April 20, 2024, 01:09:01 AM
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Topic: Harmonic Oscillator of mass m Hamiltonian expression Quantum Mechanics  (Read 4385 times)

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Offline Limbaza

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Problem: 
Consider a harmonic oscillator of mass m undergoing harmonic motion in two dimensions x and y.  The potential energy is given by
V(x,y) = (1/2)kxx2 + (1/2)kyy2.
(a)  Write down the expression for the Hamiltonian operator for such a system.
(b)  What is the general expression for the allowable energy levels of the two-dimensional harmonic oscillator?
(c)  What is the energy of the ground state (the lowest energy state)?

Hint:  The Hamiltonian operator can be written as a sum of operators.

Now I'm a bit lost on how to write the expression for the Hamiltonian. 
Is the Hamiltonian simply H = - h2/2m d2/dx2 + V(x,y)  [where V(x,y) is given above]?
Then with that Hamiltonian, solving the Schrodinger eqn is pretty straightforward to get H*psi = E*psi, now I'm a bit lost here as well to solve for the general expression for the allowable energy levels? 
I assume once I get this expression, I can plug in 0 for both x and y (the displacement of both x and y are 0) and get the lowest energy ground state?

Offline tamim83

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You are on the right track.  Remember for the 2D HO, you will need to add the extra dimension to the kinetic energy part of the hamiltonian.  For the energy expression, since you have two different force constants (an anisotropic oscillator), your energy will be a sum of two terms for each degree of freedom, namely: Enxny=hvx(nx+1/2)+hvy(ny+1/2).  Then, you should plug in nx=ny=0 to get the energy of the lowest state. 

So, you don't really need to solve the TISE for the 2D HO, you just need to be able to extend the 1D case by an extra dimension. 

Hope this helps. 

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