States of definite momentum
Suppose we have an electron in a state $\ket{\psi}$ which is described by the probability amplitude $\braket{x}{\psi}=\psi(x)$. We know that this represents a state in which the electron is spread out along the line in a certain distribution so that the probability of finding the electron in a small interval $dx$ at the location $x$ is just \begin{equation*} \prob(x,dx)=\abs{\psi(x)}^2\,dx. \end{equation*}  What can we say about the momentum of this electron? We might ask what is the probability that this electron has the momentum $p$? Let’s start out by calculating the amplitude that the state $\ket{\psi}$ is in another state $\ket{\mom p}$ which we define to be a state with the definite momentum $p$. We can find this amplitude by using our basic equation for the resolution of amplitudes, Eq. (16.19). In terms of the state $\ket{\mom p}$ \begin{equation} \label{Eq:III:16:21} \braket{\mom p}{\psi}=\int_{x=-\infty}^{+\infty} \braket{\mom p}{x}\braket{x}{\psi}\,dx. \end{equation}  And the probability that the electron will be found with the momentum $p$ should be given in terms of the absolute square of this amplitude. We have again, however, a small problem about the normalizations. In general we can only ask about the probability of finding an electron with a momentum in a small range $dp$ at the momentum $p$. The probability that the momentum is exactly some value $p$ must be zero (unless the state $\ket{\psi}$ happens to be a state of definite momentum). Only if we ask for the probability of finding the momentum in a small range $dp$ at the momentum $p$ will we get a finite probability. There are several ways the normalizations can be adjusted. We will choose one of them which we think to be the most convenient, although that may not be apparent to you just now. We take our normalizations so that the probability is related to the amplitude by \begin{equation} \label{Eq:III:16:22} \prob(p,dp)=\abs{\braket{\mom p}{\psi}}^2\,\frac{dp}{2\pi\hbar}. \end{equation}  With this definition the normalization of the amplitude $\braket{\mom p}{x}$ is determined. The amplitude $\braket{\mom p}{x}$ is, of course, just the complex conjugate of the amplitude $\braket{x}{\mom p}$, which is just the one we have written down in Eq. (16.15). With the normalization we have chosen, it turns out that the proper constant of proportionality in front of the exponential is just $1$. Namely, \begin{equation} \label{Eq:III:16:23} \braket{\mom p}{x}=\braket{x}{\mom p}\cconj=e^{-ipx/\hbar}. \end{equation}  Equation (16.21) then becomes \begin{equation} \label{Eq:III:16:24} \braket{\mom p}{\psi}=\int_{-\infty}^{+\infty} e^{-ipx/\hbar}\braket{x}{\psi}\,dx. \end{equation}  This equation together with Eq. (16.22) allows us to find the momentum distribution for any state $\ket{\psi}$. Let’s look at a particular example—for instance one in which an electron is localized in a certain region around $x=0$. Suppose we take a wave function which has the following form: \begin{equation} \label{Eq:III:16:25} \psi(x)=Ke^{-x^2/4\sigma^2}. \end{equation}  The probability distribution in $x$ for this wave function is the absolute square, or \begin{equation} \label{Eq:III:16:26} \prob(x,dx)=P(x)\,dx=K^2e^{-x^2/2\sigma^2}\,dx. \end{equation}  The probability density function $P(x)$ is the Gaussian curve shown in Fig. 16–1. Most of the probability is concentrated between $x=+\sigma$ and $x=-\sigma$. We say that the “half-width” of the curve is $\sigma$. (More precisely, $\sigma$ is equal to the root-mean-square of the coordinate $x$ for something spread out according to this distribution.)  We would normally choose the constant $K$ so that the probability density $P(x)$ is not merely proportional to the probability per unit length in $x$ of finding the electron, but has a scale such that $P(x)\,\Delta x$ is equal to the probability of finding the electron in $\Delta x$ near $x$. The constant $K$ which does this can be found by requiring that $\int_{-\infty}^{+\infty}P(x)\,dx=1$, since there must be unit probability that the electron is found somewhere. Here, we get that $K=(2\pi\sigma^2)^{-1/4}$. [We have used the fact that $\int_{-\infty}^{+\infty}e^{-t^2}\,dt=\sqrt{\pi}$; see Vol. I, footnote 40-1.] Now let’s find the distribution in momentum. Let’s let $\phi(p)$ stand for the amplitude to find the electron with the momentum $p$, \begin{equation} \label{Eq:III:16:27} \phi(p)\equiv\braket{\mom p}{\psi}. \end{equation}  Substituting Eq. (16.25) into Eq. (16.24) we get \begin{equation} \label{Eq:III:16:28} \phi(p)=\int_{-\infty}^{+\infty} e^{-ipx/\hbar}\cdot Ke^{-x^2/4\sigma^2}\,dx. \end{equation}  The integral can also be rewritten as \begin{equation} \label{Eq:III:16:29} Ke^{-p^2\sigma^2/\hbar^2}\int_{-\infty}^{+\infty} e^{-(1/4\sigma^2)(x+2ip\sigma^2/\hbar)^2}\,dx. \end{equation}  We can now make the substitution $u=x+2ip\sigma^2/\hbar$, and the integral is \begin{equation} \label{Eq:III:16:30} \int_{-\infty}^{+\infty}e^{-u^2/4\sigma^2}\,du=2\sigma\sqrt{\pi}. \end{equation}  (The mathematicians would probably object to the way we got there, but the result is, nevertheless, correct.) \begin{equation} \label{Eq:III:16:31} \phi(p)=(8\pi\sigma^2)^{1/4}e^{-p^2\sigma^2/\hbar^2}. \end{equation}  We have the interesting result that the amplitude function in $p$ has precisely the same mathematical form as the amplitude function in $x$; only the width of the Gaussian is different. We can write this as \begin{equation} \label{Eq:III:16:32} \phi(p)=(\eta^2/2\pi\hbar^2)^{-1/4}e^{-p^2/4\eta^2}, \end{equation}  where the half-width $\eta$ of the $p$-distribution function is related to the half-width $\sigma$ of the $x$-distribution by \begin{equation} \label{Eq:III:16:33} \eta=\frac{\hbar}{2\sigma}. \end{equation}  Our result says: if we make the width of the distribution in $x$ very small by making $\sigma$ small, $\eta$ becomes large and the distribution in $p$ is very much spread out. Or, conversely: if we have a narrow distribution in $p$, it must correspond to a spread-out distribution in $x$. We can, if we like, consider $\eta$ and $\sigma$ to be some measure of the uncertainty in the localization of the momentum and of the position of the electron in the state we are studying. If we call them $\Delta p$ and $\Delta x$ respectively Eq. (16.33) becomes \begin{equation} \label{Eq:III:16:34} \Delta p\,\Delta x=\frac{\hbar}{2}. \end{equation}  Interestingly enough, it is possible to prove that for any other form of a distribution in $x$ or in $p$, the product $\Delta p\,\Delta x$ cannot be smaller than the one we have found here. The Gaussian distribution gives the smallest possible value for the product of the root-mean-square widths. In general, we can say \begin{equation} \label{Eq:III:16:35} \Delta p\,\Delta x\geq\frac{\hbar}{2}. \end{equation}  This is a quantitative statement of the Heisenberg uncertainty principle, which we have discussed qualitatively many times before. We have usually made the approximate statement that the minimum value of the product $\Delta p\,\Delta x$ is of the same order as $\hbar$.