Piezoeletric tensor

8/6/2019 Piezoeletric tensor

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The strain

Definitions

In CRYSTAL, the effect of strain is defined as follows:

y The cartesian axes remain unchangedy The cartesian components of a vectorr transform as follows:

In particular, ifAis the matrix containing the cartesian components of the three elementarytranslation vectors, a1, a2 and a3

The strained matrix, A' is:

A' = (1+I%

The symmetry of the elastic tensor and Voigt notation

The tensor [IAcan be decomposed into two parts:

y an antisymmetric part, [[A, that corresponds to a pure rotation

y a symmetric part, [IA, that corresponds to a pure strain (without rotational components)

Then, if we are interested only in the strain part of the transformation, we can refer to thefollowing pure strain tensor:


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As only six components of the symmetric matrix are independent, a more compact one-indexnotation can be adopted.

Let us define, according to Voigt's convention:

In this convention the strain tensor becomes:

and can be expressed as a vector:

The ZnO case

To illustrate the effect of strain, we will use the ZnO wurtzite structure (hexagonal 6mm class).The ZnO cell parameters are:

a=3.2859 c=5.2410


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The three direct lattice vectors are labelled a1, a2, a3; EFK are, as usual, the angles between a2and a3, a1 and a3, a1 and a2, respectively. The convention adopted in CRYSTAL for the a1 and a2orientation is given in figure 1. The cartesian components ofa1 and a2 are reported. The Amatrix, is given on the right.

Figure 1: Hexagonal cell of wurtzite in the (xy) plane, according to CRYSTAL convention.

As a first example, let's apply the following strain to ZnO:

The A matrix is modified by the strain into A'


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= a

= a

E' = 900

F' = 900

K' = 1200

Figure 2: Deformation of the ZnO hexagonal cell, when the strain I isapplied.The undistorted and distorted cells are represented in light and dark

grey, respectively. The volume variation is also represented in darkgrey. The new lattice parameters are reported on the right side of thefigure.

As a second example consider:

From A', it is easily seen that = a, = c; E = F = 900

For and K' we get:

For small H values, we can disregard terms quadratic in H, so that we get for and K' (use aTaylor expansion):


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Note that (E0,III is four times H ( (E is proportional to (2H)2 and

H2, see the elastic section below).

II)

a'1=3.2717(-0.43)

E'= 900

(0.00)

a'2=3.2861(0.00)

F' = 900

(0.00)

a'3=5.2410(0.00)

K'=119.570

(-0.36)

III)

a'1=3.2575

(-0.86)

E'= 900

(0.00)

a'2=3.2861

(0.00)

F' = 900

(0.00)

a'3=5.2410

(0.00)

K'=119.140

(-0.72)

IV)

a'1=3.2861(0.00)

E'= 900

(0.00)

a'2=3.2861(0.00)

F' = 900

(0.00)

a'3=5.2410(0.00)

K'=1200

(0.00)

The elastic tensor

Definitions

The elastic constants aresecondderivatives of the energy density with respect to straincomponents:

where V is the volume of the cell and |c| is a rank 4 tensor with 3 x 3 x 3 x 3 = 81 elements(called constants).As seen before, the strain I can be expressed with a single index (i=1,6), using Voigt's notation.The elastic constants then become:


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with i,j=1,6. It is obvious from the definition that this matrix is symmetric, so the elastic tensor

has a maximum of 21 independent elements.

The elastic tensor and the symmetry

Due to the symmetry of the crystal, some of the constants are zero, or symmetry related. In orderto know which independent components must be considered, a preliminary run (a TESTGEOMrun) must be performed using the TENSORkeyword. CRYSTAL output indicates whichconstants are independent, and the coefficients relating the symmetry equivalent constants.

How to proceed

A Taylor expansion of the unit cell energy to second order in the strain components yields:

IfE(0) refers to the equilibrium configuration, the first derivatives are zero. According toequation (2), the elastic constants can then be obtained by evaluating the energy as a function ofstrain.

The energy derivatives can be evaluated numerically (CRYSTAL03) or analytically

(CRYSTAL06), with respect to the cell parameters.

When the unit cell is deformed (using the ELASTIC option), the point group may be reduced toa subgroup of the original point group (see examples below). The new point group isautomatically selected by the code. For each deformation of the unit cell, the atoms must beallowed to relax (In some cases this relaxation is not necessary, as the atoms occupy specialpositions). The optimization is performed by using the OPTGEOM option (see manual).

The ZnO case

To illustrate the calculation of an elastic tensor, we will continue to detail the ZnO wurtziteexample.

First of all, in order to know which are the independent tensor components, we run the crystalexecutable with the following geometry input:

ZnO wurtzite elastic tensor

CRYSTAL

0 0 0


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186

3.2859 5.2410

2

30 0.66666666666667 0.3333333333333 0.

8 0.66666666666667 0.3333333333333 0.38268

TENSOR

4

TESTGEOM

END

The wurtzite compounds have the 6mm hexagonal symmetry. The TENSORdirective producesthe following output:

THIS ROUTINE EVALUATES THE NON ZERO ELEMENTS

OF TENSORS OF PHYSICAL PROPERTIES UP TO ORDER 4

ORDER OF THE TENSOR : 4

| 1 A 1 E 1 B . . . |

| 1 A 1 B . . . || 1 D . . . |

| 1 C . . |

| 1 C . |

| & |

EACH LETTER CORRESPONDS TO AN INDEPENDENT COMPONENT.

INTERDEPENDENT COMPONENTS (WITH SAME LETTER) ARE RELATED BY

THE COEFFICIENT THAT PRECEDES EACH LETTER.

THE "&" SYMBOL MEANS 1/2(C11-C12) (OR 2(S11-S12)) FOR THE ORDER

4 C AND S ELASTIC TENSORS IN THE TRIGONAL OR HEXAGONAL CASES.

The number of independent terms in various tensors, in particular for the elastic and piezoelectrictensor can be found in standard textbooks, for example in Nye[1]. These tables must be usedwith some care, because the structure of the tensor depends on the adopted convention for theorientation of the cartesian axes with respect to the unit cell vectors. The output of the TENSORroutine might show permutation with respect to standard definitions[1])

There are only 5 independent components (5 different letters) to evaluate:

A: c11 = c22

B: c13 = c23

C: c44 = c55

D: c33

E: c12


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The symbol & indicates that c66 can be deduced from two independent components [c66 = ~ (c11+c12)]. This particular case happens for trigonal and hexagonal systems only.

To define the complete elastic tensor, 5 different strains must be applied. The cij constants can be

separated in two groups:

y Diagonal terms. They are directly related to second derivatives of the energy.y Off-diagonal terms. The second derivative is obtained by at least two deformations.

The diagonal terms

c11

The [ A tensor for the calculation of c11 is:

The energy expression is:

where a and b are the coefficients of a polynomial fit ofEversus H (see examples below). Then(see equation (2));

c33

The [ ] tensor for the calculation of c33 is:


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Then:

c44


NB: In Voigt's notationThe energy expression is then:

Then

The off-diagonal terms

c12



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The output ofTENSORshows that c11 = c22 (and c12 = c21 by definition), then

Then(c11 +c12) = b/V

c13

The [ A tensor for the calculation of c13 is:


Then(~c11 +~c33+c13) = b/V

Computational conditionsTo obtainEas a function of a given , the energy must be computed for various H (the strainamplitude). Particular care is required in the selection of the computational parameters and of thepoints where the energy is evaluated, in order to avoid large numerical errors in the fittingprocedure.

As regards computational conditions:


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y All the calculations were performed with the same "reference" geometry (the undistortedone) for the selection of integrals, by using the FIXINDEX option.

y The point group may reduce to a subgroup. Then internal relaxation of the atoms may benecessary. In this case, high accuracy is required in the convergence criterion of theoptimization. The following options were used to modify the default values (seeOPTGEOM option):

OPTGEOM

TOLDEG

0.00006

TOLDEX

0.00012

TOLDEE

8

ENDOPT

(default values, are less severe, namely 0.0003, 0.0012 and 7)

y Very high accuracy is also needed in the SCF step:y TOLDEE

y 10

The default value used in CRYSTAL06 for standard SCF is 5 (see manual).

The fitting

In the example below, referring to c33, 15 runs have been performed (with the strain presentedabove) with -0.035 < H< 0.035 and step 0.005.

The table included in figure 3 gives the 15 energy points that are fitted with polynomials ofvarious degrees. The value of the second derivative, b, of the fitted function at the equilibriumgeometry, is presented in Table 1 . It turns out that, in the explored interval, the first acceptablepolynomial must contain a fourth order component (H. The best value is around 1.3625. Whenthe explored interval is reduced, eliminating one or two points from each side (second and thirdrow of Table 1), the b value remains essentially the same.

H E


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-0.035-0.030-0.025-0.020-0.015

-0.010-0.0050.0000.0050.0100.0150.0200.0250.0300.035

-3705.43936297-3705.43980354-3705.44017696-3705.44048292-3705.44072111

-3705.44089138-3705.44099328-3705.44102736-3705.44099326-3705.44089111-3705.44072094-3705.44048301-3705.44017760-3705.43980510-3705.43936571

Figure 4:Energy, E(in hartree), for the 15 H values, of ZnO. Data are shown to the left;

the fitting curve is an order 3 polynomial; the energy (in mhartee) has been shifted by3705.442 hartree.

Table 1: Value of the b parameter as a function of the number of points N and order of thepolynomial P. The stability domain ofb is boxed.

N\ P

2 3 4 5 6 7

15 1.35779 1.35779 1.36276 1.36276 1.36240 1.3624013 1.35861 1.35859 1.36279 1.36275 1.36214 1.36200

11 1.35910 1.35971 1.36274 1.36243 1.36213 1.36142

This scheme must be repeated for the other four strains previously presented. Some of the strainsdecrease the number of symmetry operators, implying higher computational costs. For eachstrain here considered Table 2 gives the number of symmetry operators, the relationship betweenthe second order term b and the elastic constants and the fitted b value (as in Table 1 ).

Table 2:Number of symmetry operators, N, second order term, b, of the energy fit andrelationship between b and the elastic constants, for the five strains here considered. Theequilibrium volume, V, is 49.0065 3.

I N b


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Experimental data are available in this case for comparison and are reported in Table 3.

Table 3: Comparison of calculated and experimental elastic constants (in GPa). Variation

percentage of the HF method with respect to experiment is also reported.c11 c33 c44 c12 c13

HF 246.2 242.5 56.0 127.7 106.2

% +17.8 +12.3 +26.7 +6.4 +1.7

exp2 209.0 216.0 44.2 120.0 104.4

As expected, the Hartree-Fock hamiltonian tends to overestimate the second derivative of theenergy.

Exploiting the point symmetry

With the previous form of the I matrix the number of point symmetry operators is reduced to 4,whereas the following I matrix maintains the full point symmetry (12 operators):


Then

(c11 +~c33+c12+2 c13) = b/V

The piezoelectric tensor

Definitions

The piezoelectric constant eikl is defined, to first order in the strain[2,3], through:


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(1)

then

(2)

where Pi is the polarisation along the cartesian i axis and the derivative is evaluated at zerostrain.

As seen before, the strain I can be expressed with a single index (j=1,6), using Voigt's notation.The piezoelectric constants are then expressed as:

(3)

with i=1,3 and j=1,6; the piezoelectric tensor contains then a maximum of 3 x 6 =18 independent elements.

The piezoelectric tensor and the symmetry

Due to the symmetry of the crystal, some elements of this rank 3 tensor are null, orinterdependent (as in the elastic tensor case). The TENSORoption gives the independentconstants and the coefficients to connect the inter-related ones (see also [1]).

How to proceed

In order to avoid a multivalued problem[3,2], the polarisation is not computed directly, rather weevaluate the related phase (the Berry phase) in the reciprocal lattice vector basis. Then, theproper piezoelectric constant is deduced as follows:

(4)

where Vand ail are the undistorted unit cell volume and component along the cartesian axis i ofthe ai direct lattice vector, respectively.


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According to equation (4), the piezoelectric constants can be obtained by evaluating the phase asa function of deformations of the unit cell parameters. Note that the A matrix, contains the threetranslation vectors ordered as:

.

The "PARET" matrix, that appears in the crystal output, is the transpose ofA.

The computational scheme can be decomposed in 4 steps:

1. As in the elastic tensor case, when the unit cell is deformed , the point group is reduced toa subgroup of the original point group. Following the deformation of the unit cell,internal relaxation of the atoms may be necessary (depending on the space groupsymmetry; then use OPTGEOM to optimize the structure).

2. The Berry phase is computed for each strain, using the keyword POLARI in a properties

run.3. This phase has no physical meaning. The difference with respect to a reference (theundistorted one is usually chosen) has to be computed (properties with the keywordPIEZOBP).

4. Data are fitted and the final calculations are performed.

The computational conditions are the same as for the elastic constants. Actually, the differentstrains that must be applied in order to compute the complete piezoelectric tensor (a maximum ofsix strains), are also required for the elastic tensor, so that the piezoelectric tensor can be seen asa by-product of the calculation of the elastic tensor. Step 1 is then common with the elastic tensorcalculation. Steps 2., 3. and 4. are fast; the additional cost, with respect to the elastic tensorcalculation, is very small. Extra strains, performed in the elastic case, can be used as "crosschecks"

The ZnO case

As in the elastic case, to know which are the independent components of the [ ] tensor, we usethe following geometry input:ZnO wurtzite piezoelectric tensor

CRYSTAL

0 0 0

186

3.2859 5.2410

230 0.66666666666667 0.3333333333333 0.

8 0.66666666666667 0.3333333333333 0.38268

TENSOR

3

TESTGEOM

END

The keyword TENSORproduces the following output:THIS ROUTINE EVALUATES THE NON ZERO ELEMENTS


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| . . . . 1 A . |

| . . . 1 A . . |

| 1 B 1 B 1 C . . . |



THE COEFFICIENT THAT PRECEDES EACH LETTER.

In this case, the piezoelectric tensor is composed by only 3 independent terms:

A: e15 = e24B: e31 = e32C: e33

To get the full piezoelectric tensor 3 strains are required.

e33

The [ A tensor for the calculation of e33 is:

(5)

By applying equation (4) to this case:

(6)

Then

(7)

where a3l is the component along zof the lattice vectoral. As mentioned before, in our case thethird row of the A matrix (containing the zcomponents of the three translation vectors) is (0, 0,c).

Then


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(8)

It is to be noticed that each strain I produces three phases N, N, N and then provides threepiezoelectric constants e1i, e2i, e3i,. In the present case (Ii|I), two of them (e13 and e23) are zeroby symmetry (see the TENSORoutput); we can however compute them to check whether theyare actually null, in order to have a measure of numerical noise. Applying equation (6), we get:

(9)

and

(10)

When the elements of the A matrix are used, we get:

(11)

and

(12)


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Figure 5: Phase difference N, N, and N with respect to H! as a

function ofH

Table 4 gives the values of the phase difference N, N, and N with respect to H! along thethree reciprocal lattice vectors b1, b2, and b3.

Table 4: The phase, Ni , along the three reciprocal vectors bi, for the 15 H values considered.

H N N N

-0.035 0.000004 0.000004 -0.189461

-0.030 -0.000001 0.000010 -0.158187

-0.025 0.000008 0.000009 -0.128346

-0.020 -0.000004 0.000017 -0.099903

-0.015 0.000015 0.000013 -0.072879

-0.010 0.000012 0.000009 -0.046978

-0.005 -0.000001 0.000010 -0.022921

0.000 0.000000 0.000000 0.000000

0.005 -0.000014 0.000023 0.021157

0.010 -0.000023 0.000001 0.041188

0.015 -0.000016 -0.000002 0.0596170.020 0.000013 -0.000003 0.076505

0.025 0.000020 -0.000006 0.092376

0.030 0.000010 -0.000014 0.106365

0.035 0.000015 -0.000016 0.118843


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At this point we must evaluate the derivative of the 3 curves (presented in figure 5) at H!. Thederivatives of Nand N are null, as clearly seen from the figure. For the derivative ofN ,polynomials of various order were used (as in the evaluation of the b term for the elasticconstants). Results are reported in table 5.

Table 5: as a function of the number of points P and order of the polynomial P.

P 2 3 4 5 6 7

N

15 4.408516 4.416939 4.416938 4.410543 4.410552 4.40852413 4.411208 4.414471 4.414474 4.408827 4.408834 4.411157

11 4.412884 4.410961 4.410970 4.412133 4.412120 4.398272

The best value is:

(13)

The piezoelectric constant is then obtained from equation (8) as follows.

The prefactor is automatically calculated at the end of the PIEZOBP (and PIEZOWF)output, following the phase results:

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

PHASE PROJECTED ALONG THE RECIPROCAL LATTICE VECTORS

B1 B2 B3

PHI -0.000014 0.000023 0.021157


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PREFACTOR: |e|/(2.PI.VA)= 0.05203314E+10(C/m**3)

TTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTT PIEZOBP CPU 0.080

e31

The [ ] tensor for the calculation of e31 is:

(15)

Table 6 gives the values of the phase difference N, N, and N with respect to H!; N isplotted in figure 6.

Table 6: The phase, N, along the three reciprocal directions i , for the 15 Hvalues considered.

H N N N

-0.0175 -0.000014 0.000015 0.077271

-0.0150 -0.000011 0.000011 0.065452

-0.0125 -0.000006 0.000008 0.053811

-0.0100 -0.000010 0.000013 0.042448

-0.0075 0.000000 -0.000004 0.031564

-0.0050 -0.000008 0.000004 0.020437

-0.0025 -0.000001 0.000002 0.009939

0.0000 0.000000 0.000000 0.000000

0.0025 -0.000007 -0.000004 -0.010015

0.0050 -0.000014 0.000002 -0.019502

0.0075 0.000007 -0.000012 -0.028640

0.0100 -0.000009 0.000001 -0.037339

0.0125 0.000003 -0.000008 -0.045669

0.0150 -0.000019 0.000006 -0.05398


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0.0175 -0.000018 -0.000003 -0.061655

Let us see how to compute e31; according to equation (3) we get:

(16)

Since e31 = e32 (see TENSORoutput), we have:

(17)

In the present strain tensorI, I!I!H and I!; then:

(18)

so:

(19)

As mentioned before, in order to avoid a multivalued problem, the polarisation is not computeddirectly, rather we evaluate the related Berry phase. Then, e31 is deduced as follows:

(20)

where a31 is the component along zof the lattice vectorai. As mentioned before, in our case thethird row of the A matrix (containing the zcomponents of the three translation vectors) is (0, 0,c).

Then

(21)

Two other terms, e11 and e21, can be obtained applying the same strain ; but they are null bysymmetry (see the TENSORoutput) and will not be considered.


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Figure 6: Phase difference N withrespect to H! as a function ofH.

The derivative of the figure 6 curve at H! is:

(22)

The derivatives ofN and N are null (figures are not reported but equivalent to the I case).The piezoelectric constant is then obtained from equation (21) as follows.

e24

The [ ] tensor used for the calculation of e24 is:

(24)


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Table 7 gives the values of the phase difference, N, N, and N, with respect to H!; N and Nare plotted on figure 7.

Table 7: The projected phase, N, N, and N (along the reciprocal vectors b1 and b2), plotted forthe 15 H values considered.

H N N N

0.0000 0.000000 0.000000 0.000000

0.0025 -0.000004 -0.012387 -0.000024

-0.0025 -0.000009 0.013890 0.000670

0.0050 0.000010 -0.025240 -0.000075

-0.0050 0.000014 0.026489 0.001333

0.0075 -0.000007 -0.038484 0.000613

-0.0075 0.000006 0.039689 0.000407

0.0100 -0.000007 -0.051556 0.001110

-0.0100 0.000015 0.052273 0.002200

0.0125 -0.000006 -0.064639 0.001785

-0.0125 -0.000004 0.065635 0.0024500.0150 -0.000008 -0.077544 0.002067

-0.0150 -0.000010 0.078852 0.002046

0.0175 -0.000001 -0.090596 0.002833

-0.0175 -0.000005 0.091924 0.002834

Let us see how to compute e24; according to equation (4) we get:

(25)

where a21 is the component along y of the lattice vectoral. In our case the second row of the Amatrix (containing they components of the three translation vectors) is (-~a, a, 0).

Then


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(26)

Figure 7: Phase difference N and N with respect to H! as afunction of H.

The derivative of the figure 7 curves at H! are:

(27)

(28)


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The derivative of N is null (figure is not reported). The piezoelectric constant is then obtainedfrom equation (26) as follows:

The full piezoelectric tensor of ZnO obtained at the HF level is then (in C/m2):

(30)

Table 8: Comparison of calculated andexperimental piezoelectric constants (in C/m2).Variation percentage of the HF method withrespect to experiment is also reported.

e31 e33 c24

HF -0.543 1.203 -0.444

% +12.4 +25.3 -16.7exp5 -0.62 0.96 -0.37

Experimental data are available in this case for comparison and are reported in Table 8.

Exercises

The following exercises are classified in basic ( ) or advanced ( ) levels.

Exercise 1 ( )

Find the new direct lattice vector matrix, A', and deduce the new lattice parameters, , ,

, E', F' and K', when the following strain is applied:


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(31)

to the undistorted ZnO A matrix:

(32)

Solution

Only the zcomponent of the direct lattice vector is modified by the strain. As aconsequence,

=a =a

E' = 900F' = 900 K' = 1200

Exercise 2 ( )

Construct the first two points (the smallest H values on each side of the undistorted referencegeometry) of a ZnO input series containing the following generic strain:


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(34)

Our set of data contains 15 points (including the undistorted one) with -0.035 < Handstep 0.035.

A starting input of ZnO can be found in ZnO.d12

Note that the reference geometry, is the undistorted one (FIXINDEX keyword).

Solution

ZnO wurtzite elastic tensor

CRYSTAL

0 0 0

186

3.2859 5.2410

2

30 0.66666666666667 0.3333333333333 0.

8 0.66666666666667 0.3333333333333 0.38268

ELASTIC

-2

0. 0. 0.

0. 0. 0.0. 0. 0.005

.

.

.

END

Exercise 3 ( ) ( )

Find the new direct lattice vectors matrix, A', and deduce the new lattice parameters (as afunction of H), when the following strain is applied:

(35)


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Simplify the obtained new cell parameters just keeping terms linear in H.

Solution

=a

= 900

To the first order of delta, we have:

=a

= a = c

E =900F K =1200

Exercise 4 ( )

Run the elastic tensor symmetry analysis for BeO wurtzite, adding the TENSORkeyword in theBeO file (BeO.d12).

In the directory exercise4/out are the outputs generated by applying the following strain:

(37)


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with 15 different H values (-0.035 < H). The energy data are in exercise4/out/res.dat.

Use the shell script elastic to find, b, the second order term of the energy fitting.

Write then the relation between b and the elastic constant c33. Finally deduce the value of c33

(in hartree/ and in GPa ).

Note that the undistorted cell volume can be found in each output.

Solution

THIS ROUTINE EVALUATES THE NON ZERO ELEMENTS



| 1 A 1 E 1 B . . . |

| 1 A 1 B . . . || 1 D . . . |

| 1 C . . |

| 1 C . |

| & |



THE COEFFICIENT THAT PRECEEDS EACH LETTER.

THE "&" SYMBOL MEANS C11-C12 (OR 1/2(S11-S12)) FOR THE ORDER

4 C AND S ELASTIC TENSORS IN THE TRIGONAL OR HEXAGONAL CASES.

(38)

Then

(39)

Exercise 5 ( ) ( )

NaNO2 is an orthorhombic compound. Run its elastic tensor symmetry analysis (the basic inputforNaN02 can be found in exercise5/NaNO2.d12. How many strains are necessary to computethe full elastic tensor? Propose a set of strains, and in each case the relation between the secondorder term, b, of the energy fit, and the elastic constants.


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Solution

(40)

(41)

(42)

(43)

(44)

Exercise 6 ( )

In the directory exercise6/out are stored the outputs of the strain:

(45)


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The Berry phase has been calculated for 15 H values ranging from -0.035 to +0.035 (step of0.005) with respect to the undistorted structure (point1, H!). The phase results are stored inthe file exercise6/out/res.dat . Use the shell script piezo to plot the phase, N, along the threereciprocal directions, as a function of H. Write down the relationships between the slop of thethree curves and the three piezoelectric constants e13, e23 and e33; then deduce their values (in

C/m2

).

Solutions

Exercise 7 ( )

In the directory exercise7/out are stored the outputs of the strain:

(46)

The Berry phase has been calculated for 15 H values ranging from-0.0175 to 0.0175 with respectto the undistorted structure (point1, H!). The phase results are stored in the file

exercise7/out/res.dat . Use the shell script piezo to plot the phase, N, along the three reciprocaldirections, as a function ofH. Write the relationship between the slop of the three curves and thethree piezoelectric constants e15, e25 and e35; then deduce their values (in C/m

2).

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Piezoeletric tensor