Two-slit experiment

A somewhat simplified version

We consider a version of the experiment that allows us to use the mathematics we have learned so far. The actual experiment involves continuous states, rather than the discrete states we have described in this course.

Figure 6.3 - Simplified two-slit experiment

(I) Source

At $O$, we have a source of particles that are fired towards a screen (II) with some randomness.

(II) Screen

Most of the particles will hit the screen and be blocked, but we open thin slits in the screen at $A$, $B$ so that some of the particles get through.

Nb. We assume that the particles do not react to, and are not affected by, the screen or the slits.

(III) Detector

Here we have some sort of detector which allows us to count the number of particles arriving at various locations.

Normally, this would allow us to calculate a continuous probability distribution of all the possible locations on the detector. The simplification we consider is that, rather than a continuous detector, we have a discrete set of $N$, say, detectors spaced out at equal intervals.

We would like to calculate the probability that a particle will arrive at the $m^{\text{th}}$ detector. In fact, we shall only calculate the number of particles arriving at a particular detector. Professor Susskind notes that, in order to calculate the probability proper, we would need to consider the total probability, which means taking into account the particles that don't arrive at any of the (discrete) detectors.

One slit open

The particle starts out at the source, $O$, which we can denote by the state $\ket{O}$.

We first consider the case where only the slit at $A$ is open. If a particle fired from the source arrives at any of the detectors, then it must have passed through $A$, which we denote by the state $\ket{A}$. We say that the state has evolved from $O$ to $A$, or write $$\ket{O} \rightarrow \ket{A}$$

Let the state $\ket{n}$ represent the particle arriving at the $n^{\text{th}}$ detector. If the particle can arrive at any of the detectors, then the particle must evolve to a superposition of the possible states. That is, $$\ket{O} \rightarrow \ket{A} \rightarrow \ket{D_A}$$ where, given some $\psi_n \in \mathbb{C}$, $$\ket{D_A} = \sum_{n = 1}^{N}\psi_n\ket{n}$$

Nb. We can't demand that $\sum \psi_n^* \psi_n = 1$, since we aren't including the particles that don't arrive at any of the detectors.

The projection operator associated with arriving at the $m^{\text{th}}$ detector is simply the dyad $\ket{m}\bra{m}$, hence the average number of particles arriving at the $m^{\text{th}}$ detector is \begin{align*}\bra{D_A}\left(\ket{m}\bra{m}\right)\ket{D_A}&= \left(\sum_{k = 1}^{N}\bra{k}\psi_k^*\right)\ket{m}\bra{m}\left(\sum_{n = 1}^{N}\psi_n\ket{n}\right)\\&= \left(\sum_{k = 1}^{N}\braket{k}{m}\psi_k^*\right)\left(\sum_{n = 1}^{N}\psi_n\braket{m}{n}\right)\\&= \psi_m^*\psi_m\end{align*}

Similarly, we can describe the situation where only $B$ is open. Then, for some $\phi_n \in \mathbb{C}$, $$\ket{D_B} = \sum_{n = 1}^{N}\phi_n\ket{n}$$ such that $$\ket{O} \rightarrow \ket{B} \rightarrow \ket{D_B}$$ and the average number of particles arriving at the $m^{\text{th}}$ detector is $$\bra{D_B}\left(\ket{m}\bra{m}\right)\ket{D_B} = \phi_m^*\phi_m$$

Both slits open

Classically, with both slits open, we would expect that the average number of particles arriving at the $m^{\text{th}}$ detector is the average number arriving via $A$, plus the average number arriving from $B$, $$\psi_m^*\psi_m + \phi_m^*\phi_m$$

Figure 6.4 - Classical versus quantum outcomes

In figure 6.4, (IV) is the (continuous) result with only one of the slits open and (V) is the result predicted by classical mechanics.

(VI) shows (something like) the experimental result, described by quantum theory, which displays interference.

Nb. We need to normalise these results to describe a probability distribution.

The interference appears because, in quantum mechanics, we don't add the probabilities (averages) together, but the states themselves. Equivalently, we add the probability amplitudes.

A particle arriving at the $m^{\text{th}}$ detector could have come via $A$ or $B$, and so we need to consider the superposition of these states. That is, $$\ket{O} \rightarrow \ket{A} + \ket{B}$$

By the time evolution postulate, $$\left.\begin{matrix}\ket{A} \rightarrow \ket{D_A}\\ \ket{B} \rightarrow \ket{D_B} \end{matrix}\right\} \Rightarrow \ket{A} + \ket{B} \rightarrow \ket{D_A} + \ket{D_B} = \sum_{n=1}^{N}\left(\psi_n + \phi_n\right)\ket{n}$$

Taking the average of this state with respect to the $m^{\text{th}}$ detector now, we get \begin{align*}\sum_{k=1}^{N} \bra{k} \left(\psi_k^* + \phi_k^*\right) \ket{m}\bra{m} \sum_{n=1}^{N}\left(\psi_n + \phi_n\right)\ket{n}&= \left(\psi_m^* + \phi_m^*\right)\left(\psi_m + \phi_m\right)\\&= \psi_m^*\psi_m + \phi_m^*\phi_m + \psi_m^*\phi_m + \phi_m^*\psi_m\end{align*}

The first two terms are what we would expect classically, and the second two cross-terms are the interference terms, which result in a real number (adding a complex number to its conjugate always results in a real number).

A special case

The simplest case we can consider would be to place a detector exactly midway between the two open slits.

Figure 6.5 - Symmetric case

By the symmetry of the situation, the two amplitudes, for $A$ and $B$ would be equal, in which case the average number of particles arriving would then be $$\psi_0^*\psi_0 + \psi_0^*\psi_0 + \psi_0^*\psi_0 + \psi_0^*\psi_0 = 4\psi_0^*\psi_0$$ which is the correct experimental result.

Notice this is twice the value of the classical result, $$\psi_0^*\psi_0 + \psi_0^*\psi_0 = 2\psi_0^*\psi_0$$