- #1
damarkk
- 7
- 2
- 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
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.
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.
Last edited: