Average energies
So far we have reminded you mainly of what you already know. Now we would like to discuss a new question. How would you find the average energy of a system—say, an atom? If an atom is in a particular state of definite energy and you measure the energy, you will find a certain energy $E$. If you keep repeating the measurement on each one of a whole series of atoms which are all selected to be in the same state, all the measurements will give $E$, and the “average” of your measurements will, of course, be just $E$.  Now, however, what happens if you make the measurement on some state $\ket{\psi}$ which is not a stationary state? Since the system does not have a definite energy, one measurement would give one energy, the same measurement on another atom in the same state would give a different energy, and so on. What would you get for the average of a whole series of energy measurements?  We can answer the question by projecting the state $\ket{\psi}$ onto the set of states of definite energy. To remind you that this is a special base set, we’ll call the states $\ket{\eta_i}$. Each of the states $\ket{\eta_i}$ has a definite energy $E_i$. In this representation, \begin{equation} \label{Eq:III:20:10} \ket{\psi}=\sum_iC_i\,\ket{\eta_i}. \end{equation} When you make an energy measurement and get some number $E_i$, you have found that the system was in the state $\eta_i$. But you may get a different number for each measurement. Sometimes you will get $E_1$, sometimes $E_2$, sometimes $E_3$, and so on. The probability that you observe the energy $E_1$ is just the probability of finding the system in the state $\ket{\eta_1}$, which is, of course, just the absolute square of the amplitude $C_1=\braket{\eta_1}{\psi}$. The probability of finding each of the possible energies $E_i$ is \begin{equation} \label{Eq:III:20:11} P_i=\abs{C_i}^2. \end{equation}  How are these probabilities related to the mean value of a whole sequence of energy measurements? Let’s imagine that we get a series of measurements like this: $E_1$, $E_7$, $E_{11}$, $E_9$, $E_1$, $E_{10}$, $E_7$, $E_2$, $E_3$, $E_9$, $E_6$, $E_4$, and so on. We continue for, say, a thousand measurements. When we are finished we add all the energies and divide by one thousand. That’s what we mean by the average. There’s also a short-cut to adding all the numbers. You can count up how many times you get $E_1$, say that is $N_1$, and then count up the number of times you get $E_2$, call that $N_2$, and so on. The sum of all the energies is certainly just \begin{equation*} N_1E_1+N_2E_2+N_3E_3+\dotsb{}=\sum_iN_iE_i. \end{equation*} The average energy is this sum divided by the total number of measurements which is just the sum of all the $N_i$'s, which we can call $N$; \begin{equation} \label{Eq:III:20:12} E_{\text{av}}=\frac{\sum_iN_iE_i}{N}. \end{equation}  We are almost there. What we mean by the probability of something happening is just the number of times we expect it to happen divided by the total number of tries. The ratio $N_i/N$ should—for large $N$—be very near to $P_i$, the probability of finding the state $\ket{\eta_i}$, although it will not be exactly $P_i$ because of the statistical fluctuations. Let’s write the predicted (or “expected”) average energy as $\av{E}$; then we can say that \begin{equation} \label{Eq:III:20:13} \av{E}=\sum_iP_iE_i. \end{equation} The same arguments apply for any measurement. The average value of a measured quantity $A$ should be equal to \begin{equation*} \av{A}=\sum_iP_iA_i, \end{equation*} where $A_i$ are the various possible values of the observed quantity, and $P_i$ is the probability of getting that value.  Let’s go back to our quantum-mechanical state $\ket{\psi}$. Its average energy is \begin{equation} \label{Eq:III:20:14} \av{E}=\sum_i\abs{C_i}^2E_i=\sum_iC_i\cconj C_iE_i. \end{equation} Now watch this trickery! First, we write the sum as \begin{equation} \label{Eq:III:20:15} \sum_i\braket{\psi}{\eta_i}E_i\braket{\eta_i}{\psi}. \end{equation} Next we treat the left-hand $\bra{\psi}$ as a common “factor.” We can take this factor out of the sum, and write it as \begin{equation*} \bra{\psi}\,\biggl\{\sum_i\ket{\eta_i}E_i\braket{\eta_i}{\psi}\biggr\}. \end{equation*} This expression has the form \begin{equation*} \braket{\psi}{\phi}, \end{equation*} where $\ket{\phi}$ is some “cooked-up” state defined by \begin{equation} \label{Eq:III:20:16} \ket{\phi}=\sum_i\ket{\eta_i}E_i\braket{\eta_i}{\psi}. \end{equation} It is, in other words, the state you get if you take each base state $\ket{\eta_i}$ in the amount $E_i\braket{\eta_i}{\psi}$.  Now remember what we mean by the states $\ket{\eta_i}$. They are supposed to be the stationary states—by which we mean that for each one, \begin{equation*} \Hop\,\ket{\eta_i}=E_i\,\ket{\eta_i}. \end{equation*} Since $E_i$ is just a number, the right-hand side is the same as $\ket{\eta_i}E_i$, and the sum in Eq. (20.16) is the same as \begin{equation*} \sum_i\Hop\,\ket{\eta_i}\braket{\eta_i}{\psi}. \end{equation*} Now $i$ appears only in the famous combination that contracts to unity, so \begin{equation*} \sum_i\Hop\,\ket{\eta_i}\braket{\eta_i}{\psi}= \Hop\sum_i\ket{\eta_i}\braket{\eta_i}{\psi}=\Hop\,\ket{\psi}. \end{equation*} Magic! Equation (20.16) is the same as \begin{equation} \label{Eq:III:20:17} \ket{\phi}=\Hop\,\ket{\psi}. \end{equation} The average energy of the state $\ket{\psi}$ can be written very prettily as \begin{equation} \label{Eq:III:20:18} \av{E}=\bracket{\psi}{\Hop}{\psi}. \end{equation} To get the average energy you operate on $\ket{\psi}$ with $\Hop$, and then multiply by $\bra{\psi}$. A simple result.  Our new formula for the average energy is not only pretty. It is also useful, because now we don’t need to say anything about any particular set of base states. We don’t even have to know all of the possible energy levels. When we go to calculate, we’ll need to describe our state in terms of some set of base states, but if we know the Hamiltonian matrix $H_{ij}$ for that set we can get the average energy. Equation (20.18) says that for any set of base states $\ket{i}$, the average energy can be calculated from \begin{equation} \label{Eq:III:20:19} \av{E}=\sum_{ij}\braket{\psi}{i}\bracket{i}{\Hop}{j}\braket{j}{\psi}, \end{equation} where the amplitudes $\bracket{i}{\Hop}{j}$ are just the elements of the matrix $H_{ij}$.  Let’s check this result for the special case that the states $\ket{i}$ are the definite energy states. For them, $\Hop\,\ket{j}=E_j\,\ket{j}$, so $\bracket{i}{\Hop}{j}=E_j\,\delta_{ij}$ and \begin{equation*} \av{E}=\sum_{ij}\braket{\psi}{i}E_j\delta_{ij}\braket{j}{\psi}= \sum_iE_i\braket{\psi}{i}\braket{i}{\psi}, \end{equation*} which is right.  Equation (20.19) can, incidentally, be extended to other physical measurements which you can express as an operator. For instance, $\Lop_z$ is the operator of the $z$-component of the angular momentum $\FLPL$. The average of the $z$-component for the state $\ket{\psi}$ is \begin{equation*} \av{L_z}=\bracket{\psi}{\Lop_z}{\psi}. \end{equation*} One way to prove it is to think of some situation in which the energy is proportional to the angular momentum. Then all the arguments go through in the same way.  In summary, if a physical observable $A$ is related to a suitable quantum-mechanical operator $\Aop$, the average value of $A$ for the state $\ket{\psi}$ is given by \begin{equation} \label{Eq:III:20:20} \av{A}=\bracket{\psi}{\Aop}{\psi}. \end{equation} By this we mean that \begin{equation} \label{Eq:III:20:21} \av{A}=\braket{\psi}{\phi}, \end{equation} with \begin{equation} \label{Eq:III:20:22} \ket{\phi}=\Aop\,\ket{\psi}. \end{equation} 