Resolving state vectors
Let’s look at Eq. (8.8) again; we can think of it in the following way. Any state vector $\ket{\phi}$ can be represented as a linear combination with suitable coefficients of a set of base “vectors”—or, if you prefer, as a superposition of “unit vectors” in suitable proportions. To emphasize that the coefficients $\braket{i}{\phi}$ are just ordinary (complex) numbers, suppose we write \begin{equation} \braket{i}{\phi}=C_i.\notag \end{equation}  Then Eq. (8.8) is the same as \begin{equation} \label{Eq:III:8:10} \ket{\phi}=\sum_i\ket{i}C_i. \end{equation}  We can write a similar equation for any other state vector, say $\ket{\chi}$, with, of course, different coefficients—say $D_i$. Then we have \begin{equation} \label{Eq:III:8:11} \ket{\chi}=\sum_i\ket{i}D_i. \end{equation}  The $D_i$ are just the amplitudes $\braket{i}{\chi}$. Suppose we had started by abstracting the $\phi$ from Eq. (8.1). We would have had \begin{equation} \label{Eq:III:8:12} \bra{\chi}=\sum_i\braket{\chi}{i}\bra{i}. \end{equation}  Remembering that $\braket{\chi}{i}=\braket{i}{\chi}\cconj$, we can write this as \begin{equation} \label{Eq:III:8:13} \bra{\chi}=\sum_iD_i\cconj\,\bra{i}. \end{equation}  Now the interesting thing is that we can just multiply Eq. (8.13) and Eq. (8.10) to get back $\braket{\chi}{\phi}$. When we do that, we have to be careful of the summation indices, because they are quite distinct in the two equations. Let’s first rewrite Eq. (8.13) as \begin{equation*} \bra{\chi}=\sum_jD_j\cconj\,\bra{j}, \end{equation*}  which changes nothing. Then putting it together with Eq. (8.10), we have \begin{equation} \label{Eq:III:8:14} \braket{\chi}{\phi}=\sum_{ij}D_j\cconj\,\braket{j}{i}C_i. \end{equation}  Remember, though, that $\braket{j}{i}=\delta_{ij}$, so that in the sum we have left only the terms with $j=i$. We get \begin{equation} \label{Eq:III:8:15} \braket{\chi}{\phi}=\sum_{i}D_i\cconj\,C_i, \end{equation}  where, of course, $D_i\cconj=$ $\braket{i}{\chi}\cconj=$ $\braket{\chi}{i}$, and $C_i=\braket{i}{\phi}$. Again we see the close analogy with the dot product \begin{equation*} \FLPB\cdot\FLPA=\sum_iB_iA_i. \end{equation*}  The only difference is the complex conjugate on $D_i$. So Eq. (8.15) says that if the state vectors $\bra{\chi}$ and $\ket{\phi}$ are expanded in terms of the base vectors $\bra{i}$ or $\ket{i}$, the amplitude to go from $\phi$ to $\chi$ is given by the kind of dot product in Eq. (8.15). This equation is, of course, just Eq. (8.1) written with different symbols. So we have just gone in a circle to get used to the new symbols. We should perhaps emphasize again that while space vectors in three dimensions are described in terms of three orthogonal unit vectors, the base vectors $\ket{i}$ of the quantum mechanical states must range over the complete set applicable to any particular problem. Depending on the situation, two, or three, or five, or an infinite number of base states may be involved. We have also talked about what happens when particles go through an apparatus. If we start the particles out in a certain state $\phi$, then send them through an apparatus, and afterward make a measurement to see if they are in state $\chi$, the result is described by the amplitude \begin{equation} \label{Eq:III:8:16} \bracket{\chi}{A}{\phi}. \end{equation}  Such a symbol doesn’t have a close analog in vector algebra. (It is closer to tensor algebra, but the analogy is not particularly useful.) We saw in Chapter 5, Eq. (5.32), that we could write (8.16) as \begin{equation} \label{Eq:III:8:17} \bracket{\chi}{A}{\phi}= \sum_{ij}\braket{\chi}{i}\bracket{i}{A}{j}\braket{j}{\phi}. \end{equation}  This is just an example of the fundamental rule Eq. (8.9), used twice.  We also found that if another apparatus $B$ was added in series with $A$, then we could write \begin{equation} \label{Eq:III:8:18} \bracket{\chi}{BA}{\phi}= \sum_{ijk}\braket{\chi}{i}\bracket{i}{B}{j} \bracket{j}{A}{k}\braket{k}{\phi}. \end{equation}  Again, this comes directly from Dirac’s method of writing Eq. (8.9)—remember that we can always place a bar ($|$), which is just like the factor $1$, between $B$ and $A$. Incidentally, we can think of Eq. (8.17) in another way. Suppose we think of the particle entering apparatus $A$ in the state $\phi$ and coming out of $A$ in the state $\psi$, (“psi”). In other words, we could ask ourselves this question: Can we find a $\psi$ such that the amplitude to get from $\psi$ to $\chi$ is always identically and everywhere the same as the amplitude $\bracket{\chi}{A}{\phi}$? The answer is yes. We want Eq. (8.17) to be replaced by \begin{equation} \label{Eq:III:8:19} \braket{\chi}{\psi}= \sum_i\braket{\chi}{i}\braket{i}{\psi}. \end{equation}  We can clearly do this if \begin{equation} \label{Eq:III:8:20} \braket{i}{\psi}=\sum_j\bracket{i}{A}{j}\braket{j}{\phi}= \bracket{i}{A}{\phi}, \end{equation}  which determines $\psi$. “But it doesn’t determine $\psi$,” you say; “it only determines $\braket{i}{\psi}$.” However, $\braket{i}{\psi}$ does determine $\psi$, because if you have all the coefficients that relate $\psi$ to the base states $i$, then $\psi$ is uniquely defined. In fact, we can play with our notation and write the last term of Eq. (8.20) as \begin{equation} \label{Eq:III:8:21} \braket{i}{\psi}=\sum_j\braket{i}{j}\bracket{j}{A}{\phi}. \end{equation}  Then, since this equation is true for all $i$, we can write simply \begin{equation} \label{Eq:III:8:22} \ket{\psi}=\sum_j\ket{j}\bracket{j}{A}{\phi}. \end{equation}  Then we can say: “The state $\psi$ is what we get if we start with $\phi$ and go through the apparatus $A$.”  One final example of the tricks of the trade. We start again with Eq. (8.17). Since it is true for any $\chi$ and $\phi$, we can drop them both!  We then get1 \begin{equation} \label{Eq:III:8:23} A=\sum_{ij}\ket{i}\bracket{i}{A}{j}\bra{j}. \end{equation}  What does it mean? It means no more, no less, than what you get if you put back the $\phi$ and $\chi$. As it stands, it is an “open” equation and incomplete. If we multiply it “on the right” by $\ket{\phi}$, it becomes \begin{equation} \label{Eq:III:8:24} A\,\ket{\phi}=\sum_{ij}\ket{i}\bracket{i}{A}{j}\braket{j}{\phi}, \end{equation}  which is just Eq. (8.22) all over again. In fact, we could have just dropped the $j$’s from that equation and written \begin{equation} \label{Eq:III:8:25} \ket{\psi}=A\,\ket{\phi}. \end{equation}  The symbol $A$ is neither an amplitude, nor a vector; it is a new kind of thing called an operator. It is something which “operates on” a state to produce a new state—Eq. (8.25) says that $\ket{\psi}$ is what results if $A$ operates on $\ket{\phi}$. Again, it is still an open equation until it is completed with some bra like $\bra{\chi}$ to give \begin{equation} \label{Eq:III:8:26} \braket{\chi}{\psi}=\bracket{\chi}{A}{\phi}. \end{equation}  The operator $A$ is, of course, described completely if we give the matrix of amplitudes $\bracket{i}{A}{j}$—also written $A_{ij}$—in terms of any set of base vectors. We have really added nothing new with all of this new mathematical notation. One reason for bringing it all up was to show you the way of writing pieces of equations, because in many books you will find the equations written in the incomplete forms, and there’s no reason for you to be paralyzed when you come across them. If you prefer, you can always add the missing pieces to make an equation between numbers that will look like something more familiar. Also, as you will see, the “bra” and “ket” notation is a very convenient one. For one thing, we can from now on identify a state by giving its state vector. When we want to refer to a state of definite momentum $\FLPp$ we can say: “the state $\ket{\FLPp}$.” Or we may speak of some arbitrary state $\ket{\psi}$. For consistency we will always use the ket, writing $\ket{\psi}$, to identify a state. (It is, of course an arbitrary choice; we could equally well have chosen to use the bra, $\bra{\psi}$.)