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Curvilinear coordinates and reciprocal basis

Posted by Peeter Joot on March 9, 2014

Motivation

Here I’d like to explore some ideas from [1] where curvilinear coordinates, manifolds, and the vector derivative are introduced.

Notation

For simplicity, let’s consider the concrete example of a 2D manifold, a surface in an $n$ dimensional vector space, parameterized by two variables

\begin{aligned}\mathbf{x} = \mathbf{x}(a,b) = \mathbf{x}(u^1, u^2).\end{aligned} \hspace{\stretch{1}}(1.2.1)

Note that the indices here do not represent exponentiation. We can construct a basis for the manifold as

\begin{aligned}\mathbf{x}_i = \frac{\partial {\mathbf{x}}}{\partial {u^i}}.\end{aligned} \hspace{\stretch{1}}(1.2.2)

On the manifold we can calculate a reciprocal basis $\{\mathbf{x}^i\}$, defined by requiring, at each point on the surface

\begin{aligned}\mathbf{x}^i \cdot \mathbf{x}_j = {\delta^i}_j.\end{aligned} \hspace{\stretch{1}}(1.2.3)

Associated implicitly with this basis is a curvilinear coordinate representation defined by the projection operation

\begin{aligned}\mathbf{x} = x^i \mathbf{x}_i,\end{aligned} \hspace{\stretch{1}}(1.2.4)

(sums over mixed indexes are implied). These coordinates can be calculated by taking dot products with the reciprocal frame vectors

\begin{aligned}\mathbf{x} \cdot \mathbf{x}^i &= x^j \mathbf{x}_j \cdot \mathbf{x}^i \\ &= x^j {\delta_j}^i \\ &= x^i.\end{aligned} \hspace{\stretch{1}}(1.2.4)

Examples

Let’s pause for a couple examples that have interesting aspects.

Example: Circular coordinates on a disk

Consider an infinite disk at height $z_0$, with the origin omitted, parameterized by circular coordinates as in fig. 1.1.

Fig 1.1: Plane with circular coordinates

Points on this surface are

\begin{aligned}\mathbf{x}(r, \theta) = (r \cos\theta, r \sin\theta, z_0).\end{aligned} \hspace{\stretch{1}}(1.3.6)

The manifold basis vectors, defined by eq. 1.2.2 are

\begin{aligned}\begin{aligned}\mathbf{x}_r &= (\cos\theta, \sin\theta, 0) \\ \mathbf{x}_\theta &= r (-\sin\theta, \cos\theta, 0).\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.3.7)

By inspection, the reciprocal basis is

\begin{aligned}\begin{aligned}\mathbf{x}^r &= (\cos\theta, \sin\theta, 0) \\ \mathbf{x}^\theta &= \frac{1}{{r}} (-\sin\theta, \cos\theta, 0).\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.3.8)

The first thing to note here is that we cannot reach the points $\mathbf{x}$ of eq. 1.3.6 by linear combination of these basis vectors. Instead these basis vectors only allow us to reach other points on the surface, when already there. For example we cannot actually write

\begin{aligned}\mathbf{x} = x^r \mathbf{x}_r + x^\theta \mathbf{x}_\theta,\end{aligned} \hspace{\stretch{1}}(1.3.9)

unless $z_0 = 0$. This is why eq. 1.2.4 was described as a projective operation (and probably deserves an alternate notation). To recover the original parameterized form of the position vector on the surface, we require

\begin{aligned}\mathbf{x} = x^r \mathbf{x}_r + x^\theta \mathbf{x}_\theta + z_0 \hat{\mathbf{z}}.\end{aligned} \hspace{\stretch{1}}(1.3.10)

The coordinates $x^r, x^\theta$ follow by taking dot products

\begin{aligned}x^r &= \mathbf{x} \cdot \mathbf{x}^r \\ &= (r \cos\theta, r \sin\theta, z_0) \cdot(\cos\theta, \sin\theta, 0) \\ &= r \left( \cos^2 \theta + \sin^2 \theta \right) \\ &= r\end{aligned} \hspace{\stretch{1}}(1.0.11.11)

\begin{aligned}x^\theta &= \mathbf{x} \cdot \mathbf{x}^\theta \\ &= (r \cos\theta, r \sin\theta, z_0) \cdot\frac{1}{{r}} (-\sin\theta, \cos\theta, 0) \\ &= 0.\end{aligned} \hspace{\stretch{1}}(1.0.11.11)

Therefore, a point on the plane, relative to the origin of the plane, in this case, requires just one of the tangent plane basis vectors

\begin{aligned}\mathbf{x} = r \mathbf{x}_r.\end{aligned} \hspace{\stretch{1}}(1.0.11.11)

Example: Circumference of a circle

Now consider a circular perimeter, as illustrated in fig. 1.2, with the single variable parameterization

Fig 1.2: Circular perimeter

\begin{aligned}\mathbf{x} = r_0 \left( \cos\theta, \sin\theta \right).\end{aligned} \hspace{\stretch{1}}(1.13)

Our tangent space basis is

\begin{aligned}\mathbf{x}_\theta = r_0 \left( -\sin\theta, \cos\theta \right),\end{aligned} \hspace{\stretch{1}}(1.14)

with, by inspection, a reciprocal basis

\begin{aligned}\mathbf{x}^\theta = \frac{1}{{r_0}} \left( -\sin\theta, \cos\theta \right).\end{aligned} \hspace{\stretch{1}}(1.15)

Here we have a curious condition, since the tangent space basis vector is perpendicular to the position vector for the points on the circular surface. So, should we attempt to calculate coordinates using eq. 1.2.4, we just get zero

\begin{aligned}x^\theta &= \mathbf{x} \cdot \mathbf{x}^\theta \\ &= r_0 \left( \cos\theta, \sin\theta \right) \cdot\frac{1}{{r_0}} \left( -\sin\theta, \cos\theta \right) \\ &= 0.\end{aligned} \hspace{\stretch{1}}(1.16)

It’s perhaps notable that a coordinate representation using the tangent space basis is possible, but we need to utilize a complex geometry. Assuming

\begin{aligned}\mathbf{x} = x^\theta \mathbf{x}_\theta,\end{aligned} \hspace{\stretch{1}}(1.17)

and writing $i = \mathbf{e}_1 \mathbf{e}_2$ for the pseudoscalar, we can write

\begin{aligned}\begin{aligned}\mathbf{x} &= r_0 \mathbf{e}_1 e^{i\theta} \\ \mathbf{x}_\theta &= r_0 \mathbf{e}_2 e^{i\theta},\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.18)

so that, by inversion, the $\theta$ coordinate is

\begin{aligned}x^\theta &= \mathbf{x}\frac{1}{{\mathbf{x}_\theta}} \\ &= \left( r_0 \mathbf{e}_1 e^{i\theta} \right)\left( \frac{ e^{-i\theta } \mathbf{e}_2 }{r_0} \right) \\ &= i,\end{aligned} \hspace{\stretch{1}}(1.19)

or

\begin{aligned}\mathbf{x} = i \mathbf{x}_\theta.\end{aligned} \hspace{\stretch{1}}(1.0.20)

Example: Surface of a sphere

It is also clear that any parameterization that has radial symmetry will suffer the same issue. For example, for a radial surface in 3D with radius $r_0$ we have

\begin{aligned}\begin{aligned}\mathbf{x} &= r_0 \left( \sin\theta \cos\phi, \sin\theta \sin\phi, \cos\theta \right) \\ \mathbf{x}_\theta &= r_0 \left( \cos\theta \cos\phi, \cos\theta \sin\phi, -\sin\theta \right) \\ \mathbf{x}_\phi &= r_0 \left( -\sin\theta \sin\phi, \sin\theta \cos\phi, 0 \right) \\ \mathbf{x}^\theta &= \frac{1}{{r_0}} \left( \cos\theta \cos\phi, \cos\theta \sin\phi, -\sin\theta \right) \\ \mathbf{x}^\phi &= \frac{1}{{r_0 \sin\theta}} \left( -\sin\phi, \cos\phi, 0 \right).\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.21)

The reciprocals here were computed using the mathematica reciprocalFrameSphericalSurface.nb notebook.

Do we have a bivector parameterization of the surface using the tangent space basis? Let’s try

\begin{aligned}\mathbf{x} = x^\theta \mathbf{x}_\theta + x^\phi \mathbf{x}_\phi.\end{aligned} \hspace{\stretch{1}}(1.0.22)

Wedging with $\mathbf{x}_\theta$ and $\mathbf{x}_\phi$, and writing $i = \mathbf{e}_1 \mathbf{e}_2$, respectively yields

\begin{aligned}x^\theta &= \mathbf{x} \wedge \mathbf{x}_\phi \frac{1}{{\mathbf{x}_\theta \wedge \mathbf{x}_\phi}} \\ &= -\mathbf{e}_1 \mathbf{e}_3 \cos \phi - \mathbf{e}_2 \mathbf{e}_3 \sin \phi \\ &= \mathbf{e}_{31} e^{ i \phi}.\end{aligned} \hspace{\stretch{1}}(1.0.22)

\begin{aligned}x^\phi &= -\mathbf{x} \wedge \mathbf{x}_\theta \frac{1}{{\mathbf{x}_\theta \wedge \mathbf{x}_\phi}} \\ &= \mathbf{e}_1 \mathbf{e}_3 \cot \theta \sin \phi + i -\mathbf{e}_2 \mathbf{e}_3 \cot \theta \cos \phi \\ &= \mathbf{e}_2 \mathbf{e}_3 \cot \theta e^{i \phi} + i.\end{aligned} \hspace{\stretch{1}}(1.0.22)

However, substitution back into eq. 1.0.22 shows either pair parameterizes the radial position vector

\begin{aligned}\mathbf{x} = x^\theta \mathbf{x}_\theta = x^\phi \mathbf{x}_\phi.\end{aligned} \hspace{\stretch{1}}(1.0.25)

It is interesting that duality relationships seem to naturally arise attempting to describe points on a surface using the tangent space basis for that surface.

References

[1] A. Macdonald. Vector and Geometric Calculus. CreateSpace Independent Publishing Platform, 2012.

Nuland’s recent state dept Ukraine speech

Posted by Peeter Joot on February 21, 2014

Here is a presstv article, referring to a State department speech on Ukraine by Nuland.

In the spirit of Andrew Gavin Marshall’s podcasts, I would love to see a full translation of this speech into English from Statelish.  I imagine that the key to such a translation it would be along the following lines:

 democratic state ; modern democratic government state subservient to the USA free democracies democracies not subservient to the USA return to economic health economics subservient to the USA coordinated parallel high level diplomacy an active attempt to undermine existing government ; financing and manufacturing subversive and violent elements a tough conversation with Yanacovich I showed him lots of the blackmail material we have collected on him.  Made him realize how short his life would be if he doesn’t cooperate.  Threatened military and financial warfare. de-escalate the security situation step down so that we can install a puppet government, preferably one even more violent get Ukraine back into a conversation with Europe and with the International Monetary Fund we will fuck you over if you take on debt that will allow Russia to control you instead of taking on our debt and controls reforms that the IMF insists on are necessary for the long term economic health of the country we plan to fuck your kids and all their future progeny too foreign investment needed US exploitation is strongly desired

Gaussian quadratic form integrals and multivariable approximation of exponential integrals

Posted by Peeter Joot on January 26, 2014

[Click here for a PDF of this post with nicer formatting (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

Motivation

In [1] eq. I.2.20 is the approximation

\begin{aligned}\int d\mathbf{q} e^{-f(\mathbf{q})/\hbar} \approx e^{-f(\mathbf{a})/\hbar} \sqrt{\frac{2 \pi \hbar}{\text{Det} f''(\mathbf{a})} },\end{aligned} \hspace{\stretch{1}}(1.1.1)

where $[ f''(\mathbf{a}) ]_{ij} \equiv {\left.{{\partial^2 f/\partial q_i \partial q_j}}\right\vert}_{{\mathbf{q} = \mathbf{a}}}$. Here $\mathbf{a}$ is assumed to be an extremum of $f$. This follows from a generalization of the Gaussian integral result. Let’s derive both in detail.

Guts

First, to second order, let’s expand $f(\mathbf{q})$ around a min or max at $\mathbf{q} = \mathbf{a}$. The usual trick, presuming that one doesn’t remember the form of this generalized Taylor expansion, is to expand $g(t) = f(\mathbf{a} + t \mathbf{q})$ around $t = 0$, then evaluate at $t = 1$. We have

\begin{aligned}g'(t) &= \sum_i \frac{\partial {f(\mathbf{a} + t \mathbf{q})}}{\partial {(a_i + t q_i)}} \frac{d{{ (a_i + t q_i) }}}{dt} \\ &= \sum_i q_i \frac{\partial {f(\mathbf{a} + t \mathbf{q})}}{\partial {(a_i + t q_i)}} \\ &= \mathbf{q} \cdot \left( {\left.{{\boldsymbol{\nabla}_\mathbf{q} f(\mathbf{q})}}\right\vert}_{{\mathbf{q} = \mathbf{a} + t \mathbf{q}}} \right).\end{aligned} \hspace{\stretch{1}}(1.2.2)

The second derivative is

\begin{aligned}g''(t) = \sum_{i j} q_i q_j \frac{\partial {}}{\partial {(a_j + t q_j)}} \frac{\partial {f(\mathbf{a} + t \mathbf{q})}}{\partial {(a_i + t q_i)}},\end{aligned} \hspace{\stretch{1}}(1.2.3)

This gives

\begin{aligned}\begin{aligned}g'(0) &= \mathbf{q} \cdot \boldsymbol{\nabla}_\mathbf{q} f(\mathbf{q}) = \sum_i q_i \partial q_i f(\mathbf{q}) \\ g''(0) &= \left( \mathbf{q} \cdot \boldsymbol{\nabla}_\mathbf{q} \right)^2 f(\mathbf{q}) = \sum_{i j} q_i q_j \partial q_i \partial q_j f(\mathbf{q}).\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.2.4)

Putting these together, we have to second order in $t$ is

\begin{aligned}f(\mathbf{a} + t \mathbf{q}) \approx f(\mathbf{a}) + \sum_i q_i \partial q_i f(\mathbf{q}) \frac{t^1}{1!}+ \sum_{i j} q_i q_j \partial q_i \partial q_j f(\mathbf{q}) \frac{t^2}{2!},\end{aligned} \hspace{\stretch{1}}(1.2.5)

or

\begin{aligned}f(\mathbf{a} + \mathbf{q}) \approx f(\mathbf{a}) + \sum_i q_i {\left.{{ \left( \frac{\partial {f}}{\partial {q_i}} \right)}}\right\vert}_{\mathbf{a}}+ \frac{1}{{2}} \sum_{i j} q_i q_j {\left.{{\left( \frac{\partial^2 f}{\partial q_i \partial q_j} \right)}}\right\vert}_{\mathbf{a}}.\end{aligned} \hspace{\stretch{1}}(1.2.6)

We can put the terms up to second order in a nice tidy matrix forms

\begin{aligned}\mathbf{b} = {\left.{{\left( \boldsymbol{\nabla}_\mathbf{q} f \right)}}\right\vert}_{\mathbf{a}}\end{aligned} \hspace{\stretch{1}}(1.0.7a)

\begin{aligned}A = {\begin{bmatrix}{\left.{{ \left( \frac{\partial^2 f}{\partial q_i \partial q_j} \right)}}\right\vert}_{\mathbf{a}}\end{bmatrix}}_{i j}.\end{aligned} \hspace{\stretch{1}}(1.0.7b)

Note that eq. 1.0.7b is a real symmetric matrix, and can thus be reduced to diagonal form by an orthonormal transformation. Putting the pieces together, we have

\begin{aligned}f(\mathbf{a} + \mathbf{q}) \approx f(\mathbf{a}) + \mathbf{q}^\text{T} \mathbf{b} + \frac{1}{{2}} \mathbf{q}^\text{T} A \mathbf{q}.\end{aligned} \hspace{\stretch{1}}(1.0.8)

Integrating this, we have

\begin{aligned}\int dq_1 dq_2 \cdots dq_N \exp\left( -\left( f(\mathbf{a}) + \mathbf{q}^\text{T} \mathbf{b} + \frac{1}{{2}} \mathbf{q}^\text{T} A \mathbf{q} \right)\right)=e^{-f(\mathbf{a})}\int dq_1 dq_2 \cdots dq_N \exp\left( -\mathbf{q}^\text{T} \mathbf{b} - \frac{1}{{2}} \mathbf{q}^\text{T} A \mathbf{q} \right).\end{aligned} \hspace{\stretch{1}}(1.0.9)

Employing an orthonormal change of variables to diagonalizae the matrix

\begin{aligned}A = O^\text{T} D O,\end{aligned} \hspace{\stretch{1}}(1.0.10)

and $\mathbf{r} = O \mathbf{q}$, or $r_i = O_{ik} q_k$, the volume element after transformation is

\begin{aligned}dr_1 dr_2 \cdots dr_N &= \frac{\partial(r_1, r_2, \cdots, r_N)}{\partial(q_1, q_2, \cdots, q_N)}dq_1 dq_2 \cdots dq_N \\ &= \begin{vmatrix}O_{11} & O_{12} & \cdots & O_{1N} \\ O_{21} & O_{22} & \cdots & O_{2N} \\ \dot{v}s & \dot{v}s & \dot{v}s & \dot{v}s \\ O_{N1} & O_{N2} & \cdots & O_{NN} \\ \end{vmatrix}dq_1 dq_2 \cdots dq_N \\ &= (\text{Det} O)dq_1 dq_2 \cdots dq_N \\ &= dq_1 dq_2 \cdots dq_N \end{aligned} \hspace{\stretch{1}}(1.0.10)

Our integral is

\begin{aligned}e^{-f(\mathbf{a})}\int dq_1 dq_2 \cdots dq_N \exp\left( -\mathbf{q}^\text{T} \mathbf{b} - \frac{1}{{2}} \mathbf{q}^\text{T} A \mathbf{q} \right) &= e^{-f(\mathbf{a})}\int dr_1 dr_2 \cdots dr_N \exp\left( -\mathbf{q}^\text{T} O^\text{T} O \mathbf{b} - \frac{1}{{2}} \mathbf{q}^\text{T} O^\text{T} D O \mathbf{q} \right) \\ &= e^{-f(\mathbf{a})}\int dr_1 dr_2 \cdots dr_N \exp\left( -\mathbf{r}^\text{T} (O \mathbf{b}) - \frac{1}{{2}} \mathbf{r}^\text{T} D \mathbf{r} \right) \\ &= e^{-f(\mathbf{a})}\int dr_1 e^{ -\frac{1}{{2}} r_1^2 \lambda_1 - r_1 (O \mathbf{b})_1 }\int dr_2 e^{ -\frac{1}{{2}} r_2^2 \lambda_2 - r_2 (O \mathbf{b})_2 }\cdots \int dr_N e^{ -\frac{1}{{2}} r_N^2 \lambda_N - r_N (O \mathbf{b})_N }.\end{aligned} \hspace{\stretch{1}}(1.0.10)

We now have products of terms that are of the regular Gaussian form. One such integral is

\begin{aligned}\int e^{-a x^2/2 + J x} &= \int \exp\left(-\frac{1}{{2}} \left(\left( \sqrt{a} x - J/\sqrt{a} \right)^2- \left( J/\sqrt{a} \right)^2\right)\right) \\ &= e^{J^2/2a} \sqrt{2 \pi \int_0^\infty r dr e^{-a r^2/2}}\end{aligned} \hspace{\stretch{1}}(1.0.10)

This is just

\begin{aligned}\int e^{-a x^2/2 + J x}= e^{J^2/2a} \sqrt{ \frac{2 \pi}{a} }.\end{aligned} \hspace{\stretch{1}}(1.0.10)

Applying this to the integral of interest, writing $m_i = (O \mathbf{b})_i$

\begin{aligned}\begin{aligned}e^{-f(\mathbf{a})}\int &dq_1 dq_2 \cdots dq_N \exp\left( -\mathbf{q}^\text{T} \mathbf{b} - \frac{1}{{2}} \mathbf{q}^\text{T} A \mathbf{q} \right) \\ &=e^{-f(\mathbf{a})}e^{-m_1^2/2\lambda_1} \sqrt{ \frac{2 \pi}{\lambda_1}}e^{-m_2^2/2\lambda_2} \sqrt{ \frac{2 \pi}{\lambda_2}}\cdots e^{-m_N^2/2\lambda_N} \sqrt{ \frac{2 \pi}{\lambda_N}} \\ &=e^{-f(\mathbf{a})}\sqrt{\frac{2 \pi}{\text{Det} A}}\exp\left(-\frac{1}{{2}}\left( -m_1^2/\lambda_1 -m_2^2/\lambda_2 \cdots -m_N^2/\lambda_N \right) \right).\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.10)

This last exponential argument can be put into matrix form

\begin{aligned}-m_1^2/\lambda_1-m_2^2/\lambda_2\cdots -m_N^2/\lambda_N &= (O \mathbf{b})^\text{T} D^{-1} O \mathbf{b} \\ &= \mathbf{b}^\text{T} O^\text{T} D^{-1} O \mathbf{b} \\ &= \mathbf{b}^\text{T} A^{-1} \mathbf{b},\end{aligned} \hspace{\stretch{1}}(1.0.10)

Finally, referring back to eq. 1.0.7, we have

\begin{aligned}\int d\mathbf{q} e^{-f(\mathbf{q})} \approx e^{-f(\mathbf{a})}\sqrt{\frac{2 \pi}{\text{Det} A}}e^{-\mathbf{b}^\text{T} A^{-1} \mathbf{b}/2}.\end{aligned} \hspace{\stretch{1}}(1.0.10)

Observe that we can recover eq. 1.1.1 by noting that $\mathbf{b} = 0$ for that system was assumed (i.e. $\mathbf{a}$ was an extremum point), and by noting that the determinant scales with $1/\hbar$ since it just contains the second partials.

An afterword on notational sugar:

We didn’t need it, but it seems worth noting that we can write the Taylor expansion of eq. 1.0.8 in operator form as

\begin{aligned}f(\mathbf{a} + \mathbf{q}) = \sum_{k = 0}^\infty \frac{1}{{k!}} {\left.{{ \left( \mathbf{q} \cdot \boldsymbol{\nabla}_{\mathbf{q}'} \right)^k f(\mathbf{q}') }}\right\vert}_{{\mathbf{q}' = \mathbf{a}}}={\left.{{ e^{\mathbf{q} \cdot \boldsymbol{\nabla}_{\mathbf{q}'}} f(\mathbf{q}') }}\right\vert}_{{\mathbf{q}' = \mathbf{a}}}.\end{aligned} \hspace{\stretch{1}}(1.0.18)

References

[1] A. Zee. Quantum field theory in a nutshell. Universities Press, 2005.

Polarization angles for normal transmission and reflection

Posted by Peeter Joot on January 22, 2014

Question: Polarization angles for normal transmission and reflection ([1] pr 9.14)

For normal incidence, without assuming that the reflected and transmitted waves have the same polarization as the incident wave, prove that this must be so.

Working with coordinates as illustrated in fig. 1.1, the incident wave can be assumed to have the form

fig 1.1: Normal incidence coordinates

\begin{aligned}\tilde{\mathbf{E}}_{\mathrm{I}} = E_{\mathrm{I}} e^{i (k z - \omega t)} \hat{\mathbf{x}}\end{aligned} \hspace{\stretch{1}}(1.0.1a)

\begin{aligned}\tilde{\mathbf{B}}_{\mathrm{I}} = \frac{1}{{v}} \hat{\mathbf{z}} \times \tilde{\mathbf{E}}_{\mathrm{I}} = \frac{1}{{v}} E_{\mathrm{I}} e^{i (k z - \omega t)} \hat{\mathbf{y}}.\end{aligned} \hspace{\stretch{1}}(1.0.1b)

Assuming a polarization $\hat{\mathbf{n}} = \cos\theta \hat{\mathbf{x}} + \sin\theta \hat{\mathbf{y}}$ for the reflected wave, we have

\begin{aligned}\tilde{\mathbf{E}}_{\mathrm{R}} = E_{\mathrm{R}} e^{i (-k z - \omega t)} (\hat{\mathbf{x}} \cos\theta + \hat{\mathbf{y}} \sin\theta)\end{aligned} \hspace{\stretch{1}}(1.0.2a)

\begin{aligned}\tilde{\mathbf{B}}_{\mathrm{R}} = \frac{1}{{v}} (-\hat{\mathbf{z}}) \times \tilde{\mathbf{E}}_{\mathrm{R}} = \frac{1}{{v}} E_{\mathrm{R}} e^{i (-k z - \omega t)} (\hat{\mathbf{x}} \sin\theta - \hat{\mathbf{y}} \cos\theta).\end{aligned} \hspace{\stretch{1}}(1.0.2b)

And finally assuming a polarization $\hat{\mathbf{n}} = \cos\phi \hat{\mathbf{x}} + \sin\phi \hat{\mathbf{y}}$ for the transmitted wave, we have

\begin{aligned}\tilde{\mathbf{E}}_{\mathrm{T}} = E_{\mathrm{T}} e^{i (k' z - \omega t)} (\hat{\mathbf{x}} \cos\phi + \hat{\mathbf{y}} \sin\phi)\end{aligned} \hspace{\stretch{1}}(1.0.3a)

\begin{aligned}\tilde{\mathbf{B}}_{\mathrm{T}} = \frac{1}{{v}} \hat{\mathbf{z}} \times \tilde{\mathbf{E}}_{\mathrm{T}} = \frac{1}{{v'}} E_{\mathrm{T}} e^{i (k' z - \omega t)} (-\hat{\mathbf{x}} \sin\phi + \hat{\mathbf{y}} \cos\phi).\end{aligned} \hspace{\stretch{1}}(1.0.3b)

With no components of any of the $\tilde{\mathbf{E}}$ or $\tilde{\mathbf{B}}$ waves in the $\hat{\mathbf{z}}$ directions the boundary value conditions at $z = 0$ require the equality of the $\hat{\mathbf{x}}$ and $\hat{\mathbf{y}}$ components of

\begin{aligned}\left( \tilde{\mathbf{E}}_{\mathrm{I}} + \tilde{\mathbf{E}}_{\mathrm{R}} \right)_{x,y} = \left( \tilde{\mathbf{E}}_{\mathrm{T}} \right)_{x,y}\end{aligned} \hspace{\stretch{1}}(1.0.4a)

\begin{aligned} \left( \frac{1}{\mu} \left( \tilde{\mathbf{B}}_{\mathrm{I}} + \tilde{\mathbf{B}}_{\mathrm{R}} \right) \right)_{x,y} = \left( \frac{1}{\mu'} \tilde{\mathbf{B}}_{\mathrm{T}} \right)_{x,y}.\end{aligned} \hspace{\stretch{1}}(1.0.4b)

With $\beta = \mu v/\mu' v'$, those components are

\begin{aligned}E_{\mathrm{I}} + E_{\mathrm{R}} \cos\theta = E_{\mathrm{T}} \cos\phi \end{aligned} \hspace{\stretch{1}}(1.0.5a)

\begin{aligned}E_{\mathrm{R}} \sin\theta = E_{\mathrm{T}} \sin\phi\end{aligned} \hspace{\stretch{1}}(1.0.5b)

\begin{aligned}E_{\mathrm{R}} \sin\theta = - \beta E_{\mathrm{T}} \sin\phi\end{aligned} \hspace{\stretch{1}}(1.0.5c)

\begin{aligned}E_{\mathrm{I}} - E_{\mathrm{R}} \cos\theta = \beta E_{\mathrm{T}} \cos\phi\end{aligned} \hspace{\stretch{1}}(1.0.5d)

Equality of eq. 1.0.5b, and eq. 1.0.5c require

\begin{aligned}- \beta E_{\mathrm{T}} \sin\phi = E_{\mathrm{T}} \sin\phi,\end{aligned} \hspace{\stretch{1}}(1.0.6)

or $(\theta, \phi) \in \{(0, 0), (0, \pi), (\pi, 0), (\pi, \pi)\}$. It turns out that all of these solutions correspond to the same physical waves. Let’s look at each in turn

• $(\theta, \phi) = (0, 0)$. The system eq. 1.0.5.5 is reduced to

\begin{aligned}\begin{aligned}E_{\mathrm{I}} + E_{\mathrm{R}} &= E_{\mathrm{T}} \\ E_{\mathrm{I}} - E_{\mathrm{R}} &= \beta E_{\mathrm{T}},\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.7)

with solution

\begin{aligned}\begin{aligned}\frac{E_{\mathrm{T}}}{E_{\mathrm{I}}} &= \frac{2}{1 + \beta} \\ \frac{E_{\mathrm{R}}}{E_{\mathrm{I}}} &= \frac{1 - \beta}{1 + \beta}.\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.8)

• $(\theta, \phi) = (\pi, \pi)$. The system eq. 1.0.5.5 is reduced to

\begin{aligned}\begin{aligned}E_{\mathrm{I}} - E_{\mathrm{R}} &= -E_{\mathrm{T}} \\ E_{\mathrm{I}} + E_{\mathrm{R}} &= -\beta E_{\mathrm{T}},\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.9)

with solution

\begin{aligned}\begin{aligned}\frac{E_{\mathrm{T}}}{E_{\mathrm{I}}} &= -\frac{2}{1 + \beta} \\ \frac{E_{\mathrm{R}}}{E_{\mathrm{I}}} &= -\frac{1 - \beta}{1 + \beta}.\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.10)

Effectively the sign for the magnitude of the transmitted and reflected phasors is toggled, but the polarization vectors are also negated, with $\hat{\mathbf{n}} = -\hat{\mathbf{x}}$, and $\hat{\mathbf{n}}' = -\hat{\mathbf{x}}$. The resulting $\tilde{\mathbf{E}}_{\mathrm{R}}$ and $\tilde{\mathbf{E}}_{\mathrm{T}}$ are unchanged relative to those of the $(0,0)$ solution above.

• $(\theta, \phi) = (0, \pi)$. The system eq. 1.0.5.5 is reduced to

\begin{aligned}\begin{aligned}E_{\mathrm{I}} + E_{\mathrm{R}} &= -E_{\mathrm{T}} \\ E_{\mathrm{I}} - E_{\mathrm{R}} &= -\beta E_{\mathrm{T}},\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.11)

with solution

\begin{aligned}\begin{aligned}\frac{E_{\mathrm{T}}}{E_{\mathrm{I}}} &= -\frac{2}{1 + \beta} \\ \frac{E_{\mathrm{R}}}{E_{\mathrm{I}}} &= \frac{1 - \beta}{1 + \beta}.\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.12)

Effectively the sign for the magnitude of the transmitted phasor is toggled. The polarization vectors in this case are $\hat{\mathbf{n}} = \hat{\mathbf{x}}$, and $\hat{\mathbf{n}}' = -\hat{\mathbf{x}}$, so the transmitted phasor magnitude change of sign does not change $\tilde{\mathbf{E}}_{\mathrm{T}}$ relative to that of the $(0,0)$ solution above.

• $(\theta, \phi) = (\pi, 0)$. The system eq. 1.0.5.5 is reduced to

\begin{aligned}\begin{aligned}E_{\mathrm{I}} - E_{\mathrm{R}} &= E_{\mathrm{T}} \\ E_{\mathrm{I}} + E_{\mathrm{R}} &= \beta E_{\mathrm{T}},\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.13)

with solution

\begin{aligned}\begin{aligned}\frac{E_{\mathrm{T}}}{E_{\mathrm{I}}} &= \frac{2}{1 + \beta} \\ \frac{E_{\mathrm{R}}}{E_{\mathrm{I}}} &= -\frac{1 - \beta}{1 + \beta}.\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.14)

This time, the sign for the magnitude of the reflected phasor is toggled. The polarization vectors in this case are $\hat{\mathbf{n}} = -\hat{\mathbf{x}}$, and $\hat{\mathbf{n}}' = \hat{\mathbf{x}}$. In this final variation the reflected phasor magnitude change of sign does not change $\tilde{\mathbf{E}}_{\mathrm{R}}$ relative to that of the $(0,0)$ solution.

We see that there is only one solution for the polarization angle of the transmitted and reflected waves relative to the incident wave. Although we fixed the incident polarization with $\mathbf{E}$ along $\hat{\mathbf{x}}$, the polarization of the incident wave is maintained regardless of TE or TM labeling in this example, since our system is symmetric with respect to rotation.

References

[1] D.J. Griffith. Introduction to Electrodynamics. Prentice-Hall, 1981.

Posted by Peeter Joot on January 19, 2014

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 $k_{\mathrm{B}} T \gg \hbar \omega_{\mathrm{D}}$.

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)

One atom basis phonons in 2D

Posted by Peeter Joot on January 19, 2014

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.

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

Posted by Peeter Joot 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

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

Posted by Peeter Joot 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 Peeter Joot 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.)