States with two Bose particles
Now we would like to discuss an interesting consequence of the addition rule for Bose particles. It has to do with their behavior when there are several particles present. We begin by considering a situation in which two Bose particles are scattered from two different scatterers. We won’t worry about the details of the scattering mechanism. We are interested only in what happens to the scattered particles. Suppose we have the situation shown in Fig. 4–3. The particle $a$ is scattered into the state $1$. By a state we mean a given direction and energy, or some other given condition. The particle $b$ is scattered into the state $2$. We want to assume that the two states $1$ and $2$ are nearly the same. (What we really want to find out eventually is the amplitude that the two particles are scattered into identical directions, or states; but it is best if we think first about what happens if the states are almost the same and then work out what happens when they become identical.) Suppose that we had only particle $a$; then it would have a certain amplitude for scattering in direction $1$, say $\braket{1}{a}$. And particle $b$ alone would have the amplitude $\braket{2}{b}$ for landing in direction $2$. If the two particles are not identical, the amplitude for the two scatterings to occur at the same time is just the product \begin{equation*} \braket{1}{a}\braket{2}{b}. \end{equation*}  The probability for such an event is then \begin{equation*} \abs{\braket{1}{a}\braket{2}{b}}^2, \end{equation*}  which is also equal to \begin{equation*} \abs{\braket{1}{a}}^2\abs{\braket{2}{b}}^2. \end{equation*}  To save writing for the present arguments, we will sometimes set \begin{equation*} \braket{1}{a}=a_1,\quad \braket{2}{b}=b_2. \end{equation*}  Then the probability of the double scattering is \begin{equation*} \abs{a_1}^2\abs{b_2}^2. \end{equation*}  It could also happen that particle $b$ is scattered into direction $1$, while particle $a$ goes into direction $2$. The amplitude for this process is \begin{equation*} \braket{2}{a}\braket{1}{b}, \end{equation*}  and the probability of such an event is \begin{equation*} \abs{\braket{2}{a}\braket{1}{b}}^2= \abs{a_2}^2\abs{b_1}^2. \end{equation*}  Imagine now that we have a pair of tiny counters that pick up the two scattered particles. The probability $P_2$ that they will pick up two particles together is just the sum \begin{equation} \label{Eq:III:4:3} P_2=\abs{a_1}^2\abs{b_2}^2+\abs{a_2}^2\abs{b_1}^2. \end{equation}  Now let’s suppose that the directions $1$ and $2$ are very close together. We expect that $a$ should vary smoothly with direction, so $a_1$ and $a_2$ must approach each other as $1$ and $2$ get close together. If they are close enough, the amplitudes $a_1$ and $a_2$ will be equal. We can set $a_1=a_2$ and call them both just $a$; similarly, we set $b_1=b_2=b$. Then we get that \begin{equation} \label{Eq:III:4:4} P_2=2\abs{a}^2\abs{b}^2. \end{equation}  Now suppose, however, that $a$ and $b$ are identical Bose particles. Then the process of $a$ going into $1$ and $b$ going into $2$ cannot be distinguished from the exchanged process in which $a$ goes into $2$ and $b$ goes into $1$. In this case the amplitudes for the two different processes can interfere. The total amplitude to obtain a particle in each of the two counters is \begin{equation} \label{Eq:III:4:5} \braket{1}{a}\braket{2}{b}+\braket{2}{a}\braket{1}{b}. \end{equation}  And the probability that we get a pair is the absolute square of this amplitude, \begin{equation} \label{Eq:III:4:6} P_2=\abs{a_1b_2+a_2b_1}^2=4\abs{a}^2\abs{b}^2. \end{equation}  We have the result that it is twice as likely to find two identical Bose particles scattered into the same state as you would calculate assuming the particles were different. Although we have been considering that the two particles are observed in separate counters, this is not essential—as we can see in the following way. Let’s imagine that both the directions $1$ and $2$ would bring the particles into a single small counter which is some distance away. We will let the direction $1$ be defined by saying that it heads toward the element of area $dS_1$ of the counter. Direction $2$ heads toward the surface element $dS_2$ of the counter. (We imagine that the counter presents a surface at right angles to the line from the scatterings.) Now we cannot give a probability that a particle will go into a precise direction or to a particular point in space. Such a thing is impossible—the chance for any exact direction is zero. When we want to be so specific, we shall have to define our amplitudes so that they give the probability of arriving per unit area of a counter. Suppose that we had only particle $a$; it would have a certain amplitude for scattering in direction $1$. Let’s define $\braket{1}{a}=a_1$ to be the amplitude that $a$ will scatter into a unit area of the counter in the direction $1$. In other words, the scale of $a_1$ is chosen—we say it is “normalized” so that the probability that it will scatter into an element of area $dS_1$ is \begin{equation} \label{Eq:III:4:7} \abs{\braket{1}{a}}^2\,dS_1=\abs{a_1}^2\,dS_1. \end{equation}  If our counter has the total area $\Delta S$, and we let $dS_1$ range over this area, the total probability that the particle $a$ will be scattered into the counter is \begin{equation} \label{Eq:III:4:8} \int_{\Delta S}\abs{a_1}^2\,dS_1. \end{equation}   As before, we want to assume that the counter is sufficiently small so that the amplitude $a_1$ doesn’t vary significantly over the surface of the counter; $a_1$ is then a constant amplitude which we can call $a$. Then the probability that particle $a$ is scattered somewhere into the counter is \begin{equation} \label{Eq:III:4:9} p_a=\abs{a}^2\,\Delta S. \end{equation}  In the same way, we will have that the probability that particle $b$—when it is alone—scatters into some element of area, say $dS_2$, is \begin{equation*} \abs{b_2}^2\,dS_2. \end{equation*}  (We use $dS_2$ instead of $dS_1$ because we will later want $a$ and $b$ to go into different directions.) Again we set $b_2$ equal to the constant amplitude $b$; then the probability that particle $b$ is counted in the detector is \begin{equation} \label{Eq:III:4:10} p_b=\abs{b}^2\,\Delta S. \end{equation}  Now when both particles are present, the probability that $a$ is scattered into $dS_1$ and $b$ is scattered into $dS_2$ is \begin{equation} \label{Eq:III:4:11} \abs{a_1b_2}^2\,dS_1\,dS_2=\abs{a}^2\abs{b}^2\,dS_1\,dS_2. \end{equation}  If we want the probability that both $a$ and $b$ get into the counter, we integrate both $dS_1$ and $dS_2$ over $\Delta S$ and find that \begin{equation} \label{Eq:III:4:12} P_2=\abs{a}^2\abs{b}^2\,(\Delta S)^2. \end{equation}  We notice, incidentally, that this is just equal to $p_a\cdot p_b$, just as you would suppose assuming that the particles $a$ and $b$ act independently of each other. When the two particles are identical, however, there are two indistinguishable possibilities for each pair of surface elements $dS_1$ and $dS_2$. Particle $a$ going into $dS_2$ and particle $b$ going into $dS_1$ is indistinguishable from $a$ into $dS_1$ and $b$ into $dS_2$, so the amplitudes for these processes will interfere. (When we had two different particles above—although we did not in fact care which particle went where in the counter—we could, in principle, have found out; so there was no interference. For identical particles we cannot tell, even in principle.) We must write, then, that the probability that the two particles arrive at $dS_1$ and $dS_2$ is \begin{equation} \label{Eq:III:4:13} \abs{a_1b_2+a_2b_1}^2\,dS_1\,dS_2. \end{equation}  Now, however, when we integrate over the area of the counter, we must be careful. If we let $dS_1$ and $dS_2$ range over the whole area $\Delta S$, we would count each part of the area twice since (4.13) contains everything that can happen with any pair of surface elements $dS_1$ and $dS_2$.1 We can still do the integral that way, if we correct for the double counting by dividing the result by $2$. We get then that $P_2$ for identical Bose particles is  \begin{equation} \label{Eq:III:4:14} P_2(\text{Bose})=\tfrac{1}{2}\{4\abs{a}^2\abs{b}^2\,(\Delta S)^2\}= 2\abs{a}^2\abs{b}^2\,(\Delta S)^2. \end{equation} 
  \begin{align} P_2(\text{Bose})&=\tfrac{1}{2}\{4\abs{a}^2\abs{b}^2\,(\Delta S)^2\}\notag\\[1ex] \label{Eq:III:4:14} &=2\abs{a}^2\abs{b}^2\,(\Delta S)^2. \end{align}   Again, this is just twice what we got in Eq. (4.12) for distinguishable particles. If we imagine for a moment that we knew that the $b$ channel had already sent its particle into the particular direction, we can say that the probability that a second particle will go into the same direction is twice as great as we would have expected if we had calculated it as an independent event. It is a property of Bose particles that if there is already one particle in a condition of some kind, the probability of getting a second one in the same condition is twice as great as it would be if the first one were not already there. This fact is often stated in the following way: If there is already one Bose particle in a given state, the amplitude for putting an identical one on top of it is $\sqrt{2}$ greater than if it weren’t there. (This is not a proper way of stating the result from the physical point of view we have taken, but if it is used consistently as a rule, it will, of course, give the correct result.)