Double Slit Experiment

Prerequisites: Superposition. Full path (18 pages)

• Intuitive
• Concrete
• Abstract
• Why is it interesting?

Intuitive

When we shoot balls onto a double slit, most balls will end up somewhere in the middle. If we open just one of the slits, we will find most balls somewhere directly behind the slit. The result with both slits open is simply the addition of the two results with one slit closed.

In contrast, if we send waves like they appear in water onto a double slit the result is remarkably different. While the results with one slit closed are the same as for the balls, we see something completely different if both slits are open at the same time: an interference pattern.

Now, if we send individual electrons onto a double slit we would naively expect the same result as for balls. However, in reality, we observe an interference pattern like we did for waves.

The double slit experiment, therefore, exemplifies that we need waves to describe particles.

• Surfing and the double-slit experiment by Brian Skinner

Concrete

Take two narrow slits a distance $d$ apart and a screen at distance $L \gg d$. A particle with momentum $p$ has the de Broglie wavelength $\lambda = h/p$. In quantum mechanics each slit contributes an amplitude at a point on the screen, $\psi_1$ from slit 1 and $\psi_2$ from slit 2, and with both slits open the amplitude is their sum. The probability density of finding the particle there is (Born rule)

$$P_{12} = |\psi_1 + \psi_2|^2 = |\psi_1|^2 + |\psi_2|^2 + 2\,\mathrm{Re}\left(\psi_1^\ast \psi_2\right) = P_1 + P_2 + 2\,\mathrm{Re}\left(\psi_1^\ast \psi_2\right).$$

For classical balls only $P_1 + P_2$ appears. The last term is the interference term, and it is the whole difference between the two cases.

At angle $\theta$ from the axis, the paths from the two slits differ in length by $d\sin\theta$, so the amplitudes differ in phase by

$$\delta = \frac{2\pi d \sin\theta}{\lambda}.$$

If both slits contribute the same magnitude, $|\psi_1| = |\psi_2|$, and we write $I_0 = |\psi_1|^2$, then

$$P_{12}(\theta) = 2 I_0 (1 + \cos\delta) = 4 I_0 \cos^2\left(\frac{\pi d \sin\theta}{\lambda}\right).$$

Bright fringes sit at $d\sin\theta = n\lambda$ with integer $n$, and dark fringes halfway between, where $P_{12} = 0$ although $P_1 + P_2 \neq 0$. For small angles the position on the screen is $y \approx L\theta$, so neighbouring bright fringes are separated by

$$\Delta y \approx \frac{\lambda L}{d}.$$

Slits of finite width $a$ multiply this pattern by the single-slit envelope $\left(\sin\beta/\beta\right)^2$ with $\beta = \pi a \sin\theta/\lambda$.

Example. An electron accelerated from rest through a voltage $V$ has (non-relativistically) $p = \sqrt{2 m_e e V}$, so $\lambda = h/\sqrt{2 m_e e V} \approx 1.23\,\text{nm}/\sqrt{V/\text{volt}}$. At $V = 100\,\text{V}$ this gives $\lambda \approx 0.12\,\text{nm}$. To get fringes that one can resolve, the slit separation must therefore be very small, which is why electron interference experiments use micrometre-scale structures or biprisms instead of everyday slits.

Which-path information. Suppose a detector next to the slits ends up in state $|D_1\rangle$ if the particle passed slit 1 and $|D_2\rangle$ if it passed slit 2. The joint state is $\psi_1 |D_1\rangle + \psi_2 |D_2\rangle$, and the probability on the screen becomes

$$P = |\psi_1|^2 + |\psi_2|^2 + 2\,\mathrm{Re}\left(\psi_1^\ast \psi_2 \langle D_1 | D_2 \rangle\right).$$

If the detector states are orthogonal, so that the path can be read off in principle, the interference term vanishes and we get the classical sum $P_1 + P_2$. Partial path information ($0 < |\langle D_1|D_2\rangle| < 1$) gives fringes of reduced contrast.

• A great discussion can be found in Vol. 3 of the Feynman Lectures.

Abstract

In the Hilbert space formulation, let $|1\rangle$ and $|2\rangle$ be the (orthogonal) states of a particle that has just passed slit 1 or slit 2, and let $U$ be the unitary evolution from the slits to the screen. Write $\psi_j(x) = \langle x | U | j \rangle$. A particle prepared in $\alpha|1\rangle + \beta|2\rangle$ arrives at $x$ with probability density $|\alpha\psi_1(x) + \beta\psi_2(x)|^2$. In the path integral language, $\psi_j(x)$ is the sum of $e^{iS/\hbar}$ over all paths through slit $j$, and the total amplitude is the sum over paths through either slit.

Decoherence and complementarity. Couple the particle to an environment or detector $\mathcal{E}$ initially in $|E_0\rangle$, with the interaction

$$|j\rangle \otimes |E_0\rangle \mapsto |j\rangle \otimes |E_j\rangle, \qquad j = 1,2.$$

Tracing out $\mathcal{E}$ gives the reduced density matrix of the particle,

$$\rho = |\alpha|^2 |1\rangle\langle 1| + |\beta|^2 |2\rangle\langle 2| + \alpha\beta^\ast \langle E_2 | E_1 \rangle\, |1\rangle\langle 2| + \alpha^\ast\beta \langle E_1 | E_2 \rangle\, |2\rangle\langle 1|.$$

The screen distribution $\langle x | U\rho U^\dagger | x \rangle$ depends on the off-diagonal elements only through the interference term, which is suppressed by the factor $\langle E_1 | E_2 \rangle$. For $|\alpha| = |\beta|$ and $\psi_1$, $\psi_2$ of equal magnitude in the region of interest, the fringe visibility $V = (P_{\max} - P_{\min})/(P_{\max} + P_{\min})$ equals $|\langle E_1 | E_2 \rangle|$. The path distinguishability $D$ is defined through the best probability $p_{\text{succ}}$ of identifying the path by a measurement on $\mathcal{E}$, $D = 2p_{\text{succ}} - 1$. For two pure states with equal priors the Helstrom bound gives $p_{\text{succ}} = \frac{1}{2}\left(1 + \sqrt{1 - |\langle E_1|E_2\rangle|^2}\right)$, hence

$$D^2 + V^2 = 1$$

in this case. For mixed detector states or unequal weights one gets the inequality $D^2 + V^2 \le 1$ (Englert's duality relation). This is a quantitative form of the statement that which-path information and interference exclude each other.

Interference is second order. Write $P_A$ for the probability at a point on the screen with only the slits in the set $A$ open. The Born rule is quadratic in amplitudes, and this implies that with three slits there is no new interference beyond pairs:

$$P_{123} - P_{12} - P_{13} - P_{23} + P_1 + P_2 + P_3 = 0.$$

Two-slit interference, $P_{12} - P_1 - P_2 \neq 0$, is the lowest order term that distinguishes quantum from classical probability. Experiments with three or more slits test the vanishing of the third-order term as a check of the Born rule.

Why is it interesting?

[The Double Slit Experiment] which is impossible, absolutely impossible, to explain in any classical way, has in it the heart of quantum physics

Richard Feynman

…the most (in)famous quantum mechanical experiment of all: the double-slit experiment.Surfing and the double-slit experiment by Brian Skinner

Path to this page

1. Calculus
2. Imaginary Numbers
3. Linear Algebra
4. Functional
5. Hilbert Space
6. Legendre Transformation
7. Newton's Second Law
8. Vector Calculus
9. Newtonian Mechanics
10. Variational Calculus
11. Configuration Space
12. Euler-Lagrange Equations
13. Lagrangian Formalism
14. Phase Space
15. Hamiltonian Formalism
16. Schrödinger Equation
17. Canonical Quantum Mechanics
18. Superposition