How to evaluate the strain-induced change in reciprocal space?

[PRB147]

In real space in crystal, strain-induced change can be written as follows:
{\bf r'}=(1+\epsilon)\cdot {\bf r}
But there is no way to evaluate the strain-induced change in reciprocal space.
Can one calculate the strain-induced change in high-symmetry point in quasi-momentum space?
I check almost many books, I still can not find a way.
But Neto and his students had calculated a change in Eq.(11) in PRB Vol.80, 045401 (2009).
Would anyone here give a hint?

Thank you all!

Best wishes!

 

[read]

In real space in crystal, strain-induced change can be written as follows:
{\mathbf r'}=(1+\epsilon)\cdot {\mathbf r}
But there is no way to evaluate the strain-induced change in reciprocal space.
Can one calculate the strain-induced change in high-symmetry point in quasi-momentum space?
I check almost many books, I still can not find a way.
But Neto and his students had calculated a change in Eq.(11) in PRB Vol.80, 045401 (2009).
Would anyone here give a hint?

Thank you all!

Best wishes!

Let us denote the matrix: {\mathbf P}=(1+\epsilon). We consider that r is a column. Under the transformation P the basis vectors ({\mathbf a}', {\mathbf b}')=({\mathbf a}, {\mathbf b}) {\mathbf P}. Note, the rows. The new basis vectors will have appropriate reciprocal vectors
\begin{pmatrix} {\mathbf a}^*'\\ {\mathbf b}^*' \end{pmatrix}={\mathbf Q}\begin{pmatrix} {\mathbf a}^*\\ {\mathbf b}^* \end{pmatrix},

where \mathbf Q={\mathbf P}^{-1}.

Now we are interested how an arbitrary reciprocal vector looks like in the old reciprocal basis, but this is simply changing

a^* -> a^* h
b^* -> b^* k

in the above formula.

About the transformation in crystallography have a look in:
International Tables for Crystallography (2006). Vol. A, Chapter 5.1, pp. 78–85.

 

[PRB147]

Thank you very much, read, I will learn your explanation step by step.
By the way, the meaning of h and k is the Miller indices?

 

[read]

By the way, the meaning of h and k is the Miller indices?

Yes it is.

 

[PRB147]

Using this method, the Neto's result can not be recovered.
I am stuck in this problem for one month.

 

[read]

Using this method, the Neto's result can not be recovered.
I am stuck in this problem for one month.

There are two things here.

(i) The change of the basis from the orthogonal to the hexagonal one. Unfortunately, they do not give the matrix explicitly, but from the Fig.2 one can infers that:

A1=1/2 a1 + sqrt(3)/2 a2
B1=-1/2 a1 + sqrt(3)/2 a2,

where a1,a2 - orth. basis, A1,B1 - hexagonal. Actually this is not correct from the hex-symmetry but this is the way they used... (correct way would be to use \delta_1 and \delta_2 as a1 and a2). This matrix written by columns, as I explained before, I call P.

(i) The strain induced change of the lattice given in orth. a1,a2 system (1+e) where e is symmetric matrix with elements e11, e22 and e12.

Now you make a product (1+e) P, and then make the inverse matrix of the product. This matrix I call Q-matrix. The reciprocal lattice b1 and b2 are given by the Q-matrix elements as b1=(Q11,Q12),... in the orthogonal basis. I have got

b1= 1-e11 -e12/sqrt(3)
b2= 1//sqrt(3)- e22/sqrt(3)-e12

that corresponds to their formulas

 

[read]

A1=1/2 a1 + sqrt(3)/2 a2
B1=-1/2 a1 + sqrt(3)/2 a2,

where a1,a2 - orth. basis, A1,B1 - hexagonal. Actually this is not correct from the hex-

sorry, I have made a couple of misprints in my explanations with the basis vector notations (they does not affect the solution).

hex-notations: instead of B1 should be A2. Letters "b" stand for the reciprocal state basis vectors in this paper...

symmetry but this is the way they used... (correct way would be to use \delta_1 and \delta_2 as a1 and a2). This matrix written by columns, as I explained before, I call P.

again... instead of "a1 and a2" should be "A1 and A2", i.e I meant hex-basis

 

[PRB147]

Thank you very much! Read, Happy New year!
With my Best Wishes!