The wave function
Now that you have some idea about how things are going to look, we want to go back to the beginning and study the problem of describing the motion of an electron along a line without having to consider states connected with atoms on a lattice. We want to go back to the beginning and see what ideas we have to use if we want to describe the motion of a free particle in space. Since we are interested in the behavior of a particle, along a continuum, we will be dealing with an infinite number of possible states and, as you will see, the ideas we have developed for dealing with a finite number of states will need some technical modifications. We begin by letting the state vector $\ket{x}$ stand for a state in which a particle is located precisely at the coordinate $x$. For every value $x$ along the line—for instance $1.73$, or $9.67$, or $10.00$—there is the corresponding state. We will take these states $\ket{x}$ as our base states and, if we include all the points on the line, we will have a complete set for motion in one dimension. Now suppose we have a different kind of a state, say $\ket{\psi}$, in which an electron is distributed in some way along the line. One way of describing this state is to give all the amplitudes that the electron will be also found in each of the base states $\ket{x}$. We must give an infinite set of amplitudes, one for each value of $x$. We will write these amplitudes as $\braket{x}{\psi}$. Each of these amplitudes is a complex number and since there is one such complex number for each value of $x$, the amplitude $\braket{x}{\psi}$ is indeed just a function of $x$. We will also write it as $C(x)$, \begin{equation} \label{Eq:III:16:14} C(x)\equiv\braket{x}{\psi}. \end{equation}  We have already considered such amplitudes which vary in a continuous way with the coordinates when we talked about the variations of amplitude with time in Chapter 7. We showed there, for example, that a particle with a definite momentum should be expected to have a particular variation of its amplitude in space. If a particle has a definite momentum $p$ and a corresponding definite energy $E$, the amplitude to be found at any position $x$ would look like \begin{equation} \label{Eq:III:16:15} \braket{x}{\psi}=C(x)\propto e^{+ipx/\hbar}. \end{equation}  This equation expresses an important general principle of quantum mechanics which connects the base states corresponding to different positions in space to another system of base states—all the states of definite momentum. The definite momentum states are often more convenient than the states in $x$ for certain kinds of problems. Either set of base states is, of course, equally acceptable for a description of a quantum mechanical situation. We will come back later to the matter of the connection between them. For the moment we want to stick to our discussion of a description in terms of the states $\ket{x}$. Before proceeding, we want to make one small change in notation which we hope will not be too confusing. The function $C(x)$, defined in Eq. (16.14), will of course have a form which depends on the particular state $\ket{\psi}$ under consideration. We should indicate that in some way. We could, for example, specify which function $C(x)$ we are talking about by a subscript say, $C_\psi(x)$. Although this would be a perfectly satisfactory notation, it is a little bit cumbersome and is not the one you will find in most books. Most people simply omit the letter $C$ and use the symbol $\psi$ to define the function \begin{equation} \label{Eq:III:16:16} \psi(x)\equiv C_\psi(x)=\braket{x}{\psi}. \end{equation}  Since this is the notation used by everybody else in the world, you might as well get used to it so that you will not be frightened when you come across it somewhere else. Remember though, that we will now be using $\psi$ in two different ways. In Eq. (16.14), $\psi$ stands for a label we have given to a particular physical state of the electron. On the left-hand side of Eq. (16.16), on the other hand, the symbol $\psi$ is used to define a mathematical function of $x$ which is equal to the amplitude to be associated with each point $x$ along the line. We hope it will not be too confusing once you get accustomed to the idea. Incidentally, the function $\psi(x)$ is usually called “the wave function”—because it more often than not has the form of a complex wave in its variables. Since we have defined $\psi(x)$ to be the amplitude that an electron in the state $\psi$ will be found at the location $x$, we would like to interpret the absolute square of $\psi$ to be the probability of finding an electron at the position $x$. Unfortunately, the probability of finding a particle exactly at any particular point is zero. The electron will, in general, be smeared out in a certain region of the line, and since, in any small piece of the line, there are an infinite number of points, the probability that it will be at any one of them cannot be a finite number. We can only describe the probability of finding an electron in terms of a probability distribution2 which gives the relative probability of finding the electron at various approximate locations along the line. Let’s let $\prob(x,\Delta x)$ stand for the chance of finding the electron in a small interval $\Delta x$ located near $x$. If we go to a small enough scale in any physical situation, the probability will be varying smoothly from place to place, and the probability of finding the electron in any small finite line segment $\Delta x$ will be proportional to $\Delta x$. We can modify our definitions to take this into account.  We can think of the amplitude $\braket{x}{\psi}$ as representing a kind of “amplitude density” for all the base states $\ket{x}$ in a small region. Since the probability of finding an electron in a small interval $\Delta x$ at $x$ should be proportional to the interval $\Delta x$, we choose our definition of $\braket{x}{\psi}$ so that the following relation holds: \begin{equation*} \prob(x,\Delta x)=\abs{\braket{x}{\psi}}^2\,\Delta x. \end{equation*}  The amplitude $\braket{x}{\psi}$ is therefore proportional to the amplitude that an electron in the state $\psi$ will be found in the base state $x$ and the constant of proportionality is chosen so that the absolute square of the amplitude $\braket{x}{\psi}$ gives the probability density of finding an electron in any small region. We can write, equivalently, \begin{equation} \label{Eq:III:16:17} \prob(x,\Delta x)=\abs{\psi(x)}^2\,\Delta x. \end{equation}  We will now have to modify some of our earlier equations to make them compatible with this new definition of a probability amplitude. Suppose we have an electron in the state $\ket{\psi}$ and we want to know the amplitude for finding it in a different state $\ket{\phi}$ which may correspond to a different spread-out condition of the electron. When we were talking about a finite set of discrete states, we would have used Eq. (16.5). Before modifying our definition of the amplitudes we would have written \begin{equation} \label{Eq:III:16:18} \braket{\phi}{\psi}=\sum_{\text{all $x$}} \braket{\phi}{x}\braket{x}{\psi}. \end{equation}  Now if both of these amplitudes are normalized in the same way as we have described above, then a sum of all the states in a small region of $x$ would be equivalent to multiplying by $\Delta x$, and the sum over all values of $x$ simply becomes an integral. With our modified definitions, the correct form becomes \begin{equation} \label{Eq:III:16:19} \braket{\phi}{\psi}=\int_{\text{all $x$}} \braket{\phi}{x}\braket{x}{\psi}\,dx. \end{equation}  The amplitude $\braket{x}{\psi}$ is what we are now calling $\psi(x)$ and, in a similar way, we will choose to let the amplitude $\braket{x}{\phi}$ be represented by $\phi(x)$. Remembering that $\braket{\phi}{x}$ is the complex conjugate of $\braket{x}{\phi}$, we can write Eq. (16.19) as \begin{equation} \label{Eq:III:16:20} \braket{\phi}{\psi}=\int\phi\cconj(x)\psi(x)\,dx. \end{equation}  With our new definitions everything follows with the same formulas as before if you always replace a summation sign by an integral over $x$. We should mention one qualification to what we have been saying. Any suitable set of base states must be complete if it is to be used for an adequate description of what is going on. For an electron in one dimension it is not really sufficient to specify only the base states $\ket{x}$, because for each of these states the electron may have a spin which is either up or down. One way of getting a complete set is to take two sets of states in $x$, one for up spin and the other for down spin. We will, however, not worry about such complications for the time being.