FOlks,
I am having difficulty understanding from image 144.jpg eqn 1.3.25, how Sakurai identified this as a square matrix by column matrix.
Can anyone shed light on this? THanks
Well, I wouldn't use the language "identified this as a square matrix by column matrix." Sakurai appears to have identified Eq. 1.3.25 as an application of the rule for multiplication of a column vector by a square matrix. The final result, as we can see, is a bracket (a scalar). To go from the first line of 1.3.25 to the second, all Sakurai did was insert the identity in the form of
$\displaystyle \displaystyle I=\sum_{a''}|a''\rangle\langle a''|.$
That's certainly a square matrix.
I guess I'm not entirely sure what your question is. Perhaps my comments could help to clarify your question? Are you asking why Sakurai can write down Eq. 1.3.25? Or are you asking how he recognized 1.3.25 as, partly, the multiplication of a square matrix with a column vector? Or something else?
Take a look at this thread. That might help explain a few things. Let me know if you have other questions.
THis an interesting thread particularly
You say that x and y are column vectors, ie the kets alpha and gamma are column vectors in eqn 1.3.26.
But I still dont see how these arrive to be column matrices...so I guess will have to take it as given for the moment :-)
Pardon my ignorance on this!
Typically, kets are defined as column vectors. (Incidentally, in your last post, you use the phrases "column vector" and "column matrix". Are you meaning the same thing by those two terms?) Bras are something for which you have to deduce their structure such that the inner product makes sense - they are vectors in the dual space.
Ok, that is good information. I also see that an inner product is a complex number therefore in matrix representation this has to be in the form
(1*N)(N*N)(N*1) =(1*1) to get a number and this equivalent to
$\displaystyle
\langle a|\hat X|b \rangle
$
Additionally
(1*N)(N*1)=(1*1) equivalent to $\displaystyle \langle a| b \rangle$
(N*1)(1*N)=(N*N) equivalent to $\displaystyle \langle b| a \rangle$
Hope I have them right :-)