quantum mechanics - Energy of system in eigenstate of Hamiltonian

I know how to find the spectrum of the Hamiltonian to get the allowed energies for a system. If the spectrum is quantized, I can get definite values for each energy level. But when the system is in one of the corresponding eigenstates, does it actually have exactly that value? How does Heisenberg uncertainty come into play here?

Answer

Heisenberg uncertainty is, in its general form as Robertson-Schrödinger uncertainty, given by

$$ \sigma_A \sigma_B \geq \frac{1}{2} \langle [A,B] \rangle $$

for any two observables $A,B$. Now, in an eigenstate $\lvert a \rangle$ of $A$, $\sigma_A$ vanishes, but also $\langle [A,B]\rangle = \langle a \rvert(AB - BA) \lvert a \rangle = \langle a \rvert( a B - B a )\lvert a \rangle = a \langle a \rvert( B - B )\lvert a \rangle = 0$, so the uncertainty relation is always obeyed.

Since you comment about Griffith's statement that a free particle never has definite energy, look at this:

A free particle has the Hamiltonian

$$ H = \frac{p^2}{m}$$

which has eigenfunctions $f_p(x) = \mathrm{e}^{\mathrm{i}px}$ if we go to the position representation where $p = \partial_x$.

Now, $f_p(x)$ is not square-integrable, and hence not a physical state, since physical states are described by wavefunctions in $L^2(\mathbb{R})$. Therefore, the eigenfunctions of the Hamiltonian are not true eigenvectors inside the Hilbert space, but only lie in a larger space that contains objects that are not normalizable. This observation leads one naturally to consider the idea of a Gel'fand triple.

Therefore, no physical state is an eigenstate of this Hamiltonian, and free particles never have definite energy. This has nothing to do with the uncertainty principle, and everything to do with the form of the Hamiltonian.