Quantum Harmonic Oscillator, what is #E_0#?

[damarkk]

Homework Statement

Quantum Mechanics, Quantum Harmonic Oscillator in 2D

Relevant Equations

$H_0 = \hbar \omega a^{\dagger}a$

Hello to everyone. I'm sorry for the foolish question.

The text is

An harmonic oscillator in two dimension isothropic of masses m and frequency $\omega$ is described by hamiltonian

H0=hbarωax†ax+ℏωay†ay

and there is a perturbation described by $H'=\alpha x y$.

1. Find the energy for the first three eigenstates (fundamental, first and second excitated states) for non-perturbed oscillator and compute their degeneration.

My attempt.
=
There are one fundamental state $|0_x 0_y \rangle$ with energy $E_0=E_{0x}+E_{0y}=\frac{\hbar \omega}{2}+\frac{\hbar \omega}{2}=\hbar \omega$.

The first level has $E_1 = 2\hbar \omega$ and degeneration equal to two because the correspondent states are $|0_x 1_y \rangle$, $|1_x 0_y \rangle$.

The second level has $E_2 = 3\hbar \omega$ energyn and degeneration equal to three. The correspondent states are $|2_x \0_y \rangle$, $|1_x \1_y \rangle$, $|0_x \2_y \rangle$.

My question is: through Schrödinger Equation for eigenstates, $H_0 |n_x n_y \rangle = E_0|n_x n_y \rangle = \hbar \omega (n_x+n_y+1) |n_x n_y \rangle$.
But for fundamental states $n_x = n_y = 0$ in this hamiltonian formula, because is without $1/2$ term on x and y segment. Is then $E_0 = 0$? I find this strange, because fundamental level has of course a physical energy. Of course we have to $E_0 = \hbar \omega$, but where are the addendum $+1$ in this hamiltonian?

This is not error in exercise text.


P.S.
I'm sorry but I don't understand why Latex is not formatted.

 

[berkeman]

Homework Statement: Quantum Mechanics, Quantum Harmonic Oscillator in 2D
Relevant Equations: $H_0 = \hbar \omega a^{\dagger}a$

I'm sorry but I don't understand why Latex is not formatted.

I tried to fix it up for you. Use double-# delimiters for in-line LaTeX here at PF, and double-$ delimiters for stand-alone lines of LaTeX. See the LaTeX Guide link below the Edit window.

 

[damarkk]

I tried to fix it up for you. Use double-# delimiters for in-line LaTeX here at PF, and double-$ delimiters for stand-alone lines of LaTeX. See the LaTeX Guide link below the Edit window.

Thank you, sir. Sorry for this mistaken.

 

[vela]

If you shift the energy by a constant amount, does it matter?

 

[damarkk]

It doesn't matter obviously. But this is not my question.
If you have hamiltonian like $H = \hbar \omega N_x +\hbar \omega N_y$ and if $N_x |n_x n_y \rangle = n_x |n_x n_y \rangle$, $N_y |n_x n_y \rangle = n_y |n_x n_y \rangle$, then $H |n_x n_y \rangle = \hbar \omega (n_x + n_y) |n_x n_y \rangle = 0$ if $|n_x n_y \rangle = |0_x 0_y \rangle$.

And this is my point: if $H |0_x 0_y \rangle = 0$, how I can say that $E_0 = \hbar \omega$?

Note that by definition we must to have $H |n_x n_y \rangle = E_N |n_x n_y \rangle$.

 

[DrClaude]

And this is my point: if $H |0_x 0_y \rangle = 0$, how I can say that $E_0 = \hbar \omega$?

You can't. As the Hamiltonian is set up, there is no zero-point energy (*). This is perfectly legitimate, as the zero of energy is arbitrary.

(*) If you were to draw the corresponding potential $V(x)$, you would find that the minimum at $V(0)$ is not zero but negative, such that the ground state is exactly at $E_0=0$.