Quantum harmonic oscl. half space potential

AI Thread Summary
When the quantum harmonic oscillator potential is defined over [0..infinity], the energy eigenstates still follow the pattern of hw(n + 1/2), but are confined to a half-space due to the presence of an infinite wall. This wall necessitates a node at the origin, meaning only the solutions with a node survive. Consequently, the energy levels and wavefunctions are modified to reflect these new boundary conditions, leading to shifts in the energy levels. The overall behavior of the oscillator is influenced by this half-space potential, while the fundamental formula for energy eigenstates remains unchanged. The confinement alters the oscillator's characteristics significantly.
YZer
Messages
3
Reaction score
0
I'm trying to figure out what happens to the energy eigenstates when the quantum harmonic oscillator potential is over [0..infinity] rather then
[-infinity..infinity]. Originaly its hw(n +1/2)...

Thanks
 
Physics news on Phys.org
The half space SHO is what you get when you superpose the SHO with an infinite wall. Since a wavefunction cannot penetrate an infinite wall, you have to have a node there. So of all the regular SHO solutions, only the ones with a node at the origin survive. Their corresponding energies are the only allowed energies.
 
for the question! When the quantum harmonic oscillator potential is over [0..infinity], instead of [-infinity..infinity], the energy eigenstates will still follow the same pattern of hw(n + 1/2), where h is Planck's constant and w is the frequency of the oscillator. However, the main difference is that the energy eigenstates will now be confined to a half-space, rather than being able to exist in both positive and negative regions. This is because the potential in the region from 0 to infinity acts as a barrier, not allowing the oscillator to exist in the negative region. This will result in a change in the energy levels and wavefunctions of the oscillator, as the boundary conditions are now different. The energy levels will be shifted and the wavefunctions will be modified to reflect the confined nature of the oscillator. Overall, the behavior of the oscillator will be affected by the presence of the half-space potential, but the general formula for the energy eigenstates will remain the same.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top