For a given Hamiltonian, what is the purpose of finding the eigenvalues and eigenvectors? What is the physical meaning of each?
3 Answers
If you're dealing with the time-independent Schrödinger equation (TISE), then you get the following eigenvalue problem: $$H\Psi = E\Psi$$
where we define $E$ to be the energy of the state.
Thus, its eigenvalues corresponds to the energy of your state. And, the eigenvector (or eigenstate) correspond to the state with said energy.
If you're interested in the energy of your system, you're interested in solving the eigenvalue problem involving your Hamiltonian.
When you have found the eigenvalues $E_n$ and eigenvectors $|\psi_n\rangle$ of the Hamiltonian, then you can build the most general solution of the time-dependent Schrödinger equation: $$|\psi(t)\rangle = \sum_n c_n|\psi_n\rangle e^{-iE_nt/\hbar}$$
The physical meaning is:
- A state $|\psi_n\rangle e^{-iE_nt/\hbar}$ is a stationary state, i.e. its shape doesn't change with time $t$.
- The eigenvalues $E_n$ are the energies of these stationary states.
If you have an eigenstate of $\hat H$, then there is no fluctuation in the possible outcome: $$ \Delta E=\langle \hat H^2\rangle -\langle \hat H\rangle^2 =0 $$ The energy of the state $\psi_E(x)$ so that $\hat H\psi_E(x)=E\psi_E(x)$ is thus always the same (no fluctuation) and you can use this energy $E$ to “name” the state.
Thus it makes sense to say “this state has energy E”, just like it makes sense to refer to me as user ZeroTheHero: my handle is “the eigenvalue” of the operator “username”.
Of course it’s much preferable to name states using properties that are physically useful and constant in time, which is why one uses eigenvalues and eigenvectors of operators which commute with $\hat H$ since such operators share with $H$ a common set of eigenvectors. Hence, for hydrogen, we can talk about a state having energy $E_n$ AND angular momentum $\ell$ since $\hat H$ and $L^2$ have a common set of eigenvectors.