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test-typst/.gitignore
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test-typst/.gitignore
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main.pdf
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@@ -1,5 +1,7 @@
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#import "@preview/starter-journal-article:0.4.0": article, author-meta
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#import "@preview/tablem:0.2.0": tablem
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#import "@preview/physica:0.9.4": pdv
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#set par.line(numbering: "1")
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// TODO: fix indent of first line
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#show figure.caption: it => {
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@@ -187,20 +189,226 @@ The 18 negligible-polar phonons correspond to 14 irreducible representations of
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Phonons belonging to A#sub[1] and B#sub[1] representations vibration along z-axis and are non-degenerate,
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while phonons belonging to E#sub[1] and E#sub[2] representations vibrate in plane and are doubly degenerate.
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Phonons belonging to B#sub[1] representation are Raman-inactive, as their Raman tensors vanish.
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// TODO: 调整英语
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In contrast, phonons belonging to other representations are Raman-active,
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the Raman tensors of them have non-zero components,
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indicating that these phonons might be visible in Raman experiment under appropriate polarization configurations.
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// 各个模式的声子可以使用怎样的偏振光看到（即拉曼张量的非零分量）可以联合考虑 C6v 和 C2v 的表示来判断，如表所示。
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// TODO: 翻译
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and the non-zero components of Raman tensor of each phonon
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could be determined by considering further in the C#sub[2v] point group @table-rep.
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These Raman-active phonons might be visible in Raman experiment under appropriate polarization configurations.
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However, the actual visibility of each phonon depends on the magnitudes of its Raman tensor components,
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which cannot be computed solely from symmetry analysis.
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// TODO: 画个表
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// TODO: 完善表格
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#figure({
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let m2(content) = table.cell(colspan: 2, content);
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table(columns: 7, align: center + horizon, inset: (x: 3pt, y: 5pt),
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[*Rep in C6v*], [A#sub[1]], [B#sub[1]], m2[E#sub[1]], m2[E#sub[2]],
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[*Rep in C2v*], [A#sub[1]], [B#sub[1]], [B#sub[2]], [B#sub[1]], [A#sub[2]], [A#sub[1]],
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[*Vib Direction*], [z], [z], [x], [y], [x], [y],
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[*Raman Tensor*],
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[$mat(a, , ; , a, ; , , b)$], [$0$], [$mat( , , a; , , ; a, , ;)$], [$mat( , , ; , , a; , a, ;)$],
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[$mat( , a, ; a, , ; , , ;)$], [$mat( a, , ; , -a, ; , , ;)$],
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[*Raman Intensity*],
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[$mat(a^2, , ; , a^2, ; , , b^2)$], [$0$], m2[$mat( , , a^2; , , a^2; a^2, a^2, ;)$],
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m2[$mat( a^2, a^2, ; a^2, a^2, ; , , ;)$]
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)},
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caption: [Rep],
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placement: none,
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)<table-rep>
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Here we propose a method to estimate the magnitudes of the Raman tensors of these phonons.
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// TODO: 写出来这个方法，并验证。
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/*
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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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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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Furthermore, we list predicted modes and their Raman tensors, in @table-predmode.
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Frequency could be estimated by, how many atoms are moving towards its neighbor.
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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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[C A], [$mat(,a_2+eta_2,-a_1-eta_1;a_2+eta_2,,;-a_1-eta_1,,;)$],
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[$mat(a_4+eta_4,,;,-a_4-eta_4,-a_3-eta_3;,-a_3-eta_3,;)$], [$mat(-a_5-eta_5,,;,-a_5-eta_5,;,,-a_6-eta_6;)$],
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[Si A], [$mat(,a_2,a_1;a_2,,;a_1,,;)$], [$mat(a_4,,;,-a_4,a_3;,a_3,;)$], [$mat(a_5,,;,a_5,;,,a_6;)$],
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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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[$mat(-a_4-eta_4-zeta_4,,;,a_4+eta_4+zeta_4,-a_3-eta_3-zeta_3;,-a_3-eta_3-zeta_3,;)$],
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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 B1], [$mat(,a_2+epsilon_2,a_1+epsilon_1;a_2+epsilon_2,,;a_1+epsilon_1,,;)$],
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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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[$mat(a_5+epsilon_5,,;,a_5+epsilon_5,;,,a_6+epsilon_6;)$],
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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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[$mat(-a_4-eta_4,,;,a_4+eta_4,-a_3-eta_3;,-a_3-eta_3,;)$], [$mat(-a_5-eta_5,,;,-a_5-eta_5,;,,-a_6-eta_6;)$],
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[Si C], [$mat(,-a_2,a_1;-a_2,,;a_1,,;)$], [$mat(-a_4,,;,a_4,a_3;,a_3,;)$], [$mat(a_5,,;,a_5,;,,a_6;)$],
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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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[$mat(a_4+eta_4+zeta_4,,;,-a_4-eta_4-zeta_4,-a_3-eta_3-zeta_3;,-a_3-eta_3-zeta_3,;)$],
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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 B2], [$mat(,-a_2-epsilon_2,a_1+epsilon_1;-a_2-epsilon_2,,;a_1+epsilon_1,,;)$],
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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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[$mat(a_5+epsilon_5,,;,a_5+epsilon_5,;,,a_6+epsilon_6;)$],
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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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// Raman Tensor for A1: line1 xz/yz; line2 zz
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#page(flipped: true)[#figure({
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let m2(content) = table.cell(colspan: 2, content);
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let m3(content) = table.cell(colspan: 3, content);
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table(columns: 4, align: center + horizon, inset: (x: 3pt, y: 5pt),
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[*Representation in C#sub[6v]*], m3[A1],
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[*Representation in C#sub[2v]*], m3[A1],
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[*x*], m2[0.5], [1],
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[*Relative Vibration Direction*], [$++--++--$], [$+--++--+$], [$+-+-+-+-$],
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[*Vibration Direction*], m3[z],
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[*Raman Tensor Predicted*], [$2(-epsilon_5+zeta_5)$ #linebreak() $2(-epsilon_6+zeta_6)$],
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[$2(epsilon_5+zeta_5)$ #linebreak() $2(epsilon_6+zeta_6)$],
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[$-4(2a_5+eta_5+epsilon_5+zeta_5)$ #linebreak() $-4(2a_6+eta_6+epsilon_6+zeta_6)$],
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[*Raman Intensity Predicted*], m2[weak], [strong],
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[*Raman Tensor Calculated*],
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[0.10 #linebreak() -1.33], [-1.68 #linebreak() 1.34], [7.68 #linebreak() -21.65],
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[*Atom-pair that Move Relatively In-plane*], [4], [0], [4],
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[*Atom-pair that Move Relatively Out-plane*], [0], [4], [4],
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[*Predicted Frequency*], [low], [medium], [high],
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[*Calculated Frequency*], [591.90], [812.87], [933.80],
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)},
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caption: [Predicted modes and their "Raman tensor"],
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placement: none,
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)<table-predmode>]
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/*
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这里应该有办法来估计。下面是我总结的规律：
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按照我们规定的 ABCB 层序，并将拉曼张量的大小归结为键长的变化的话：
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* 对于 E2 表示（AC层运动方向必须相反，B1/B2层运动方向必须相反，因此只讨论A和B1层）
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@@ -216,6 +424,21 @@ Here we propose a method to estimate the magnitudes of the Raman tensors of thes
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* 对于其它键，根据对称性由上面的结果直接写出。
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* 写出各个模式的拉曼张量（上面的线性组合）。即可以直接看到结果。
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*/
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//
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// 考虑 A 层的竖着的键。面内的振动模式对应的拉曼张量落在两个空间中。
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// 在第一个空间中，它的形式为：
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//
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// $
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// mat(a, , ; , -a, ; , , ;)
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// mat(, a, ; a, , ; , , ;)
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// $
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//
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// 在第二个空间中，它的形式为：
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//
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// $
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// mat(, , a; , , ; a, , ;)
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// mat(, , ; , , a; , a, ;)
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// $
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// 我们计算了拉曼活性声子的频率及拉曼张量，并与实验对比，如表如图所示。
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// 其中有几个声子的拉曼活性较弱，有几个比较强。强的都可以在实验上看到；但弱的能否看到则取决于它是否恰好位于强模式的附近。
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