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陈浩南
2025-07-21 16:14:52 +08:00
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paper/appendix/predict/default.typ
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@@ -11,7 +11,7 @@
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。
若模式的该拉曼张量只包含小量（$epsilon$, $eta$ and $zeta$），
说明该模式中，原子振动导致的拉曼效应互相抵消了大部分，该模式的拉曼活性较弱；
而若该拉曼张量包含较大的常数项，说明该模式的拉曼效应较强。
而若该拉曼张量包含较大的常数项（$a$），说明该模式的拉曼效应较强。
The center principle of our approximation is to assign the Raman tensor
(i.e., change of polarizability caused by atomic displacement)
@@ -33,7 +33,7 @@ Otherwise, if the Raman tensor contains quantities with large absolute value ($a
In this section, AB#sub[1]CB#sub[2] instead of ABCB was used to denote the four bilayers in 4H-SiC primative cell
to clearly distinguish the two B layers.
This is because the local environment of the two B layers is mirror symmetric rather than translationally symmetric,
This is because the local environment of the two B layers is mirror symmetric with each other,
thus their Raman tensors are not equal.
=== Raman tensor of Si atoms in A and C layers
@@ -42,8 +42,8 @@ This is because the local environment of the two B layers is mirror symmetric ra
根据前文，我们知道，当这两个原子同步地沿 x 正方向振动时，它们属于 E1(C6v) B2(C2v) 表示，拉曼张量可以写为：
We first derive the Raman tensor of Si atoms in A and C layer vibrating along x direction.
When the two atoms vibrate synchronously in the positive x direction,
they belong to the representation of E#sub[1] of C#sub[6v] or B#sub[2] of C#sub[2v].
The vibration where the two atoms vibrate synchronously along the positive x direction,
belongs to the representation of E#sub[1] of C#sub[6v] and B#sub[2] of C#sub[2v].
Thus, their Raman tensor can be written as:
$ mat(,,2a_1;,,;2a_1,,;) $
@@ -56,9 +56,9 @@ where $a_i (i = 1 "to" 6)$ are unknown constants.
它们属于 E#sub[2] of C#sub[6v] or A#sub[2] of C#sub[2v] 表示，
拉曼张量可以写为：
When the Si atom in A layer vibrates in the positive x direction
while the Si atom in C layer vibrates in the negative x direction,
they belong to the representation of E#sub[2] of C#sub[6v] or A#sub[2] of C#sub[2v].
For the vibration where the Si atom in A layer vibrates towards the positive x direction
and the Si atom in C layer vibrates towards the negative x direction,
it belongs to the representation of E#sub[2] of C#sub[6v] and A#sub[2] of C#sub[2v].
Thus, their Raman tensor can be written as:
$ mat(,2a_2,;2a_2,,;,,;) $
@@ -72,33 +72,31 @@ $ mat(,a_2,a_1;a_2,,;a_1,,;), mat(,-a_2,a_1;-a_2,,;a_1,,;), $
接下来讨论 A/C 层 Si 原子沿 y 方向振动时的拉曼张量，使用相似的方法可以得到：
The Raman tensors of Si atoms in A and C layers vibrating along positive y direction
can be obtained using the method, which gives:
can be obtained using the same method, which gives:
$ mat(a_4,,;,-a_4,a_3;,a_3,;), mat(-a_4,,;,a_4,a_3;,a_3,;) $
$\{a_1, a_2, a_3, a_4\}$ 之间并不独立。为了确定它们之间的关系，我们考虑将体系绕 z 轴旋转 120 度，
同一个点的坐标旋转前后分别为 $r$ 和 $C_3 r$，其中 $r$ 是一个列向量，$C_3$ 是一个旋转矩阵：
$\{a_1, a_2, a_3, a_4\}$ 之间并不独立。为了确定它们之间的关系，我们考虑 A 层中 Si 原子沿 b 轴正方向振动所导致的拉曼张量（记为 $alpha'$）。
一方面，它可以看作由向 x 和 y 正方向振动的拉曼张量（记为 $alpha_x$ 和 $alpha_y$）通过线性组合得到；
另一方面，它也可以看作由 $alpha_x$ 通过将体系绕 z 轴旋转 $120 degree$ 得到。
因此：
where $\{a_3, a_4\}$ are not independent of $\{a_1, a_2\}$.
To determine the relationship between $\{a_1, a_2\}$ and $\{a_3, a_4\}$,
the system is rotated by $120 degree$ around the z axis.
The point located at $r$ before rotation should be at $C_3 r$ after rotation,
where $r$ is a column vector and $C_3$ is a rotation matrix:
the Raman tensor of Si atoms in A layer vibrating along the positive b-axis direction
(denoted as $alpha'$) was considered.
On one hand,
it can be expressed as a linear combination of the Raman tensors vibrating in the positive x and y directions
(denoted as $alpha_x$ and $alpha_y$, respectively);
on the other hand,
it can also be obtained from $alpha_x$ by rotating the system by $120 degree$.
Thus:
$ C_3 = mat(-1/2,-sqrt(3)/2,0;sqrt(3)/2,-1/2,0;0,0,1;) $
记 A 层中 Si 原子沿与 x 轴夹角 $120 degree$ 的方向振动所导致的拉曼张量为 $alpha'$，
沿 x 和 y 正方向振动的拉曼张量则为 $alpha_x$ 和 $alpha_y$。
一方面，$alpha'$ 可以从 $alpha_x$ 出发，将体系旋转 $120 degree$ 得到；
另一方面，$alpha'$ 也可以由 $alpha_x$ 和 $alpha_y$ 通过线性组合得到：
The Raman tensor of Si atoms in A layer vibrating along the direction at an angle of $120 degree$ with the x axis
is denoted as $alpha'$,
while those vibrating in the positive x and y directions are denoted as $alpha_x$ and $alpha_y$, respectively.
On one hand, $alpha'$ can be obtained from $alpha_x$ by rotating the system by $120 degree$;
on the other hand, $alpha'$ can also be expressed as a linear combination of $alpha_x$ and $alpha_y$:
$ alpha' = C_3 alpha_x C_3^T = -1/2 alpha_x + sqrt(3)/2 alpha_y $
$
alpha' = C_3 alpha_x C_3^T = -1/2 alpha_x + sqrt(3)/2 alpha_y, \
"where" C_3 = mat(-1/2,-sqrt(3)/2,;sqrt(3)/2,-1/2,;,,1;),
alpha_x = mat(,a_2,a_1;a_2,,;a_1,,;), alpha_y = mat(a_4,,;,-a_4,a_3;,a_3,;)
$
化简可得：
@@ -109,8 +107,7 @@ $ a_3 = a_1, a_4 = a_2 $
与 x 和 y 方向的情况类似，可以推导出 A/C 层 C 原子沿 z 方向振动时的拉曼张量。总结如下。
The Raman tensors of C atoms in A and C layers vibrating along z direction can be derived similarly.
Thus the results are summarized as follows:
The results are summarized as follows:
#figure(
table(columns: 6, align: center + horizon, inset: (x: 3pt, y: 5pt),
@@ -141,7 +138,7 @@ $ mat(,a'_2,a'_1;a'_2,,;a'_1,,;), mat(,-a'_2,a'_1;-a'_2,,;a'_1,,;) $
因此可以推测它们的拉曼张量只有较小的不同，即：
Because the local environment of Si atoms in B layer is very similar to that in A layer (as shown in @figure-same),
we can assume that their Raman tensors differ only by small quantities, i.e.,
we can assume that their Raman tensors differ only by quantities with small absolute values, i.e.,
$
a'_1 = a_1 + epsilon_1, abs(epsilon_1) << abs(a_1), \
@@ -152,7 +149,7 @@ $
由此可以写出 B 层 Si 原子沿 x 方向振动时的拉曼张量：
Thus, the Raman tensor of Si atoms in B#sub[1] and B#sub[2] layers vibrating along x direction
Thus, the Raman tensor of Si atoms in B#sub[1] and B#sub[2] layers vibrating along positive x direction
can be written as:
$
@@ -185,72 +182,95 @@ The Raman tensors of Si atoms in B#sub[1] and B#sub[2] layers vibrating along ot
=== Raman tensor of C atoms
考虑 A/C 层 C 原子的拉曼张量，使用与 Si 原子类似的方法，可以得到：
C 原子的拉曼张量，使用与 Si 原子类似的方法，可以得到。
例如 A 层 C 原子的拉曼张量可以写为：
The Raman tensors of C atoms in A and C layers can be obtained using a similar method:
The Raman tensors of C atoms can be obtained using a similar method as that of Si atoms.
For example, the Raman tensor of C atom in A layer can be written as:
$
mat(,b_2,b_1;b_2,,;b_1,,;), mat(,-b_2,b_1;-b_2,,;b_1,,;)
$
#figure(
table(columns: 6, align: center + horizon, inset: (x: 3pt, y: 5pt),
table.cell(colspan: 3)[*Vibration Direction*], [x], [y], [z],
table.cell(rowspan: 2)[*Raman tensor #linebreak() of atoms*], table.cell(rowspan: 2)[A layer],
[C],
[$alpha_"Cx" = mat(,b_2,b_1;b_2,,;b_1,,;)$],
[$alpha_"Cy" = mat(b_2,,;,-b_2,b_1;,b_1,;)$],
[$alpha_"Cz" = mat(b_5,,;,b_5,;,,b_6;)$],
[Si],
[$alpha_"Six" = mat(,a_2,a_1;a_2,,;a_1,,;)$],
[$alpha_"Siy" = mat(a_2,,;,-a_2,a_1;,a_1,;)$],
[$alpha_"Siz" = mat(a_5,,;,a_5,;,,a_6;)$],
),
placement: none,
)
我们需要估计 ${b_1, b_2}$ 与 ${a_1, a_2}$ 之间的关系。
考虑 A 层 C 原子的环境，它可以由 A 层 Si 原子通过以下操作得到：
考虑 A 层 C 原子的环境，它可以由 A 层 Si 原子通过以下三步操作得到：
先沿基平面取镜像，然后反转电荷，再调整原子质量等其它因素，如图所示。
我们分两步来推导这个过程中拉曼张量的变化。
我们分别考虑这些过程中拉曼张量的变化。
The relationship between $\{b_1, b_2\}$ and $\{a_1, a_2\}$ needs to be estimated.
The environment of C atoms in A layer can be obtained from that of Si atoms in A layer
by first taking a mirror image along the basal plane,
then reversing the charge and adjusting the atomic mass and other factors,
as shown in @figure-sitoc.
The change of Raman tensor during this process could be derived following these two steps.
The environment of C atoms in A layer can be obtained from that of Si atoms in A layer with the following three steps:
first taking a mirror image along the basal plane, then reversing the charge,
and finally adjusting the atomic mass and other factors, as shown in @fig-sitoc.
The change of Raman tensor during these processes would be discussed separately.
#include "fig-sitoc.typ"
记翻转前后的 Si 原子拉曼张量为 $alpha$ 和 $alpha'$。
考虑在外场 $E$ 作用下，Si 原子沿 x 方向振动导致的系统能量变化为：
第一步中，记翻转后的拉曼张量为 $alpha'_"Six"$、$alpha'_"Siy"$ 和 $alpha'_"Siz"$。
对于 x 和 y 方向振动的拉曼张量，只需要将群元素 $sigma_"h" = op("diag") (1, 1, -1)$ 作用上去即可；
对于 z 方向振动的拉曼张量，还需要乘以 $-1$，因为在这个过程中振动的方向发生了改变。
The Raman tensor of the Si atom in A layer before and after taking the mirror image
was denoted as $alpha$ and $alpha'$, respectively.
Before the mirror image,
the system energy change caused by the vibration of this atom along x direction
under an external electric field $E$ is:
In the first step,
the Raman tensors after taking the mirror image were denoted as $alpha'_"Six"$, $alpha'_"Siy"$ and $alpha'_"Siz"$.
For $alpha'_"Six"$ and $alpha'_"Siy"$, they are connected with $alpha_"Six"$ and $alpha_"Siy"$
by the group element $sigma_"h" = op("diag") (1, 1, -1)$.
For $alpha'_"Siz"$, an additional factor of $-1$ is needed,
because the direction of vibration has changed during this process.
$ Delta E = E^T alpha E, #[where] alpha = mat(,a_2,a_1;a_2,,;a_1,,;) $
$
alpha'_"Six" = sigma_"h" alpha_"Six" sigma_"h"^T = mat(,a_2,-a_1;a_2,,;-a_1,,;), \
alpha'_"Siy" = sigma_"h" alpha_"Siy" sigma_"h"^T = mat(a_2,,;,-a_2,-a_1;,-a_1,;), \
alpha'_"Siz" = -sigma_"h" alpha_"Siz" sigma_"h"^T = mat(-a_5,,;,-a_5,;,,-a_6;),
$
若在翻转的过程中，将电场同样翻转，则总能量不变。因此：
翻转电荷的过程不会导致拉曼张量的变化。
这可以通过考虑在外场 $E$ 作用下的能量变化来得知。
记电荷翻转前后的拉曼张量分别为 $alpha$ 和 $alpha'$。
若在翻转电荷的过程中，外加电场同样翻转，则总能量不变。
因此 $E^T alpha E = (-E)^T alpha' (-E)$，因此 $alpha = alpha'$。
When the electric field is also flipped during the mirror image,
the total energy does not change, i.e.,
In the second step (reversing the charge), the Raman tensor does not change.
This can be derived by considering the energy caused by an external electric field $E$.
The Raman tensors before and after charge reversal were denoted as $alpha$ and $alpha'$.
When the direction of the external electric field is also reversed during the charge reversal,
the total energy does not change, i.e., $E^T alpha E = (-E)^T alpha' (-E)$.
Thus, we have $alpha = alpha'$.
$ E^T alpha E = Delta E = (sigma E)^T alpha' (sigma E), #[where] sigma = mat(1,0,0;0,1,0;0,0,-1;) $
整理可得：
第三步中，我们假定原子质量和其它因素的变化对拉曼张量的影响较小，
即 $alpha'_"Six"$、$alpha'_"Siy"$ 和 $alpha'_"Siz"$ 与 $alpha_"Six"$、$alpha_"Siy"$ 和 $alpha_"Siz"$ 之间仅有较小的差异。
因此：
In the third step, we assume that the change in atomic mass and other factors has a small effect on the Raman tensor,
i.e., $alpha'_"Six"$, $alpha'_"Siy"$ and $alpha'_"Siz"$ differ from $alpha_"Six"$, $alpha_"Siy"$ and $alpha_"Siz"$
only by small quantities, respectively.
Thus:
$ alpha' = mat(,a_2,-a_1;a_2,,;-a_1,,;) $
$
b_1 = -a_1 - zeta_1, abs(zeta_1) << abs(a_1), \
b_2 = a_2 + zeta_2, abs(zeta_2) << abs(a_2), \
b_5 = -a_5 - zeta_5, abs(zeta_5) << abs(a_5), \
b_6 = -a_6 - zeta_6, abs(zeta_6) << abs(a_6),
$
记电荷反转后，Si 原子拉曼张量为 $alpha''$。
若将外加电场方向同时反转，则能量不变。即：
=== Summary
// TODO: 能量写明是 Delta E
我们将各个原子的拉曼张量总结于 @table-singleatom，用它推测了各个模式的拉曼张量并与第一性原理计算对比，
结果如 @table-predmode 所示。
The Raman tensor of Si atom after charge reversal was denoted as $alpha''$.
The energy does not change when the direction of the external electric field is also reversed, i.e.,
$ E^T alpha' E = (sigma' E)^T alpha'' (sigma' E), #[where] sigma' = -1 $
Thus:
$ alpha'' = alpha' $
使用类似的方法，得到 Si 原子沿其它方向和 C 原子沿各个方向的拉曼张量，以及各个模式的拉曼张量。
Similarily, we can write out the Raman tensors of Si atoms virbrating along other directions
and the Raman tensors of C atoms,
and thus the Raman tensors of all phonon modes.
The results are summarized in @table-singleatom and @table-predmode.
We summarized the Raman tensors of each atom in @table-singleatom.
The result was used to predict the Raman tensors of each mode and comparing with first-principles calculations,
which is shown in @table-predmode.
#include "table-singleatom.typ"
#include "table-predmode.typ"

2
paper/appendix/predict/fig-sitoc.typ
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@@ -4,4 +4,4 @@
where the nearest neighbors are exactly the same, and only half of the next nearest neighbors are different.
],
placement: none,
)<figure-sitoc>
)<fig-sitoc>

24
paper/appendix/predict/table-singleatom.typ
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@@ -5,36 +5,36 @@
table.cell(rowspan: 8)[*Raman tensor #linebreak() of atoms*],
r2[A layer],
[C],
[$mat(,b_2,b_1;b_2,,;b_1,,;)$],
[$mat(b_2,,;,-b_2,b_1;,b_1,;)$],
[$mat(b_5,,;,b_5,;,,b_6;)$],
[$mat(,a_2+zeta_2,-a_1-zeta_1;a_2+zeta_2,,;-a_1-zeta_1,,;)$],
[$mat(a_2+zeta_2,,;,-a_2-zeta_2,-a_1-zeta_1;,-a_1-zeta_1,;)$],
[$mat(-a_5-zeta_5,,;,-a_5-zeta_5,;,,-a_6-zeta_6;)$],
[Si],
[$mat(,a_2,a_1;a_2,,;a_1,,;)$],
[$mat(a_2,,;,-a_2,a_1;,a_1,;)$],
[$mat(a_5,,;,a_5,;,,a_6;)$],
r2[B#sub[1] layer],
[C],
[$mat(,-b_2-eta_2,b_1+eta_1;-b_2-eta_2,,;b_1+eta_1,,;)$],
[$mat(-b_2-eta_2,,;,b_2+eta_2,b_1+eta_1;,b_1+eta_1,;)$],
[$mat(b_5+eta_5,,;,b_5+eta_5,;,,b_6+eta_6;)$],
[$mat(,-a_2-zeta_2-eta_2,-a_1-zeta_1+eta_1;-a_2-zeta_2-eta_2,,;-a_1-zeta_1+eta_1,,;)$],
[$mat(-a_2-zeta_2-eta_2,,;,a_2+zeta_2+eta_2,-a_1-zeta_1+eta_1;,-a_1-zeta_1+eta_1,;)$],
[$mat(-a_5-zeta_5+eta_5,,;,-a_5-zeta_5+eta_5,;,,-a_6-zeta_6+eta_6;)$],
[Si],
[$mat(,a_2+epsilon_2,a_1+epsilon_1;a_2+epsilon_2,,;a_1+epsilon_1,,;)$],
[$mat(a_2+epsilon_2,,;,-a_2-epsilon_2,a_1+epsilon_1;,a_1+epsilon_1,;)$],
[$mat(a_5+epsilon_5,,;,a_5+epsilon_5,;,,a_6+epsilon_6;)$],
r2[C layer],
[C],
[$mat(,-b_2,b_1;-b_2,,;b_1,,;)$],
[$mat(-b_2,,;,b_2,b_1;,b_1,;)$],
[$mat(b_5,,;,b_5,;,,b_6;)$],
[$mat(,-a_2-zeta_2,-a_1-zeta_1;-a_2-zeta_2,,;-a_1-zeta_1,,;)$],
[$mat(-a_2-zeta_2,,;,a_2+zeta_2,-a_1-zeta_1;,-a_1-zeta_1,;)$],
[$mat(-a_5-zeta_5,,;,-a_5-zeta_5,;,,-a_6-zeta_6;)$],
[Si],
[$mat(,-a_2,a_1;-a_2,,;a_1,,;)$],
[$mat(-a_2,,;,a_2,a_1;,a_1,;)$],
[$mat(a_5,,;,a_5,;,,a_6;)$],
r2[B#sub[2] layer],
[C],
[$mat(,b_2+eta_2,b_1+eta_1;b_2+eta_2,,;b_1+eta_1,,;)$],
[$mat(b_2+eta_2,,;,-b_2-eta_2,b_1+eta_1;,b_1+eta_1,;)$],
[$mat(b_5+eta_5,,;,b_5+eta_5,;,,b_6+eta_6;)$],
[$mat(,a_2+zeta_2+eta_2,-a_1-zeta_1+eta_1;a_2+zeta_2+eta_2,,;-a_1-zeta_1+eta_1,,;)$],
[$mat(a_2+zeta_2+eta_2,,;,-a_2-zeta_2-eta_2,-a_1-zeta_1+eta_1;,-a_1-zeta_1+eta_1,;)$],
[$mat(-a_5-zeta_5+eta_5,,;,-a_5-zeta_5+eta_5,;,,-a_6-zeta_6+eta_6;)$],
[Si],
[$mat(,-a_2-epsilon_2,a_1+epsilon_1;-a_2-epsilon_2,,;a_1+epsilon_1,,;)$],
[$mat(-a_2-epsilon_2,,;,a_2+epsilon_2,a_1+epsilon_1;,a_1+epsilon_1,;)$],