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Quadratic Debye

Posted by peeterjoot on January 19, 2014

[Click here for a PDF of this post with nicer formatting]

Question: Quadratic Debye phonons (2013 midterm pr B2)

Assume a quadratic dispersion relation for the longitudinal and transverse modes

\begin{aligned}\omega = \left\{\begin{array}{l}b_{\mathrm{L}} q^2 \\ b_{\mathrm{T}} q^2\end{array}\right..\end{aligned} \hspace{\stretch{1}}(1.2)

Part a

Find the density of states.

Part b

Find the Debye frequency.

Part c

In terms of k_{\mathrm{B}} \Theta = \hbar \omega_{\mathrm{D}}, and

\begin{aligned}\mathcal{I} = \int_0^\infty \frac{y^{5/2} e^{y} dy}{\left( { e^y - 1} \right)^2 },\end{aligned} \hspace{\stretch{1}}(1.2)

find the specific heat for k_{\mathrm{B}} T \ll \hbar \omega_{\mathrm{D}}.

Part d

Find the specific heat for k_{\mathrm{B}} T \gg \hbar \omega_{\mathrm{D}}.

Answer

Part a

Working straight from the definition

\begin{aligned}Z(\omega) &= \frac{V}{(2 \pi)^3 } \sum_{L, T} \int \frac{df_\omega}{ \left\lvert { \boldsymbol{\nabla}_\mathbf{q} \omega } \right\rvert } \\ &= \frac{V}{(2 \pi)^3 } \left( { {\left.{{\frac{4 \pi q^2}{2 b_{\mathrm{L}} q} }}\right\vert}_{{\mathrm{L}}} + {\left.{{\frac{2 \times 4 \pi q^2}{2 b_{\mathrm{T}} q} }}\right\vert}_{{\mathrm{T}}} } \right) \\ &= \frac{V}{4 \pi^2 } \left( { \frac{q_{\mathrm{L}}}{b_{\mathrm{L}}} + \frac{2 q_{\mathrm{T}}}{b_{\mathrm{T}}} } \right).\end{aligned} \hspace{\stretch{1}}(1.3)

With q_{\mathrm{L}} = \sqrt{\omega/b_{\mathrm{L}}} and q_{\mathrm{T}} = \sqrt{\omega/b_{\mathrm{T}}}, this is

\begin{aligned}Z(\omega) = \frac{V}{4 \pi^2 } \left( { \frac{1}{b_{\mathrm{L}}^{3/2}} + \frac{2}{b_{\mathrm{T}}^{3/2}} } \right)\sqrt{\omega}\end{aligned} \hspace{\stretch{1}}(1.4)

Part b

The Debye frequency was given implicitly by

\begin{aligned}\int_0^{\omega_{\mathrm{D}}} Z(\omega) d\omega = 3 r N,\end{aligned} \hspace{\stretch{1}}(1.5)

which gives

\begin{aligned}3 r N=\frac{2}{3} \frac{V}{4 \pi^2 } \left( { \frac{1}{b_{\mathrm{L}}^{3/2}} + \frac{2}{b_{\mathrm{T}}^{3/2}} } \right)\omega_{\mathrm{D}}^{3/2}=\frac{V}{6 \pi^2 } \left( { \frac{1}{b_{\mathrm{L}}^{3/2}} + \frac{2}{b_{\mathrm{T}}^{3/2}} } \right)\omega_{\mathrm{D}}^{3/2}\end{aligned} \hspace{\stretch{1}}(1.6)

Part c

Assuming a Bose distribution and ignoring the zero point energy, which has no temperature dependence, the specific heat, the temperature derivative of the energy density, is

\begin{aligned}C_{\mathrm{V}} &= \frac{d}{d T} \frac{1}{{V}} \int Z(\omega) \frac{\hbar \omega}{ e^{\hbar \omega/ k_{\mathrm{B}} T } - 1} d\omega \\ &= \frac{1}{{V}} \frac{d}{d T} \int Z(\omega) \frac{\hbar \omega}{ \hbar \omega/ k_{\mathrm{B}} T + \frac{1}{{2}}( \hbar \omega/k_{\mathrm{B}} T)^2 + \cdots } d\omega \\ &\approx \frac{1}{{V}} \frac{d}{d T} \int Z(\omega) k_{\mathrm{B}} T d\omega \\ &= \frac{1}{{V}} k_{\mathrm{B}} 3 r N.\end{aligned} \hspace{\stretch{1}}(1.7)

Part d

First note that the density of states can be written

\begin{aligned}Z(\omega) = \frac{9 r N}{ 2 \omega_{\mathrm{D}}^{3/2} } \omega^{1/2},\end{aligned} \hspace{\stretch{1}}(1.8)

for a specific heat of

\begin{aligned}C_{\mathrm{V}} &= \frac{d}{d T} \frac{1}{{V}} \int_0^\infty \frac{9 r N}{ 2 \omega_{\mathrm{D}}^{3/2} } \omega^{1/2} \frac{\hbar \omega}{ e^{\hbar \omega/ k_{\mathrm{B}} T } - 1} d\omega \\ &= \frac{9 r N}{ 2 V \omega_{\mathrm{D}}^{3/2} } \int_0^\infty d\omega \omega^{1/2} \frac{d}{d T} \frac{\hbar \omega}{ e^{\hbar \omega/ k_{\mathrm{B}} T } - 1} \\ &= \frac{9 r N}{ 2 V \omega_{\mathrm{D}}^{3/2} } \int_0^\infty d\omega \omega^{1/2} \frac{-\hbar \omega}{ \left( {e^{\hbar \omega/ k_{\mathrm{B}} T } - 1} \right)^2 }  e^{\hbar \omega/k_{\mathrm{B}} T} \hbar \omega/k_{\mathrm{B}} \left( {-\frac{1}{{T^2}}} \right) \\ &= \frac{9 r N k_{\mathrm{B}} }{ 2 V \omega_{\mathrm{D}}^{3/2} } \left( { \frac{ k_{\mathrm{B}} T}{\hbar} } \right)^{3/2}\int_0^\infty d \frac{\hbar \omega}{k_{\mathrm{B}} T} \left( {\frac{\hbar \omega}{k_{\mathrm{B}} T}} \right)^{1/2} \frac{1}{ \left( {e^{\hbar \omega/ k_{\mathrm{B}} T } - 1} \right)^2 }  e^{\hbar \omega/k_{\mathrm{B}} T} \left( { \frac{\hbar \omega}{k_{\mathrm{B}} T} } \right)^2 \\ &= \frac{9 r N k_{\mathrm{B}} }{ 2 V \omega_{\mathrm{D}}^{3/2} } \left( { \frac{ k_{\mathrm{B}} T}{\hbar} } \right)^{3/2}\int_0^\infty dy \frac{y^{5/2} e^y }{ \left( {e^y - 1} \right)^2 } \\& = \frac{9 r N k_{\mathrm{B}} }{ 2 V } \left( { \frac{ T}{\Theta} } \right)^{3/2} \mathcal{I}.\end{aligned} \hspace{\stretch{1}}(1.9)

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One atom basis phonons in 2D

Posted by peeterjoot on January 19, 2014

[Click here for a PDF of this post with nicer formatting]

Let’s tackle a problem like the 2D problem of the final exam, but first more generally. Instead of a square lattice consider the lattice with the geometry illustrated in fig. 1.1.

Fig 1.1: Oblique one atom basis

Here, \mathbf{a} and \mathbf{b} are the vector differences between the equilibrium positions separating the masses along the K_1 and K_2 interaction directions respectively. The equilibrium spacing for the cross coupling harmonic forces are

\begin{aligned}\begin{aligned}\mathbf{r} &= (\mathbf{b} + \mathbf{a})/2 \\ \mathbf{s} &= (\mathbf{b} - \mathbf{a})/2.\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.1)

Based on previous calculations, we can write the equations of motion by inspection

\begin{aligned}\begin{aligned}m \dot{d}{\mathbf{u}}_\mathbf{n} = &-K_1 \text{Proj}_{\hat{\mathbf{a}}} \sum_\pm \left( { \mathbf{u}_\mathbf{n} - \mathbf{u}_{\mathbf{n} \pm(1, 0)}} \right)^2 \\ &-K_2 \text{Proj}_{\hat{\mathbf{b}}} \sum_\pm \left( { \mathbf{u}_\mathbf{n} - \mathbf{u}_{\mathbf{n} \pm(0, 1)}} \right)^2 \\ &-K_3 \text{Proj}_{\hat{\mathbf{r}}} \sum_\pm \left( { \mathbf{u}_\mathbf{n} - \mathbf{u}_{\mathbf{n} \pm(1, 1)}} \right)^2 \\ &-K_4 \text{Proj}_{\hat{\mathbf{s}}} \sum_\pm \left( { \mathbf{u}_\mathbf{n} - \mathbf{u}_{\mathbf{n} \pm(1, -1)}} \right)^2.\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.2)

Inserting the trial solution

\begin{aligned}\mathbf{u}_\mathbf{n} = \frac{1}{{\sqrt{m}}} \boldsymbol{\epsilon}(\mathbf{q}) e^{i( \mathbf{r}_\mathbf{n} \cdot \mathbf{q} - \omega t) },\end{aligned} \hspace{\stretch{1}}(1.3)

and using the matrix form for the projection operators, we have

\begin{aligned}\begin{aligned}\omega^2 \boldsymbol{\epsilon} &=\frac{K_1}{m} \hat{\mathbf{a}} \hat{\mathbf{a}}^\text{T} \boldsymbol{\epsilon}\sum_\pm\left( { 1 - e^{\pm i \mathbf{a} \cdot \mathbf{q}} } \right) \\ & +\frac{K_2}{m} \hat{\mathbf{b}} \hat{\mathbf{b}}^\text{T} \boldsymbol{\epsilon}\sum_\pm\left( { 1 - e^{\pm i \mathbf{b} \cdot \mathbf{q}} } \right) \\ & +\frac{K_3}{m} \hat{\mathbf{b}} \hat{\mathbf{b}}^\text{T} \boldsymbol{\epsilon}\sum_\pm\left( { 1 - e^{\pm i (\mathbf{b} + \mathbf{a}) \cdot \mathbf{q}} } \right) \\ & +\frac{K_3}{m} \hat{\mathbf{b}} \hat{\mathbf{b}}^\text{T} \boldsymbol{\epsilon}\sum_\pm\left( { 1 - e^{\pm i (\mathbf{b} - \mathbf{a}) \cdot \mathbf{q}} } \right) \\ &=\frac{4 K_1}{m} \hat{\mathbf{a}} \hat{\mathbf{a}}^\text{T} \boldsymbol{\epsilon} \sin^2\left( { \mathbf{a} \cdot \mathbf{q}/2 } \right)+\frac{4 K_2}{m} \hat{\mathbf{b}} \hat{\mathbf{b}}^\text{T} \boldsymbol{\epsilon} \sin^2\left( { \mathbf{b} \cdot \mathbf{q}/2 } \right) \\ &+\frac{4 K_3}{m} \hat{\mathbf{r}} \hat{\mathbf{r}}^\text{T} \boldsymbol{\epsilon} \sin^2\left( { (\mathbf{b} + \mathbf{a}) \cdot \mathbf{q}/2 } \right)+\frac{4 K_4}{m} \hat{\mathbf{s}} \hat{\mathbf{s}}^\text{T} \boldsymbol{\epsilon} \sin^2\left( { (\mathbf{b} - \mathbf{a}) \cdot \mathbf{q}/2 } \right).\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.4)

This fully specifies our eigenvalue problem. Writing

\begin{aligned}\begin{aligned}S_1 &= \sin^2\left( { \mathbf{a} \cdot \mathbf{q}/2 } \right) \\ S_2 &= \sin^2\left( { \mathbf{b} \cdot \mathbf{q}/2 } \right) \\ S_3 &= \sin^2\left( { (\mathbf{b} + \mathbf{a}) \cdot \mathbf{q}/2 } \right) \\ S_4 &= \sin^2\left( { (\mathbf{b} - \mathbf{a}) \cdot \mathbf{q}/2 } \right)\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.5.5)

\begin{aligned}\boxed{A = \frac{4}{m}\left( { K_1 S_1 \hat{\mathbf{a}} \hat{\mathbf{a}}^\text{T} + K_2 S_2 \hat{\mathbf{b}} \hat{\mathbf{b}}^\text{T} + K_3 S_3 \hat{\mathbf{r}} \hat{\mathbf{r}}^\text{T} + K_4 S_4 \hat{\mathbf{s}} \hat{\mathbf{s}}^\text{T}} \right),}\end{aligned} \hspace{\stretch{1}}(1.0.5.5)

we wish to solve

\begin{aligned}A \boldsymbol{\epsilon} = \omega^2 \boldsymbol{\epsilon} = \lambda \boldsymbol{\epsilon}.\end{aligned} \hspace{\stretch{1}}(1.0.6)

Neglecting the specifics of the matrix at hand, consider a generic two by two matrix

\begin{aligned}A = \begin{bmatrix}a & b \\ c & d\end{bmatrix},\end{aligned} \hspace{\stretch{1}}(1.0.6)

for which the characteristic equation is

\begin{aligned}0 &= \begin{vmatrix}\lambda - a & - b \\ -c & \lambda -d \end{vmatrix} \\ &= (\lambda - a)(\lambda - d) - b c \\ &= \lambda^2 - (a + d) \lambda + a d - b c \\ &= \lambda^2 - (Tr A) \lambda + \left\lvert {A} \right\rvert \\ &= \left( {\lambda - \frac{Tr A}{2}} \right)^2- \left( {\frac{Tr A}{2}} \right)^2 + \left\lvert {A} \right\rvert.\end{aligned} \hspace{\stretch{1}}(1.0.6)

So our angular frequencies are given by

\begin{aligned}\omega^2 = \frac{1}{{2}} \left( { Tr A \pm \sqrt{ \left(Tr A\right)^2 - 4 \left\lvert {A} \right\rvert }} \right).\end{aligned} \hspace{\stretch{1}}(1.0.6)

The square root can be simplified slightly

\begin{aligned}\left( {Tr A} \right)^2 - 4 \left\lvert {A} \right\rvert \\ &= (a + d)^2 -4 (a d - b c) \\ &= a^2 + d^2 + 2 a d - 4 a d + 4 b c \\ &= (a - d)^2 + 4 b c,\end{aligned} \hspace{\stretch{1}}(1.0.6)

so that, finally, the dispersion relation is

\begin{aligned}\boxed{\omega^2 = \frac{1}{{2}} \left( { d + a \pm \sqrt{ (d - a)^2 + 4 b c } } \right),}\end{aligned} \hspace{\stretch{1}}(1.0.6)

Our eigenvectors will be given by

\begin{aligned}0 = (\lambda - a) \boldsymbol{\epsilon}_1 - b\boldsymbol{\epsilon}_2,\end{aligned} \hspace{\stretch{1}}(1.0.6)

or

\begin{aligned}\boldsymbol{\epsilon}_1 \propto \frac{b}{\lambda - a}\boldsymbol{\epsilon}_2.\end{aligned} \hspace{\stretch{1}}(1.0.6)

So, our eigenvectors, the vectoral components of our atomic displacements, are

\begin{aligned}\boldsymbol{\epsilon} \propto\begin{bmatrix}b \\ \omega^2 - a\end{bmatrix},\end{aligned} \hspace{\stretch{1}}(1.0.6)

or

\begin{aligned}\boxed{\boldsymbol{\epsilon} \propto\begin{bmatrix}2 b \\ d - a \pm \sqrt{ (d - a)^2 + 4 b c }\end{bmatrix}.}\end{aligned} \hspace{\stretch{1}}(1.0.6)

Square lattice

There is not too much to gain by expanding out the projection operators explicitly in general. However, let’s do this for the specific case of a square lattice (as on the exam problem). In that case, our projection operators are

\begin{aligned}\hat{\mathbf{a}} \hat{\mathbf{a}}^\text{T} = \begin{bmatrix}1 \\ 0\end{bmatrix}\begin{bmatrix}1 & 0\end{bmatrix}=\begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(1.0.16a)

\begin{aligned}\hat{\mathbf{b}} \hat{\mathbf{b}}^\text{T} = \begin{bmatrix}0\\ 1 \end{bmatrix}\begin{bmatrix}0 &1 \end{bmatrix}=\begin{bmatrix}0 & 0 \\ 0 & 1\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(1.0.16b)

\begin{aligned}\hat{\mathbf{r}} \hat{\mathbf{r}}^\text{T} = \frac{1}{{2}}\begin{bmatrix}1 \\ 1 \end{bmatrix}\begin{bmatrix}1 &1 \end{bmatrix}=\frac{1}{{2}}\begin{bmatrix}1 & 1 \\ 1 & 1\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(1.0.16c)

\begin{aligned}\hat{\mathbf{s}} \hat{\mathbf{s}}^\text{T} = \frac{1}{{2}}\begin{bmatrix}-1 \\ 1 \end{bmatrix}\begin{bmatrix}-1 &1 \end{bmatrix}=\frac{1}{{2}}\begin{bmatrix}1 & -1 \\ -1 & 1\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(1.0.16d)

\begin{aligned}\begin{aligned}S_1 &= \sin^2\left( { \mathbf{a} \cdot \mathbf{q} } \right) \\ S_2 &= \sin^2\left( { \mathbf{b} \cdot \mathbf{q} } \right) \\ S_3 &= \sin^2\left( { (\mathbf{b} + \mathbf{a}) \cdot \mathbf{q} } \right) \\ S_4 &= \sin^2\left( { (\mathbf{b} - \mathbf{a}) \cdot \mathbf{q} } \right),\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.16d)

Our matrix is

\begin{aligned}A = \frac{2}{m}\begin{bmatrix}2 K_1 S_1 + K_3 S_3 + K_4 S_4 & K_3 S_3 - K_4 S_4 \\ K_3 S_3 - K_4 S_4 & 2 K_2 S_2 + K_3 S_3 + K_4 S_4\end{bmatrix},\end{aligned} \hspace{\stretch{1}}(1.0.16d)

where, specifically, the squared sines for this geometry are

\begin{aligned}S_1 = \sin^2 \left( { \mathbf{a} \cdot \mathbf{q}/2 } \right) = \sin^2 \left( { a q_x/2} \right)\end{aligned} \hspace{\stretch{1}}(1.0.19a)

\begin{aligned}S_2 = \sin^2 \left( { \mathbf{b} \cdot \mathbf{q}/2 } \right) = \sin^2 \left( { a q_y/2} \right)\end{aligned} \hspace{\stretch{1}}(1.0.19b)

\begin{aligned}S_3 = \sin^2 \left( { (\mathbf{b} + \mathbf{a}) \cdot \mathbf{q}/2 } \right) = \sin^2 \left( { a (q_x + q_y)/2} \right)\end{aligned} \hspace{\stretch{1}}(1.0.19c)

\begin{aligned}S_4 = \sin^2 \left( { (\mathbf{b} - \mathbf{a}) \cdot \mathbf{q}/2 } \right) = \sin^2 \left( { a (q_y - q_x)/2} \right).\end{aligned} \hspace{\stretch{1}}(1.0.19d)

Using eq. 1.0.6, the dispersion relation and eigenvectors are

\begin{aligned}\omega^2 = \frac{2}{m} \left( { \sum_i K_i S_i \pm \sqrt{ (K_2 S_2 - K_1 S_1)^2 + (K_3 S_3 - K_4 S_4)^2 } } \right)\end{aligned} \hspace{\stretch{1}}(1.0.20.20)

\begin{aligned}\boldsymbol{\epsilon} \propto\begin{bmatrix}K_3 S_3 - K_4 S_4 \\ K_2 S_2 - K_1 S_1 \pm \sqrt{ (K_2 S_2 - K_1 S_1)^2 + (K_3 S_3 - K_4 S_4)^2 } \end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(1.0.20.20)

This calculation is confirmed in oneAtomBasisPhononSquareLatticeEigensystem.nb. Mathematica calculates an alternate form (equivalent to using a zero dot product for the second row), of

\begin{aligned}\boldsymbol{\epsilon} \propto\begin{bmatrix}K_1 S_1 - K_2 S_2 \pm \sqrt{ (K_2 S_2 - K_1 S_1)^2 + (K_3 S_3 - K_4 S_4)^2 } \\ K_3 S_3 - K_4 S_4 \end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(1.0.20.20)

Either way, we see that K_3 S_3 - K_4 S_4 = 0 leads to only horizontal or vertical motion.

With the exam criteria

In the specific case that we had on the exam where K_1 = K_2 and K_3 = K_4, these are

\begin{aligned}\omega^2 = \frac{2}{m} \left( { K_1 (S_1 + S_2) + K_3(S_3 + S_4) \pm \sqrt{ K_1^2 (S_2 - S_1)^2 + K_3^2 (S_3 - S_4)^2 } } \right)\end{aligned} \hspace{\stretch{1}}(1.0.22.22)

\begin{aligned}\boldsymbol{\epsilon} \propto\begin{bmatrix}K_3 \left( { S_3 - S_4 } \right) \\ K_1 \left( { (S_1 - S_2) \pm \sqrt{ (S_2 - S_1)^2 + \left( \frac{K_3}{K_1} \right)^2 (S_3 - S_4)^2 } } \right)\end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(1.0.22.22)

For horizontal and vertical motion we need S_3 = S_4, or for a 2 \pi \times \text{integer} difference in the absolute values of the sine arguments

\begin{aligned}\pm ( a (q_x + q_y) /2 ) = a (q_y - q_y) /2 + 2 \pi n.\end{aligned} \hspace{\stretch{1}}(1.0.22.22)

That is, one of

\begin{aligned}\begin{aligned}q_x &= \frac{2 \pi}{a} n \\ q_y &= \frac{2 \pi}{a} n\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.22.22)

In the first BZ, that is one of q_x = 0 or q_y = 0.

System in rotated coordinates

On the exam, where we were asked to solve for motion along the cross directions explicitly, there was a strong hint to consider a rotated (by \pi/4) coordinate system.

The rotated the lattice basis vectors are \mathbf{a} = a \mathbf{e}_1, \mathbf{b} = a \mathbf{e}_2, and the projection matrices. Writing \hat{\mathbf{r}} = \mathbf{f}_1 and \hat{\mathbf{s}} = \mathbf{f}_2, where \mathbf{f}_1 = (\mathbf{e}_1 + \mathbf{e}_2)/\sqrt{2}, \mathbf{f}_2 = (\mathbf{e}_2 - \mathbf{e}_1)/\sqrt{2}, or \mathbf{e}_1 = (\mathbf{f}_1 - \mathbf{f}_2)/\sqrt{2}, \mathbf{e}_2 = (\mathbf{f}_1 + \mathbf{f}_2)/\sqrt{2}. In the \{\mathbf{f}_1, \mathbf{f}_2\} basis the projection matrices are

\begin{aligned}\hat{\mathbf{a}} \hat{\mathbf{a}}^\text{T} = \frac{1}{{2}}\begin{bmatrix}1 \\ -1\end{bmatrix}\begin{bmatrix}1 & -1\end{bmatrix}= \frac{1}{{2}} \begin{bmatrix}1 & -1 \\ -1 & 1\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(1.0.25a)

\begin{aligned}\hat{\mathbf{b}} \hat{\mathbf{b}}^\text{T} = \frac{1}{{2}}\begin{bmatrix}1 \\ 1\end{bmatrix}\begin{bmatrix}1 & 1\end{bmatrix}= \frac{1}{{2}} \begin{bmatrix}1 & 1 \\ 1 & 1\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(1.0.25b)

\begin{aligned}\hat{\mathbf{r}} \hat{\mathbf{r}}^\text{T} = \begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(1.0.25c)

\begin{aligned}\hat{\mathbf{s}} \hat{\mathbf{s}}^\text{T} = \begin{bmatrix}0 & 0 \\ 0 & 1\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(1.0.25d)

The dot products that show up in the squared sines are

\begin{aligned}\mathbf{a} \cdot \mathbf{q}=a \frac{1}{{\sqrt{2}}} (\mathbf{f}_1 - \mathbf{f}_2) \cdot (\mathbf{f}_1 k_u + \mathbf{f}_2 k_v)=\frac{a}{\sqrt{2}} (k_u - k_v)\end{aligned} \hspace{\stretch{1}}(1.0.26a)

\begin{aligned}\mathbf{b} \cdot \mathbf{q}=a \frac{1}{{\sqrt{2}}} (\mathbf{f}_1 + \mathbf{f}_2) \cdot (\mathbf{f}_1 k_u + \mathbf{f}_2 k_v)=\frac{a}{\sqrt{2}} (k_u + k_v)\end{aligned} \hspace{\stretch{1}}(1.0.26b)

\begin{aligned}(\mathbf{a} + \mathbf{b}) \cdot \mathbf{q} = \sqrt{2} a k_u \end{aligned} \hspace{\stretch{1}}(1.0.26c)

\begin{aligned}(\mathbf{b} - \mathbf{a}) \cdot \mathbf{q} = \sqrt{2} a k_v \end{aligned} \hspace{\stretch{1}}(1.0.26d)

So that in this basis

\begin{aligned}\begin{aligned}S_1 &= \sin^2 \left( { \frac{a}{\sqrt{2}} (k_u - k_v) } \right) \\ S_2 &= \sin^2 \left( { \frac{a}{\sqrt{2}} (k_u + k_v) } \right) \\ S_3 &= \sin^2 \left( { \sqrt{2} a k_u } \right) \\ S_4 &= \sin^2 \left( { \sqrt{2} a k_v } \right)\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.26d)

With the rotated projection operators eq. 1.0.5.5 takes the form

\begin{aligned}A = \frac{2}{m}\begin{bmatrix}K_1 S_1 + K_2 S_2 + 2 K_3 S_3 & K_2 S_2 - K_1 S_1 \\ K_2 S_2 - K_1 S_1 & K_1 S_1 + K_2 S_2 + 2 K_4 S_4\end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(1.0.26d)

This clearly differs from eq. 1.0.16d, and results in a different expression for the eigenvectors, but the same as eq. 1.0.20.20 for the angular frequencies.

\begin{aligned}\boldsymbol{\epsilon} \propto\begin{bmatrix}K_2 S_2 - K_1 S_1 \\ K_4 S_4 - K_3 S_3 \mp \sqrt{ (K_2 S_2 - K_1 S_1)^2 + (K_3 S_3 - K_4 S_4)^2 }\end{bmatrix},\end{aligned} \hspace{\stretch{1}}(1.0.26d)

or, equivalently

\begin{aligned}\boldsymbol{\epsilon} \propto\begin{bmatrix}K_4 S_4 - K_3 S_3 \mp \sqrt{ (K_2 S_2 - K_1 S_1)^2 + (K_3 S_3 - K_4 S_4)^2 } \\ K_1 S_1 - K_2 S_2 \\ \end{bmatrix},\end{aligned} \hspace{\stretch{1}}(1.0.26d)

For the K_1 = K_2 and K_3 = K_4 case of the exam, this is

\begin{aligned}\boldsymbol{\epsilon} \propto\begin{bmatrix}K_1 (S_2 - S_1 ) \\ K_3 \left( { S_4 - S_3 \mp \sqrt{ \left( \frac{K_1}{K_3} \right)^2 (S_2 - S_1)^2 + (S_3 - S_4)^2 } } \right)\end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(1.0.26d)

Similar to the horizontal coordinate system, we see that we have motion along the diagonals when

\begin{aligned}\pm \frac{a}{\sqrt{2}} (k_u - k_v) = \frac{a}{\sqrt{2}} (k_u + k_v) + 2 \pi n,\end{aligned} \hspace{\stretch{1}}(1.0.26d)

or one of

\begin{aligned}\begin{aligned}k_u &= \sqrt{2} \frac{\pi}{a} n \\ k_v &= \sqrt{2} \frac{\pi}{a} n\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.26d)

Stability?

The exam asked why the cross coupling is required for stability. Clearly we have more complex interaction. The constant \omega surfaces will also be more complex. However, I still don’t have a good intuition what exactly was sought after for that part of the question.

Numerical computations

A Manipulate allowing for choice of the spring constants and lattice orientation, as shown in fig. 1.2, is available in phy487/oneAtomBasisPhonon.nb. This interface also provides a numerical calculation of the distribution relation as shown in fig. 1.3, and provides an animation of the normal modes for any given selection of \mathbf{q} and \omega(\mathbf{q}) (not shown).

Fig 1.2: 2D Single atom basis Manipulate interface

Fig 1.3: Sample distribution relation for 2D single atom basis.

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Two body harmonic oscillator in 3D, and 2D diamond lattice vibrations

Posted by peeterjoot on January 7, 2014

Abridged harmonic oscillator notes

[This is an abbreviation of more extensive PDF notes associated with the latter part of this post.]

Motivation and summary of harmonic oscillator background

After having had some trouble on a non-1D harmonic oscillator lattice problem on the exam, I attempted such a problem with enough time available to consider it properly. I found it helpful to consider first just two masses interacting harmonically in 3D, each displaced from an equilibrium position.

The Lagrangian that described this most naturally was found to be

\begin{aligned}\mathcal{L} = \frac{1}{2} m_1 \left( \dot{\mathbf{r}}_1 \right)^2+\frac{1}{2} m_2 \left( \dot{\mathbf{r}}_2 \right)^2- \frac{K}{2} \left( \left\lvert {\mathbf{r}_2 - \mathbf{r}_1} \right\rvert - a \right)^2.\end{aligned} \hspace{\stretch{1}}(2.1)

This was solved in absolute and displacement coordinates, and then I moved on to consider a linear expansion of the harmonic potential about the equilibrium point, a problem closer to the exam problem (albeit still considering only two masses). The equilibrium points were described with vectors \mathbf{a}_1, \mathbf{a}_2 as in fig. 2.1, where \Delta \mathbf{a} = \left\lvert {\Delta \mathbf{a}} \right\rvert (\cos \theta_1, \cos\theta_2, \cos\theta_3).

fig 2.1: Direction cosines relative to equilibrium position difference vector

 

Using such coordinates, and generalizing, it was found that the complete Lagrangian, to second order about the equilibrium positions, is

\begin{aligned}\mathcal{L} = \sum_j \frac{m_i}{2} \dot{u}_{ij}^2 -\frac{K}{2} \sum_{i j} \cos\theta_i \cos\theta_j \left( u_{2 i} - u_{1 i} \right)\left( u_{2 j} - u_{1 j} \right).\end{aligned} \hspace{\stretch{1}}(2.2)

Evaluating the Euler-Lagrange equations, the equations of motion for the displacements were found to be

\begin{aligned}\begin{aligned}m_1 \ddot{\mathbf{u}}_1 &= K \widehat{\Delta \mathbf{a}} \left( \widehat{\Delta \mathbf{a}} \cdot \Delta \mathbf{u} \right) \\ m_2 \ddot{\mathbf{u}}_2 &= -K \widehat{\Delta \mathbf{a}} \left( \widehat{\Delta \mathbf{a}} \cdot \Delta \mathbf{u} \right),\end{aligned}\end{aligned} \hspace{\stretch{1}}(2.3)

or

\begin{aligned}\boxed{\begin{aligned}\mu \Delta \ddot{\mathbf{u}} &= -K \widehat{\Delta \mathbf{a}} \left( \widehat{\Delta \mathbf{a}} \cdot \Delta \mathbf{u} \right) \\ m_1 \ddot{\mathbf{u}}_1 + m_2 \ddot{\mathbf{u}}_2 &= 0.\end{aligned}}\end{aligned} \hspace{\stretch{1}}(2.4)

Observe that on the RHS above we have a projection operator, so we could also write

\begin{aligned}\mu \Delta \ddot{\mathbf{u}} = -K \text{Proj}_{\widehat{\Delta \mathbf{a}}} \Delta \mathbf{u}.\end{aligned} \hspace{\stretch{1}}(2.5)

We see that the equations of motion for the displacements of a system of harmonic oscillators has a rather pleasant expression in terms of projection operators, where we have projections onto the unit vectors between each pair of equilibrium position.

A number of harmonically coupled masses

Now let’s consider masses at lattice points indexed by a lattice vector \mathbf{n}, as illustrated in fig. 2.2.

fig 2.2: Masses harmonically coupled in a lattice

 

With a coupling constant of K_{\mathbf{n} \mathbf{m}} between lattice points indexed \mathbf{n} and \mathbf{m} (located at \mathbf{a}_\mathbf{n} and \mathbf{a}_\mathbf{m} respectively), and direction cosines for the equilibrium direction vector between those points given by

\begin{aligned}\mathbf{a}_\mathbf{n} - \mathbf{a}_\mathbf{m} = \Delta \mathbf{a}_{\mathbf{n} \mathbf{m}}= \left\lvert {\Delta \mathbf{a}_{\mathbf{n} \mathbf{m}}} \right\rvert (\cos \theta_{\mathbf{n} \mathbf{m} 1},\cos \theta_{\mathbf{n} \mathbf{m} 2},\cos \theta_{\mathbf{n} \mathbf{m} 3}),\end{aligned} \hspace{\stretch{1}}(2.6)

the Lagrangian is

\begin{aligned}\mathcal{L} = \sum_{\mathbf{n}, i} \frac{m_\mathbf{n}}{2} \dot{u}_{\mathbf{n} i}^2-\frac{1}{2} \sum_{\mathbf{n} \ne \mathbf{m}, i, j} \frac{K_{\mathbf{n} \mathbf{m}}}{2} \cos\theta_{\mathbf{n} \mathbf{m} i}\cos\theta_{\mathbf{n} \mathbf{m} j}\left( u_{\mathbf{n} i} - u_{\mathbf{m} i} \right)\left( u_{\mathbf{n} j} - u_{\mathbf{m} j} \right)\end{aligned} \hspace{\stretch{1}}(2.7)

Evaluating the Euler-Lagrange equations for the mass at index \mathbf{n} we have

\begin{aligned}\frac{d}{dt} \frac{\partial {\mathcal{L}}}{\partial {\dot{u}_{\mathbf{n} k}}} =m_\mathbf{n} \ddot{u}_{\mathbf{n} k},\end{aligned} \hspace{\stretch{1}}(2.8)

and

\begin{aligned}\frac{\partial {\mathcal{L}}}{\partial {u_{\mathbf{n} k}}} &= -\sum_{\mathbf{m}, i, j}\frac{K_{\mathbf{n} \mathbf{m}}}{2} \cos\theta_{\mathbf{n} \mathbf{m} i}\cos\theta_{\mathbf{n} \mathbf{m} j}\left(\delta_{i k}\left( u_{\mathbf{n} j} - u_{\mathbf{m} j} \right)+\left( u_{\mathbf{n} i} - u_{\mathbf{m} i} \right)\delta_{j k}\right) \\ &= -\sum_{\mathbf{m}, i}K_{\mathbf{n} \mathbf{m}}\cos\theta_{\mathbf{n} \mathbf{m} k}\cos\theta_{\mathbf{n} \mathbf{m} i}\left( u_{\mathbf{n} i} - u_{\mathbf{m} i} \right) \\ &= -\sum_{\mathbf{m}}K_{\mathbf{n} \mathbf{m}}\cos\theta_{\mathbf{n} \mathbf{m} k}\widehat{\Delta \mathbf{a}} \cdot \Delta \mathbf{u}_{\mathbf{n} \mathbf{m}},\end{aligned} \hspace{\stretch{1}}(2.9)

where \Delta \mathbf{u}_{\mathbf{n} \mathbf{m}} = \mathbf{u}_\mathbf{n} - \mathbf{u}_\mathbf{m}. Equating both, we have in vector form

\begin{aligned}m_\mathbf{n} \ddot{\mathbf{u}}_\mathbf{n} = -\sum_{\mathbf{m}}K_{\mathbf{n} \mathbf{m}}\widehat{\Delta \mathbf{a}}\left( \widehat{\Delta \mathbf{a}} \cdot \Delta \mathbf{u}_{\mathbf{n} \mathbf{m}} \right),\end{aligned} \hspace{\stretch{1}}(2.10)

or

\begin{aligned}\boxed{m_\mathbf{n} \ddot{\mathbf{u}}_\mathbf{n} = -\sum_{\mathbf{m}}K_{\mathbf{n} \mathbf{m}}\text{Proj}_{ \widehat{\Delta \mathbf{a}} } \Delta \mathbf{u}_{\mathbf{n} \mathbf{m}},}\end{aligned} \hspace{\stretch{1}}(2.11)

This is an intuitively pleasing result. We have displacement and the direction of the lattice separations in the mix, but not the magnitude of the lattice separation itself.

Two atom basis, 2D diamond lattice

As a concrete application of the previously calculated equilibrium harmonic oscillator result, let’s consider a two atom basis diamond lattice where the horizontal length is a and vertical height is b.

Indexing for the primitive unit cells is illustrated in fig. 2.3.

fig 2.3: Primitive unit cells for rectangular lattice

 

Let’s write

\begin{aligned}\begin{aligned}\mathbf{r} &= a (\cos\theta, \sin\theta) = a \hat{\mathbf{r}} \\ \mathbf{s} &= a (-\cos\theta, \sin\theta) = a \hat{\mathbf{s}} \\ \mathbf{n} &= (n_1, n_2) \\ \mathbf{r}_\mathbf{n} &= n_1 \mathbf{r} + n_2 \mathbf{s},\end{aligned}\end{aligned} \hspace{\stretch{1}}(2.12)

For mass m_\alpha, \alpha \in \{1, 2\} assume a trial solution of the form

\begin{aligned}\mathbf{u}_{\mathbf{n},\alpha} = \frac{\boldsymbol{\epsilon}_\alpha(\mathbf{q})}{\sqrt{m_\alpha}} e^{i \mathbf{r}_n \cdot \mathbf{q} - \omega t}.\end{aligned} \hspace{\stretch{1}}(2.13)

The equations of motion for the two particles are

\begin{aligned}\begin{aligned}m_1 \ddot{\mathbf{u}}_{\mathbf{n}, 1} &= - K_1 \text{Proj}_{\hat{\mathbf{x}}} \left(  \mathbf{u}_{\mathbf{n}, 1} - \mathbf{u}_{\mathbf{n} - (0,1), 2}  \right)- K_1 \text{Proj}_{\hat{\mathbf{x}}} \left(  \mathbf{u}_{\mathbf{n}, 1} - \mathbf{u}_{\mathbf{n} - (1,0), 2}  \right) \\  & \quad- K_2 \text{Proj}_{\hat{\mathbf{y}}} \left(  \mathbf{u}_{\mathbf{n}, 1} - \mathbf{u}_{\mathbf{n}, 2}  \right)- K_2 \text{Proj}_{\hat{\mathbf{y}}} \left(  \mathbf{u}_{\mathbf{n}, 1} - \mathbf{u}_{\mathbf{n} - (1,1), 2}  \right) \\  & \quad- K_3 \sum_\pm\text{Proj}_{\hat{\mathbf{r}}} \left(  \mathbf{u}_{\mathbf{n}, 1} - \mathbf{u}_{\mathbf{n} \pm (1,0), 1}  \right)- K_4 \sum_\pm\text{Proj}_{\hat{\mathbf{s}}} \left(  \mathbf{u}_{\mathbf{n}, 1} - \mathbf{u}_{\mathbf{n} \pm (0,1), 1}  \right)\end{aligned}\end{aligned} \hspace{\stretch{1}}(2.0.14.14)

\begin{aligned}\begin{aligned}m_2 \ddot{\mathbf{u}}_{\mathbf{n}, 2} &= - K_1 \text{Proj}_{\hat{\mathbf{x}}} \left(  \mathbf{u}_{\mathbf{n}, 2} - \mathbf{u}_{\mathbf{n} + (1,0), 1}  \right)- K_1 \text{Proj}_{\hat{\mathbf{x}}} \left(  \mathbf{u}_{\mathbf{n}, 2} - \mathbf{u}_{\mathbf{n} + (0,1), 1}  \right)\\  &\quad- K_2 \text{Proj}_{\hat{\mathbf{y}}} \left(  \mathbf{u}_{\mathbf{n}, 2} - \mathbf{u}_{\mathbf{n}, 1}  \right)- K_2 \text{Proj}_{\hat{\mathbf{y}}} \left(  \mathbf{u}_{\mathbf{n}, 2} - \mathbf{u}_{\mathbf{n} + (1,1), 1}  \right)\\  &\quad- K_3 \sum_\pm\text{Proj}_{\hat{\mathbf{r}}} \left(  \mathbf{u}_{\mathbf{n}, 2} - \mathbf{u}_{\mathbf{n} \pm (1,0), 2}  \right)- K_4 \sum_\pm\text{Proj}_{\hat{\mathbf{s}}} \left(  \mathbf{u}_{\mathbf{n}, 2} - \mathbf{u}_{\mathbf{n} \pm (0,1), 2}  \right)\end{aligned}\end{aligned} \hspace{\stretch{1}}(2.0.14.14)

Insertion of the trial solution gives

\begin{aligned}\begin{aligned} \omega^2 \sqrt{m_1} \boldsymbol{\epsilon}_1&= K_1 \text{Proj}_{\hat{\mathbf{x}}} \left(  \frac{\boldsymbol{\epsilon}_1}{\sqrt{m_1}} - \frac{\boldsymbol{\epsilon}_2}{\sqrt{m_2}} e^{ - i \mathbf{s} \cdot \mathbf{q} }  \right)+ K_1 \text{Proj}_{\hat{\mathbf{x}}} \left(  \frac{\boldsymbol{\epsilon}_1}{\sqrt{m_1}} - \frac{\boldsymbol{\epsilon}_2}{\sqrt{m_2}} e^{ - i \mathbf{r} \cdot \mathbf{q} }  \right) \\  &\quad+ K_2 \text{Proj}_{\hat{\mathbf{y}}} \left(  \frac{\boldsymbol{\epsilon}_1}{\sqrt{m_1}} - \frac{\boldsymbol{\epsilon}_2}{\sqrt{m_2}}  \right)+ K_2 \text{Proj}_{\hat{\mathbf{y}}} \left(  \frac{\boldsymbol{\epsilon}_1}{\sqrt{m_1}} - \frac{\boldsymbol{\epsilon}_2}{\sqrt{m_2}} e^{ - i (\mathbf{r} + \mathbf{s}) \cdot \mathbf{q} }  \right) \\  &\quad+ K_3 \left(  \text{Proj}_{\hat{\mathbf{r}}} \frac{\boldsymbol{\epsilon}_1}{\sqrt{m_1}}  \right)\sum_\pm\left(  1 - e^{ \pm i \mathbf{r} \cdot \mathbf{q} }  \right)+ K_4 \left(  \text{Proj}_{\hat{\mathbf{s}}} \frac{\boldsymbol{\epsilon}_1}{\sqrt{m_1}}  \right)\sum_\pm\left(  1 - e^{ \pm i \mathbf{s} \cdot \mathbf{q} }  \right)\end{aligned}\end{aligned} \hspace{\stretch{1}}(2.0.15.15)

\begin{aligned}\begin{aligned}\omega^2 \sqrt{m_2} \boldsymbol{\epsilon}_2&=K_1 \text{Proj}_{\hat{\mathbf{x}}} \left(  \frac{\boldsymbol{\epsilon}_2}{\sqrt{m_2}} - \frac{\boldsymbol{\epsilon}_1}{\sqrt{m_1}} e^{ + i \mathbf{r} \cdot \mathbf{q} }  \right)+ K_1 \text{Proj}_{\hat{\mathbf{x}}} \left(  \frac{\boldsymbol{\epsilon}_2}{\sqrt{m_2}} - \frac{\boldsymbol{\epsilon}_1}{\sqrt{m_1}} e^{ + i \mathbf{s} \cdot \mathbf{q} }  \right)\\  &\quad+ K_2 \text{Proj}_{\hat{\mathbf{y}}} \left(  \frac{\boldsymbol{\epsilon}_2}{\sqrt{m_2}} - \frac{\boldsymbol{\epsilon}_1}{\sqrt{m_1}}  \right)+ K_2 \text{Proj}_{\hat{\mathbf{y}}} \left(  \frac{\boldsymbol{\epsilon}_2}{\sqrt{m_2}} - \frac{\boldsymbol{\epsilon}_1}{\sqrt{m_1}} e^{ + i (\mathbf{r} + \mathbf{s}) \cdot \mathbf{q} }  \right) \\  &\quad+ K_3 \left(  \text{Proj}_{\hat{\mathbf{r}}} \frac{\boldsymbol{\epsilon}_2}{\sqrt{m_2}}  \right)\sum_\pm\left(  1 - e^{ \pm i \mathbf{r} \cdot \mathbf{q} }  \right)+ K_4 \left(  \text{Proj}_{\hat{\mathbf{s}}} \frac{\boldsymbol{\epsilon}_2}{\sqrt{m_2}}  \right)\sum_\pm\left(  1 - e^{ \pm i \mathbf{s} \cdot \mathbf{q} }  \right)\end{aligned}\end{aligned} \hspace{\stretch{1}}(2.0.15.15)

Regrouping, and using the matrix form \text{Proj}_{\hat{\mathbf{u}}} = \hat{\mathbf{u}} \hat{\mathbf{u}}^\text{T} for the projection operators, this is

\begin{aligned}\left(\omega^2 - \frac{2}{m_1} \left(  K_1 \hat{\mathbf{x}} \hat{\mathbf{x}}^T + K_2 \hat{\mathbf{y}} \hat{\mathbf{y}}^T + 2 K_3 \hat{\mathbf{r}} \hat{\mathbf{r}}^T \sin^2 (\mathbf{r} \cdot \mathbf{q}/2) + 2 K_4 \hat{\mathbf{s}} \hat{\mathbf{s}}^T \sin^2 (\mathbf{s} \cdot \mathbf{q}/2)  \right)\right)\boldsymbol{\epsilon}_1 = -\left(  K_1 \hat{\mathbf{r}} \hat{\mathbf{r}}^\text{T} \left(  e^{ - i \mathbf{s} \cdot \mathbf{q} } + e^{ - i \mathbf{r} \cdot \mathbf{q} }  \right) + K_2 \hat{\mathbf{s}} \hat{\mathbf{s}}^\text{T} \left(  1 + e^{ - i (\mathbf{r} + \mathbf{s}) \cdot \mathbf{q} }  \right)  \right)\frac{\boldsymbol{\epsilon}_2}{\sqrt{ m_1 m_2 }}\end{aligned} \hspace{\stretch{1}}(2.0.16.16)

\begin{aligned}\left( \omega^2 - \frac{2}{m_2} \left(  K_1 \hat{\mathbf{x}} \hat{\mathbf{x}}^T + K_2 \hat{\mathbf{y}} \hat{\mathbf{y}}^T + 2 K_3 \hat{\mathbf{r}} \hat{\mathbf{r}}^T \sin^2 (\mathbf{r} \cdot \mathbf{q}/2)+ 2 K_4 \hat{\mathbf{s}} \hat{\mathbf{s}}^T \sin^2 (\mathbf{s} \cdot \mathbf{q}/2) \right) \right)\boldsymbol{\epsilon}_2 = -\left( K_1    \hat{\mathbf{r}} \hat{\mathbf{r}}^\text{T}   \left(    e^{ i \mathbf{s} \cdot \mathbf{q} }   +   e^{ i \mathbf{r} \cdot \mathbf{q} }    \right) +   K_2    \hat{\mathbf{s}} \hat{\mathbf{s}}^\text{T}   \left(    1   +   e^{ i (\mathbf{r} + \mathbf{s}) \cdot \mathbf{q} }    \right) \right)\frac{\boldsymbol{\epsilon}_1}{\sqrt{ m_1 m_2 }}\end{aligned} \hspace{\stretch{1}}(2.0.16.16)

As a single matrix equation, this is

\begin{aligned}A = K_1 \hat{\mathbf{x}} \hat{\mathbf{x}}^T + K_2 \hat{\mathbf{y}} \hat{\mathbf{y}}^T + 2 K_3 \hat{\mathbf{r}} \hat{\mathbf{r}}^T \sin^2 (\mathbf{r} \cdot \mathbf{q}/2)+ 2 K_4 \hat{\mathbf{s}} \hat{\mathbf{s}}^T \sin^2 (\mathbf{s} \cdot \mathbf{q}/2)\end{aligned} \hspace{\stretch{1}}(2.0.17.17)

\begin{aligned}B = e^{ i (\mathbf{r} + \mathbf{s}) \cdot \mathbf{q}/2 }\left( {   K_1    \hat{\mathbf{r}} \hat{\mathbf{r}}^\text{T}   \cos\left(  (\mathbf{r} - \mathbf{s}) \cdot \mathbf{q}/2  \right)+   K_2    \hat{\mathbf{s}} \hat{\mathbf{s}}^\text{T}   \cos\left(  (\mathbf{r} + \mathbf{s}) \cdot \mathbf{q}/2  \right)} \right)\end{aligned} \hspace{\stretch{1}}(2.0.17.17)

\begin{aligned}0 =\begin{bmatrix}\omega^2 - \frac{2 A}{m_1} & \frac{B^{*}}{\sqrt{m_1 m_2}} \\ \frac{B}{\sqrt{m_1 m_2}} & \omega^2 - \frac{2 A}{m_2} \end{bmatrix}\begin{bmatrix}\boldsymbol{\epsilon}_1 \\ \boldsymbol{\epsilon}_2\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(2.0.17.17)

Observe that this is an eigenvalue problem E \mathbf{e} = \omega^2 \mathbf{e} for matrix

\begin{aligned}E = \begin{bmatrix}\frac{2 A}{m_1} & -\frac{B^{*}}{\sqrt{m_1 m_2}} \\ -\frac{B}{\sqrt{m_1 m_2}} & \frac{2 A}{m_2} \end{bmatrix},\end{aligned} \hspace{\stretch{1}}(2.0.17.17)

and eigenvalues \omega^2.

To be explicit lets put the A and B functions in explicit matrix form. The orthogonal projectors have a simple form

\begin{aligned}\text{Proj}_{\hat{\mathbf{x}}} = \hat{\mathbf{x}} \hat{\mathbf{x}}^\text{T}= \begin{bmatrix}1 \\ 0\end{bmatrix}\begin{bmatrix}1 & 0\end{bmatrix}=\begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(2.0.19a)

\begin{aligned}\text{Proj}_{\hat{\mathbf{y}}} = \hat{\mathbf{y}} \hat{\mathbf{y}}^\text{T}= \begin{bmatrix}0 \\ 1\end{bmatrix}\begin{bmatrix}0 & 1\end{bmatrix}=\begin{bmatrix}0 & 0 \\ 0 & 1\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(2.0.19b)

For the \hat{\mathbf{r}} and \hat{\mathbf{s}} projection operators, we can use half angle formulations

\begin{aligned}\text{Proj}_{\hat{\mathbf{r}}} = \hat{\mathbf{r}} \hat{\mathbf{r}}^\text{T}= \begin{bmatrix}\cos\theta \\ \sin\theta\end{bmatrix}\begin{bmatrix}\cos\theta & \sin\theta\end{bmatrix}=\begin{bmatrix}\cos^2\theta & \cos\theta \sin\theta \\ \cos\theta \sin\theta & \sin^2 \theta\end{bmatrix}=\frac{1}{2}\begin{bmatrix}1 + \cos \left(  2 \theta  \right) & \sin \left(  2 \theta  \right) \\ \sin \left(  2 \theta  \right) & 1 - \cos \left(  2 \theta  \right)\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(2.0.20.20)

\begin{aligned}\text{Proj}_{\hat{\mathbf{s}}} = \hat{\mathbf{s}} \hat{\mathbf{s}}^\text{T}= \begin{bmatrix}-\cos\theta \\ \sin\theta\end{bmatrix}\begin{bmatrix}-\cos\theta & \sin\theta\end{bmatrix}=\begin{bmatrix}\cos^2\theta & -\cos\theta \sin\theta \\ -\cos\theta \sin\theta & \sin^2 \theta\end{bmatrix}=\frac{1}{2}\begin{bmatrix}1 + \cos \left(  2 \theta  \right) & -\sin \left(  2 \theta  \right) \\ -\sin \left(  2 \theta  \right) & 1 - \cos \left(  2 \theta  \right)\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(2.0.20.20)

After some manipulation, and the following helper functions

\begin{aligned}\begin{aligned}\alpha_\pm &= K_3 \sin^2 (\mathbf{r} \cdot \mathbf{q}/2) \pm K_4 \sin^2 (\mathbf{s} \cdot \mathbf{q}/2) \\ \beta_\pm &= K_1 \cos\left(  (\mathbf{r} - \mathbf{s}) \cdot \mathbf{q}/2  \right) \pm K_2 \cos\left(  (\mathbf{r} + \mathbf{s}) \cdot \mathbf{q}/2  \right),\end{aligned}\end{aligned} \hspace{\stretch{1}}(2.0.20.20)

the block matrices of eq. 2.0.17.17 take the form

\begin{aligned}A = \begin{bmatrix}K_1 + \alpha_+ (1 + \cos\left(  2 \theta  \right)) & \alpha_- \sin\left(  2 \theta  \right) \\ \alpha_- \sin\left(  2 \theta  \right) & K_2 + \alpha_+ (1 - \cos\left(  2 \theta  \right))\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(2.0.22.22)

\begin{aligned}B =    e^{ i (\mathbf{r} + \mathbf{s}) \cdot \mathbf{q}/2 }\begin{bmatrix}   \beta_+ (1 + \cos \left(  2 \theta  \right)) & \beta_- \sin \left(  2 \theta  \right) \\    \beta_- \sin \left(  2 \theta  \right) & \beta_+( 1 -\cos \left(  2 \theta  \right))\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(2.0.22.22)

A final bit of simplification for B possible, noting that \mathbf{r} + \mathbf{s} = 2 a (0, \sin\theta ), and \mathbf{r} - \mathbf{s} = 2 a(\cos\theta, 0), so

\begin{aligned}\beta_\pm = K_1 \cos\left(  a \cos\theta q_x  \right) \pm K_2 \cos\left(  a \sin\theta q_y  \right),\end{aligned} \hspace{\stretch{1}}(2.0.22.22)

and

\begin{aligned}B =    e^{ i a \sin\theta q_y }\begin{bmatrix}   \beta_+ (1 + \cos \left(  2 \theta  \right)) & \beta_- \sin \left(  2 \theta  \right) \\    \beta_- \sin \left(  2 \theta  \right) & \beta_+( 1 -\cos \left(  2 \theta  \right))\end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(2.0.22.22)

It isn’t particularly illuminating to expand out the determinant for such a system, even though it can be done symbolically without too much programming. However, what is easy after formulating the matrix for this system, is actually solving it. This is done, and animated, in twoAtomBasisRectangularLatticeDispersionRelation.cdf

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(Markham) Markville Mall Sony store: an interesting variation of warranty fraud.

Posted by peeterjoot on December 18, 2013

I mailed the following response to the Markville mall’s Sony store, where one of their employees attempted a new variety of warranty fraud:

I discovered when I attempted to exchange the headphones that you completely misrepresented yourself when I made the purchase. The item that you scanned showed up at a lower price than the sticker price. Instead of offering me a chance to buy that item at that price you padded the price back up to the sticker price (or rather close to it, short a few dollars), by adding in an extended warranty.

You did this despite the fact that I said I would never voluntarily purchase one of these extended warranties. Since you portrayed this as something that was “for free”, I did not object. However, you completely missed your chance at honesty, because I should have been offered the sale price. You stated that you lowered the price “so that you could offer me the warranty without costing me anything”, however, that lowered price was already the listed sale price in your system. This was something that you did not disclose.

That is, in my opinion, undeniably fraud.

Since you were so careful to make sure that both you and your colleague were represented on the bill, I can only assume that you are on commission. It would be very hard to argue that this was not a blatant attempt to pad your commission, at my expense. I shudder to consider how many other customers have been exploited in this way.

I’ll never shop at the Sony Store again. You have lost my patronage, and I’ll not hesitate to tell anybody who is considering a Sony Store purchase to be very careful at your store, to avoid this new interesting variation of warranty fraud.

The mechanism of the fraud attempt used here was that I was sold a pair of headphones that were on sale, but the salesman padded the price up to the sticker price including a “free” extended warranty.  He blatantly told me that he was reducing the price for me so that he could include the extended warranty for free.  In the end this made it appear that I got $5 off the sticker price, and got an extended warranty to boot.

I only discovered this because as well as attempting to defraud me, they also gave me the wrong headphones (I’d asked for a noise cancelling model).  I had not yet noticed this, however, the Sony Store now provides a service (a rather nice innovation) where they will provide you with a soft copy of your receipt if you provide them your email address.  Because of this, they had my contact info to proactively inform me about the incorrect boxe of headphones that I’d been given.  When I attempted the exchange, it was at a point when the staff member who did this transaction was not there, so I was able to discover what actually occurred.

At the point of return, I was offered a reduction in price for the warranty should I desire it, but it still would have meant paying for it.  This new “cheaper warranty” that I never wanted in the first place would still have cost me (not saving me money in a too good to be true fashion as it originally appeared), so I turned that down.  I then discovered that I’d also have to pay more for the correct headphones.  It was only $5 more, but by this time I was completely fed up.  Reflecting now, I was also very annoyed at myself for having fallen for this trick.  I just returned them completely.

It is always interesting to learn of new fraud techniques. This is a new one that I had not seen before. Taking advantage of an unlisted sales price to sell additional undesired content, so that commissions could be padded. Because the sales price was on their system, the salesman did not require any management approval to override the system with a lower price, and was able to make it appear that he had “lowered the price” for me so that I could get something for free.  It’s actually very clever.

Could this have been an honest mistake?  I doubt it very much.

EDIT:

The Sony salesman who I had dealt with contacted me after this, stating:

I am sorry for whatever misunderstanding happened yesterday.  Can I please call you & explain you the situation & see how can I make your experience with us more conferrable.  Please let me know on what phone number I can reach you on. I am sorry once again & I believe you will give me a chance to explain & I will do my best to solve this issue.

My response was that he needed to resolve that with his management, not me. If that was done, then I’d be willing to talk to them (not him).

They (management through him) eventually offered me a deal on the item that I’d returned. It wasn’t really my intent to be fishing for a deal, and I’d continued shopping after all this for an alternate set of phones to buy from somewhere else. However, the timing and the offer were both good since I hadn’t yet found a replacement item I was happy with, and I ended up accepting that offer.

Picking up the set they’d set aside for me provided a chance to talk to his manager, who hadn’t been given the complete story. Despite that, after talking to the manager, I’m no longer certain that the salesman understood exactly why I objected to the transaction. This may not have been an instance of fraud as it initially seemed to be, instead it could have been a blunder by a fairly new staff member, confused by the wrong price showing up after the scan, and tried to “fix it” in a way that he naively thought would be satisfactory. I don’t think I’ll ever know for sure.

Posted in Incoherent ramblings | Tagged: , , , | 1 Comment »

Pre-final-exam update of notes for PHY487 (condensed matter physics)

Posted by peeterjoot on December 2, 2013

Here’s an update of my class notes from Winter 2013, University of Toronto Condensed Matter Physics course (PHY487H1F), taught by Prof. Stephen Julian.  This includes notes for all the examinable lectures (i.e. excluding superconductivity).  I’ll post at least one more update later, probably after the exam, including notes from the final lecture, and my problem set 10 solution.

NOTE: This v.4 update of these notes is still really big (~18M).  Some of my mathematica generated 3d images appear to result in very large pdfs.

Changelog for this update (relative to the the first, and second, and third Changelogs) :

December 02, 2013 Lecture 23, Superconductivity

December 01, 2013 Lecture 22, Intro to semiconductor physics

December 01, 2013 Lecture 21, Electron-phonon scattering

November 26, 2013 Problem Set 9, Electron band structure, density of states, and effective mass

November 22, 2013 Lecture 20, Electric current (cont.)

November 20, 2013 Problem Set 8, Tight Binding.

November 18, 2013 Lecture 19, Electrical transport (cont.)

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Signs, signs, everywhere there’s signs. Do this, don’t do that, can’t you read the sign.

Posted by peeterjoot on November 28, 2013

Here’s a sign that was recently installed in one of the men’s washrooms at work

IMG_1032

and a close up of it

IMG_1033

Yes, there is a set of men’s showers in the X1 location of the building, which is consistent with the graphic of the shower in the sign.  I couldn’t, however, for the life of me, figure out why somebody who was at the washroom sink would have to be reminded that the lab has shower facilities.  Nor could I figure out why we now had a sign that appeared to be instructing those at the sink to pray in the shower.

I’ve pointed it out this sign to a few people now when I was at the sink, but nobody else appeared to be able to decipher it.  Sofia, who is more wise in the ways of the world, told me what this is about: part of the Islamic prayer ritual involves washing one’s feet.

Sure enough, I remember that there was a guy who used to wash his feet in the bathroom sink in the summer.  He always wore sandals, and I thought his washing was because he thought his feet smelled (I wondered why he didn’t switch to socks and shoes if that was the case, and now feel silly for not just asking him).

So, it seems that this sign was likely commissioned and installed just for this one individual.  Somebody objected to his feet washing.  That objector was probably also completely ignorant like me, and likely didn’t know that this was part of his religious ritual.  I’d guess that foot washing objector still doesn’t know that.

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“fun” bug in 64-bit perl 5.10.0 bigint sprintf

Posted by peeterjoot on November 19, 2013

Here’s a rather unexpected bug with perl sprintf

#! /usr/bin/perl

use strict;
use warnings;
use bigint ;

my $a = hex( "0x0A0000001D05A820" ) ;
printf( "0x%016X\n", $a ) ;
printf( "%d\n", $a ) ;
printf( "$a\n" ) ;

The %X printf produces a value where the least significant 0x20 is lost:

$ ./bigint
0x0A0000001D05A800
720575940866189312
720575940866189344

Observe that the loss occurs in the printf and not the hex() call, since 720575940866189344 == 0x0A0000001D05A820.

This bug appears to be fixed in some version of perl <= 5.16.2. Oh, the joys of using ancient operating system versions so that we can support customers on many of the ancient deployments that they seem to like to run on.

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second update of phy487 notes (Condensed Matter Physics)

Posted by peeterjoot on November 16, 2013

Here’s an update of my (incomplete) lecture notes for the Winter 2013, University of Toronto Condensed Matter Physics course (PHY487H1F), taught by Prof. Stephen Julian. This makes updates to these notes since the first, and second versions posted.

NOTE: This v.3 update of these notes is still really big (~16M).  Some of my mathematica generated 3d images appear to result in very large pdfs.

This set of notes includes the following these additions (not many of which were posted separately for this course)

November 15, 2013 Semiconductors

November 13, 2013 Thomas-Fermi screening, and nearly free electron model

November 11, 2013 3 dimensional band structures, Fermi surfaces of real metals

November 05, 2013 Fourier coefficient integral for periodic function

November 04, 2013 Tight binding model

November 04, 2013 Fermi properties, free electron specific heat, bulk modulus

November 01, 2013 Nearly free electron model, periodic potential (cont.)

October 29, 2013 Free electron model of metals

October 28, 2013 Electrons in a periodic lattice

October 23, 2013 Huygens diffraction

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Political correctness crap in IBM HR circles

Posted by peeterjoot on November 12, 2013

Apparently IBM has fired so many people that they have invented (or started using) the term RA (Resource Action) instead of fired, used like so:

Bob was RAed.

There has been another round of RAs.

If you don’t want to be RAed, make sure your PBC (personal business committment) document is filled in and sounds good, since there’s a new earnings report due out tomorrow.

This is supposed to sound nicer, but I think it’s the opposite.  To me this sounds like the individual is now a resource, and is be moved around like an entry in some accounting table.  RA means that, unfortunately, they ended up in the delete column.  It’s very impersonal.  Perhaps this is just to make the firing manager feel better, since that manager cannot do anything about the firing quotas when they occur.

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Remembering cultural insanity and propagandization

Posted by peeterjoot on November 11, 2013

It’s Remembrance day, and we are once again barraged with the heroism of war and sacrifice

warProp1

We get the cuddly manly shots of veterans posing on the field

warProp2

these last two pictures taken from the Markham community paper, where we have our poor veteran Stan, who recalls the “guts blood and luck at Juno Beach” quoted

we got to the beach, and there were dead people all over … You didn’t stop to pick someone up to even [to sic] check if they were still alive.  You just kept moving, otherwise you would have been shot yourself.

I surely sympathize for Stan.  This is truly horrifying, and must have been traumatic.  It’s something that you’d want to forget, and would haunt you for the rest of your life.  However, our unfortunate Stan was sold a narrative.

We must fight the evil villain.

warProp5

People are dying and it is our duty to protect.

warProp6

Kids are doing their part

warProp4

Women are doing their part

warProp8

It is brave and honourable to do your part

warProp3

Even the whisky makers are doing their part!

warProp7

The list of selling points goes on and on.   “Be ashamed of your fear of death.  It’s the right thing to do. … “  It was a vicious and evil sales pitch.  This is a narrative that was backed by hordes of propaganda.

A lot of profit was made by this war.  This is true of all war.

In a sane world, what would we be remembering?  Follow the money.  We should remember those that profited from the war.  We should remember those that directly or indirectly sold armaments to both sides.  Most importantly, we should remember those that bankrolled the war on all sides.  Stan was not on the winning side.  He was on the loosing side.  The winning side resides in corporate and banking boardrooms.  This is what we need to remember whenever the drums of war start beating.

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