2 The Double Slit Experiment

2.1 Particles and Waves

In classical mechanics, you have learned about two distinct things: particles and waves. In this chapter we will explore the ‘double-slit experiment’. This shows that the elementary constituents of matter (electrons, photons, any of the elementary particles) exhibit characteristics of both particles and waves, known as ‘particle-wave duality’. The experiment also demonstrates that the laws of nature are fundamentally probabilistic.

First, a quick reminder about the classical mechanics of particles and waves moving in one dimension.

• A particle has a definite position and momentum \((x(t),p(t))\) at each time \(t\). The time evolution of position and momentum is found by solving Hamilton’s equations, \[\dot{x} = \frac{\partial H}{\partial p} \qquad \dot{p} = - \frac{\partial H}{\partial x} \, ,\] where \(H\) is the Hamiltonian function. For a particle of mass \(m\) moving in a potential \(V(x)\) the Hamiltonian is \[H = \frac{p^2}{2m} +V(x) \, .\]

• A wave is described by an amplitude \(\psi(x,t)\). This might be the displacement of a string, or a component of the electromagnetic field, as a function of position \(x\) and time \(t\). The amplitude is a solution of a partial differential equation, such as the wave equation \[\frac{\partial^2\psi}{\partial t^2} - v^2 \frac{\partial^2\psi}{\partial x^2} = 0 \, ,\] where \(v\) is the velocity.

Experiments probing microscopic distances show that the elementary constituents of matter exhibit characteristics of both particles and waves. To see this concretely, we are going to examine the ‘double-slit experiment’. Although it is very often presented as a thought experiment, it has actually been performed it in the laboratory, for instance with electrons. A more extensive account can be found in Volume III of the Feynman Lectures.

For arguments sake, we will imagine performing this experiment with electrons. We will first explain the outcome of the experiment according to classical mechanics, assuming that electrons are particles and then waves. We will then explain how electrons actually behave in nature.

2.2 Double-Slit : Particles

Let us first suppose that electrons are particles. There is a source emitting these particles at a uniform rate in random directions towards a screen with two small slits \(S_1\) and \(S_2\). The particles that pass through one of the slits arrive one at a time at a detector \(D\) on the other side of the screen.

By averaging over a long period of time, the detector measures the rate that particles arrive per unit area, as a function of the vertical direction \(x\). We call this the ‘intensity’. Suppose that:

• The intensity measured with only \(S_1\) open is \(I_1(x)\).

• The intensity measured with only \(S_2\) open is \(I_2(x)\).

• The intensity measured with both \(S_1\) and \(S_2\) open is \(I(x)\).

Since particles arrive at the detector one at a time and must pass through either \(S_1\) or \(S_2\), the intensities add up \[I(x) = I_1(x)+I_2(x) \, . \label{eq:intensity-particles}\] The result is thus simply the appearance of one or two peaks, depending on the separation of the slits and the distance to the screen.

2.3 Double-Slit : Waves

Now suppose instead that electrons are waves with amplitude \(\psi(x,t)\). There is a source emitting waves uniformly towards a screen with two small slits \(S_1\) and \(S_2\). The waves pass through the slits and arrive continuously at a detector \(D\).

Suppose that

• The amplitude at the detector with only \(S_1\) open is \(\psi_1(x)\).

• The amplitude at the detector with only \(S_2\) open is \(\psi_2(x)\).

• The amplitude at the detector with both \(S_1\) and \(S_2\) open is \(\psi(x)\).

Let us assume that the wave amplitude obeys a linear partial differential equation. Then the principle of superposition means that \[\psi(x) = \psi_1(x) + \psi_2(x) \, .\] Note that we are ignoring the dependence on time \(t\), which is not important in the argument that follows.

By averaging over a long period of time, the detector measures the rate that energy is deposited per unit area, as a function of the vertical direction \(x\). We again call this the ‘intensity’. The energy carried by a wave is proportional to the modulus squared of the amplitude. Ignoring the constant of proportionality, \[I_1(x) = |\psi_1(x)|^2\,, \qquad I_2(x) = |\psi_2(x)|^2\,, \qquad I(x) = |\psi(x)|^2\,,\] are the intensities measured by the detector with only \(S_1\), only \(S_2\), and both \(S_1\) and \(S_2\) open respectively. They are related by \[\begin{aligned} I(x) & = |\psi(x)|^2 \\ & = |\psi_1(x) + \psi_2(x)|^2 \\ & = |\psi_1(x)|^2 + |\psi_2(x)|^2 + 2 \mathrm{Re}(\psi_1(x)\overline{\psi_2(x)})\\ & = I_1(x) + I_2(x) + \sqrt{I_1(x)I_2(x)} \cos(\delta(x))\,, \label{eq:intensity-waves} \end{aligned}\] where \(\delta(x)\) is the relative phase of \(\psi_1(x)\) and \(\psi_2(x)\). The additional term compared to the result for particles is known as the ‘interference’ term. It generates the kind of interference pattern illustrated in the figure above.

Let us compute the interference pattern more explicitly using a series of approximations. For familiarity, let us suppose that the amplitude \(\psi(x,t)\) obeys the wave equation with velocity \(v\). We further assume the slits are very thin, so that we are effectively dealing with two pointlike sources, producing the two waves \[\psi_1(x,t) = C \, e^{i ( k r_1 - \omega t)}\,, \quad \psi_2(x,t) = C \, e^{i ( k r_2 - \omega t)} \, ,\] where

• \(C\) is a normalisation constant that is unimportant in what follows,

• \(k = \omega / v\) where \(\omega\) is the angular frequency of the wave,

• \(r_1\), \(r_2\) are the distances from the slits \(S_1\), \(S_2\) to a point on the detector at height \(x\).

With these approximations, the total amplitude is \[\psi(x,t) = C(e^{ikr_1} +e^{ikr_2} ) e^{-i\omega t}\] and therefore the intensity at a point \(x\) on the screen will be given by \[\begin{aligned} I(x) \label{e:interference} & = C^2 | \, e^{ikr_1} +e^{ikr_2} \, |^2 \\ & = 2C^2 \left( 1+ \cos( k(r_1-r_2 ) ) \right) \\ & = 4 C^2 \cos^2\left(\frac{k}{2}(r_1-r_2 ) \right) \, . \end{aligned}\] There is constructive interference when \(k(r_1-r_2) = 2 n \pi\) and destructive interference when \(k(r_1-r_2) = (2n+1) \pi\) where \(n\in \mathbb{Z}\), so the intensity will clearly display an interference pattern.

To determine the intensity function \(I(x)\) explicitly is tricky since the distances \(r_1\), \(r_2\) are complicated functions of \(x\). The two right-angled triangles give \[\label{e:r1r2} r_1^2 = L^2 + \left(x-\frac{a}{2}\right)^2\,,\quad r_2^2 = L^2 + \left(x+\frac{a}{2}\right)^2\,.\] Taking a square root to obtain \(r_1\) and \(r_2\), we can then subtract these two expressions to get the phase we have in \(\eqref{e:interference}\). To get some more insight into that expression, it is useful to expand \(r_1\) and \(r_2\) for small values of \(a\), that is, for \(a\ll x\). Using a Taylor expansion we get \[r_1 = \sqrt{L^2 + x^2} - \frac{a x}{2\sqrt{L^2+x^2}} + {\cal O}(a^2)\,,\] and similar for \(r_2\). Subtracting the two and assuming furthermore that \(x\ll L\), we get the simple expression \(r_2-r_1=a x /L\), and using this in \(\eqref{e:interference}\) thus gives \[I(x) \approx 4 C^2 \cos^2\left( \frac{ka}{2L} x \right) \, .\]

The result is thus quite different from that in the previous section: waves give a characteristic intensity interference pattern, with an approximate separation between maxima as given above. To plot the full pattern, just use \(\eqref{e:r1r2}\) directly.

2.4 Double-Slit: the real world

Let us summarise our results for particles and waves and compare to the experimental result for electrons.

Particles

1. Particles arrive one at a time.

2. There is no interference, \(I = I_1+I_2\).

Waves

1. Waves arrive continuously.

2. There is interference, \(I = I_1+I_2+ 2\sqrt{I_1 I_2} \cos \delta\).

However, when the experiment is performed with real-world electrons, the result is neither of these two. Despite the fact that we can set up the experiment such that we can really register the arrival of electrons on the screen one-by-one, there nevertheless is an interference pattern. So we find that:

Real world electrons

1. Electrons arrive one at a time.

2. There is interference, \(I = I_1+I_2+ 2\sqrt{I_1 I_2} \cos \delta\).

So electrons exhibit characteristics of both particles and waves. They arrive one by one like particles, but the rate that particles arrive exhibits interference like a wave. This behaviour cannot be explained in classical mechanics.

The double experiment is, however, compatible with the following statements:

• Electrons arrive at the detector one at a time.

• Each of them is described by an amplitude \(\psi(x,t)\).

• The modulus squared \(|\psi(x,t)|^2\) is the probability distribution for the detector to find an electron at position \(x\) at time \(t\).

• The amplitude \(\psi(x,t)\) obeys a linear partial differential equation to ensure the principle of superposition.

We must abandon the idea that an electron has a definite position \(x(t)\) at all times. Instead, we may only predict the probability to measure the position \(x\) in a certain region. The same conclusion applies to momentum, energy and any other measurable quantity. The laws of nature are inherently probabilistic!

2.5 Problems

1. Approximations:

   We have discussed several approximations to arrive at the wave interference pattern \[I(x) = 4 C^2 \cos^2\left(\frac{k}{2}(r_1-r_2 ) \right) \, .\] There is one hidden assumption which we have not mentioned explicitly, which is related to the normalisation constant \(C\). Can you spot it? What would the effect be of relaxing this assumption?

   Solution ▶ Strictly speaking, the intensity of the wave drops as a function of the distance from the source, unless you are in one dimension. In two dimensions, you would have \(C \sim 1/r\). This means that as you get further away from the axis, and the distance to the slit gets larger, the intensity drops.

2. PROBLEMS CLASS 1
   Finite-width single slit Another approximation we have made is that the slits are infinitesimally thin, which of course in the real world would not allow for electrons to pass through. We can do better than that.

   Consider first a single slit of width \(b\) as in the figure below, centered on the \(x\)-axis.

   

   You can view this as a superposition of a continuum of sources, so that the wave is given by \[\label{e:psicont} \psi(x,t) = C \int_{s=-b/2}^{b/2} e^{i k r(x,s) - i\omega t} {\rm d}s\,.\] Express \(r(x,s)\) as the sum of \(r_0\) plus a correction, then perform this integral. Show that the intensity \(|\psi(x,t)|^2\) takes the form \[|\psi(x,t)|^2 = \tilde{C}^2 \frac{\sin^2\beta}{\beta^2} \,,\quad \beta = \frac{bkx}{2L}\,.\] Give a qualitative plot of \(|\psi(x,t)|^2\) versus \(x\).

   Solution ▶ Start by writing \[r_0^2 = L^2 + x^2\,,\quad r^2(x,s) = L^2 + (x-s)^2\,.\] Expanding to first order in \(x/L\) gives \[r_0(x) = L + {\cal O}((x/L)^2)\,,\quad r(x,s) = L\sqrt{(s/L)^2+1} - \frac{L}{\sqrt{(s/L)^2+1}} (x/L) + {\cal O}((x/L)^2)\,.\] Now also expand \(r(x,s)\) to first order in \(s/L\) to get \[r(x,s) = L - L (s/L) (x/L) + {\cal O}((x/L)^2, (s/L)^2)\,.\] A Taylor series expansion to lowest order in \(s/L\) and \(x/L\) thus yields \[r(x,s) = r_0 - \frac{sx}{L} + {\cal O}((x/L)^2, (s/L)^2 )\,.\] This means that the amplitude at the screen is \[\begin{aligned} \psi(x) &\approx C \int_{-b/2}^{b/2} e^{ikr(x,s) - i\omega t} {\rm d}s = C e^{ikr_0 - i\omega t} \int_{-b/2}^{b/2} e^{-iksx/L} {\rm d}s\\[1ex] &= C e^{ikr_0-i\omega t} \frac{iL}{kx} e^{-iksx/L}\Big|_{s=-b/2}^{b/2} = C e^{ikr_0-i\omega t} \frac{2L}{kx} \sin\left(\frac{kbx}{2L}\right)\,. \end{aligned}\] The intensity is the norm-squared, so \[|\psi(x)|^2 = |C|^2 b^2 \frac{\sin^2\beta}{\beta^2}\,,\quad \beta = \frac{kbx}{2L}\,.\] A plot for \(kb/2L=1\) is given below.

   

3. Finite-width double slit:

   Now consider a situation with finite-width slits, as in the figure below. Instead of integrating the contributions from \(-b/2\) to \(b/2\), we should now integrate \(\eqref{e:psicont}\) over \((a-b)/2\) to \((a+b)/2\), and then add the contribution from the second slit.

   

   Perform these integrals. Show that the result can be written in the form of the product of two factors, one which is the interference pattern of infinitesimally thin double slits separated by a distance \(a\), and one the interference pattern of a single slit of finite width \(b\).

   Solution ▶ Instead of doing the integral over \(-b/2 < s < b/2\), we now have two integrals, one ranging over \((a-b)/2 < s < (a+b)/2\) and the other ranging over \(-(a+b)/2 < s < -(a-b)/2\). The integrand is the same. Working out the 4 exponential terms, you get, up to the factors irrelevant for \(|\psi|^2\), \[\psi(x) \propto \left(\frac{2L}{kx} \sin\beta\right) \cos\alpha\,,\] where the first factor is the single-slit result and the second factor the double-slit. We can write the intensity as \[|\psi(x)|^2 \propto \frac{\sin^2\beta}{\beta^2} \cos^2(\alpha)\,,\quad \beta = \frac{kbx}{2L}\,,\quad \alpha=\frac{kax}{2L}\,.\] Plots for separation \(a\) equal to \(5\) times the slit size \(b\) are given below.