Number of k-cells for a 1D lattice in reciprocal space

In summary, the conversation is about finding the number of allowed values of k in a 1D lattice in reciprocal space. The initial solution provided is incorrect, and the speaker is asking for guidance on how to determine the correct values. The second speaker explains that the allowed values of k depend on the size of the lattice, and gives an analogy to the Wigner-Seitz cell in real space. They also mention that k represents a point in momentum space and its values can be determined by the size of the reciprocal lattice.
  • #1
patric44
296
39
Homework Statement
find the numer of k-cells for a 1d lattice in reciprocal space ?
Relevant Equations
attached
hi guys
our solid state professor sent us a work sheet that included this example :
example6.jpg

i solved it not sure its correct tho :
kcell2.jpg

is it that simple , or this is not the right approach for it ?
 
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  • #2
patric44 said:
Homework Statement:: find the numer of k-cells for a 1d lattice in reciprocal space ?
Relevant Equations:: attached

hi guys
our solid state professor sent us a work sheet that included this example :
View attachment 260534
i solved it not sure its correct tho :
View attachment 260537
is it that simple , or this is not the right approach for it ?
It is the right approach, but the result is not correct.
What are the allowed values of k? What is the length of each cell?
 
  • #3
ehild said:
It is the right approach, but the result is not correct.
What are the allowed values of k? What is the length of each cell?
i did something wrong , i guess the length of the brillion zone is
$$ \frac{2π}{a} = \frac{2π}{3*10^-10} = 2.094*E10 A^- $$
$$dK =\frac{2.094E10}{11} =1.9*10^9 A^-$$
the allowed values of k i guess would be multiples of dk ,1dk,2dk ...
 
Last edited:
  • #4
patric44 said:
i did something wrong , i guess the length of the brillion zone is
$$ \frac{2π}{a} = \frac{2π}{3*10^-10} = 2.094*E10 A^- $$
$$dK =\frac{2.094E10}{11} =1.9*10^9 A^-$$
the allowed values of k i guess would be multiples of dk ,1dk,2dk ...
It is better for the length, but not the allowed values. Is k=0 allowed? And negative k values? And what about the magnitude? Is 1/A° really 10^9 A°^-?
 
  • #5
ehild said:
It is better for the length, but not the allowed values. Is k=0 allowed? And negative k values? And what about the magnitude? Is 1/A° really 10^9 A°^-?
what is the way for knowing which k is allowed ? a little hint
2π/3*10^-10 = 2.094*10^10 and since "a" has units of A so this value has units of is A^-1
so k = 2.094*10^10/11 = 1.9*10^9 still has units of A^-1 ?
what am i doing wrong
 
  • #6
patric44 said:
what is the way for knowing which k is allowed ? a little hint
2π/3*10^-10 = 2.094*10^10 and since "a" has units of A so this value has units of is A^-1
so k = 2.094*10^10/11 = 1.9*10^9 still has units of A^-1 ?
what am i doing wrong
You converted Angstroms to meters. 1 A° =10-10 m. What is 2π/(3A°) then? Always write out the units!
 
  • #7
ehild said:
You converted Angstroms to meters. 1 A° =10-10 m. What is 2π/(3A°) then? Always write out the units!
Sorry My bad :), what does he mean by allowed value of k is it just multiples of dk like I just wrote or something else
How would I know which values is allowed say 0 or negative like you mentioned
 
  • #8
k could have values of = 2nπ/a but if i let that be k would exceed the length of the first brillon zone ?
 
  • #9
patric44 said:
k could have values of = 2nπ/a but if i let that be k would exceed the length of the first brillon zone ?
No, the allowed k values depend on the size of the lattice. You remembered that in your first post.
There are as many allowed k values in the first Brillouin Zone, as Brawais cells in the real crystal, that is 11 in this problem. You know the size of the cells in the reciprocal lattice (what did you get at last?) : the Brillouin Zone is the Wigner-Seitz cell in that space. If the lattice points in the real crystal are at x= 0, +-a, +-2a... those of the reciprocal lattice are at n2pi/a. The Wigner Seitz cell in he real lattice is symmetric to x=0, the edges of the first one are at -a/2 and a/2. The First Brillouin Zone in the reciprocal lattice is analogous to that.
k means a point in the momentum space (reciprocal space) as x means a point in the real space. k is the wave vector of a wave traveling in the crystal. Two waves are identical if their phase differ by a reciprocal cell vector. You certainly learned about it!
 
  • #10
ehild said:
No, the allowed k values depend on the size of the lattice. You remembered that in your first post.
There are as many allowed k values in the first Brillouin Zone, as Brawais cells in the real crystal, that is 11 in this problem. You know the size of the cells in the reciprocal lattice (what did you get at last?) : the Brillouin Zone is the Wigner-Seitz cell in that space. If the lattice points in the real crystal are at x= 0, +-a, +-2a... those of the reciprocal lattice are at n2pi/a. The Wigner Seitz cell in he real lattice is symmetric to x=0, the edges of the first one are at -a/2 and a/2. The First Brillouin Zone in the reciprocal lattice is analogous to that.
k means a point in the momentum space (reciprocal space) as x means a point in the real space. k is the wave vector of a wave traveling in the crystal. Two waves are identical if their phase differ by a reciprocal cell vector. You certainly learned about it!
thanks for helping , so there are only 11 allowed values of k with an interval between them = 2π/Na = 0.19A.
 
  • #11
patric44 said:
thanks for helping , so there are only 11 allowed values of k with an interval between them = 2π/Na = 0.19A.
Yes, there are 11 allowed values of k in the first Brillouin zone (what is its size?) The problem asks you to give these k values.
 
  • #12
i would say k = 0.19 , 0.38 ,0.57 ,0.76 ,... up to 2.09 as an integer multiples of dk
 
  • #13
patric44 said:
i would say k = 0.19 , 0.38 ,0.57 ,0.76 ,... up to 2.09 as an integer multiples of dk
The Brillouin zone is a Wigner-Seitz cell, symmetric to k=0. Is k=0 also allowed? And what is the unit of k? It is not A as you wrote.
 
  • #14
ehild said:
The Brillouin zone is a Wigner-Seitz cell, symmetric to k=0. Is k=0 also allowed? And what is the unit of k? It is not A as you wrote.
Screenshot_2020-04-13-14-56-42.jpg

A^-1
 
  • #15
patric44 said:
Nice! :cool:
Do not forget, that k is a wave-vector, and the wave can travel forward and backward, so k can be both + and -
 
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Likes patric44 and jim mcnamara
  • #16
Thank you so much
 
  • #17
patric44 said:
Thank you so much
you are welcome.
 

1. What is a k-cell in reciprocal space?

A k-cell in reciprocal space is a unit cell that represents a specific region of the reciprocal lattice. It is defined by a set of reciprocal lattice vectors and is used to describe the periodicity of a crystal in reciprocal space.

2. How is the number of k-cells determined for a 1D lattice in reciprocal space?

The number of k-cells for a 1D lattice in reciprocal space is determined by the length of the reciprocal lattice vector. It is equal to the length of the reciprocal lattice vector divided by the length of the first reciprocal lattice vector.

3. What factors affect the number of k-cells for a 1D lattice in reciprocal space?

The number of k-cells for a 1D lattice in reciprocal space is affected by the length of the reciprocal lattice vector, the length of the first reciprocal lattice vector, and the dimensionality of the lattice. It can also be affected by the symmetry of the lattice and any potential defects or disorder present.

4. How does the number of k-cells for a 1D lattice in reciprocal space relate to the number of k-points in the Brillouin zone?

The number of k-cells for a 1D lattice in reciprocal space is equal to the number of k-points in the Brillouin zone. This is because each k-cell represents a unique k-point in the Brillouin zone, and the number of k-cells is directly proportional to the number of k-points.

5. Can the number of k-cells for a 1D lattice in reciprocal space be calculated for a non-periodic system?

No, the number of k-cells for a 1D lattice in reciprocal space can only be calculated for a periodic system. Non-periodic systems do not have a well-defined reciprocal lattice, and therefore the concept of k-cells does not apply.

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