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#import "@preview/physica:0.9.5": pdv, super-T-as-transpose
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#show: super-T-as-transpose
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= Appendix
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#include "predict/default.typ"
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#include "predmode.typ"
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The center principle is to assign the Raman tensor (i.e., change of polarizability caused by atomic displacement)
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to each atom in the unit cell.
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This including the following steps:
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- Write out the change of polarizability caused by displacement of Si atom in A and C layer,
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Where unknown non-zero components are denoted by $a_1$, $a_2$, $a_5$, $a_6$.
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For example, when we move the Si atom in A layer slightly towards the x+ direction in $d$ distance,
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the change of polarizability should be $mat(,a_2,a_1;a_2,,;a_1,,)d$.
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This could be done by conclusion above.
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- The Si atom in B layer have similar local environment as the A and C layer, with only a little difference.
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We denote these difference by $epsilon_1$, $epsilon_2$, $epsilon_5$, $epsilon_6$,
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and the absolute value of $epsilon_i$ should be much smaller than $a_i$.
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For example, when we move the Si atom in B layer slightly towards the x+ direction in $d$ distance,
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the change of polarizability should be $mat(,a_2+epsilon_2,a_1+epsilon_1;a_2+epsilon_2,,;a_1+epsilon_1,,)d$.
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- The local environment of C atom in A layer is similar to the Si atom in A layer with charge reversed and
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the system reversed along xy plane.
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We denote these difference by $eta_1$, $eta_2$, $eta_5$, $eta_6$,
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and the absolute value of $epsilon_i$ should be much smaller than $a_i$.
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For example, when we move the C atom in A layer slightly towards the x+ direction in $d$ distance,
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the change of polarizability should be $mat(,a_2+eta_2,-a_1-eta_1;a_2+eta_2,,;-a_1-eta_1,,)d$.
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- Similar to the case in Si atoms, we derive the change of polarizability
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caused by moving C atom in B layer slightly towards the x+ direction in $d$ distance,
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which should be $mat(,-a_2-eta_2-zeta_2,-a_1-eta_1-zeta_1;-a_2-eta_2-zeta_2,,;-a_1-eta_1-zeta_1,,)d$.
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Lets assign Raman tensor onto each atom.
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That is, Raman tensor is derivative of the polarizability with respect to the atomic displacement:
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$
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alpha = pdv(chi, u)
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$
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where $u$ should be the displacement of the atom corresponding to a phonon mode.
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But, even when $u$ is *NOT* the displacement of a phonon
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(for example, lets only slightly move Si atom in A layer, keeping other atoms fixed),
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the (high-frequency) polarizability is still well-defined,
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and the will still cause a change in the polarizability.
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Even more, the group representation theory is still applicable in this condition:
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the only thing that matters is, when applying $g$ to the system,
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the tensor transformed into $g^(-1) alpha g$ or $g alpha g^(-1)$,
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no matter $alpha$ is Raman tensor or something else, or it is related to a phonon or not.
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Thus, we can, in principle, "assign" Raman tensor of a phonon, to each atom.
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This "assign" is unique since both the atom movement and all phonons have 24 dimensions.
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Next, we consider what these single-atom-caused "Raman tensors" looks like.
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For example, what happens if we move the Si atom in A layer slightly along the x+ direction?
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Consider also move the Si atom in C layer slightly, along x+ or x- direction.
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How about the Raman tensor caused by the both two atoms?
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In first case, this is B2 representation in E1 representation. Thus the Raman tensor should be something like:
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$
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mat(,,2a_1;,,;2a_1,,;)
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$
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In the second case, it is A2 in E2. It turns out:
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$
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mat(,2a_2,;2a_2,,;,,;)
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$
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The average of these two tensors should be the s"Raman tensor" cause by move only the Si atom in A layer,
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slightly towards x+ direction.
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$
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mat(,a_2,a_1;a_2,,;a_1,,;)
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$
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The difference should be the "Raman tensor" of the second atom.
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$
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mat(,-a_2,a_1;-a_2,,;a_1,,;)
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$
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// This approach applied relied on the fact that, all Si atom in 4H-SiC is "distinguishable" by the symmetry operations.
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// I mean, what will happen if we have two Si atoms in A layer?
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// Apparently, we could not extract the "Raman tensor" of only one of the two atoms.
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// This is the case for the 6H-SiC.
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// Hence, we will provide a more general approach to estimate the "Raman tensor" of a single atom.
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Consider the Si atom in the B1 layer.
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It lives in an environment quite similar to the A layer.
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Thus, the "Raman tensor" caused by it should be similar to the one caused by the A layer:
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$
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mat(,a_2+epsilon_2,a_1+epsilon_1;a_2+epsilon_2,,;a_1+epsilon_1,,;)
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$
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Similar to the Si atom in B2 layer:
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$
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mat(,-a_2-epsilon_2,a_1+epsilon_1;-a_2-epsilon_2,,;a_1+epsilon_1,,;)
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$
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Same approach applied for Si atom vibrate in y direction.
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When we move both Si atoms in A and C layer in y+ direction,
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it is B1 in E1, thus the "Raman tensor" should be:
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$
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mat(,,;,,2a_3;,2a_3,;)
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$
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And if we move Si in A layer towards y+ but Si in C layer towards y-,
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it is A2 in E2:
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$
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mat(2a_4,,;,-2a_4,;,,;)
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$
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Thus we get the "Raman tensor" of Si atom in A layer sololy move towards y+ direction:
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$
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mat(a_4,,;,-a_4,a_3;,a_3,;)
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$
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and the "Raman tensor" of Si atom in C layer towards y+ direction:
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$
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mat(-a_4,,;,a_4,a_3;,a_3,;)
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$
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Same applied for the Si atom in B layer:
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$
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mat(a_4+epsilon_4,,;,-a_4-epsilon_4,a_3+epsilon_3;,a_3+epsilon_3,;)
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$
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$
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mat(-a_4-epsilon_4,,;,a_4+epsilon_4,a_3+epsilon_3;,a_3+epsilon_3,;)
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$
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Before consider z-direction, it is important to note that, $a_1$ $a_2$ $a_3$ $a_4$ are not independent.
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Consider vibration along x+ direction (lets say the distance is $d$).
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System energy caused by external electric field and vibration is:
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$
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E^T (mat(,,2a_1;,,;2a_1,,) d) E
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$
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Apply C#sub[3] to atom vibration and external field, energy should not change. We got:
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$
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(mat(-1/2,-sqrt(3)/2,;sqrt(3)/2,-1/2,;,,1)E)^T ( mat(,,2a_1;,,;2a_1,,)(-1/2 d) + mat(,,;,,2a_3;,2a_3,)(sqrt(3)/2 d) )
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(mat(-1/2,-sqrt(3)/2,;sqrt(3)/2,-1/2,;,,1)E)
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$
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It is equal to:
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$
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E^T (mat(,,1/2 a_1 + 3/2 a_3;,,sqrt(3)/2 a_1 - sqrt(3)/2 a_3;1/2 a_1 + 3/2 a_3,sqrt(3)/2 a_1 - sqrt(3)/2 a_3,) d) E
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$
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Thus:
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$
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1/2 a_1 + 3/2 a_3 = 2a_1 #linebreak()
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sqrt(3)/2 a_1 - sqrt(3)/2 a_3 = 0
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$
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Thus $a_1 = a_3$.
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Apply the same method, we get $abs(a_2) = abs(a_4)$.
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Since we have not define the sign of $a_4$, we could take $a_2 = a_4$.
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Same for $epsilon$.
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Now consider what if we move the Si atom in A layer along z+ direction.
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If we move the Si atom in C layer along z+ direction, it is A1:
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$
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mat(2a_5,,;,2a_5,;,,2a_6;)
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$
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If we move the Si atom in C layer along z- direction, it is B1:
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$
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0
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$
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Thus we get the "Raman tensor" of Si atom in A or C layer towards z+ direction:
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$
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mat(a_5,,;,a_5,;,,a_6;)
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$
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Lets consider the C atom in A layer.
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It should be somehow similar to the Si atom in A layer, but with a negative sign in some places,
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and then add or subtract some little value.
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Actually, the "transformation" of Si atom in A layer to C atom in A layer applied in the following steps:
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- reverse charge.
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- reverse system along xy plane.
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First we consider the first step.
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Taking the define of electricity tenser:
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$
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P = chi E
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$
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Lets reverse charge of the system, say we now have electricity tensor $chi'$. We get:
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$
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-P = chi'(-E)
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$
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Thus we get $chi' = chi$, the first step does not change the electricity tensor, nor the "Raman tensor".
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Now we consider the second step.
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For electricity tensor, it will become:
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$
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mat(1,,;,1,;,,-1) chi mat(1,,;,1,;,,-1)
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$
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For $u$, when it is along x or y direction, it will not change. When it is along z direction, it will become $-u$.
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So in conclusion, Raman tensor of C atom in A layer could be estimated from the Raman tensor of Si atom in A layer, by:
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- for movement alone x and y direction, xz yz should be applied a negative sign.
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- for movement alone z direction, xx xy yy zz should be applied a negative sign.
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Export "Raman tensor" of C atom in C layer from C atom in A layer, in the same way.
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Now consider the C atom in B1 layer.
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Is it similar to the C atom in A layer, just like that for Si atom?
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No. It turns out to be similar to the C atom in C layer.
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We summarize these stuff into @table-singleatom.
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Until now, we only consider the "Raman tensor" caused by single atom or atoms move in the same amplitudes.
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However, that is not the case in real phonon.
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- In some A1 modes, only Si or C atom moves. If we take the magnitude of eigenvector as 1,
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then amplitude of each atom is $1/(4sqrt(m_#text[Si]))$ or $1/(4sqrt(m_#text[C]))$.
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- In other cases, the amplitude of Si and C are in the ration of $m_#text[C] : m_#text[Si]$.
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thus the amplitude of Si atom is $1/2 sqrt(1/(m_#text[Si]+m_#text[Si]^2/m_#text[C]))$, so do the C atom.
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Furthermore, we list predicted modes and their Raman tensors, in @table-predmode.
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- $a$: Raman tensor of Si atom in A layer, large value.
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- $epsilon$: Difference of Raman tensors of Si atom in A and B1 layer, small value.
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- $eta$: Difference of Raman tensors of C and Si atom in A layer, small value.
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- $zeta$: Difference of Raman tensors of C atoms in A and B layer, small value.
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#page(flipped: true)[#figure({
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table(columns: 4, align: center + horizon, inset: (x: 3pt, y: 5pt),
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[*Move Direction*], [x], [y], [z],
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[Si A], [$mat(,a_2,a_1;a_2,,;a_1,,;)$], [$mat(a_2,,;,-a_2,a_1;,a_1,;)$], [$mat(a_5,,;,a_5,;,,a_6;)$],
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[C A], [$mat(,a_2+eta_2,-a_1-eta_1;a_2+eta_2,,;-a_1-eta_1,,;)$],
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[$mat(a_2+eta_2,,;,-a_2-eta_2,-a_1-eta_1;,-a_1-eta_1,;)$], [$mat(-a_5-eta_5,,;,-a_5-eta_5,;,,-a_6-eta_6;)$],
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[Si B1], [$mat(,a_2+epsilon_2,a_1+epsilon_1;a_2+epsilon_2,,;a_1+epsilon_1,,;)$],
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[$mat(a_2+epsilon_2,,;,-a_2-epsilon_2,a_1+epsilon_1;,a_1+epsilon_1,;)$],
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[$mat(a_5+epsilon_5,,;,a_5+epsilon_5,;,,a_6+epsilon_6;)$],
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[C, B1], [$mat(,-a_2-eta_2-zeta_2,-a_1-eta_1-zeta_1;-a_2-eta_2-zeta_2,,;-a_1-eta_1-zeta_1,,;)$],
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[$mat(-a_2-eta_2-zeta_2,,;,a_2+eta_2+zeta_2,-a_1-eta_1-zeta_1;,-a_1-eta_1-zeta_1,;)$],
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[$mat(-a_5-eta_5-zeta_5,,;,-a_5-eta_5-zeta_5,;,,-a_6-eta_6-zeta_6;)$],
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[Si C], [$mat(,-a_2,a_1;-a_2,,;a_1,,;)$], [$mat(-a_2,,;,a_2,a_1;,a_1,;)$], [$mat(a_5,,;,a_5,;,,a_6;)$],
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[C, C], [$mat(,-a_2-eta_2,-a_1-eta_1;-a_2-eta_2,,;-a_1-eta_1,,;)$],
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[$mat(-a_2-eta_2,,;,a_2+eta_2,-a_1-eta_1;,-a_1-eta_1,;)$], [$mat(-a_5-eta_5,,;,-a_5-eta_5,;,,-a_6-eta_6;)$],
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[Si B2], [$mat(,-a_2-epsilon_2,a_1+epsilon_1;-a_2-epsilon_2,,;a_1+epsilon_1,,;)$],
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[$mat(-a_2-epsilon_2,,;,a_2+epsilon_2,a_1+epsilon_1;,a_1+epsilon_1,;)$],
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[$mat(a_5+epsilon_5,,;,a_5+epsilon_5,;,,a_6+epsilon_6;)$],
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[C, B2], [$mat(,a_2+eta_2+zeta_2,-a_1-eta_1-zeta_1;a_2+eta_2+zeta_2,,;-a_1-eta_1-zeta_1,,;)$],
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[$mat(a_2+eta_2+zeta_2,,;,-a_2-eta_2-zeta_2,-a_1-eta_1-zeta_1;,-a_1-eta_1-zeta_1,;)$],
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[$mat(-a_5-eta_5-zeta_5,,;,-a_5-eta_5-zeta_5,;,,-a_6-eta_6-zeta_6;)$],
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)},
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caption: ["Raman tensor" caused by single atom],
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placement: none,
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)<table-singleatom>]
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273
paper/appendix/predict/default.typ
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273
paper/appendix/predict/default.typ
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@@ -0,0 +1,273 @@
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#import "@preview/physica:0.9.5": pdv, super-T-as-transpose
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#show: super-T-as-transpose
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== Approximation of Raman tensor of 4H-SiC
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近似的核心思路。
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我们的近似方法基于这样一个原则：将拉曼张量（即原子位移引起的极化率变化）分配给单位晶胞中的每个原子。
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由于同种原子的局部环境相似、只有次近邻才有不同，因此我们将同种原子导致的拉曼效应的差视为一个小量。
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将一个模式中所有参与振动的原子的贡献相加，就可以得到该模式的拉曼张量。
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The center principle of our approximation is to assign the Raman tensor
|
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(i.e., change of polarizability caused by atomic displacement)
|
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to each atom in the unit cell.
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Because the local environment of the same type of atom is similar and only different for the next nearest neighbors,
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we consider the difference in Raman effect caused by the same type of atom as a small quantity.
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The Raman tensor of a phonon mode can be obtained
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by summing the contributions of all atoms participating in the vibration.
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推导 A/C 层 Si 原子沿 x 方向振动时的拉曼张量。
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我们首先推导 A/C 层 Si 原子沿 x 方向振动时的拉曼张量。
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根据前文，我们知道，当这两个原子同向和反向振动时，它们分别属于 E1(C6v) B2(C2v) 和 E2(C6v) A2(C2v) 表示，因此它们的拉曼张量分别为：
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We first derive the Raman tensor of Si atoms in A/C layer vibrating along x direction.
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When the two atoms vibrate in the same direction and opposite direction,
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they belong to the representation of E#sub[1] of C#sub[6v] or B#sub[2] of C#sub[2v]
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and E#sub[2] of C#sub[6v] or A#sub[2] of C#sub[2v], respectively.
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Thus, their Raman tensors are in the form of:
|
||||
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$ mat(,,2a_1;,,;2a_1,,;), mat(,2a_2,;2a_2,,;,,;), $
|
||||
|
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其中 $a_1$ 和 $a_2$ 是两个未知的常数。
|
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where $a_1$ and $a_2$ are two constants with unknown values.
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||||
|
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将上述结果相加或相减，得到 A 层和 C 层 Si 原子沿 x 方向振动时的拉曼张量分别为：
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||||
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||||
By adding or subtracting the above results,
|
||||
we get the Raman tensors of Si atoms in A and C layers vibrating along x direction:
|
||||
|
||||
$ mat(,a_2,a_1;a_2,,;a_1,,;), mat(,-a_2,a_1;-a_2,,;-a_1,,;), $
|
||||
|
||||
近似给出 B 层 Si 原子沿 x 方向振动时的拉曼张量。
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||||
|
||||
注意到 B 层 Si 原子与 A 层 Si 原子
|
||||
|
||||
#include "fig-same.typ"
|
||||
|
||||
|
||||
The center principle is to assign the Raman tensor (i.e., change of polarizability caused by atomic displacement)
|
||||
to each atom in the unit cell.
|
||||
This including the following steps:
|
||||
- Write out the change of polarizability caused by displacement of Si atom in A and C layer,
|
||||
Where unknown non-zero components are denoted by $a_1$, $a_2$, $a_5$, $a_6$.
|
||||
For example, when we move the Si atom in A layer slightly towards the x+ direction in $d$ distance,
|
||||
the change of polarizability should be $mat(,a_2,a_1;a_2,,;a_1,,)d$.
|
||||
This could be done by conclusion above.
|
||||
- The Si atom in B layer have similar local environment as the A and C layer, with only a little difference.
|
||||
We denote these difference by $epsilon_1$, $epsilon_2$, $epsilon_5$, $epsilon_6$,
|
||||
and the absolute value of $epsilon_i$ should be much smaller than $a_i$.
|
||||
For example, when we move the Si atom in B layer slightly towards the x+ direction in $d$ distance,
|
||||
the change of polarizability should be $mat(,a_2+epsilon_2,a_1+epsilon_1;a_2+epsilon_2,,;a_1+epsilon_1,,)d$.
|
||||
- The local environment of C atom in A layer is similar to the Si atom in A layer with charge reversed and
|
||||
the system reversed along xy plane.
|
||||
We denote these difference by $eta_1$, $eta_2$, $eta_5$, $eta_6$,
|
||||
and the absolute value of $epsilon_i$ should be much smaller than $a_i$.
|
||||
For example, when we move the C atom in A layer slightly towards the x+ direction in $d$ distance,
|
||||
the change of polarizability should be $mat(,a_2+eta_2,-a_1-eta_1;a_2+eta_2,,;-a_1-eta_1,,)d$.
|
||||
- Similar to the case in Si atoms, we derive the change of polarizability
|
||||
caused by moving C atom in B layer slightly towards the x+ direction in $d$ distance,
|
||||
which should be $mat(,-a_2-eta_2-zeta_2,-a_1-eta_1-zeta_1;-a_2-eta_2-zeta_2,,;-a_1-eta_1-zeta_1,,)d$.
|
||||
|
||||
Lets assign Raman tensor onto each atom.
|
||||
That is, Raman tensor is derivative of the polarizability with respect to the atomic displacement:
|
||||
$
|
||||
alpha = pdv(chi, u)
|
||||
$
|
||||
where $u$ should be the displacement of the atom corresponding to a phonon mode.
|
||||
But, even when $u$ is *NOT* the displacement of a phonon
|
||||
(for example, lets only slightly move Si atom in A layer, keeping other atoms fixed),
|
||||
the (high-frequency) polarizability is still well-defined,
|
||||
and the will still cause a change in the polarizability.
|
||||
Even more, the group representation theory is still applicable in this condition:
|
||||
the only thing that matters is, when applying $g$ to the system,
|
||||
the tensor transformed into $g^(-1) alpha g$ or $g alpha g^(-1)$,
|
||||
no matter $alpha$ is Raman tensor or something else, or it is related to a phonon or not.
|
||||
|
||||
Thus, we can, in principle, "assign" Raman tensor of a phonon, to each atom.
|
||||
This "assign" is unique since both the atom movement and all phonons have 24 dimensions.
|
||||
|
||||
Next, we consider what these single-atom-caused "Raman tensors" looks like.
|
||||
For example, what happens if we move the Si atom in A layer slightly along the x+ direction?
|
||||
Consider also move the Si atom in C layer slightly, along x+ or x- direction.
|
||||
How about the Raman tensor caused by the both two atoms?
|
||||
In first case, this is B2 representation in E1 representation. Thus the Raman tensor should be something like:
|
||||
$
|
||||
mat(,,2a_1;,,;2a_1,,;)
|
||||
$
|
||||
In the second case, it is A2 in E2. It turns out:
|
||||
$
|
||||
mat(,2a_2,;2a_2,,;,,;)
|
||||
$
|
||||
The average of these two tensors should be the s"Raman tensor" cause by move only the Si atom in A layer,
|
||||
slightly towards x+ direction.
|
||||
$
|
||||
mat(,a_2,a_1;a_2,,;a_1,,;)
|
||||
$
|
||||
The difference should be the "Raman tensor" of the second atom.
|
||||
$
|
||||
mat(,-a_2,a_1;-a_2,,;a_1,,;)
|
||||
$
|
||||
|
||||
// This approach applied relied on the fact that, all Si atom in 4H-SiC is "distinguishable" by the symmetry operations.
|
||||
// I mean, what will happen if we have two Si atoms in A layer?
|
||||
// Apparently, we could not extract the "Raman tensor" of only one of the two atoms.
|
||||
// This is the case for the 6H-SiC.
|
||||
// Hence, we will provide a more general approach to estimate the "Raman tensor" of a single atom.
|
||||
|
||||
Consider the Si atom in the B1 layer.
|
||||
It lives in an environment quite similar to the A layer.
|
||||
Thus, the "Raman tensor" caused by it should be similar to the one caused by the A layer:
|
||||
$
|
||||
mat(,a_2+epsilon_2,a_1+epsilon_1;a_2+epsilon_2,,;a_1+epsilon_1,,;)
|
||||
$
|
||||
Similar to the Si atom in B2 layer:
|
||||
$
|
||||
mat(,-a_2-epsilon_2,a_1+epsilon_1;-a_2-epsilon_2,,;a_1+epsilon_1,,;)
|
||||
$
|
||||
|
||||
Same approach applied for Si atom vibrate in y direction.
|
||||
When we move both Si atoms in A and C layer in y+ direction,
|
||||
it is B1 in E1, thus the "Raman tensor" should be:
|
||||
$
|
||||
mat(,,;,,2a_3;,2a_3,;)
|
||||
$
|
||||
And if we move Si in A layer towards y+ but Si in C layer towards y-,
|
||||
it is A2 in E2:
|
||||
$
|
||||
mat(2a_4,,;,-2a_4,;,,;)
|
||||
$
|
||||
Thus we get the "Raman tensor" of Si atom in A layer sololy move towards y+ direction:
|
||||
$
|
||||
mat(a_4,,;,-a_4,a_3;,a_3,;)
|
||||
$
|
||||
and the "Raman tensor" of Si atom in C layer towards y+ direction:
|
||||
$
|
||||
mat(-a_4,,;,a_4,a_3;,a_3,;)
|
||||
$
|
||||
Same applied for the Si atom in B layer:
|
||||
$
|
||||
mat(a_4+epsilon_4,,;,-a_4-epsilon_4,a_3+epsilon_3;,a_3+epsilon_3,;)
|
||||
$
|
||||
$
|
||||
mat(-a_4-epsilon_4,,;,a_4+epsilon_4,a_3+epsilon_3;,a_3+epsilon_3,;)
|
||||
$
|
||||
|
||||
Before consider z-direction, it is important to note that, $a_1$ $a_2$ $a_3$ $a_4$ are not independent.
|
||||
Consider vibration along x+ direction (lets say the distance is $d$).
|
||||
System energy caused by external electric field and vibration is:
|
||||
$
|
||||
E^T (mat(,,2a_1;,,;2a_1,,) d) E
|
||||
$
|
||||
Apply C#sub[3] to atom vibration and external field, energy should not change. We got:
|
||||
$
|
||||
(mat(-1/2,-sqrt(3)/2,;sqrt(3)/2,-1/2,;,,1)E)^T ( mat(,,2a_1;,,;2a_1,,)(-1/2 d) + mat(,,;,,2a_3;,2a_3,)(sqrt(3)/2 d) )
|
||||
(mat(-1/2,-sqrt(3)/2,;sqrt(3)/2,-1/2,;,,1)E)
|
||||
$
|
||||
It is equal to:
|
||||
$
|
||||
E^T (mat(,,1/2 a_1 + 3/2 a_3;,,sqrt(3)/2 a_1 - sqrt(3)/2 a_3;1/2 a_1 + 3/2 a_3,sqrt(3)/2 a_1 - sqrt(3)/2 a_3,) d) E
|
||||
$
|
||||
Thus:
|
||||
$
|
||||
1/2 a_1 + 3/2 a_3 = 2a_1 #linebreak()
|
||||
sqrt(3)/2 a_1 - sqrt(3)/2 a_3 = 0
|
||||
$
|
||||
Thus $a_1 = a_3$.
|
||||
Apply the same method, we get $abs(a_2) = abs(a_4)$.
|
||||
Since we have not define the sign of $a_4$, we could take $a_2 = a_4$.
|
||||
Same for $epsilon$.
|
||||
|
||||
Now consider what if we move the Si atom in A layer along z+ direction.
|
||||
If we move the Si atom in C layer along z+ direction, it is A1:
|
||||
$
|
||||
mat(2a_5,,;,2a_5,;,,2a_6;)
|
||||
$
|
||||
If we move the Si atom in C layer along z- direction, it is B1:
|
||||
$
|
||||
0
|
||||
$
|
||||
Thus we get the "Raman tensor" of Si atom in A or C layer towards z+ direction:
|
||||
$
|
||||
mat(a_5,,;,a_5,;,,a_6;)
|
||||
$
|
||||
|
||||
Lets consider the C atom in A layer.
|
||||
It should be somehow similar to the Si atom in A layer, but with a negative sign in some places,
|
||||
and then add or subtract some little value.
|
||||
Actually, the "transformation" of Si atom in A layer to C atom in A layer applied in the following steps:
|
||||
- reverse charge.
|
||||
- reverse system along xy plane.
|
||||
First we consider the first step.
|
||||
Taking the define of electricity tenser:
|
||||
$
|
||||
P = chi E
|
||||
$
|
||||
Lets reverse charge of the system, say we now have electricity tensor $chi'$. We get:
|
||||
$
|
||||
-P = chi'(-E)
|
||||
$
|
||||
Thus we get $chi' = chi$, the first step does not change the electricity tensor, nor the "Raman tensor".
|
||||
|
||||
Now we consider the second step.
|
||||
For electricity tensor, it will become:
|
||||
$
|
||||
mat(1,,;,1,;,,-1) chi mat(1,,;,1,;,,-1)
|
||||
$
|
||||
For $u$, when it is along x or y direction, it will not change. When it is along z direction, it will become $-u$.
|
||||
|
||||
So in conclusion, Raman tensor of C atom in A layer could be estimated from the Raman tensor of Si atom in A layer, by:
|
||||
- for movement alone x and y direction, xz yz should be applied a negative sign.
|
||||
- for movement alone z direction, xx xy yy zz should be applied a negative sign.
|
||||
|
||||
Export "Raman tensor" of C atom in C layer from C atom in A layer, in the same way.
|
||||
|
||||
Now consider the C atom in B1 layer.
|
||||
Is it similar to the C atom in A layer, just like that for Si atom?
|
||||
No. It turns out to be similar to the C atom in C layer.
|
||||
|
||||
We summarize these stuff into @table-singleatom.
|
||||
|
||||
Until now, we only consider the "Raman tensor" caused by single atom or atoms move in the same amplitudes.
|
||||
However, that is not the case in real phonon.
|
||||
- In some A1 modes, only Si or C atom moves. If we take the magnitude of eigenvector as 1,
|
||||
then amplitude of each atom is $1/(4sqrt(m_#text[Si]))$ or $1/(4sqrt(m_#text[C]))$.
|
||||
- In other cases, the amplitude of Si and C are in the ration of $m_#text[C] : m_#text[Si]$.
|
||||
thus the amplitude of Si atom is $1/2 sqrt(1/(m_#text[Si]+m_#text[Si]^2/m_#text[C]))$, so do the C atom.
|
||||
|
||||
|
||||
Furthermore, we list predicted modes and their Raman tensors, in @table-predmode.
|
||||
|
||||
- $a$: Raman tensor of Si atom in A layer, large value.
|
||||
- $epsilon$: Difference of Raman tensors of Si atom in A and B1 layer, small value.
|
||||
- $eta$: Difference of Raman tensors of C and Si atom in A layer, small value.
|
||||
- $zeta$: Difference of Raman tensors of C atoms in A and B layer, small value.
|
||||
|
||||
|
||||
#page(flipped: true)[#figure({
|
||||
table(columns: 4, align: center + horizon, inset: (x: 3pt, y: 5pt),
|
||||
[*Move Direction*], [x], [y], [z],
|
||||
[Si A], [$mat(,a_2,a_1;a_2,,;a_1,,;)$], [$mat(a_2,,;,-a_2,a_1;,a_1,;)$], [$mat(a_5,,;,a_5,;,,a_6;)$],
|
||||
[C A], [$mat(,a_2+eta_2,-a_1-eta_1;a_2+eta_2,,;-a_1-eta_1,,;)$],
|
||||
[$mat(a_2+eta_2,,;,-a_2-eta_2,-a_1-eta_1;,-a_1-eta_1,;)$], [$mat(-a_5-eta_5,,;,-a_5-eta_5,;,,-a_6-eta_6;)$],
|
||||
[Si B1], [$mat(,a_2+epsilon_2,a_1+epsilon_1;a_2+epsilon_2,,;a_1+epsilon_1,,;)$],
|
||||
[$mat(a_2+epsilon_2,,;,-a_2-epsilon_2,a_1+epsilon_1;,a_1+epsilon_1,;)$],
|
||||
[$mat(a_5+epsilon_5,,;,a_5+epsilon_5,;,,a_6+epsilon_6;)$],
|
||||
[C, B1], [$mat(,-a_2-eta_2-zeta_2,-a_1-eta_1-zeta_1;-a_2-eta_2-zeta_2,,;-a_1-eta_1-zeta_1,,;)$],
|
||||
[$mat(-a_2-eta_2-zeta_2,,;,a_2+eta_2+zeta_2,-a_1-eta_1-zeta_1;,-a_1-eta_1-zeta_1,;)$],
|
||||
[$mat(-a_5-eta_5-zeta_5,,;,-a_5-eta_5-zeta_5,;,,-a_6-eta_6-zeta_6;)$],
|
||||
[Si C], [$mat(,-a_2,a_1;-a_2,,;a_1,,;)$], [$mat(-a_2,,;,a_2,a_1;,a_1,;)$], [$mat(a_5,,;,a_5,;,,a_6;)$],
|
||||
[C, C], [$mat(,-a_2-eta_2,-a_1-eta_1;-a_2-eta_2,,;-a_1-eta_1,,;)$],
|
||||
[$mat(-a_2-eta_2,,;,a_2+eta_2,-a_1-eta_1;,-a_1-eta_1,;)$], [$mat(-a_5-eta_5,,;,-a_5-eta_5,;,,-a_6-eta_6;)$],
|
||||
[Si B2], [$mat(,-a_2-epsilon_2,a_1+epsilon_1;-a_2-epsilon_2,,;a_1+epsilon_1,,;)$],
|
||||
[$mat(-a_2-epsilon_2,,;,a_2+epsilon_2,a_1+epsilon_1;,a_1+epsilon_1,;)$],
|
||||
[$mat(a_5+epsilon_5,,;,a_5+epsilon_5,;,,a_6+epsilon_6;)$],
|
||||
[C, B2], [$mat(,a_2+eta_2+zeta_2,-a_1-eta_1-zeta_1;a_2+eta_2+zeta_2,,;-a_1-eta_1-zeta_1,,;)$],
|
||||
[$mat(a_2+eta_2+zeta_2,,;,-a_2-eta_2-zeta_2,-a_1-eta_1-zeta_1;,-a_1-eta_1-zeta_1,;)$],
|
||||
[$mat(-a_5-eta_5-zeta_5,,;,-a_5-eta_5-zeta_5,;,,-a_6-eta_6-zeta_6;)$],
|
||||
)},
|
||||
caption: ["Raman tensor" caused by single atom],
|
||||
placement: none,
|
||||
)<table-singleatom>]
|
||||
5
paper/appendix/predict/fig-same.typ
Normal file
5
paper/appendix/predict/fig-same.typ
Normal file
@@ -0,0 +1,5 @@
|
||||
#figure(
|
||||
image("/画图/AB相似/embed.svg"),
|
||||
caption: [Light incidence configurations in our Raman experiments.],
|
||||
placement: none,
|
||||
)<figure-same>
|
||||
@@ -49,6 +49,10 @@
|
||||
// 增加标题的间距
|
||||
#show heading: set block(below: 1em)
|
||||
|
||||
// 公式编号
|
||||
|
||||
#set math.equation(numbering: "(1)")
|
||||
|
||||
#include "introduction.typ"
|
||||
#include "method/default.typ"
|
||||
#include "result/default.typ"
|
||||
|
||||
@@ -75,7 +75,7 @@
|
||||
- [x] 中文
|
||||
- [x] 英文
|
||||
- [x] 调整语言
|
||||
- [ ] 填充分离 Si C 结果的数据，对称分布小量
|
||||
- [ ] 填充最终结果
|
||||
- [ ] 模式的表格
|
||||
- [x] 大致内容
|
||||
- [ ] 调整、填充数据
|
||||
@@ -93,6 +93,13 @@
|
||||
- [ ] 复杂替位原子
|
||||
- [ ] 面缺陷（BPD）
|
||||
- [ ] 附录
|
||||
- [/] 推导细节
|
||||
- [x] 总述
|
||||
- [x] A/C 层 Si 面内
|
||||
- [ ] 画图
|
||||
- [ ] B 层 Si 面内
|
||||
- [ ] 其它情况
|
||||
- [ ] 总结表格
|
||||
- [ ] 分离 Si
|
||||
- [ ] 推导细节
|
||||
- [ ] 杂项
|
||||
|
||||
12
画图/AB相似/atomA.csv
Normal file
12
画图/AB相似/atomA.csv
Normal file
@@ -0,0 +1,12 @@
|
||||
type,x,y,z,radius
|
||||
0,3.08813,0.00000,10.10781,1.18
|
||||
0,0.00000,0.00000,10.10781,1.18
|
||||
0,1.54407,2.67440,10.10781,1.18
|
||||
1,1.54407,0.89147, 9.48201,0.77
|
||||
0,1.54407,0.89147, 7.58104,1.18
|
||||
1,1.54407,2.67440, 6.94819,0.77
|
||||
1,3.08813,0.00000, 6.94819,0.77
|
||||
1,0.00000,0.00000, 6.94819,0.77
|
||||
0,3.08813,0.00000, 5.05312,1.18
|
||||
0,0.00000,0.00000, 5.05312,1.18
|
||||
0,1.54407,2.67440, 5.05312,1.18
|
||||
|
12
画图/AB相似/atomB.csv
Normal file
12
画图/AB相似/atomB.csv
Normal file
@@ -0,0 +1,12 @@
|
||||
type,x,y,z,radius
|
||||
0, 1.54407, 0.89147,7.58104,1.18
|
||||
0,-1.54407, 0.89147,7.58104,1.18
|
||||
0,-0.00000,-1.78294,7.58104,1.18
|
||||
1, 0.00000, 0.00000,6.94819,0.77
|
||||
0, 0.00000, 0.00000,5.05312,1.18
|
||||
1, 0.00000, 1.78294,4.42732,0.77
|
||||
1,-1.54407,-0.89147,4.42732,0.77
|
||||
1, 1.54407,-0.89147,4.42732,0.77
|
||||
0, 0.00000, 1.78294,2.52635,1.18
|
||||
0,-1.54407,-0.89147,2.52635,1.18
|
||||
0, 1.54407,-0.89147,2.52635,1.18
|
||||
|
11
画图/AB相似/bondA.csv
Normal file
11
画图/AB相似/bondA.csv
Normal file
@@ -0,0 +1,11 @@
|
||||
x0,y0,z0,x1,y1,z1
|
||||
3.08813,0.00000,10.10781,1.54407,0.89147, 9.48201
|
||||
0.00000,0.00000,10.10781,1.54407,0.89147, 9.48201
|
||||
1.54407,2.67440,10.10781,1.54407,0.89147, 9.48201
|
||||
1.54407,0.89147, 9.48201,1.54407,0.89147, 7.58104
|
||||
1.54407,0.89147, 7.58104,1.54407,2.67440, 6.94819
|
||||
1.54407,0.89147, 7.58104,3.08813,0.00000, 6.94819
|
||||
1.54407,0.89147, 7.58104,0.00000,0.00000, 6.94819
|
||||
1.54407,2.67440, 6.94819,1.54407,2.67440, 5.05312
|
||||
3.08813,0.00000, 6.94819,3.08813,0.00000, 5.05312
|
||||
0.00000,0.00000, 6.94819,0.00000,0.00000, 5.05312
|
||||
|
11
画图/AB相似/bondB.csv
Normal file
11
画图/AB相似/bondB.csv
Normal file
@@ -0,0 +1,11 @@
|
||||
x0,y0,z0,x1,y1,z1
|
||||
1.54407, 0.89147,7.58104, 0.00000, 0.00000,6.94819
|
||||
-1.54407, 0.89147,7.58104, 0.00000, 0.00000,6.94819
|
||||
-0.00000,-1.78294,7.58104, 0.00000, 0.00000,6.94819
|
||||
0.00000, 0.00000,6.94819, 0.00000, 0.00000,5.05312
|
||||
0.00000, 0.00000,5.05312, 0.00000, 1.78294,4.42732
|
||||
0.00000, 0.00000,5.05312,-1.54407,-0.89147,4.42732
|
||||
0.00000, 0.00000,5.05312, 1.54407,-0.89147,4.42732
|
||||
0.00000, 1.78294,4.42732, 0.00000, 1.78294,2.52635
|
||||
-1.54407,-0.89147,4.42732,-1.54407,-0.89147,2.52635
|
||||
1.54407,-0.89147,4.42732, 1.54407,-0.89147,2.52635
|
||||
|
BIN
画图/AB相似/embed.svg
LFS
Normal file
BIN
画图/AB相似/embed.svg
LFS
Normal file
Binary file not shown.
BIN
画图/AB相似/main.png
LFS
Normal file
BIN
画图/AB相似/main.png
LFS
Normal file
Binary file not shown.
BIN
画图/AB相似/main.pvsm
LFS
Normal file
BIN
画图/AB相似/main.pvsm
LFS
Normal file
Binary file not shown.
BIN
画图/AB相似/main.svg
LFS
Normal file
BIN
画图/AB相似/main.svg
LFS
Normal file
Binary file not shown.
Reference in New Issue
Block a user