,” I say? Let me expand on that. Here is the key fact: the Schrödinger evolution is linear. In other words, the evolution of $a \mathbf v_1 + b \mathbf v_2$ is just $a \mathbf v_1(t) + b \mathbf v_2(t)$. It is because we have linear evolution that we can consider any input vector to be a linear-combination of our basis eigenvectors, and then we can calculate the output of those vectors simply by knowing what happens to each eigenvector independently. In the case of a time-independent Hamiltonian in QM, each eigenvector’s time evolution is multiplied by a phase rotating with the speed of the energy, $|n(t)\rangle = e^{-i E_n t/\hbar} |n\rangle.$

Fermions then make this essential

Eventually things will get even more intriguing, because you’ll have fermions. Fermions are a topic in many-particle quantum mechanics and often require much more sophisticated approaches like “second quantization” to describe fully. But the basic idea is that you’ll solve for a Hamiltonian and then weakly-interacting fermions will start occupying those levels independently of each other.

This is actually, you will learn, the quantum basis for the periodic table of the elements. So if you look at how many elements are in each row, you'll find these mysterious numbers $2, 8, 18, 32$. The quantum behavior of electrons, which are fermions, suggests that each expansion is due to new higher-angular-momentum states being unlocked, so you should really look at the differences $2, 6, 10, 14$. Furthermore it says that you should divide by a factor of 2 due to a spin degeneracy: this is secretly really $1, 3, 5, 7.$ Now that sounds much more simple. (If you just divided by 2 you might see that each number is double a square number; it turns out every square number is a sum of consecutive odds.)

Indeed you will prove that for an arbitrary spherically symmetric potential there is a relationship between energy levels and angular momentum, where there is a level number $n=1,2,\dots$ and some total angular momentum $\langle L^2\rangle = \hbar^2 \ell (\ell + 1)$ for some integer $\ell$, and some projection of that angular momentum along some arbitrary axis $\langle L_z\rangle = \hbar m$ for some number $|m| \le \ell.$ So the first expansion corresponds to these new states $n=2, \ell=1$ having spin $+\hbar, 0, -\hbar$ in one particular direction; those states then reappear for $n=3, 4, 5\dots$, the second expansion corresponds to these new states $n=3, \ell=2$ having spin $2\hbar, \hbar, 0, -\hbar, -2\hbar$ along that particular direction, reappearing for all later shells; and so on. That $\ell = 0, 1, \dots n-1$ also emerges for inverse square laws, although you may notice from the periodic table that the angular momentum contributes to the energy, so these “electron shells” do not fill in order of n directly.

Anyway, this relationship is just a result of spherical symmetry and the inverse-square attraction of the nucleus and nothing else, so as long as the electrons with respect to each other kind of “blur out” into a spherically symmetric inverse-square mean field approximation (the shielding effect), it is also approximately true even if you already have multiple electrons and you are adding another electron to the mix. So this is the quantum origin of the periodic table.