# Quantum anharmonic (cubic) oscillator.

1. Apr 20, 2005

### Opi_Phys

Hello,

I tried to consider one-dimensional anharmonic (cubic) oscillator in second quantization formalism. And I found that energy difference between the first excited state and the ground state (as well as energy difference between the second excited stated and the the first one) as function of number of used basis functions show no convergence on this number. Should it be? The more basis functions I use the less is energy difference between the first excited and ground states. Should it be?

In more details, I start form Hamiltonian in x-p representation:
H = p^2/2 + 1/2(k + a*x)*x^2, where "a" is anharmonicity.

Then I use transition from x and p operator to creation and annihilation operators and obtain the following Hamiltonian:
H = epsilon1 * ap * am + V + Delta * (am + ap) * (am + ap) * (am + ap),
where ap - creation operator, and am - annihilation operator.
epsilon, V, Delta - some constants.

Then I rewrite Hamiltonian in the eigen states of Harmonic oscillators. It means that I multiply previous Hamiltonian (from left and right side) on unitary operator:
H_new = I * H_old * I,
where I = cket(0)*bra(0) + cket(1)*bra(1) + ... + cket(n)*bra(n).

Then new Hamiltonian I write in matrix form:
sum_{i,j=1}^{n} cket(i) * H(i,j) * bra(j)

And finally I diagonalize matrix H(i,j) to find energy levels of the system.
And I see no convergence of result as a function of "n"...

2. Apr 20, 2005

### vanesch

Staff Emeritus
I could be wrong, never tried this. But isn't the fact that you have classically no energy bound from below a problem ? For x = very large negative number, and p small, H can be as low as anything, no ? So what's your "ground" state ?

Intuitively, I'd think that a smooth blob in psi, very very far to the left on the x-axis, must have an expectation value for H which is VERY NEGATIVE...
Now, <psi | H | psi> is always larger than E0 of the ground state. So your ground state goes through the floor, no ?

cheers,
Patrick.