In linear algebra all the world’s a matrix (even vectors). Everyone (except me in the last post) numbers matrix elements by the following subscript convention — the row always comes first, then the columns (mnemonic Roman Catholic). Similarly matrix size is always written a x b where a is the number of rows and b the number of columns. Vectors in quantum mechanics are written both ways, as column vectors 1 x n, or as row vectors (n x 1).
Vectors aren’t usually called matrices, but matrices they are when it comes to multiplication. Vectors can be multiplied by a matrix (or multiply a matrix) using the usual matrix multiplication rules. That’s one reason the example in LASGFQM – VI was so tedious — I wanted to show how matrices of different sizes could be multiplied together. The order of the matrices is crucial. The first matrix A must have the same number of columns that the second matrix (B) has rows — otherwise it just doesn’t work. The product matrix has the number of rows of matrix A and the columns of matrix B.
So it is possible to form A B where A is 3 x 4 and B is 4 x 5 giving a 3 x 5 matrix, but B A makes no sense. If you get stuck use the Hubbard method of writing them out (see the last post). Here is a 3 x 3 matrix (A) multiplying a 3 x 1 matrix (vector B)
A11 A12 A13 A11*B11 + A12 B21 + A13 * B31 — this is a single number
A21 A22 A23 A21*B11 + A22*B21 + A23* B31 — ditto
A31 A32 A33 etc.
AB is just another 3 x 1 vector. So the matrix just transforms one 3 dimensional vector into another
You should draw a similar diagram and see why B A is impossible. What about
C (3 x 1) times D (3 x 3)? You get CD a 3 x 1 matrix (row vector) back .
D11 D12 D13
D21 D22 D23
D31 D32 D33
C11 C12 C13 What is CD12?
Suppose we get concrete and make B into a column vector of the following type
A11 A12 A13 A11
A21 A22 A23 A21
A31 A32 A33 A31
The first time I saw this, I didn’t understand it. I thought the mathematicians were going back to the old Cartesian system of standard orthonormal vectors. They weren’t doing this at all. Recall that we’re in a vector space and the column vector is really the 3 coefficients multiplying the 3 basis vectors (which are not specified). So you don’t have to mess around with choosing a basis, the result is true for ALL bases of a 3 dimensional vector space. The power of abstraction. The first column of A shows what the first basis vector goes to (in general), the second column shows what the second basis goes to. Back in LSQFQM – IV, it was explained why any linear transformation (call it T) of a basis vector (call it C1) to another vector space must look like this
T(C1) = t11 * D1 + t12 * D2 + . .. for however many basis vectors vector space D has.
Well, in the above example we’re going from a 3 dimensional vector space to another, and the first row of matrix A tells us what basis vector #1 is going to. This is why every linear transformation can be represented by a matrix and every matrix represents a linear transformation. Sometimes abstraction saves a lot of legwork.
A more geometric way to look at all this is to regard an n x n matrix multiplying an n x 1 vector as moving it around in n dimensional space (keeping one end fixed at the origin — see below). So
1 0 0
0 1 0
0 0 2
just multiplies the third basis vector by 2 leaving the other two alone.
The notation is consistent. Recall that any linear transformation must leave the zero vector unchanged (see LSQFQM – I for a proof). Given the rules for multiplying a matrix times a vector, this happens with a column vector which is all zeros.
The geometrically inclined can start thinking about what the possible linear transformations can do to three dimensional space (leaving the origin fixed). Rotations about the origin are one possibility, expansion or contraction along a single basis vector are two more, projections down to a 2 dimensional plane or a 1 dimensional line are two more. There are others (particularly when we’re in a vector space with complex numbers for coefficients — e.g. all of quantum mechanics).
Up next time, eigenvectors, adjoints, and (hopefully) Hermitian operators. That will be about it. The point of these posts (which are far more extensive than I thought they would be when I started out) is to show you how natural the language of linear algebra is, once you see what’s going on under the hood. It is not to teach quantum mechanics, which I’m still learning to see how it is used in chemistry. QM is far from natural (although it describes the submicroscopic world — whether it can ever describe the world we live in is another question), but, if these posts are any good at all, you should be able to understand the language in which QM is expressed.