The two-slit interference pattern
Now we would like to consider a matter which was discussed in some detail in Chapter 1. This time we will do it with the full glory of the amplitude idea to show you how it works out. We take the same experiment shown in Fig. 3–1, but now with the addition of a light source behind the two holes, as shown in Fig. 3–3. In Chapter 1, we discovered the following interesting result. If we looked behind slit $1$ and saw a photon scattered from there, then the distribution obtained for the electrons at $x$ in coincidence with these photons was the same as though slit $2$ were closed. The total distribution for electrons that had been “seen” at either slit $1$ or slit $2$ was the sum of the separate distributions and was completely different from the distribution with the light turned off. This was true at least if we used light of short enough wavelength. If the wavelength was made longer so we could not be sure at which hole the scattering had occurred, the distribution became more like the one with the light turned off. Let’s examine what is happening by using our new notation and the principles of combining amplitudes. To simplify the writing, we can again let $\phi_1$ stand for the amplitude that the electron will arrive at $x$ by way of hole $1$, that is, \begin{equation*} \phi_1=\braket{x}{1}\braket{1}{s}. \end{equation*}  Similarly, we’ll let $\phi_2$ stand for the amplitude that the electron gets to the detector by way of hole $2$: \begin{equation*} \phi_2=\braket{x}{2}\braket{2}{s}. \end{equation*}  These are the amplitudes to go through the two holes and arrive at $x$ if there is no light. Now if there is light, we ask ourselves the question: What is the amplitude for the process in which the electron starts at $s$ and a photon is liberated by the light source $L$, ending with the electron at $x$ and a photon seen behind slit $1$? Suppose that we observe the photon behind slit $1$ by means of a detector $D_1$, as shown in Fig. 3–3, and use a similar detector $D_2$ to count photons scattered behind hole $2$. There will be an amplitude for a photon to arrive at $D_1$ and an electron at $x$, and also an amplitude for a photon to arrive at $D_2$ and an electron at $x$. Let’s try to calculate them. Although we don’t have the correct mathematical formula for all the factors that go into this calculation, you will see the spirit of it in the following discussion. First, there is the amplitude $\braket{1}{s}$ that an electron goes from the source to hole $1$. Then we can suppose that there is a certain amplitude that while the electron is at hole $1$ it scatters a photon into the detector $D_1$. Let us represent this amplitude by $a$. Then there is the amplitude $\braket{x}{1}$ that the electron goes from slit $1$ to the electron detector at $x$. The amplitude that the electron goes from $s$ to $x$ via slit $1$ and scatters a photon into $D_1$ is then \begin{equation*} \braket{x}{1}\,a\,\braket{1}{s}. \end{equation*}  Or, in our previous notation, it is just $a\phi_1$. There is also some amplitude that an electron going through slit $2$ will scatter a photon into counter $D_1$. You say, “That’s impossible; how can it scatter into counter $D_1$ if it is only looking at hole $1$?”  If the wavelength is long enough, there are diffraction effects, and it is certainly possible. If the apparatus is built well and if we use photons of short wavelength, then the amplitude that a photon will be scattered into detector $1$, from an electron at $2$ is very small. But to keep the discussion general we want to take into account that there is always some such amplitude, which we will call $b$. Then the amplitude that an electron goes via slit $2$ and scatters a photon into $D_1$ is \begin{equation*} \braket{x}{2}\,b\,\braket{2}{s}=b\phi_2. \end{equation*}   The amplitude to find the electron at $x$ and the photon in $D_1$ is the sum of two terms, one for each possible path for the electron. Each term is in turn made up of two factors: first, that the electron went through a hole, and second, that the photon is scattered by such an electron into detector $1$; we have \begin{equation} \label{Eq:III:3:8} \biggl\langle \begin{subarray}{l} \displaystyle \text{electron at $x$}\\[1ex] \displaystyle \text{photon at $D_1$} \end{subarray} \!\! \biggm| \! \begin{subarray}{l} \displaystyle \text{electron from $s$}\\[1ex] \displaystyle \text{photon from $L$} \end{subarray} \biggr\rangle= a\phi_1\!+b\phi_2. \end{equation}  We can get a similar expression when the photon is found in the other detector $D_2$. If we assume for simplicity that the system is symmetrical, then $a$ is also the amplitude for a photon in $D_2$ when an electron passes through hole $2$, and $b$ is the amplitude for a photon in $D_2$ when the electron passes through hole $1$. The corresponding total amplitude for a photon at $D_2$ and an electron at $x$ is \begin{equation} \label{Eq:III:3:9} \biggl\langle \begin{subarray}{l} \displaystyle \text{electron at $x$}\\[1ex] \displaystyle \text{photon at $D_2$} \end{subarray} \!\! \biggm| \! \begin{subarray}{l} \displaystyle \text{electron from $s$}\\[1ex] \displaystyle \text{photon from $L$} \end{subarray} \biggr\rangle= a\phi_2\!+b\phi_1. \end{equation}  Now we are finished. We can easily calculate the probability for various situations. Suppose that we want to know with what probability we get a count in $D_1$ and an electron at $x$. That will be the absolute square of the amplitude given in Eq. (3.8), namely, just $\abs{a\phi_1+b\phi_2}^2$. Let’s look more carefully at this expression. First of all, if $b$ is zero—which is the way we would like to design the apparatus—then the answer is simply $\abs{\phi_1}^2$ diminished in total amplitude by the factor $\abs{a}^2$. This is the probability distribution that you would get if there were only one hole—as shown in the graph of Fig. 3–4(a). On the other hand, if the wavelength is very long, the scattering behind hole $2$ into $D_1$ may be just about the same as for hole $1$. Although there may be some phases involved in $a$ and $b$, we can ask about a simple case in which the two phases are equal. If $a$ is practically equal to $b$, then the total probability becomes $\abs{\phi_1+\phi_2}^2$ multiplied by $\abs{a}^2$, since the common factor $a$ can be taken out. This, however, is just the probability distribution we would have gotten without the photons at all. Therefore, in the case that the wavelength is very long—and the photon detection ineffective—you return to the original distribution curve which shows interference effects, as shown in Fig. 3–4(b). In the case that the detection is partially effective, there is an interference between a lot of $\phi_1$ and a little of $\phi_2$, and you will get an intermediate distribution such as is sketched in Fig. 3–4(c). Needless to say, if we look for coincidence counts of photons at $D_2$ and electrons at $x$, we will get the same kinds of results. If you remember the discussion in Chapter 1, you will see that these results give a quantitative description of what was described there. Now we would like to emphasize an important point so that you will avoid a common error. Suppose that you only want the amplitude that the electron arrives at $x$, regardless of whether the photon was counted at $D_1$ or $D_2$. Should you add the amplitudes given in Eqs. (3.8) and (3.9)? No! You must never add amplitudes for different and distinct final states. Once the photon is accepted by one of the photon counters, we can always determine which alternative occurred if we want, without any further disturbance to the system. Each alternative has a probability completely independent of the other. To repeat, do not add amplitudes for different final conditions, where by “final” we mean at that moment the probability is desired—that is, when the experiment is “finished.” You do add the amplitudes for the different indistinguishable alternatives inside the experiment, before the complete process is finished. At the end of the process you may say that you “don’t want to look at the photon.” That’s your business, but you still do not add the amplitudes. Nature does not know what you are looking at, and she behaves the way she is going to behave whether you bother to take down the data or not. So here we must not add the amplitudes. We first square the amplitudes for all possible different final events and then sum. The correct result for an electron at $x$ and a photon at either $D_1$ or $D_2$ is \begin{gather} \biggl\lvert \biggl\langle \begin{subarray}{l} \displaystyle \text{e at $x$}\\[1ex] \displaystyle \text{ph at $D_1$} \end{subarray}\!\! \biggm|\! \begin{subarray}{l} \displaystyle \text{e from $s$}\\[1ex] \displaystyle \text{ph from $L$} \end{subarray} \biggr\rangle \biggr\rvert^2\!\!+ \biggl\lvert \biggl\langle \begin{subarray}{l} \displaystyle \text{e at $x$}\\[1ex] \displaystyle \text{ph at $D_2$} \end{subarray}\!\! \biggm|\! \begin{subarray}{l} \displaystyle \text{e from $s$}\\[1ex] \displaystyle \text{ph from $L$} \end{subarray} \biggr\rangle \biggr\rvert^2\notag\\[2ex] \label{Eq:III:3:10} =\abs{a\phi_1\!+b\phi_2}^2+\;\abs{a\phi_2\!+b\phi_1}^2. \end{gather} 