The average energy of an atom
Suppose we want the average energy of an atom in a state described by a wave function $\psi(\FLPr)$; How do we find it? Let’s first think of a one-dimensional situation with a state $\ket{\psi}$ defined by the amplitude $\braket{x}{\psi}=\psi(x)$. We are asking for the special case of Eq. (20.19) applied to the coordinate representation. Following our usual procedure, we replace the states $\ket{i}$ and $\ket{j}$ by $\ket{x}$ and $\ket{x'}$, and change the sums to integrals. We get \begin{equation} \label{Eq:III:20:23} \av{E}=\!\iint\braket{\psi}{x}\bracket{x}{\Hop}{x'} \braket{x'}{\psi}\,dx\,dx'\!. \end{equation} This integral can, if we wish, be written in the following way: \begin{equation} \label{Eq:III:20:24} \int\braket{\psi}{x}\braket{x}{\phi}\,dx, \end{equation} with \begin{equation} \label{Eq:III:20:25} \braket{x}{\phi}=\int\bracket{x}{\Hop}{x'}\braket{x'}{\psi}\,dx'. \end{equation} The integral over $x'$ in (20.25) is the same one we had in Chapter 16—see Eq. (16.50) and Eq. (16.52)—and is equal to \begin{equation*} -\frac{\hbar^2}{2m}\,\frac{d^2}{dx^2}\,\psi(x)+V(x)\psi(x). \end{equation*} We can therefore write \begin{equation} \label{Eq:III:20:26} \braket{x}{\phi}=\biggl\{\! -\frac{\hbar^2}{2m}\,\frac{d^2}{dx^2}+V(x)\biggr\}\psi(x). \end{equation}  Remember that $\braket{\psi}{x}=$ $\braket{x}{\psi}\cconj=$ $\psi\cconj(x)$; using this equality, the average energy in Eq. (20.23) can be written as \begin{equation} \label{Eq:III:20:27} \av{E}=\!\int\!\psi\cconj(x)\biggl\{\! -\frac{\hbar^2}{2m}\frac{d^2}{dx^2}\!+\!V(x)\!\biggr\}\psi(x)\,dx. \end{equation} Given a wave function $\psi(x)$, you can get the average energy by doing this integral. You can begin to see how we can go back and forth from the state-vector ideas to the wave-function ideas.  The quantity in the braces of Eq. (20.27) is an algebraic operator.1 We will write it as $\Hcalop$ \begin{equation*} \Hcalop=-\frac{\hbar^2}{2m}\,\frac{d^2}{dx^2}+V(x). \end{equation*} With this notation Eq. (20.23) becomes \begin{equation} \label{Eq:III:20:28} \av{E}=\int\psi\cconj(x)\Hcalop\psi(x)\,dx. \end{equation}  The algebraic operator $\Hcalop$ defined here is, of course, not identical to the quantum-mechanical operator $\Hop$. The new operator works on a function of position $\psi(x)=\braket{x}{\psi}$ to give a new function of $x$, $\phi(x)=\braket{x}{\phi}$; while $\Hop$ operates on a state vector $\ket{\psi}$ to give another state vector $\ket{\phi}$, without implying the coordinate representation or any particular representation at all. Nor is $\Hcalop$ strictly the same as $\Hop$ even in the coordinate representation. If we choose to work in the coordinate representation, we would interpret $\Hop$ in terms of a matrix $\bracket{x}{\Hop}{x'}$ which depends somehow on the two “indices” $x$ and $x'$; that is, we expect—according to Eq. (20.25)—that $\braket{x}{\phi}$ is related to all the amplitudes $\braket{x}{\psi}$ by an integration. On the other hand, we find that $\Hcalop$ is a differential operator. We have already worked out in Section 16–5 the connection between $\bracket{x}{\Hop}{x'}$ and the algebraic operator $\Hcalop$.  We should make one qualification on our results. We have been assuming that the amplitude $\psi(x)=\braket{x}{\psi}$ is normalized. By this we mean that the scale has been chosen so that \begin{equation*} \int\abs{\psi(x)}^2\,dx=1; \end{equation*} so the probability of finding the electron somewhere is unity. If you should choose to work with a $\psi(x)$ which is not normalized you should write \begin{equation} \label{Eq:III:20:29} \av{E}=\frac{\int\psi\cconj(x)\Hcalop\psi(x)\,dx} {\int\psi\cconj(x)\psi(x)\,dx}. \end{equation} It’s the same thing.  Notice the similarity in form between Eq. (20.28) and Eq. (20.18). These two ways of writing the same result appear often when you work with the $x$-representation. You can go from the first form to the second with any $\Aop$ which is a local operator, where a local operator is one which in the integral \begin{equation*} \int\bracket{x}{\Aop}{x'}\braket{x'}{\psi}\,dx' \end{equation*} can be written as $\Acalop\psi(x)$, where $\Acalop$ is a differential algebraic operator. There are, however, operators for which this is not true. For them you must work with the basic equations in (20.21) and (20.22).  You can easily extend the derivation to three dimensions. The result is that2 \begin{equation} \label{Eq:III:20:30} \av{E}=\int\psi\cconj(\FLPr)\Hcalop\psi(\FLPr)\,dV, \end{equation} with \begin{equation} \label{Eq:III:20:31} \Hcalop=-\frac{\hbar^2}{2m}\,\nabla^2+V(\FLPr), \end{equation} and with the understanding that \begin{equation} \label{Eq:III:20:32} \int\abs{\psi}^2\,dV=1. \end{equation} The same equations can be extended to systems with several electrons in a fairly obvious way, but we won’t bother to write down the results.  With Eq. (20.30) we can calculate the average energy of an atomic state even without knowing its energy levels. All we need is the wave function. It’s an important law. We’ll tell you about one interesting application. Suppose you want to know the ground-state energy of some system—say the helium atom, but it’s too hard to solve Schrödinger’s equation for the wave function, because there are too many variables. Suppose, however, that you take a guess at the wave function—pick any function you like—and calculate the average energy. That is, you use Eq. (20.29)—generalized to three dimensions-to find what the average energy would be if the atom were really in the state described by this wave function. This energy will certainly be higher than the ground-state energy which is the lowest possible energy the atom can have.3 Now pick another function and calculate its average energy. If it is lower than your first choice you are getting closer to the true ground-state energy. If you keep on trying all sorts of artificial states you will be able to get lower and lower energies, which come closer and closer to the ground-state energy. If you are clever, you will try some functions which have a few adjustable parameters. When you calculate the energy it will be expressed in terms of these parameters. By varying the parameters to give the lowest possible energy, you are trying out a whole class of functions at once. Eventually you will find that it is harder and harder to get lower energies and you will begin to be convinced that you are fairly close to the lowest possible energy. The helium atom has been solved in just this way—not by solving a differential equation, but by making up a special function with a lot of adjustable parameters which are eventually chosen to give the lowest possible value for the average energy. 