Particle confined in 3D box - quantum states

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Homework Help Overview

The discussion revolves around the quantum mechanics of a particle confined in a three-dimensional box, specifically focusing on the interpretation of quantum states and energy levels. Participants are exploring the relationship between quantum numbers and energy levels in both one-dimensional and three-dimensional confinement scenarios.

Discussion Character

  • Conceptual clarification, Assumption checking, Mixed

Approaches and Questions Raised

  • Participants are questioning the values that quantum numbers (nx, ny, nz) can take for a given energy level in a 3D box. There is also discussion about the implications of integer solutions to energy equations and the physical relevance of approximations in quantum calculations.

Discussion Status

The discussion is ongoing, with participants providing insights and raising questions about the physical interpretation of the results. Some have suggested reconsidering the precision of the given energy values and their implications for possible quantum states.

Contextual Notes

There are concerns regarding the accuracy of the provided data and units, as well as the physical feasibility of certain energy levels based on the calculations presented. Participants are also noting the limitations imposed by the precision of the energy values discussed.

greg997
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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
 
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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.
 
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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:

##\ ##
 
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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.
 
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