Exercise 2.1

$\def\i{\mathbf {i}}\def\ket#1{|{#1}\rangle}\def\bra#1{\langle{#1}|}\def\braket#1#2{\langle{#1}|{#2}\rangle}\mathbf{Exercise\ 2.1}$

Let the direction $\ket v$ of polaroid $B$'s preferred axis be given as a function of $\theta$, $\ket v = \cos\theta\ket\to + \sin\theta\ket\uparrow$, and suppose that the polaroids $A$ and $C$ remain horizontally and vertically polarized as in the experiment of Section 2.1. What fraction of photons reach the screen? Assume that each photon generated by the laser pointer has random polarization.

page revision: 1, last edited: 12 Nov 2011 04:40

The fraction $0.5$ of photons that pass $A$ and are in state $\ket{\rightarrow}$. The projection of $\ket{\rightarrow}$ onto the preferred axis of $B$ is $\cos \theta$ so $(\cos \theta)^2$ of the photons pass $B$ and are in state $\ket{v}$. The projection of $\ket{v}$ onto the axis $\ket{\uparrow}$ of $C$ is $\sin \theta$. Thus, overall $0.5 (\cos \theta)^2 (\sin \theta)^2$ of the photons reach the screen.

ReplyOptionsThe question seemed ambiguous to me. "Assume that each photon generated by the laser pointer has random polarization" - does that mean 50% $\ket{\rightarrow}$ and 50% $\ket{\uparrow}$, or does it mean a uniformly distributed random superposition? But either way, it's true that a fraction 0.5 of photons pass A and are then in state $\ket{\rightarrow}$, so the answer given is correct.

ReplyOptions