Answer to Question #202898 in Quantum Mechanics for Swagata Ghosh

Question #202898

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


1
Expert's answer
2021-06-03T18:26:25-0400

"H\u03c8n=En\u03c8n"

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{\u210f2d^2}{2mdx^2}"

Equations in Schrödinger (SE)

"\\frac{\u2212\u210f2d^2}{2mdx^2}\u03c8n(x){}=E_n\u03c8_n(x)"

"\u03c8_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^2\u03c0^2\u210f^2}{2ma^2}"

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



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