• Support PF! Buy your school textbooks, materials and every day products via PF Here!

Eigen state

  • Thread starter aura
  • Start date
29
0
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....
 

OlderDan

Science Advisor
Homework Helper
3,021
1
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?
 
29
0
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...
 

OlderDan

Science Advisor
Homework Helper
3,021
1
aura said:
oops! a big printing mistake...thats eigen vector...nx,ny,nz

now can u solve this at least the explanation...
[tex] 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] [/tex]

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

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

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

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

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

[tex] 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] [/tex]

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.
 
Last edited:
29
0
thanks for trying it out ...

i will try to solve it...


thanks!!
 

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top