Quantum anharmonic (cubic) oscillator.

[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"...

 

[vanesch]

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.

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.

 

[R0man]

Hello,

Thank you for sharing your research on the quantum anharmonic (cubic) oscillator. It seems like you have done a thorough analysis using the second quantization formalism. To answer your question, the lack of convergence in the energy difference between the first excited state and the ground state, as well as the second excited state and the first excited state, is not unexpected. This is because the anharmonic oscillator is a non-linear system, and thus, its energy levels are not equally spaced like in a harmonic oscillator. Moreover, the use of more basis functions will lead to a more accurate representation of the system, resulting in a smaller energy difference between the states. Therefore, it is expected that the energy difference will decrease as the number of basis functions increases.

However, it is important to note that the convergence of the energy levels is not the only factor to consider in the accuracy of the results. Other factors such as the truncation error and the choice of basis functions can also affect the accuracy. Therefore, it is important to carefully analyze and compare the results obtained from different basis sets and consider the overall accuracy of the results rather than just the convergence of energy levels.

Overall, your research on the quantum anharmonic (cubic) oscillator is interesting and provides valuable insights into the behavior of this system. Thank you for sharing your findings.