Quantum Harmonic Oscillator, what is #E_0#?

damarkk
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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, becuase 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.
 
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damarkk said:
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.
 
berkeman said:
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.
 
If you shift the energy by a constant amount, does it matter?
 
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##.
 
damarkk said:
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##.
 
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