Question

Consider H cap=(p_x) ^2/2+x²/2+x×(p_x) , find out the eigen energy state of Hamiltonian

Answer 1

$Hψn=Enψn$

Where H is the operator Hamiltonion, the corresponding value is a collection of wave (eigen)functions.

Consider the basic quantum system of a single particle in a 1D box with infinite walls and zero potential within the box, if you're unfamiliar with your own values.

The system operator in Hamilton is:

$H=\frac{ℏ2d^2}{2mdx^2}$

Equations in Schrödinger (SE)

$\frac{−ℏ2d^2}{2mdx^2}ψn(x){}=E_nψ_n(x)$

$ψ_n(x)=C sin \frac{n\pi x}{a}$

The length of the box is n=1,2,3,... and a.

$E_n=\frac{n^2π^2ℏ^2}{2ma^2}$

Where n=1,2,3, − the SE's own functions and the En's own values.