# Hamiltonian operator

Is there any difference between Hamiltonian operator and E? Or do we describe H as an operation that is performed over (psi) to give us E as a function of (psi)??

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stevendaryl
Staff Emeritus
Is there any difference between Hamiltonian operator and E? Or do we describe H as an operation that is performed over (psi) to give us E as a function of (psi)??
Usually, $H$ is an operator, and $E$ is a real number, its eigenvalue. For example, for a free particle,

$H = -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2}$

When applied to $\psi(x) = e^{ikx}$ you get:

$H \psi = \frac{\hbar^2 k^2}{2m} \psi$

So for this particular $\psi$, $E = \frac{\hbar^2 k^2}{2m}$, which is a real number.

Orodruin
Staff Emeritus