Particle Mass m in Cube Box: Density of Eigenstates

AI Thread Summary
The discussion focuses on determining the density of eigenstates for a particle of mass m in a cubic box, with the energy spectrum defined by E=(k^2)/(2m) = 2(π^2)(nx^2 + ny^2 + nz^2)/(mL^2). Participants note that the n values represent the eigenstates, which can be counted to find combinations leading to the same energy level. They highlight the need to calculate the number of degenerate states for each energy and suggest that the density of states can be derived by dividing these counts by an energy interval. The conversation emphasizes the potential utility of visualizing the problem in terms of a spherical energy surface to simplify calculations for larger n values. Overall, the goal is to establish a method for calculating the density of states in this quantum mechanical scenario.
aura
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Consider particle of mass m in a cubic box of length L which has energy spectrum given by E=(k sqr)/2m =2 (pi sqr) (nx sqr+ ny sqr +nz sqr)/m (L sqr).what will be the density of states (eigen states per unit energy interval)

k is boltzman const..nx,ny,nz are unit vectors in resp. directions...
 
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aura said:
Consider particle of mass m in a cubic box of length L which has energy spectrum given by E=(k sqr)/2m =2 (pi sqr) (nx sqr+ ny sqr +nz sqr)/m (L sqr).what will be the density of states (eigen states per unit energy interval)

k is boltzman const..nx,ny,nz are unit vectors in resp. directions...

Aren't the n values the eigenstates with n_j = 1,2,3...?

This looks to be a counting problem to find the number of combinations of the three n values leading to the same sum of squares. Clearly, for the lowest energy there is only one. After that, what are the possibilities, and what happens to the difference between energy levels as the n values increase?
 
OlderDan said:
Aren't the n values the eigenstates with n_j = 1,2,3...?

This looks to be a counting problem to find the number of combinations of the three n values leading to the same sum of squares. Clearly, for the lowest energy there is only one. After that, what are the possibilities, and what happens to the difference between energy levels as the n values increase?

oops! a big printing mistake...thats eigen vector...nx,ny,nz

now can u solve this at least the explanation...
 
aura said:
oops! a big printing mistake...thats eigen vector...nx,ny,nz

now can u solve this at least the explanation...
E_{n_x,n_y,n_z} = \left[\frac{2\ \pi^2}{mL^2}\right] \left[n_x^2+n_y^2+n_z^2\right]

E_{1,1,1} = \left[\frac{2\ \pi^2}{mL^2}\right] \left[3\right]

E_{2,1,1} = E_{1,2,1} = E_{1,1,2} = \left[\frac{2\ \pi^2}{mL^2}\right] \left[6\right]

E_{2,2,1} = E_{2,1,2} = E_{1,2,2} = \left[\frac{2\ \pi^2}{mL^2}\right] \left[9\right]

E_{3,1,1} = E_{1,3,1} = E_{1,1,3} = \left[\frac{2\ \pi^2}{mL^2}\right] \left[11\right]

E_{2,2,2} = \left[\frac{2\ \pi^2}{mL^2}\right] \left[12\right]

E_{1,2,3} = E_{1,3,2} = E_{2,1,3} = E_{2,3,1} = E_{3,1,2} = E_{3,2,1} = \left[\frac{2\ \pi^2}{mL^2}\right] \left[14\right]

etc, etc.

I believe you are supposed to be figuring out all possible energies and how many degenerate states there are for each energy, and then divide the number of states by some energy interval to find the density. Unless I have missed some, the density is a bit erratic for these low numbered states. For larger n, perhaps you can come up with a general expression for how many states there are between some energy E and and a slightly higher level to come up with a number of states per unit energy interval. The sum of squares is suggestive that thinking in terms of the number of states contained within a spherical energy surface might prove helpful.
 
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thanks for trying it out ...

i will try to solve it...


thanks!
 
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