Particle confined in 3D box - quantum states

  • Thread starter Thread starter greg997
  • Start date Start date
  • Tags Tags
    Quantum states
AI Thread Summary
The discussion centers on the confusion surrounding quantum states of a particle confined in a 3D box, particularly regarding the values of quantum numbers nx, ny, and nz for a given energy level. It is clarified that while n=3 indicates a specific quantum state in 1D confinement, in 3D, there can be multiple combinations of nx, ny, and nz that correspond to the same energy level. The participants highlight the importance of considering the approximate nature of physical measurements, suggesting that strict adherence to integer values may not be practical. Additionally, concerns are raised about the accuracy of the provided energy values, which are noted to have limited precision. The conclusion drawn is that the interpretation of energy levels may need to account for these approximations and the inherent complexities of quantum mechanics.
greg997
Messages
105
Reaction score
2
Homework Statement
Find quantum numbers for each of three possible quantum states
Relevant Equations
E= (( h^2)/(8mD) ) (nx^2_ny^2, nz^2)
Hi Everyone.
I hope someone can point me in right direction. I am struggling to work this out . If it was 1d confinement the calculated n number would be the energy level. So for example n= 3, means that quantum number is n= 3 and there is 3 possible quantum states. Is that correct?

With 3D box i am getting confused what values nx , ny, nx can have for the E given.
1.jpg
 
Physics news on Phys.org
I asked Wolfram as shown
1714694120082.png

There seems no integers to satisfy the relation. In neighbor, (2,3,27) satisfies
a^2+b^2+c^2=582 and (2,2,24),(6,8,22) satisfies
a^2+b^2+c^2=584 where I excluded (0,10,22) which includes physically prohibited 0.
 
Last edited:
Thank you for that. What would be the conclusion, interpretation of such solution? - No such energy levels exists?
 
I am surprised that, with 1 digit of precision in the given variables, the search is limited to ##\sum n^2 = 583\quad ## :rolleyes:

##\ ##
 
  • Like
Likes Steve4Physics
greg997 said:
Thank you for that. What would be the conclusion, interpretation of such solution? - No such energy levels exists?

In a narrow sense of mathematics, you are right. But in physics almost all the numbers in calculation is approximate. I am afraid that thinking of integer 583 just is not practical.
 
greg997 said:
If it was 1d confinement the calculated n number would be the energy level. So for example n= 3, means that quantum number is n= 3 and there is 3 possible quantum states. Is that correct?
Not quite. Given a particular value of energy, if you found n=3 that would mean the system is in the n=3 state. There is only one such state for a simple '1D particle in a box' and there are an infinite number of other states (each with its own unique energy value).

greg997 said:
With 3D box i am getting confused what values nx , ny, nx can have for the E given.
View attachment 344371
The box is about the size of a (large) atom.
The mass is about ##10^{15}## times bigger than the mass of a large atom.
The value of energy is remakably small.
So I’m wondering if you have the correct data/units?

And, as others have pointed out, an energy of ##5 \times 10^{-37}## J is precise to only 1 significant figure. This would mean the energy is between ##4.5 \times 10^{-37}## J and ##5.5 \times 10^{-37}## J. Similarly for other values.
 
Last edited:
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged 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