diff --git a/paper/appendix/default.typ b/paper/appendix/default.typ
index aeb17bb..26c2b91 100644
--- a/paper/appendix/default.typ
+++ b/paper/appendix/default.typ
@@ -1,230 +1,4 @@
-#import "@preview/physica:0.9.5": pdv, super-T-as-transpose
-#show: super-T-as-transpose
-
 = Appendix
 
+#include "predict/default.typ"
 #include "predmode.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>]
diff --git a/paper/appendix/predict/default.typ b/paper/appendix/predict/default.typ
new file mode 100644
index 0000000..926078e
--- /dev/null
+++ b/paper/appendix/predict/default.typ
@@ -0,0 +1,273 @@
+#import "@preview/physica:0.9.5": pdv, super-T-as-transpose
+#show: super-T-as-transpose
+
+== Approximation of Raman tensor of 4H-SiC
+
+近似的核心思路。
+
+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。
+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。
+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。
+
+The center principle of our approximation is to assign the Raman tensor
+    (i.e., change of polarizability caused by atomic displacement)
+  to each atom in the unit cell.
+Because the local environment of the same type of atom is similar and only different for the next nearest neighbors,
+  we consider the difference in Raman effect caused by the same type of atom as a small quantity.
+The Raman tensor of a phonon mode can be obtained
+  by summing the contributions of all atoms participating in the vibration.
+
+推导 A/C 层 Si 原子沿 x 方向振动时的拉曼张量。
+
+我们首先推导 A/C 层 Si 原子沿 x 方向振动时的拉曼张量。
+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：
+
+We first derive the Raman tensor of Si atoms in A/C layer vibrating along x direction.
+When the two atoms vibrate in the same direction and opposite direction,
+  they belong to the representation of E#sub[1] of C#sub[6v] or B#sub[2] of C#sub[2v]
+    and E#sub[2] of C#sub[6v] or A#sub[2] of C#sub[2v], respectively.
+Thus, their Raman tensors are in the form of:
+
+$ mat(,,2a_1;,,;2a_1,,;), mat(,2a_2,;2a_2,,;,,;), $
+
+其中 $a_1$ 和 $a_2$ 是两个未知的常数。
+
+where $a_1$ and $a_2$ are two constants with unknown values.
+
+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：
+
+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。
+
+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>]
diff --git a/paper/appendix/predict/fig-same.typ b/paper/appendix/predict/fig-same.typ
new file mode 100644
index 0000000..df1fbd4
--- /dev/null
+++ b/paper/appendix/predict/fig-same.typ
@@ -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>
diff --git a/paper/main.typ b/paper/main.typ
index 3288d16..c8a685a 100644
--- a/paper/main.typ
+++ b/paper/main.typ
@@ -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"
diff --git a/paper/todo.typ b/paper/todo.typ
index fc80554..b8ba596 100644
--- a/paper/todo.typ
+++ b/paper/todo.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
diff --git a/画图/AB相似/atomA.csv b/画图/AB相似/atomA.csv
new file mode 100644
index 0000000..b02e540
--- /dev/null
+++ b/画图/AB相似/atomA.csv
@@ -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
diff --git a/画图/AB相似/atomB.csv b/画图/AB相似/atomB.csv
new file mode 100644
index 0000000..fe739cd
--- /dev/null
+++ b/画图/AB相似/atomB.csv
@@ -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
diff --git a/画图/AB相似/bondA.csv b/画图/AB相似/bondA.csv
new file mode 100644
index 0000000..86c8d4c
--- /dev/null
+++ b/画图/AB相似/bondA.csv
@@ -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
diff --git a/画图/AB相似/bondB.csv b/画图/AB相似/bondB.csv
new file mode 100644
index 0000000..c53bff1
--- /dev/null
+++ b/画图/AB相似/bondB.csv
@@ -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
diff --git a/画图/AB相似/embed.svg b/画图/AB相似/embed.svg
new file mode 100644
index 0000000..e573b16
--- /dev/null
+++ b/画图/AB相似/embed.svg
@@ -0,0 +1,3 @@
+version https://git-lfs.github.com/spec/v1
+oid sha256:0c31a371ccaf22b9b12682d29a24c99de72465abcf5558884d298a50667a7374
+size 332902
diff --git a/画图/AB相似/main.png b/画图/AB相似/main.png
new file mode 100644
index 0000000..bcb20da
--- /dev/null
+++ b/画图/AB相似/main.png
@@ -0,0 +1,3 @@
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+oid sha256:0ef5a754e4f5ba784d33c7196bd38adba936885a294d6c23d7f9440d0aeb4ccd
+size 240978
diff --git a/画图/AB相似/main.pvsm b/画图/AB相似/main.pvsm
new file mode 100644
index 0000000..b8cdbd4
--- /dev/null
+++ b/画图/AB相似/main.pvsm
@@ -0,0 +1,3 @@
+version https://git-lfs.github.com/spec/v1
+oid sha256:4bede7b191e43899b2d50b60abe30a3460069ed00ac8bfe0a4aba78a42252b06
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diff --git a/画图/AB相似/main.svg b/画图/AB相似/main.svg
new file mode 100644
index 0000000..311b2ce
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