Generalization to $\boldsymbol{N}$-state systems
We have finished with all the two-state systems we wanted to talk about. In the following chapters we will go on to study systems with more states. The extension to $N$-state systems of the ideas we have worked out for two states is pretty straightforward. It goes like this.  If a system has $N$ distinct states, we can represent any state $\ket{\psi(t)}$ as a linear combination of any set of base states $\ket{i}$, where $i=1$, $2$, $3$, $\ldots$, $N$; \begin{equation} \label{Eq:III:11:57} \ket{\psi(t)}=\sum_{\text{all $i$}}\ket{i}C_i(t). \end{equation} The coefficients $C_i(t)$ are the amplitudes $\braket{i}{\psi(t)}$. The behavior of the amplitudes $C_i$ with time is governed by the equations \begin{equation} \label{Eq:III:11:58} i\hbar\,\ddt{C_i(t)}{t}=\sum_jH_{ij}C_j, \end{equation} where the energy matrix $H_{ij}$ describes the physics of the problem. It looks the same as for two states. Only now, both $i$ and $j$ must range over all $N$ base states, and the energy matrix $H_{ij}$—or, if you prefer, the Hamiltonian—is an $N$ by $N$ matrix with $N^2$ numbers. As before, $H_{ij}\cconj=H_{ji}$—so long as particles are conserved—and the diagonal elements $H_{ii}$ are real numbers.  We have found a general solution for the $C$’s of a two-state system when the energy matrix is constant (doesn’t depend on $t$). It is also not difficult to solve Eq. (11.58) for an $N$-state system when $H$ is not time dependent. Again, we begin by looking for a possible solution in which the amplitudes all have the same time dependence. We try \begin{equation} \label{Eq:III:11:59} C_i=a_ie^{-(i/\hbar)Et}. \end{equation} When these $C_i$’s are substituted into (11.58), the derivatives $dC_i(t)/dt$ become just $(-i/\hbar)EC_i$. Canceling the common exponential factor from all terms, we get \begin{equation} \label{Eq:III:11:60} Ea_i=\sum_jH_{ij}a_j. \end{equation} This is a set of $N$ linear algebraic equations for the $N$ unknowns $a_1$, $a_2$, $\ldots$, $a_N$, and there is a solution only if you are lucky—only if the determinant of the coefficients of all the $a$'s is zero. But it’s not necessary to be that sophisticated; you can just start to solve the equations any way you want, and you will find that they can be solved only for certain values of $E$. (Remember that $E$ is the only adjustable thing we have in the equations.)  If you want to be formal, however, you can write Eq. (11.60) as \begin{equation} \label{Eq:III:11:61} \sum_j(H_{ij}-\delta_{ij}E)a_j=0. \end{equation} Then you can use the rule—if you know it—that these equations will have a solution only for those values of $E$ for which \begin{equation} \label{Eq:III:11:62} \Det\,(H_{ij}-\delta_{ij}E)=0. \end{equation} Each term of the determinant is just $H_{ij}$, except that $E$ is subtracted from every diagonal element. That is, (11.62) means just \begin{equation} \label{Eq:III:11:63} \Det \!\! % ebook remove \begin{pmatrix} H_{11}\!-\!E & H_{12} & H_{13} & \dots\\[1ex] H_{21} & H_{22}\!-\!E & H_{23} & \dots\\[1ex] H_{31} & H_{32} & H_{33}\!-\!E & \dots\\[1ex] \dots & \dots & \dots & \dots \end{pmatrix} \!\!=0.% ebook remove % ebook insert: =0. \end{equation} This is, of course, just a special way of writing an algebraic equation for $E$ which is the sum of a bunch of products of all the terms taken a certain way. These products will give all the powers of $E$ up to $E^N$.  So we have an $N$th order polynomial equal to zero, and there are, in general, $N$ roots. (We must remember, however, that some of them may be multiple roots—meaning that two or more roots are equal.) Let’s call the $N$ roots \begin{equation} \label{Eq:III:11:64} E_{\slI},E_{\slII},E_{\slIII},\dotsc,E_{\bldn},\dotsc, E_{\bldN}. \end{equation} (We will use $\bldn$ to represent the $n$th Roman numeral, so that $\bldn$ takes on the values $\slI$, $\slII$, $\ldots$, $\bldN$.)  It may be that some of these energies are equal—say $E_{\slII}=E_{\slIII}$—but we will still choose to call them by different names.  The equations (11.60)—or (11.61)—have one solution for each value of $E$. If you put any one of the $E$'s—say $E_{\bldn}$—into (11.60) and solve for the $a_i$, you get a set which belongs to the energy $E_{\bldn}$. We will call this set $a_i(\bldn)$.  Using these $a_i(\bldn)$ in Eq. (11.59), we have the amplitudes $C_i(\bldn)$ that the definite energy states are in the base state $\ket{i}$. Letting $\ket{\bldn}$ stand for the state vector of the definite energy state at $t=0$, we can write \begin{equation*} C_i(\bldn)=\braket{i}{\bldn}e^{-(i/\hbar)E_{\bldn}t}, \end{equation*} with \begin{equation} \label{Eq:III:11:65} \braket{i}{\bldn}=a_i(\bldn). \end{equation} The complete definite energy state $\ket{\psi_{\bldn}(t)}$ can then be written as \begin{equation*} \ket{\psi_{\bldn}(t)}=\sum_i\ket{i}a_i(\bldn)e^{-(i/\hbar)E_{\bldn}t}, \end{equation*} or \begin{equation} \label{Eq:III:11:66} \ket{\psi_{\bldn}(t)}=\ket{\bldn}e^{-(i/\hbar)E_{\bldn}t}. \end{equation} The state vectors $\ket{\bldn}$ describe the configuration of the definite energy states, but have the time dependence factored out. Then they are constant vectors which can be used as a new base set if we wish.  Each of the states $\ket{\bldn}$ has the property—as you can easily show—that when operated on by the Hamiltonian operator $\Hop$ it gives just $E_{\bldn}$ times the same state: \begin{equation} \label{Eq:III:11:67} \Hop\,\ket{\bldn}=E_{\bldn}\,\ket{\bldn}. \end{equation}  The energy $E_{\bldn}$ is, then, a number which is a characteristic of the Hamiltonian operator $\Hop$. As we have seen, a Hamiltonian will, in general, have several characteristic energies. In the mathematician’s world these would be called the “characteristic values” of the matrix $H_{ij}$. Physicists usually call them the “eigenvalues” of $\Hop$. (“Eigen” is the German word for “characteristic” or “proper.”) With each eigenvalue of $\Hop$—in other words, for each energy—there is the state of definite energy, which we have called the “stationary state.” Physicists usually call the states $\ket{\bldn}$ “the eigenstates of $\Hop$.” Each eigenstate corresponds to a particular eigenvalue $E_{\bldn}$.  Now, generally, the states $\ket{\bldn}$—of which there are $N$—can also be used as a base set. For this to be true, all of the states must be orthogonal, meaning that for any two of them, say $\ket{\bldn}$ and $\ket{\bldm}$, \begin{equation} \label{Eq:III:11:68} \braket{\bldn}{\bldm}=0. \end{equation} This will be true automatically if all the energies are different. Also, we can multiply all the $a_i(\bldn)$ by a suitable factor so that all the states are normalized—by which we mean that \begin{equation} \label{Eq:III:11:69} \braket{\bldn}{\bldn}=1 \end{equation} for all $\bldn$.  When it happens that Eq. (11.63) accidentally has two (or more) roots with the same energy, there are some minor complications. First, there are still two different sets of $a_i$'s which go with the two equal energies, but the states they give may not be orthogonal. Suppose you go through the normal procedure and find two stationary states with equal energies—let’s call them $\ket{\mu}$ and $\ket{\nu}$. Then it will not necessarily be so that they are orthogonal—if you are unlucky, \begin{equation*} \braket{\mu}{\nu}\neq0. \end{equation*} It is, however, always true that you can cook up two new states, which we will call $\ket{\mu'}$ and $\ket{\nu'}$, that have the same energies and are also orthogonal, so that \begin{equation} \label{Eq:III:11:70} \braket{\mu'}{\nu'}=0. \end{equation} You can do this by making $\ket{\mu'}$ and $\ket{\nu'}$ a suitable linear combination of $\ket{\mu}$ and $\ket{\nu}$, with the coefficients chosen to make it come out so that Eq. (11.70) is true. It is always convenient to do this. We will generally assume that this has been done so that we can always assume that our proper energy states $\ket{\bldn}$ are all orthogonal.  We would like, for fun, to prove that when two of the stationary states have different energies they are indeed orthogonal. For the state $\ket{\bldn}$ with the energy $E_{\bldn}$, we have that \begin{equation} \label{Eq:III:11:71} \Hop\,\ket{\bldn}=E_{\bldn}\,\ket{\bldn}. \end{equation} This operator equation really means that there is an equation between numbers. Filling the missing parts, it means the same as \begin{equation} \label{Eq:III:11:72} \sum_j\bracket{i}{\Hop}{j}\braket{j}{\bldn}= E_{\bldn}\braket{i}{\bldn}. \end{equation} If we take the complex conjugate of this equation, we get \begin{equation} \label{Eq:III:11:73} \sum_j\bracket{i}{\Hop}{j}\cconj\braket{j}{\bldn}\cconj= E_{\bldn}\cconj\braket{i}{\bldn}\cconj. \end{equation} Remember now that the complex conjugate of an amplitude is the reverse amplitude, so (11.73) can be rewritten as \begin{equation} \label{Eq:III:11:74} \sum_j\braket{\bldn}{j}\bracket{j}{\Hop}{i}= E_{\bldn}\cconj\braket{\bldn}{i}. \end{equation} Since this equation is valid for any $i$, its “short form” is \begin{equation} \label{Eq:III:11:75} \bra{\bldn}\,\Hop=E_{\bldn}\cconj\bra{\bldn}, \end{equation} which is called the adjoint to Eq. (11.71).  Now we can easily prove that $E_{\bldn}$ is a real number. We multiply Eq. (11.71) by $\bra{\bldn}$ to get \begin{equation} \label{Eq:III:11:76} \bracket{\bldn}{\Hop}{\bldn}=E_{\bldn}, \end{equation} since $\braket{\bldn}{\bldn}=1$. Then we multiply Eq. (11.75) on the right by $\ket{\bldn}$ to get \begin{equation} \label{Eq:III:11:77} \bracket{\bldn}{\Hop}{\bldn}=E_{\bldn}\cconj. \end{equation} Comparing (11.76) with (11.77) it is clear that \begin{equation} \label{Eq:III:11:78} E_{\bldn}=E_{\bldn}\cconj, \end{equation} which means that $E_{\bldn}$ is real. We can erase the star on $E_{\bldn}$ in Eq. (11.75).  Finally we are ready to show that the different energy states are orthogonal. Let $\ket{\bldn}$ and $\ket{\bldm}$ be any two of the definite energy base states. Using Eq. (11.75) for the state $\bldm$, and multiplying it by $\ket{\bldn}$, we get that \begin{equation*} \bracket{\bldm}{\Hop}{\bldn}=E_{\bldm}\braket{\bldm}{\bldn}. \end{equation*} But if we multiply (11.71) by $\bra{\bldm}$, we get \begin{equation*} \bracket{\bldm}{\Hop}{\bldn}=E_{\bldn}\braket{\bldm}{\bldn}. \end{equation*} Since the left sides of these two equations are equal, the right sides are, also: \begin{equation} \label{Eq:III:11:79} E_{\bldm}\braket{\bldm}{\bldn}=E_{\bldn}\braket{\bldm}{\bldn}. \end{equation} If $E_{\bldm}=E_{\bldn}$ the equation does not tell us anything. But if the energies of the two states $\ket{\bldm}$ and $\ket{\bldn}$ are different ($E_{\bldm}\neq E_{\bldn}$), Eq. (11.79) says that $\braket{\bldm}{\bldn}$ must be zero, as we wanted to prove. The two states are necessarily orthogonal so long as $E_{\bldn}$ and $E_{\bldm}$ are numerically different. 