# Making light of quantum mechanics

April 30, 2020.

### Contents

1. Introduction
2. Fun with photons
3. Quantum mechanics

## 2. Fun with photons

Classically, light consists of intertwined electric and magnetic fields, separately wobbling up and down, but at right angles to each other. In quantum mechanics, this wave can also be viewed as a particle called the photon. The photon’s electric field wobbles up and down in some direction called the polarisation.

Let’s set up a coordinate system with the $z$-axis coinciding with the path of the photon, and $x$ and $y$ perpendicular to it. The polarization makes an angle $\theta$ with respect to the $x$-axis, as picture below. A polariser is a slit (in real life, a series of aligned slits) which allows only certain polarisations to pass through. Surprisingly, experiments with polarisers that you can do at home reveal the basic features of quantum mechanics!

### 2.1. Flashlights and sunglasses

Let’s grab a polariser and put it at some angle $\theta_P$. If a single photon comes along with polarisation $\theta$, what happens? Intuitively, the polariser should act like a filter which only lets the angle $\theta_P$ through; anything else is blocked. Although it’s hard to produce a single photon with fixed polarisation, it’s easy to produce many photons with random polarisations, simply by turning on a flashlight. If $\theta$ is randomly chosen, most of the time $\theta \neq \theta_P$, so we might expect a polariser to block the light from a flashlight. You can test this prediction with a pair of polarising sunglasses. You’ll find that they don’t block the flashlight. It would be crazy to make polarised sunglasses otherwise!

Let’s see how many photons do get blocked. Consider a chunk of the beam of light as it leaves the flashlight. It has some area $A$, a width $w$, and a density of photons $n$, so the total number of photons is $N = Awn$. If the beam is monochromatic (one colour), each photon carries the same amount of energy, $\epsilon$, so the total amount of energy is

$E = N\epsilon = Awn\epsilon.$

We can measure the intensity of light using a light meter, or even an app on your phone. It doesn’t measure the total energy in a chunk, or even the power (energy per unit time) delivered by the beam, but rather, the power per unit are:

$I = \frac{E}{t A} = \frac{wn\epsilon}{t} = nc\epsilon.$

Here, we used the fact that $w/t = c$ is the speed of light. Note that the intensity is proportional to the density of photons, $n$, and two constants, $c$ and $\epsilon$. Using a light meter app, you can check that after the light passes through the subglasses, the intensity is reduced by half, and hence the density of photons is reduced by half:

$I' = n'c\epsilon = \frac{1}{2}I = \frac{1}{2}nc\epsilon \quad \Longrightarrow \quad n' = \frac{1}{2}n.$

In other words, even though the photons have random polarisations $\theta$, half of them are let through. How could this possibly happen?

### 2.2. Quantum polarisers

To explain what’s happening, let’s revisit our original hypothesis: only light with $\theta = \theta_P$ is let through. Another way of thinking about this is as follows. When light hits the slit, the polariser “measures” the angle $\theta$, and decides to let the light through if $\theta = \theta_P$. Otherwise, it simply absorbs the photon. If this view is correct, then in a beam of randomly polarised photons, most get blocked.

Experimentally, we see that only half the photons are blocked, so we need to update our model of what the polariser does to a photon. To help us formulate a model, let’s do some more experiments, but now with multiple polarisers. To start off with, apply a polariser with $\theta_1$ to a flashlight beam of intensity $I$. The intensity is halved to $I_1 = I/2$, and all the light that comes out has polarisation angle $\theta_1$. To check this, we can apply another polariser with angle $\theta_2 = \theta_1$. The intensity after the second polariser is exactly the same, $I_2 = I_1$, so all the photons are allowed through!

This suggests two rules for polarisers:

P1. Any photons emerging from a polariser with angle $\theta_P$ have polarisation angle $\theta_P$.

P2’. A polariser with angle $\theta_P$ always admits a photon with angle $\theta_P$.

What about photons with a different angle? Using (P1), we can test this very precisely. Fix the angle of the first polariser at $\theta_1 = 0$ for simplicity, and let the angle $\theta_2 = \theta$ of the second polariser vary. We can measure the resultant intensity $I_2(\theta)$. At $\theta = 0$, we know from (P2’) that the intensity is unchanged, $I_2 = I_1$. But as we rotate the second polariser away from the $x$-axis, the intensity decreases, until it disappears altogether at $\theta = 90^\circ$. When the polarisers are at right angles, no light gets through at all!

Fitting a curve to the measurements, one finds that the intensity as a function of angle is

$I_2(\theta) = I_1 \cos^2\theta.$

This is called Malus’s law after Étienne-Louis Malus (1775–1812). If $N_1$ photons come out of the first filter with horizontal polarisation, only $N_2 = N_1 \cos^2\theta$ make it through the second, so it seems as if the polariser makes a random decision to let a photon through, with probability

$p = \frac{N_2}{N_1} = \cos^2\theta.$

Note that $\theta = \theta_2 - \theta_1 = \Delta\theta$ is the relative angle between photon and polariser, this suggests a general rule (P2), which includes (P2’) as a special case:

P2. If a photon and polariser have relative angle $\Delta\theta$, the photon passes through with

$p(\Delta \theta) = \cos^2(\Delta \theta).$

We have based our rules on experiments using two filters, but they more conventionally introduced using three filters and the famous Stern-Gerlach experiment (1922). You can predict the outcome of this experiment using rules (P1) and (P2) in Exercise 1. In Exercise 2, you can see why intensity is halved when a flashlight beam is polarised.

Exercise 1 (Stern-Gerlach). The original Stern-Gerlach experiment was performed with electrons. Here, you can work out what happens in an analogous experiment for photons. As above, consider a flashlight of intensity $I$ illuminating two filters. If these are at right angles, they block all incoming light and $I_2 =0$.

(a) Suppose we insert a filter in between the first and the second, oriented at $45^\circ$ relative to each. Using rules (P1) and (P2), what is the resultant intensity?

(b) Confirm your result experimentally. A couple of minutes at the pharmacy (in the sunglasses section) should suffice.

Exercise 2 (halving intensity). We still haven’t explained why the unpolarised flashlight beam is halved in intensity. Recall that, in a beam of unpolarised light, the angle of individual photons $\theta$ is random. To halve the intensity, we need the average value of $p(\theta) = \cos^2\theta$, for $\theta = 0$ to $\theta = 360^\circ$, to be $1/2$.

(a) Use the double angle formula to show that

$\cos^2\theta - \frac{1}{2} = \frac{1}{2}\cos(2x).$

(b) Conclude that $p(\theta)$ is symmetric around the line $y =1/2$.

(c) The average value $p_{\text{avg}}$ is the area underneath the curve $p(\theta)$ compared to the area beneath the curve with $p = 1$. Use (a) and (b) to argue that $p_{\text{avg}} = 1/2$.

### 2.3. Splitting light

When we have two “crossed” filters, i.e. filters at right angles, all incoming light is blocked. But what if we combine these crossed slits into a single polariser? Our rules don’t apply in this case, so we need to do some more experiments. Since sunglasses are polarised in a single direction only, this is a little harder to perform at home, so I’ll just tell you what happens and we can continue theory-building from there.

If you shine a flashlight on the cross, and compare the intensity before and afterwards, you find no difference. Every photon gets through! On top of that, the polarisations are split evenly between the two directions. We can learn even more by polarising the light first. For concreteness, orient the cross so that the first slit is at $\theta_x = 0$ and the second at $\theta_y = 90^\circ$. If an incoming beam is polarised at angle $\theta$, it will obey Malus’s law is true for both slits. In other words, it will have horizontal orientation $\theta_x = 0$ with probability

$p_x = \cos^2\theta$

and vertical orientation $\theta_y = 90^\circ$ with probability

$p_y = \cos^2(\theta - 90^\circ) = \sin^2\theta.$

Every photon gets through since these probabilities add up to $1$:

$p_x + p_y = \sin^2\theta + \cos^2\theta = 1.$

There is a nice geometric way to interpret this. Recall that, if $\theta$ labels an angle on the unit circle, then the $x$ and $y$ coordinates of the point are

$x(\theta) = \cos\theta, \quad y(\theta) = \sin \theta.$

To get probabilities, we just square coordinates on the unit circle. It’s useful to

## 3. Quantum mechanics

### 3.1. States and measurements

Let’s recap. To explain why you can see through polarised sunglasses, we have to assume that polarisers make a fundamentally random decision about what to do with an incoming photon. The universe, even at a microscopic level, is probabilistic. Maybe this seems natural to you, but this was shocking for physicists in the early 20th century, used to thinking of the world in classical terms. It was too much for Einstein’s sensibilities. He famously wrote

God does not play dice.

Ironically, in his efforts to discredit quantum mechanics, Einstein invented entanglement, perhaps the most useful and essentially quantum phenomenon, as we will see below. Even when Einstein was wrong, he was right!

### 3.3. Entanglement

• qm: polarisers/filters, stern-gerlach, superposition, uncertainty, entanglement, qkd, quantum computing, measurement problem, parallel universes

### Extra

Procedurally, we might imagine it as follows: the polariser measures the photon’s polarisation $\theta$, checks it against the angle of the slit $\theta_P$, and if the two are different, “rejects” the photon. We’ll call this a “classical” polariser for reasons we explain below. A beam of light will consist of many photons with random polarisation angles $\theta$. Since $\theta$ can be chosen from anywhere on the circle, most photons in the beam won’t have $\theta = \theta_P$, and if the polariser is classical, will be blocked. You can perform a simple experiment to see what happens.

It’s not too hard to account for this when we remember that, with a single slit at angle $\theta_P$, half the photons went through (and picked up polarisation $\theta_P$) and the other half were absorbed by the filter. It seems that, instead of being absorbed, the photons are now getting spat out with the second polarisation $\theta_P + 90^\circ$. And the same thing is true if we think about things from the perspective of the second filter, so this interpretation is consistent!

Once again, we can do better Recall that if we place the angle $\theta$ on the unit circle, the $x$ and $y$ coordinates are

$x = \cos\theta, \quad y = \sin \theta.$

It looks like every photon that comes along is getting polarised into one of the two directions, with half assigned to the first polarisation and half to the second.

Now introduce a second, ordinary polariser, align it with the first of the crossed slits, and measure the intensity $I_1$. We can similarly measure the intensity $I_2$ when the polariser is aligned with the second of the crossed slits. Each of these intensities is precisely half the original intensity:

$I' = I = 2I_1 = 2I_2.$