# Proportion of photons passing through three polarizers

I've started recently reading book "Quantum Computing, A Gentle Introduction". After each chapter there are exercises for self study. For some of them there are answers, for some not. So far I've been reading "Feynman lectures on physics" and I've decided to make those exercises.

I've worked on answer on my own, and when I look at one provided by the author I am not sure about my thinking. This question is generally follow up for my previous one: Calculating amplitude for chain of polaroids, where I am presenting my thinking, but as pointed out by @ACuriousMind my notation (as well as thinking) might be wrong.

Exercise 2.1:

Let the direction |v⟩ of polaroid B's preferred axis be given as a function of θ, |v⟩=cosθ|→⟩+sinθ|↑⟩, 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.

And this is the image question is referring to:

The answer is $0.5(cosθ)^{2}(sinθ)^{2}$

• Do you know the derivation of the exact same solution in the classical set up of a light source and three linear polarizes using Malus law? Think in that direction. Commented May 5, 2016 at 18:32
• @RajathKrishnaR I am not sure about it. Also, taking into account that this is one of the first questions in the book and this law was never even remotely mentioned I wouldn't expect that. Commented May 5, 2016 at 18:36

I am not able to understand your notation and even though your question is not clear enough I will try to answer it based on my understanding of what the question is.

Question: Consider three linear polarizers $P_1, P_2 and P_3$ kept one after the other in front of a photon source (unpolarized). The axis of $P_1$ is vertical, that of $P_2$ is at at an angle $\theta$ with respect to $P_1$ and that of $P_3$ is horizontal (i.e $90^o$ with respect to the axis of $P_1$). What proportion of photon passing through $P_1$ will come out of $P_3$?

This question is important experimentally as well as is an important pedagogical method to teach the importance of superposition and its consequences. Consider unpolarized photons passing through $P_1$. As all photos have random polarizations(which is when we call the source to be unpolarized) the proportion of photons that pass through P_1 is $\frac{1}{2}$. After passing through $P_1$ all photons are in the same state $|up>$ or "vertically polarized". Let is associate $P_2$ with a basis and call it $|v>$ (along the polarization axis) and $|h>$ (perpendicular to the polarization axis).

Now, the state $|up>$ can be decomposed with respect to this ${|v>,|h>}$ basis as: $$|up>=|v><v|up> + |h><h|up>$$ $$=|v>cos(\theta)+|h>sin(\theta).$$

Now, the probability the photon has a state $|v>$ after passing through the second polarizer (in fact all photons that pass through $P_2$ are in state $|v>$ as others are blocked by the polarizer) is given by $$|<v|up>|^2=cos^2(\theta)$$. Now, the photons that passed through P_2 are in the state |v>.

Let us associate another basis with $P_3$. $|m>$ along the direction of polarization axis and $|n>$ perpendicular to it. Now, the state $|v>$ can again be decomposed with respect to this new basis. $$|v>=|m><m|v> + |n><n|v>$$ $$=|m>cos(90^o-\theta) + |n>sin(90^o-\theta) \quad\quad (why?)$$.

Again, the probability that the photon is in state $|m>$ is $$|<m|v>|^2=cos^2(90^o-\theta)=sin^2(\theta).$$

Thus, the proportion of photon which passes through the whole setup is $$P=\frac{1}{2}*cos^2(\theta)*sin^2(\theta).$$

• Let me know whether I answered your question. Anyone who wants to make the answer better by adding pictures for this setup along with the basis mentioned are welcome . Commented May 6, 2016 at 4:38
• Thank you very much for your answer. Well, as for the question - this is exact copy form the book I'm using, so I didn't make it up. Commented May 6, 2016 at 12:38

Let the initial intensity of the light be $I_0$. After the first polarizer, the intensity will be $I_1=\frac12 I_0$. This is because the light was initially unpolarized.

Now the light is horizontally polarized. To figure out the intensity through the second polarizer, remember that the intensity is proportional to the amplitude of the electromagnetic wave $I\propto E_0^2$. So all we need to do is project the polarization vector of the horizontally polarized light onto the diagonal one which makes an angle $\theta$ with the horizontal. This gives us a $\cos^2$ term in the intensity. The next polarizer will make an angle of $\pi/2-\theta$ with the diagonally polarized light. When plugging this into another cosine and projecting onto the vertical axis the cosine goes to a sine and we are left with the final form:

$$\frac{I}{I_0}=\frac12 \cos^2\theta \sin^2\theta.$$

I left out some details intentionally in hopes that you will think about it and come to a better understanding of what is happening. Just remember that intensity is proportional to the amplitude squared and to project those amplitudes onto the respective axis to simulate passage through a polarizer.

Think of the problem as three polaroid interactions: Since each pass through a polaroid creates a photon with known polarization state that is then used as input for the next polaroid, the probability of reaching the screen can be written as a product of the probability of passing through each polaroid. Thus Ptotal = P1 * P2 * P3

P1 is the probability of an unpolarized photon passing through horizontal (|1>)polarizer A. Think of it like flipping a coin: for a random distribution, the probability of 1 out of 2 outcomes is P1 = 1/2.

P2 is the probability of a horizontally (|1>) polarized photon from polaroid A passing through polaroid B whose preferred axis |v> = cos(theta)|1> + sin(theta)|0>. This probability is the square of |v>'s |1> coefficient: for coefficient cos(theta), P2 = (cos(theta))squared.

P3 is the probability of a B (cos(theta)|1> + sin(theta)|0>) polarized photon passing through vertical (|0>) polarizer C. This probability is the square of |v>'s |0> coefficient: for coefficient sin(theta) , P3 = (sin(theta))squared.

Thus the probability of reaching the screen is Ptotal = P1 * P2 * P3 = (1/2) * (cos(theta))squared * (sin(theta))squared.