Physics 402 Test 2

I pledge that I have given nor received unauthorized assistance during the completion of this work.

Signature: width10cm height1pt depth0pt

Multiple Choice Questions (Circle your choice, 4 points apiece).

1. The Stern-Gerlach experiment in quantum physics demonstrates the quantization of spin. Sample data are shown in the figure. The conclusion is that

   (a)	the electron is a fermion and can have spin up or down	(d)	The electron can only have spin down
   (b)	the electron has no spin	(e)	the electron is a fermion and can have spin 3/2, 1/2, -1/2, -3/2
   (c)	the electron can only have spin up

   
   
   
   

2. Consider a mechanical model of the proton where the spin is due to its rotation. Assume the proton to be a uniform, solid sphere and derive the equatorial velocity. Assume $m_p = 1.67\times 10^{-24}~grams$ and $r_p = 10^{-13}~cm$.

   (a)	$1.58\times 10^{10}~cm/s$	(d)	$1.87\times 10^{10}~cm/s$
   (b)	$3.00\times 10^{10}~cm/s$	(e)	$7.88\times 10^{10}~cm/s$
   (c)	$3.94\times 10^{10}~cm/s$

   
   

3. In the Zeeman effect, the energy of a spectral line is found to be changed in a magnetic field. What is the amount of the energy change?

   (a)	$eB/2m_e c$	(d)	$2\mu_B B$
   (b)	$eB/m_e c$	(e)	$\mu_B B/2$
   (c)	$\mu_B B$

   
   
   
   

4. A plane wave solution of the electromagnetic wave equation is
   
   $${\bf E} = E_{0y} \cos (\omega t - kx + \alpha) \hat j + E_{0z} \cos (\omega t - kx + \beta) \hat k$$
   
   
   Under what conditions is this light circularly polarized?

   (a)	$\beta - \alpha = \pm \pi/2$	(d)	$\beta - \alpha = \pm \pi/2$ and $E_{0y} = 2E_{0z}$
   (b)	$\beta - \alpha = \pm \pi/2$ and $E_{0y} = E_{0z}$	(e)	$\beta = \alpha$ and $E_{0y} = 2E_{0z}$
   (c)	$\beta = \alpha$ and $E_{0y} = E_{0z}$

Problems. Clearly show all work for full credit.

1. (20 pts.)

Consider the correlation factor $a$ in the Aspect experiment where a transmitted photon counts as $+1$ and a reflected photon counts as $-1$. For a polarizer set to the angle $\theta_1$ and an ensemble of photons with a single polarization angle $\chi$, show the classical model predicts $$\langle a (\theta_1,\chi) \rangle = \cos 2(\theta_1 - \chi) \quad$$ Recall that Malus's Law states the intensity of polarized light transmitted through a polarizer is $$I_{trans} = I_{inc} \cos^2 \beta$$ where $I_{trans}$ is the transmitted intensity, $I_{inc}$ is the incident intensity, and $\beta$ is the angle between the plane of polarization of the incident light and the angle of the polarizer.

2. (32 pts.)

Using the expressions for $\hat L^2$ and $\hat L_{\pm}$ shown below operate with $\hat L^2$ on the following uncoupled angular momentum state for two $p$ electrons $$\vert 2011\rangle = \sqrt{ 1 \over 6}~ \vert 11\rangle_1 \ve... ...~ +~ \sqrt{1 \over 6}~ \vert 1 -1 \rangle_1 \vert 11\rangle_2$$ to verify the $lm$ entry for the $\vert lml_1 l_2\rangle$ eigenstate. $$\hat L_2 = \hat L_1^2 + \hat L_2^2 + 2\hat L_{1z} \hat L_{2z} + \hat L_{1+} \hat L_{2-} + \hat L_{1-} \hat L_{2+}$$ $$\hat L_\pm \vert lm \rangle = \hbar \sqrt{l(l+1) - m(m\pm 1)} ~ \vert lm \rangle$$

3. (32 pts.)

We used the Coulomb potential to describe the interaction between the proton and electron in our initial, simple model of the hydrogen atom. We used perturbation theory to incorporate the spin-orbit interaction into the model using $$H_{so}^\prime = {1 \over 2m_e^2 c^2} \vec L \cdot \vec S {1 \over r} {dV(r) \over dr}$$ where $V(r)$ is the potential energy between the electron and the proton. Suppose now the proton and electron are part of some larger molecule and we can now more accurately describe the interaction using a harmonic oscillator potential. Recall that the harmonic oscillator potential energy is $V(r) = kr^2/2$ and the energy levels of an unperturbed oscillator are $E_n = (n + 1/2) \hbar \omega_0$. What is the effect of the spin-orbit interaction on the spectrum of harmonic oscillator states? In other words, what energy shifts does this perturbation cause? You might find some of the following expressions useful. $$\hat L_+ = \hat L_x + i\hat L_y \qquad \hat L_- = \hat L_x -... ...- \over 2} \qquad \hat L_x = {\hat L_+ - \hat L_- \over 2i}$$ $$\hat L^2 = (\hat {\bf ~L}_1 + \hat {\bf ~L}_2)\cdot (\hat ... ...at L_{2z} + \hat L_{1+} \hat L_{2-} + \hat L_{1-} \hat L_{2+}$$ $$\hat L \cdot \hat S = \hat L_x \hat S_x + \hat L_y \hat S_y ... ... L_+ \hat S_- + \hat L_- \hat S_+ \over 2} + \hat L_z \hat S_z$$ $$\vec B = -{\vec p \over mc}\times \hat E \qquad \vec \mu_l ... ...er 2 m_e c} \vec L \qquad \vec \mu_s = {e \over m_e c} \vec S$$

Table of Constants

Speed of light	$c$	$2.9979\times 10^8 ~m/s$
Boltzmann's constant	$k_B$	$1.381\times 10^{-23}~J/K$
		$8.62\times 10^{-5}~eV/k$
Planck's constant	$h$	$6.621 \times 10^{-34}~J-s$
		$4.1357\times 10^{-15}~eV-s$
	$\hbar$	$1.0546\times 10^{-34}~J-s$
		$6.5821\times 10^{-16}~eV-s$
	$\hbar c$	$197~MeV-fm$
	$\hbar c$	$1970~eV-{\rm\AA}$
Electron charge	$e$	$1.6\times 10^{-19}~C$
Electron mass	$m_e$	$9.11\times 10^{-31}~kg$
		$0.511~MeV/c^2$
Proton mass	$m_p$	$1.67\times 10^{-27}kg$
		$938~MeV/c^2$
Neutron mass	$m_n$	$1.68\times 10^{-27}~kg$
		$939~MeV/c^2$
atomic mass unit	$u$	$1.66\times 10^{-27}~kg$
		$931.5~MeV/c^2$
Fine structure constant	$e^2$	$\hbar c /137$

Rotational Inertias

Hoop about central axis	$MR^2$
Annular ring about central axis	${1\over 2}M(R_1^2 + R_2^2)$
Solid disk about central axis	${1\over 2}MR^2$
Solid disk about central diameter	${1\over 3}MR^2$
Solid sphere about any diameter	${2\over 5} MR^2$
Thin spherical shell about any diameter	${2\over 3} MR^2$

Trigonometric Identities

$\sin^2\theta + \cos^2 \theta = 1$	$\sec^2 \theta - \tan^2 \theta = 1$
$\sin 2\theta =2\sin\theta\cos\theta$	$\cos 2\theta = 2 \cos^2\theta - 1$
$\cos 2\theta = \cos^2\theta - \sin^2\theta$	$\cos 2\theta = 1 - 2 \sin^2\theta$
$\sin(\alpha\pm \beta) = \sin\alpha \cos\beta \pm \cos\alpha \sin\beta$	$\cos(\alpha\pm \beta) = \cos\alpha \cos\beta \mp \sin\alpha \sin\beta$
$\sin\alpha + \sin\beta = 2\sin{\alpha+\beta\over 2}\cos{\alpha-\beta\over 2}$	$\sin\alpha - \sin\beta = 2\cos{\alpha+\beta\over 2}\sin{\alpha-\beta\over 2}$
$\cos\alpha+\cos\beta = 2\cos{\alpha+\beta\over 2}\cos{\alpha-\beta\over 2}$	$\cos\alpha-\cos\beta = 2\sin{\alpha+\beta\over 2}\sin{\eta-\alpha\over 2}$