Energy eigenvalue of Hamiltonian operator

Question

Hamiltonian is the total energy of the system. Then, is its eigenvalue $E$ also total energy of the system? What is the difference between them? Both of them are energy.

Answer 1

For bound states:

$$\mathcal{H}\psi_n=E_n\psi_n$$

Where $\mathcal{H}$ is the Hamiltonion operator, $\psi_n$ a set of wave (eigen)functions and $E_n$ the corresponding eigenvalues.

If you're not familiar with eigenvalues, consider the simple quantum system of a single particle in a 1D box with infinite potential walls and zero potential inside the box.

The Hamiltonian operator of the system is:

$$\mathcal{H}=-\frac{\hbar^2}{2m}\frac{\operatorname{d^2} }{\operatorname{d}x^2}$$

The Schrödinger equation (SE) becomes:

$$-\frac{\hbar^2}{2m}\frac{\operatorname{d^2} }{\operatorname{d}x^2}\psi_n(x)=E_n\psi_n(x)$$

Solving, as per the link above, we get:

$$\psi_n(x)=C\sin\Big(\frac{n\pi x}{a}\Big)$$ Where $n=1,2,3,...$ and $a$ is the length of the box. The eigenvalues (allowed energy levels) compute (as per the same link) to:

$$E_n=\frac{n^2\pi^2\hbar^2}{2ma^2}$$ Where $n=1,2,3,...$

$\psi_n(x)$ are the eigenfunctions of the SE and $E_n$ are its eigenvalues.

Answer 2

You need to review operators. if $\mathcal{H}$ is an Hamiltonian, and $\phi(t,x)$ is some wave vector, then

$\mathcal{H}\phi=\sum a_i\phi_i$

So, the operator is what you act with (operate) on a vector to change it to another vector, often represented as a sum of base vecotrs as I have written.

In the special case:

$\mathcal{H}\phi=E_i\phi_i$

$\phi$ is such a base vector, and $E$ the eigenvalue. Now we need to consider on the physical interpretation of the operator (this is more complicated to define), and in this case we say E is the energy - a scalar.