SHO in 2D - ground state energy

You can try solving it and see what happens. However, I think it would be easier to solve in cartesian coordinates, since the potential is already expressed in terms of x,y, and z. In summary, the conversation discusses a two-dimensional harmonic oscillator described by a potential with a positive constant. The problem is to find the ground-state energy of the oscillator. The suggested method is to use a change of coordinates to get rid of the cross term in the Hamiltonian. The resulting Hamiltonian can be seen as two harmonic oscillators with different frequencies. The conversation then discusses the eigenstates of a modified Hamiltonian with a non-spherical potential. It is suggested that the ground state will have a skewed symmetry and that there may be mixing between different
  • #1
nhanle
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0

Homework Statement


A two-dimensional harmonic oscillator is described by a potential of the form
V(x,y) = 1/2 m [tex] \omega^{2}[/tex](x[tex]^{2}[/tex]+y[tex]^{2}[/tex] + [tex]\alpha (x-y)^{2}[/tex]
where [tex]\alpha[/tex] is a positive constant.

Homework Equations


Find the ground-state energy of the oscillator

The Attempt at a Solution


I have tried to plug in the energy of SHO for each dimension x,y; yielding E = h_bar [tex]\omega[/tex](nx+1/2) + h_bar [tex]\omega[/tex](ny+1/2)
which method should I use to solve the third term i.e. [tex]\alpha (x-y)^{2}[/tex]?
 
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  • #2
You can only use [itex]E=\hbar\omega(n+1/2)[/itex] if the Hamiltonian is of the form

[tex]\hat{H} = \frac{\hat{p}^2}{2m} + \frac{1}{2} m\omega^2\hat{x}^2[/tex]

In this problem, the xy cross term messes things up, so you can't just look at it as two harmonic oscillators with the same frequency. You want to find a change of coordinates that will get rid of the cross term.
 
  • #3
Thank you, after your suggestion

here is my attempt

x' = x + y
y' = x - y

So the Hamiltonian becomes

[tex]\hat{H}= \frac{\hat{p'^2}}{2m} + m\omega^2( \frac{1}{2} (\hat{x'^2}) + \frac{(1+ 2\alpha)}{2}(\hat{y'^2}))[/tex]

where [tex]\hat{p'}^2 = (\hat{p_{x'}})^2 + (\hat{p_{y'}})^2 [/tex]

Question:
- Is it similar to the 2D SHO with the perturbed term [tex]\hat{H^{1}}} = \alpha\:\hat{y'}^2[/tex]?
 
Last edited:
  • #4
Shouldn't there be a factor of sqrt(2) in your coordinate transformation?

You don't need to treat it as a perturbation. Your Hamiltonian is now that of two harmonic oscillators of different frequencies:

[tex]\hat{H}= \left[\frac{\hat{p}'_x^2}{2m} + \frac{1}{2} m\omega^2 \hat{x}'^2\right] + \left[\frac{\hat{p}'_y^2}{2m} + \frac{1}{2} m\omega'^2 \hat{y}'^2\right][/tex]

where [itex]\omega'^2 = \omega^2(1+2\alpha)[/itex].
 
  • #5
wow, thank you so much.

Just to extend this arguments further. If the question was about the eigenstates of the modified Hamiltonian
[tex]\hat{H} = \frac{\hat{p}^2}{2m} + m\omega^2 ( x^2 + y^2 + \alpha (y-x)^2 + z^2)[/tex]

, can I assume:

1) the ground state |n,l,m> has its symmetry skewed in the direction of (x-y)

2) the n =2 state has mixing between 211 and 21-1 that split the energy

Is it correct assumption? Just want to know so that I won't get lost when trying to do the calculation
 
  • #6
nhanle said:
wow, thank you so much.

Just to extend this arguments further. If the question was about the eigenstates of the modified Hamiltonian
[tex]\hat{H} = \frac{\hat{p}^2}{2m} + m\omega^2 ( x^2 + y^2 + \alpha (y-x)^2 + z^2)[/tex]

, can I assume:

1) the ground state |n,l,m> has its symmetry skewed in the direction of (x-y)
Yes, the spherical symmetry is broken by the (x-y) term.

Did you mean |nx,ny,nz>? Or did you solve the 3D harmonic oscillator, with the non-spherically symmetric potential, in spherical coordinates?
2) the n =2 state has mixing between 211 and 21-1 that split the energy

Is it correct assumption? Just want to know so that I won't get lost when trying to do the calculation
I don't know.
 

What is SHO in 2D?

SHO in 2D refers to the Simple Harmonic Oscillator in two dimensions. It is a mathematical model used to describe the motion of a particle or system under the influence of a restoring force that is directly proportional to its displacement from equilibrium.

What is the ground state energy in SHO in 2D?

The ground state energy in SHO in 2D is the lowest possible energy that a particle or system can have in this state. It is also known as the zero-point energy and is equal to 1/2 of the energy quantum, or Planck's constant, multiplied by the oscillator frequency.

How is the ground state energy calculated in SHO in 2D?

The ground state energy in SHO in 2D can be calculated using the formula E0 = (n + 1/2)ħω, where n is the quantum number and ħ is the reduced Planck's constant. The value of n for the ground state is always 0, making the ground state energy equal to 1/2 of the energy quantum multiplied by the oscillator frequency.

What is the significance of the ground state energy in SHO in 2D?

The ground state energy in SHO in 2D is significant because it represents the lowest possible energy that a particle or system can have in this state. It serves as a starting point for calculating higher energy levels and is essential for understanding the behavior of particles and systems in this model.

How does the ground state energy in SHO in 2D vary with different parameters?

The ground state energy in SHO in 2D can vary depending on the values of parameters such as the oscillator frequency, mass of the particle, and reduced Planck's constant. However, it will always be 1/2 of the energy quantum multiplied by the oscillator frequency, regardless of the specific values of these parameters.

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