Normalization of the states in $\boldsymbol{x}$
We return now to the discussion of the modifications of our basic equations which are required when we are dealing with a continuum of base states. When we have a finite number of discrete states, a fundamental condition which must be satisfied by the set of base states is \begin{equation} \label{Eq:III:16:36} \braket{i}{j}=\delta_{ij}. \end{equation}  If a particle is in one base state, the amplitude to be in another base state is $0$. By choosing a suitable normalization, we have defined the amplitude $\braket{i}{i}$ to be $1$. These two conditions are described by Eq. (16.36). We want now to see how this relation must be modified when we use the base states $\ket{x}$ of a particle on a line. If the particle is known to be in one of the base states $\ket{x}$, what is the amplitude that it will be in another base state $\ket{x'}$? If $x$ and $x'$ are two different locations along the line, then the amplitude $\braket{x}{x'}$ is certainly $0$, so that is consistent with Eq. (16.36). But if $x$ and $x'$ are equal, the amplitude $\braket{x}{x'}$ will not be $1$, because of the same old normalization problem. To see how we have to patch things up, we go back to Eq. (16.19), and apply this equation to the special case in which the state $\ket{\phi}$ is just the base state $\ket{x'}$. We would have then \begin{equation} \label{Eq:III:16:37} \braket{x'}{\psi}=\int\braket{x'}{x}\psi(x)\,dx. \end{equation}  Now the amplitude $\braket{x}{\psi}$ is just what we have been calling the function $\psi(x)$. Similarly the amplitude $\braket{x'}{\psi}$, since it refers to the same state $\ket{\psi}$, is the same function of the variable $x'$, namely $\psi(x')$. We can, therefore, rewrite Eq. (16.37) as \begin{equation} \label{Eq:III:16:38} \psi(x')=\int\braket{x'}{x}\psi(x)\,dx. \end{equation}  This equation must be true for any state $\ket{\psi}$ and, therefore, for any arbitrary function $\psi(x)$. This requirement should completely determine the nature of the amplitude $\braket{x}{x'}$—which is, of course, just a function that depends on $x$ and $x'$. Our problem now is to find a function $f(x,x')$, which when multiplied into $\psi(x)$, and integrated over all $x$ gives just the quantity $\psi(x')$. It turns out that there is no mathematical function which will do this!  At least nothing like what we ordinarily mean by a “function.” Suppose we pick $x'$ to be the special number $0$ and define the amplitude $\braket{0}{x}$ to be some function of $x$, let’s say $f(x)$. Then Eq. (16.38) would read as follows: \begin{equation} \label{Eq:III:16:39} \psi(0)=\int f(x)\psi(x)\,dx. \end{equation}  What kind of function $f(x)$ could possibly satisfy this equation? Since the integral must not depend on what values $\psi(x)$ takes for values of $x$ other than $0$, $f(x)$ must clearly be $0$ for all values of $x$ except $0$. But if $f(x)$ is $0$ everywhere, the integral will be $0$, too, and Eq. (16.39) will not be satisfied. So we have an impossible situation: we wish a function to be $0$ everywhere but at a point, and still to give a finite integral. Since we can’t find a function that does this, the easiest way out is just to say that the function $f(x)$ is defined by Eq. (16.39). Namely, $f(x)$ is that function which makes (16.39) correct. The function which does this was first invented by Dirac and carries his name. We write it $\delta(x)$. All we are saying is that the function $\delta(x)$ has the strange property that if it is substituted for $f(x)$ in the Eq. (16.39), the integral picks out the value that $\psi(x)$ takes on when $x$ is equal $0$; and, since the integral must be independent of $\psi(x)$ for all values of $x$ other than $0$, the function $\delta(x)$ must be $0$ everywhere except at $x=0$. Summarizing, we write \begin{equation} \label{Eq:III:16:40} \braket{0}{x}=\delta(x), \end{equation}  where $\delta(x)$ is defined by \begin{equation} \label{Eq:III:16:41} \psi(0)=\int\delta(x)\psi(x)\,dx. \end{equation} Notice what happens if we use the special function “$1$” for the function $\psi$ in Eq. (16.41). Then we have the result \begin{equation} \label{Eq:III:16:42} 1=\int\delta(x)\,dx. \end{equation}  That is, the function $\delta(x)$ has the property that it is $0$ everywhere except at $x=0$ but has a finite integral equal to unity. We must imagine that the function $\delta(x)$ has such a fantastic infinity at one point that the total area comes out equal to one. One way of imagining what the Dirac $\delta$-function is like is to think of a sequence of rectangles—or any other peaked function you care to—which gets narrower and narrower and higher and higher, always keeping a unit area, as sketched in Fig. 16–2. The integral of this function from $-\infty$ to $+\infty$ is always $1$. If you multiply it by any function $\psi(x)$ and integrate the product, you get something which is approximately the value of the function at $x=0$, the approximation getting better and better as you use the narrower and narrower rectangles. You can if you wish, imagine the $\delta$-function in terms of this kind of limiting process. The only important thing, however, is that the $\delta$-function is defined so that Eq. (16.41) is true for every possible function $\psi(x)$. That uniquely defines the $\delta$-function. Its properties are then as we have described. If we change the argument of the $\delta$-function from $x$ to $x-x'$, the corresponding relations are \begin{gather} \delta(x-x')=0,\quad x'\neq x,\notag\\[2ex] \label{Eq:III:16:43} \int\delta(x-x')\psi(x)\,dx=\psi(x'). \end{gather}  If we use $\delta(x-x')$ for the amplitude $\braket{x}{x'}$ in Eq. (16.38), that equation is satisfied. Our result then is that for our base states in $x$, the condition corresponding to (16.36) is \begin{equation} \label{Eq:III:16:44} \braket{x'}{x}=\delta(x-x'). \end{equation}  We have now completed the necessary modifications of our basic equations which are necessary for dealing with the continuum of base states corresponding to the points along a line. The extension to three dimensions is fairly obvious; first we replace the coordinate $x$ by the vector $\FLPr$. Then integrals over $x$ become replaced by integrals over $x$, $y$, and $z$. In other words, they become volume integrals. Finally, the one-dimensional $\delta$-function must be replaced by just the product of three $\delta$-functions, one in $x$, one in $y$, and the other in $z$, $\delta(x-x')\,\delta(y-y')\,\delta(z-z')$. Putting everything together we get the following set of equations for the amplitudes for a particle in three dimensions: \begin{gather} \label{Eq:III:16:45} \braket{\phi}{\psi}=\int \braket{\phi}{\FLPr}\braket{\FLPr}{\psi}\, dV,\\[1.5ex] \label{Eq:III:16:46} \begin{aligned} \braket{\FLPr}{\psi}&=\psi(\FLPr),\\[1.5ex] \braket{\FLPr}{\phi}&=\phi(\FLPr), \end{aligned}\\[1.5ex] \label{Eq:III:16:47} \braket{\phi}{\psi}=\int \phi\cconj(\FLPr)\psi(\FLPr)\,dV,\\[1.5ex] \label{Eq:III:16:48} \braket{\FLPr'}{\FLPr}=\delta(x-x')\,\delta(y-y')\,\delta(z-z'). \end{gather}  What happens when there is more than one particle? We will tell you about how to handle two particles and you will easily see what you must do if you want to deal with a larger number. Suppose there are two particles, which we can call particle No. $1$ and particle No. $2$. What shall we use for the base states? One perfectly good set can be described by saying that particle $1$ is at $x_1$ and particle $2$ is at $x_2$, which we can write as $\ket{x_1,x_2}$. Notice that describing the position of only one particle does not define a base state. Each base state must define the condition of the entire system. You must not think that each particle moves independently as a wave in three dimensions. Any physical state $\ket{\psi}$ can be defined by giving all of the amplitudes $\braket{x_1,x_2}{\psi}$ to find the two particles at $x_1$ and $x_2$. This generalized amplitude is therefore a function of the two sets of coordinates $x_1$ and $x_2$. You see that such a function is not a wave in the sense of an oscillation that moves along in three dimensions. Neither is it generally simply a product of two individual waves, one for each particle. It is, in general, some kind of a wave in the six dimensions defined by $x_1$ and $x_2$. If there are two particles in nature which are interacting, there is no way of describing what happens to one of the particles by trying to write down a wave function for it alone. The famous paradoxes that we considered in earlier chapters—where the measurements made on one particle were claimed to be able to tell what was going to happen to another particle, or were able to destroy an interference—have caused people all sorts of trouble because they have tried to think of the wave function of one particle alone, rather than the correct wave function in the coordinates of both particles. The complete description can be given correctly only in terms of functions of the coordinates of both particles.