The Double-Slit Experiment

We will base this discussion almost entirely on the content provided by Professor David Tong (Tong (n.d.)).

The double slit set-up is a simple but iconic experiment that vividly demonstrates some of the stranger aspects of quantum mechanics. To set the scene, we first describe two experiments from classical physics. The set-up in both is the same: there is a wall, with two openings in it. These are the eponymous double slits. Behind the wall is a detector. You stand in front of the wall and throw things. The only difference between the two experiments lies in what you throw.

First, we throw particles. These are boring classical particles, like tennis balls. You throw randomly in the direction of the wall. Some simply bounce off the wall, while others get lucky and go through the slits. Then you measure the distribution of balls seen in the detector.

In this case, things are nice and simple. Suppose that you close the bottom slit, keeping only the top slit open, and then measure the distribution of balls reaching the detector, call this $P_{1}$. Similarly, if you close the top slit, but keep the bottom slit open then you measure the same distribution just shifted downwards, call this $P\left(2\right)$. Now open both slits. What you see is the sum of the previous distributions $P_{12}=P_{1}+P_{2}$, so far, so good.

Our second experiment is the same, but now with waves rather than particles. For our purposes, these can be water waves, sound waves or light waves. For any kind of wave, there is a new ingredient that dramatically changes the outcome: this is interference. Here we give a broad brush description of the physics.

First, we open one slit and not the other. The waves are diffracted at the slit and propagate out radially towards the detector where we can measure their intensity. The end result is very similar to that of a particle: a distribution spread out, peaked around the position of the first slit.

The real difference comes when we open both slits. The waves pass through along two paths and, when they recombine on the other side, add constructively or destructively. Crucially, at some positions the intensity when both slits are open can be less than when any single slit is open. This happens, for example, at positions where the two waves cancel themselves out completely so the intensity on the screen vanishes. An interference pattern of this kind can be seen vividly using lasers.

With this classical intuition in place, we now turn to the quantum world. We will perform the double slit experiment one last time, now with quantum particles, say electrons. For appropriately sized slits, the result is a combination of the two classical experiments described above. First, when we detect an electron it appears just like a classical particle would appear. That means that each time we play the game, the electron appears as a point on the detector, rather than something continuous and smeared out like a wave. However, after playing the game many times, the probability distribution seen on the screen agrees with that of a classical wave, exhibiting an interference pattern.

Perhaps the most striking aspect of this result is that there are certain places on the screen where fewer electrons are detected when both slits are open than when any one slit is open. There’s no way to explain this from a classical worldview where the particle follows a trajectory that goes either through the first slit or through the second. It must be that each particle somehow knows about the presence of both slits.

We can understand this behaviour from what little we know about quantum mechanics. First, when one slit is open, the particle is described by a wavefunction $\psi_{1}\left(x\right)$ and the probability distribution that we see on the screen is

$P_{1}\left(x\right)=\left|\psi_{1}\left(x\right)\right|^{2}$.

Similarly, with the second slit open and the first closed, we have $P_{2}\left(x\right)=\left|\psi_{2}\left(x\right)\right|^{2}$. But when both slits are open, the principle of superposition tells us that it is the wavefunctions that add, not the probabilities. This means that we have $\psi_{12}\left(x\right)=\psi_{1}\left(x\right)+\psi_{2}\left(x\right)$ and so the new probability distribution is

We see that the cross-terms mean that $P_{12}\neq P_{1}+P_{2}$. These cross-terms are responsible for the interference pattern.

The conclusion that each electron necessarily passes through both slits is, to say the least, a little surprising. One could try to find a clever way to evade it, perhaps by setting up some ingenious apparatus near one of the slits that will tell us for sure which slit the particle travelled through. It turns out that any attempt to do this will necessarily disturb the state of the particle. For example, if you know for sure that the particle passed through slit 1 (or, equivalently, that it did not pass through slit 2) then the wavefunction collapses to $\psi_{12}\left(x\right)\rightarrow\psi_{1}\left(x\right)$, and the interference pattern disappears.

Quantum mysteries are often phrased as the question of whether an electron is a particle or a wave. The answer that is usually given is: both. A much better answer is: neither. Electrons, like all quantum objects, exist in a regime where our everyday experience is of little use. The idea that such objects can be described in terms of things that we can find on the beach is, to put it mildly, a little immature.

A quantum particle is, instead, described by a wavefunction. The right way to proceed is not to search for some classical metaphor that will capture some aspect of the quantum particle, but instead to better understand the rules that govern these wavefunctions.

I would like to emphasize this experiment once again. Notice that, individually, each electron is detected on the screen as a localized event, that is, as a point-like detection. This is its particle-like characteristics. It is not possible to observe the interference pattern from a single electron.

The pattern emerges statistically when we repeat the experiment many times, since the distribution of electrons on the screen follows a probability distribution that exhibits the characteristic features of wave interference. This is its wave-like characteristic.

But notice that the mathematical description in terms of a wave applies to the wave function, which we use to calculate probabilities, such as the probability of finding the electron at a given position, rather than as a direct description of the motion of the electron itself.

We should also keep clearly in mind that this does not mean that the electron “knows” that it is being observed or responds to a consciousness. What matters is the physical interaction involved in obtaining information about the system. In order to determine which slit the electron passed through, it is necessary to interact with the electron in a way that provides information about its path. When we obtain such information, we can determine its position with respect to the measurement being performed. Mathematically, this interaction changes the quantum state of the system. In terms of wave-function collapse, it is as if we are starting the experiment again: the state is no longer the same as it was before the measurement. The change does not occur because we are “observing” the electron, but because we are physically interacting with it.

The underlying mathematics¶

Let us now resume the discussion, drawing on the work of Professor Cresser (Cresser (2001)) to discuss the topic with a bit more mathematics.

The argument to be presented here is not meant to be rigorous, but more to suggest a way of thinking about quantum mechanics that can be made more precise, and very general in that it can be applied to any physical system. The experimental set up will be as before. Particles, which have all passed through exactly the same preparation procedure and hence are presumably all in the same state, are produced at a source $S$ . These particles are then incident on a screen in which there are two narrow slits, 1 and 2, and beyond this screen is a further screen which the particles will ultimately strike, the point at which they hit this screen being registered in some way.

Since the particles are all prepared in the same way, they will presumably all be associated with the same wave function. The detailed form of the wave function will depend on many things, not the least of which is that the particles are all produced at the source S . We will introduce the notation for the wave function $\Psi_{s}\left(x\right)$ as the probability amplitude of finding the particle at $x$ given that it originated at the source $S$^[1] . Let us move the discussion forward and denote $\Psi_{s}\left(x\right)=\left\langle x|s\right\rangle$.

We will not be concerning ourselves with the possible time dependence of the wave function here – in effect we are assuming some kind of steady state. This wave function will propagate through space and at the positions of the two slits will have values $\Psi_{s}\left(n\right)=\left\langle n|s\right\rangle$, $n=1,2$. These waves will then pass through these slits on their way towards the observation screen.

The task then is to determine the amplitude of the waves at a point $x$ on the observation screen due to waves that originated at each of the slits. To do that we will first of all suppose that, if the amplitude of the wave incident on slit 1 is unity^[2], then the resulting wave amplitude at a position $x$ on the screen is $\Psi_{1}\left(x\right)=\left\langle x|1\right\rangle$.

But since the amplitude of the wave at slit 1 is $\Psi_{s}\left(1\right)=\left\langle 1|s\right\rangle$ then it appears reasonable to suppose that we scale up the amplitude of the wave from slit 1 incident on the observation screen at $x$ by the same factor, i.e. $\Psi_{s}\left(1\right)$, so that the amplitude of the wave incident on the screen at $x$ that emerged from the slit 1 will be $\Psi_{s}\left(1\right)\Psi_{1}\left(x\right)$. Of course, this is an assumption, but it is what is observed for, say, light waves passing through a slit – if the amplitude of a light wave incident on a slit is doubled, then the amplitude of the wave that emerges from the slit is also doubled. Light waves are linear. Probability amplitudes are hence also assumed to be linear.

We have a similar case for the second slit. Consequently, the total amplitude of the wave at x will be

where $\Psi_{s}\left(x\right)$ is the amplitude of waves at $x$ that originated from the source $S$ . This then is the probability amplitude of observing a particle at $x$ given that it originated from the source S , i.e. by the Born interpretation $\left|\Psi_{s}\left(x\right)\right|^{2}dx$ is the probability density of observing a particle in the region $x$ to $x+dx$. This result is the same as the one presented before, but here we have expressed this total probability amplitude as a sum of two contributions of the form:

Probability amplitude for the particle to pass from the source $S$ to point $x$ through slit $n$ $=$ Probability amplitude of observing the particle at slit $n$ given that it originated from the source S $\times$ Probability amplitude of observing the particle at slit $n$ given that it originated from the source $S$^[3].

Remembering what we had:

So, evidently: $\psi_{n}=\Psi_{s}\left(n\right)\Psi_{n}\left(x\right)$. We can now better understand the adopted vector notation:

• $\left|s\right\rangle \equiv$ state of the particle when at the position of the source $S$;

• $\left|n\right\rangle \equiv$ state of the particle when at the position of slit $n$;

• $\left|z\right\rangle \equiv$ state of the particle when at the position $x$;

So $\Psi_{s}\left(x\right)\left\langle x|s\right\rangle$is the probability amplitude of finding the particle at $x$, i.e. in the state $\left|x\right\rangle$, given that it was initially at the source $S$ , i.e. in the state $\left|s\right\rangle$. Since $\left\langle \cdot|\cdot\right\rangle$ is a scalar, we can rearrange:

We can do something radical, which is to ’cancel’ the $\left\langle x\right|$ because the expression above will be true for all values of the variable $x$, i.e. it does not hold true for just one value of $x$. If we denote the coefficients by $c_{n}=\left\langle n|s\right\rangle$, we can also write:

That is, we have a state vector $\left|\psi\right\rangle \equiv\left|s\right\rangle$, written in the vectors $\left\{ \left|1\right\rangle ,\left|2\right\rangle \right\}$. If they form a complete orthonormal basis, note that the probability of finding the particle at slit $n$ is $\left|\left\langle n|\psi\right\rangle \right|^{2}=\left|c_{n}\right|^{2}$. But after a which-slit measurement, if the particle is found at slit $n$, the wave function becomes $\left|\psi\right\rangle =\left|n\right\rangle$, such that the probability of finding the particle at $x$ is

This is significantly different from what we would observe if we considered only the probability of the particle being identified at $x$, without the intermediate measurement:

The new description is a very potent one, and constitutes the central feature of quantum mechanics, the idea that a system, in some sense, can be described simultaneously in a number of different physical states, otherwise known as a ‘superposition of states’.

It is this sort of result that lies at the heart of quantum mechanics – the idea that a system in a certain state can behave as if it is in oa number of other states, and to do so in a way that is probabilistic in nature.

We will show now that the result which we have ‘derived’ in the case of the two slit experiment can also be shown (and much more rigorously) to be an intrinsic part of the Stern-Gerlach experiment.

It is interesting to note that, in fact, $\left\langle x|s\right\rangle =\left\langle x|1\right\rangle \left\langle 1|s\right\rangle +\left\langle x|2\right\rangle \left\langle 2|s\right\rangle$ , corresponds to a statistical description of the behavior of nature through the wave function (or state vector), which is supported by circumstantial evidence, though neither the wave function itself nor the probability amplitudes are directly observed.