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Posts Tagged ‘boundary value conditions’

Polarization angles for normal transmission and reflection

Posted by peeterjoot 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.

phy456 Problem set 4, problem 2 notes

Posted by peeterjoot on October 12, 2011

[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.)]

Problem 2.

I was deceived by an incorrect result in Mathematica, which led me to believe that the second order energy perturbation was zero (whereas part (c) of the problem asked if it was greater or lesser than zero). I started starting writing this up to show my reasoning, but our Professor quickly provided an example after class showing how this zero must be wrong, and I didn’t have to show him any of this.

Setup

Recall first the one dimensional particle in a box. Within the box we have to solve

\begin{aligned}\frac{P^2}{2m} \psi = E\psi\end{aligned} \hspace{\stretch{1}}(1.1)

and find

\begin{aligned}\psi \sim e^{\frac{i}{\hbar} \sqrt{2 m E} x} \end{aligned} \hspace{\stretch{1}}(1.2)

With

\begin{aligned}k = \frac{\sqrt{2 m E}}{\hbar}\end{aligned} \hspace{\stretch{1}}(1.3)

our general state, involving terms of each sign, takes the form

\begin{aligned}\psi = A e^{ i k x } +B e^{ -i k x }\end{aligned} \hspace{\stretch{1}}(1.4)

Inserting boundary conditions gives us

\begin{aligned}\begin{bmatrix}\psi(-L/2) \\ \psi(L/2)\end{bmatrix}\begin{bmatrix}e^{ -i k \frac{L}{2} } +e^{ i k \frac{L}{2} } \\ e^{ i k \frac{L}{2} } +e^{ -i k \frac{L}{2} }\end{bmatrix}\begin{bmatrix}A \\ B\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(1.5)

The determinant is zero

\begin{aligned}e^{-i k L} - e^{i k L} = 0,\end{aligned} \hspace{\stretch{1}}(1.6)

which provides our constraint on $k$

\begin{aligned}e^{2 i k L} = 1.\end{aligned} \hspace{\stretch{1}}(1.7)

We require $2 k L = 2 \pi n$ for any integer $n$, or

\begin{aligned}k = \frac{\pi n}{L}.\end{aligned} \hspace{\stretch{1}}(1.8)

This quantizes the energy, and inverting 1.3 gives us

\begin{aligned}E = \frac{1}{{2m}} \left( \frac{\hbar \pi n }{L} \right)^2.\end{aligned} \hspace{\stretch{1}}(1.9)

To complete the task of matching boundary value conditions we cheat and recall that the particular linear combinations that we need to match the boundary constraint of zero at $\pm L/2$ were sums and differences yielding cosines and sines respectively. Since

\begin{aligned}{\left.{{\sin\left( \frac{\pi n x }{L} \right) }}\right\vert}_{{x = \pm L/2}} = \pm \sin\left(\frac{\pi n}{2}\right)\end{aligned} \hspace{\stretch{1}}(1.10)

So sines are the wave functions for $n = 2, 4, ...$ since $\sin(n \pi) = 0$ for integer $n$. Similarly

\begin{aligned}{\left.{{\cos\left( \frac{\pi n x }{L} \right) }}\right\vert}_{{x = \pm L/2}} = \cos\left(\frac{\pi n}{2}\right).\end{aligned} \hspace{\stretch{1}}(1.11)

Cosine becomes zero at $\pi/2, 3\pi/2, \cdots$, so our wave function is the cosine for $n = 1, 3, 5, \cdots$.

Normalizing gives us

\begin{aligned}\psi_n(x) = \sqrt{\frac{2}{L}}\left\{\begin{array}{l l}\cos\left(\frac{\pi n x}{L}\right) & \quad n = 1, 3, 5, \cdots \\ \sin\left(\frac{\pi n x}{L}\right) & \quad n = 2, 4, 6, \cdots \end{array}\right.\end{aligned} \hspace{\stretch{1}}(1.12)

Two non-interacting particles. Three lowest energy levels and degeneracies

Forming the Hamiltonian for two particles in the box without interaction, we have within the box

\begin{aligned}H = \frac{P_1^2}{2m} +\frac{P_2^2}{2m} \end{aligned} \hspace{\stretch{1}}(1.13)

we can apply separation of variables, and it becomes clear that our wave functions have the form

\begin{aligned}\psi_{nm}(x_1, x_2) = \psi_n(x_1) \psi_m(x_2)\end{aligned} \hspace{\stretch{1}}(1.14)

Plugging in

\begin{aligned}H \psi = E \psi,\end{aligned} \hspace{\stretch{1}}(1.15)

supplies the energy levels for the two particle wavefunction, giving

\begin{aligned}\begin{aligned}H \psi_{nm} &= \frac{\hbar^2}{2m}\left(\left(\frac{\pi n}{L}\right)^2+\left(\frac{\pi m}{L}\right)^2\right)\psi_{nm} \\ &= \frac{1}{2m} \left(\frac{\hbar \pi}{L}\right)^2 ( n^2 + m^2 ) \psi_{nm}\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.16)

Letting $n, m$ each range over $[1,3]$ for example we find

\begin{aligned}\begin{array}{l l l}n & m & n^2 + m^2 \\ 1 & 1 & 2 \\ 1 & 2 & 5 \\ 1 & 3 & 10 \\ 2 & 1 & 5 \\ 2 & 2 & 8 \\ 2 & 3 & 13 \\ 3 & 1 & 10 \\ 3 & 2 & 13 \\ 3 & 3 & 18\end{array}\end{aligned} \hspace{\stretch{1}}(1.17)

It’s clear that our lowest energy levels are

\begin{aligned}\frac{1}{m} \left(\frac{\hbar \pi}{L}\right)^2 \\ \frac{5}{2m} \left(\frac{\hbar \pi}{L}\right)^2 \\ \frac{4}{m} \left(\frac{\hbar \pi}{L}\right)^2 \end{aligned}

with degeneracies $1, 2, 1$ respectively.

Ground state energy with interaction perturbation to first order.

With $c_0$ positive and an interaction potential of the form

\begin{aligned}U(X_1, X_2) = - c_0 \delta(X_1 - X_2)\end{aligned} \hspace{\stretch{1}}(1.18)

The second order perturbation of the ground state energy is

\begin{aligned}E = E_{11}^{(0)} + H_{11;11}' + \sum_{nm \ne 11} \frac{{\left\lvert{H_{11;11}' }\right\rvert}^2}{E_{11} - E_{nm}}\end{aligned} \hspace{\stretch{1}}(1.19)

where

\begin{aligned}E_{11}^{(0)} = \frac{1}{m} \left(\frac{\hbar \pi}{L}\right)^2,\end{aligned} \hspace{\stretch{1}}(1.20)

and

\begin{aligned}H_{nm;ab}' = -c_0 {\langle {\psi_{nm}} \rvert} \delta(X_1 - X_2) {\lvert {\psi_{ab}} \rangle}\end{aligned} \hspace{\stretch{1}}(1.21)

to proceed, we need to expand the matrix element

\begin{aligned}{\langle {\psi_{nm}} \rvert} \delta(X_1 - X_2) {\lvert {\psi_{ab}} \rangle}&=\int dx_1 dx_2 dy_1 dy_2\left\langle{{\psi_{nm}}} \vert {{x_1 x_2}}\right\rangle {\langle {x_1 x_2} \rvert} \delta(X_1 - X_2) {\lvert {y_1 y_2 } \rangle} \left\langle{{y_1 y_2}} \vert {{\psi_{ab}}}\right\rangle \\ &=\int dx_1 dx_2 dy_1 dy_2\left\langle{{\psi_{nm}}} \vert {{x_1 x_2}}\right\rangle \delta(x_1 - x_2) \delta^2(\mathbf{x} - \mathbf{y}) \left\langle{{y_1 y_2}} \vert {{\psi_{ab}}}\right\rangle \\ &=\int dx_1 dx_2 \left\langle{{\psi_{nm}}} \vert {{x_1 x_2}}\right\rangle \delta(x_1 - x_2) \left\langle{{x_1 x_2}} \vert {{\psi_{ab}}}\right\rangle \\ &=\int_{-L/2}^{L/2} dx\psi_{nm}(x, x)\psi_{ab}(x, x)\end{aligned}

So, for our first order calculation we need

\begin{aligned}H_{11; 11}' &= - c_0\int_{-L/2}^{L/2} dx\psi_{11}(x, x)\psi_{11}(x, x) \\ &=\frac{4}{L^2}\int_{-L/2}^{L/2} dx\cos^4( \pi x /L ) \\ &=- \frac{3 c_0}{2 L}\end{aligned}

For the second order perturbation of the energy, it is clear that this will reduce the first order approximation for each matrix element that is non-zero.

Attempting that calculation with \href{https://github.com/peeterjoot/physicsplay/blob/796c8e3739ae1a9ca26270a0e91384afba45661d/notes/phy456/problem\

This worksheet can be seen to be giving misleading results, by evaluating

\begin{aligned}\int_{-\frac{L}{2}}^{\frac{L}{2}} \left(\frac{2}{L}\right)^2 \cos ^2\left(\frac{\pi x}{L}\right) \cos ^2\left(\frac{3 \pi x}{L}\right) \, dx = \frac{1}{L}\end{aligned} \hspace{\stretch{1}}(1.22)

Yet, the FullSimplify gives

\begin{aligned}\text{FullSimplify}\left[\int_{-\frac{L}{2}}^{\frac{L}{2}} \text{Cos}\left[\frac{\pi x}{L}\right]^2 \left(\frac{2}{L}\right)^2 \text{Cos}\left[\frac{(2 n+1) \pi x}{L}\right] \text{Cos}\left[\frac{(2 m+1) \pi x}{L}\right] \, dx,\{m,n\}\in \text{Integers}\right] = 0\end{aligned} \hspace{\stretch{1}}(1.23)