The experiment 2 is the famously known double slit experiment. It was first realised in 1927 by Thomas Young. The double slit shows how the light (and sometimes matter) can act at the same time as a wave and a particle, from a classical point of view. The goal here is to get a better understanding on what happens when photons pass through a double slit and learn the basics of Dirac notation to solve mathematically what is happening.
The next simulation lets you have a play with the parameters to see how they affect the final pattern. You can change the number of slit(N), their width(b), their distance(d) and the wavelenght of the laser(L) by moving the pink dots on the control panel.
The prove you that animation is not lying, we built the setup in a lab.
These are the data we got from our lab setup. The left plot has two curve, for either slit covered. In the following section, we will explain what we did in the lab, what the data shows, what is the explaination for all of this and the maths behind the theorical curves.
Curious of what an optical setup looks like? Here are some pictures of the double slit and the single slit setup (same setup, we just change the slit).
Here is the representation of the double slit the setup you see in the picture. All of the equipement is shown, but the scaling is not representative
We are using a laser with a wavelenght of 808nm. There's a beam splitter and a half-wave plate. Toghter, they make sure the polarization of the beam is diagonal. The beam then goes to a lens converging the beam on the double (or single) slit. Then the beam reaches the detector.
Our detector is actually a single photon detector. The recreate a full pattern, it has to be mount on a linear moving plate. There is also a bunch of filter and long pass filter in front of it the ensure not saturating the detector.
You can download all our data here:
The file has 4 columns. The first two need to be ignore in this experiment. The third one is the distance travelled by the actuator in $\mu$m. The last column is the number of count on the detector for 2 second of acquisition.
In the graphic for the single slit, we use a same double slit where we blocked one slit at the time with tape. We see the two curves with a small gap of 175 $\mu$m. It correspond to the gap between both slit (around 200 $\mu$m).
In the second plot, we can easily distinguish 3 main peaks, which show interference. The number of peaks and their size are define by the dimension of the double slit, you might get 4 or 5 or more peak with other dimension. The important part is that the central peak is the highest and the two other has the same intensity. If not, it means the slit is not exactly perpendicular to the beam. A final check we can do is to compare this plot with theory. Using the equations givin in the theory section, we can verify how many peak we should be able to see. That theorical curve is the gray dash line. We see a little offset in the intensity and in the size of the pattern, which canbe due to the lenses placement are the accuracy of the linear actuator.
In this section, we will cover a new concept in the Dirac notation to add to what we already learn from experiement 1 and find the equation for the interference pattern without touching any integral.
To truly understand the double slit experience, we need to let go of classic physics and jump into quantum mechanics, which mean accepting that photons are probabilistics. We called the function that describe photon's state a wave-function. It combined different parameters, time, position and spin. The wave-function is generally denote $\Psi$.
If we think about a single slit first and we shoot a beam of light through it. Light will be acting like a wave. When it passes the slit, it will diffract. If we put a screen or a detector after the slit, we will see the diffraction pattern. Now, if we do the same thing, but with a double slit, the light will have a 50-50 chance to go through either of the slit. Every photon that goes thorugh slit A will interfere with the one from slit B and create an interference pattern. Actually, there's a little lie in this statement. If we shoot only one photon at the time, we will still see interference. When we say there's a 50-50 chance to go through either slit, we actually means that the photon will go through both slit at the same time. Which means the photon can interfere with itself. It might be hard to concive that the photon goes through the two slits at the same time, but that is quantum mechanics.
The equation of a photon passing through only one slit, called diffraction is defined as:
$D =\Big(\frac{sin(\alpha)}{\alpha} \Big)^2 \text{ , with } \alpha = \pi \frac{asin(\theta)}{\lambda}$
Where a is the width of the slit, theta is the angle from a point on the screen to the slit and $\lambda$ is the wave-lenght of the photons. On the right is an example of a pattern with random parameters.
When you add the second slit, you will obtain an interference pattern inside of of the envelop create by the diffraciton pattern. A general form of the interference equation is:
$I = \Big(\frac{sin(N\beta)}{Nsin(\beta)} \Big)^2 \text{ , with } \beta = \pi \frac{dsin(\theta)}{\lambda}$
WhereN is the number of slit, d is the distance between the slits. To get the pattern, you need to multiply interference with diffraction. The result with random paramters is shown on the left.
This experiment won't be much more difficult than the previous one form a mathematical point of view. We only need the add one concept, whcich is the phase of photon.
As we know, photons oscillate in space as they move. They might go from left to right, or from up to down or any angle. We call the phase the place were they are in their motion. That place is reallyimportant when we are talking about intefrence. Photons will not interfere the same way depending on the phase. Let's say two photons oscillate in the same direction (i.e. left to right). If they are both at the same "spot" in their interference (so the same phase) their interaction will be constructive.If one is at the left "spot" and the other at the right one, the interaction will be destructive.
The phase can be represented as an exponential. Exemple:
In this equation, "i" is the compexe number. If you don't know what it means, you can almost ignore it. Just know that "i" square equal minus 1. "k" is a called the wavenumber, which equal $2\pi/\lambda$. Finally, "d" is the distance the photon as travelled.
The last thing we need to learn is when two photons take different path. We will need to divide them into to place and recombined if they do recombined. Let's say a beam start at a position $\ket{S}$ and half the photon goes to to point $\ket{A}$ and the other hlaf to point $\ket{B}$.
The fraction $1/\sqrt{2}$ comes normalized the equation. If point A and B are not at the same distance, we could have add the phase shift in regard to both distance they have travelled,respectively a and b.
Full approach for the double slit
1. Let's start with the initial position $\ket{S}$. Because their is no polarizer, we can loose the polarization state.
2. Let's split the beam into two path, one that reach A, the other B. Both path are at the same distance from the laser, distance we will call l.
3. Now, both path will get to position $\ket{X}$. This position will move to form thehole pattern, which means the distance both photons will travel won't stay the same. If X is exactly in the middle, they will both be the same, if X is higher, a will be shorter, and so on. Let's scale "a" the distance from slit A and "b" the distance from slit B.
Finally, the bring $\ket{A}$ and $\ket{B}$ together, we need to do the reverse operation as when they split. Now multiply by $\sqrt{2}$
4. We now have found our last state. To get the interference pattern, wejuste need totake the complex conjugate of this last equation. As a reminder, this operation mean to multiply the equation by itself, by changing all the $\ket{}$ to $\bra{}$ and all the i becomes -i.
To resolve this, we need to think about it just a little. First, let's start with the $\braket{X|X}$. In the last experiment, we said we need to take the scalair product between the left-side and the right-side. In this case, the scalaire product between anything and itself always gives 1. Lucky us, we won't see any other product scalaire for any experiement that does not result as 1.
Now, for the exponential, the trick is whenever you multiply 2 exponential, you can add the exponant together.
For those who don't already know, $e^{0} =1$. If you are not used to trigonometric identity, you can stop here. This is our final answer. But trigonometry can be more eye appealing. Let's use the identity: $2cos(a) = e^{ia} + e^{-ia}$
5. This last equation is the intererence pattern. We need to multiply it by the diffraction pattern to get the final result.
