poolwin2001 said:
How are operators represented by matrix method?
The relationship between linear transformations and matrices is quite simple.
Suppose U and V are vector spaces, and that
T:U\rightarrow V
is linear. Any vector in either of the two vector spaces can be expressed as a linear combination of basis vectors. We can use this, and the linearity of T, to express y=Tx in two different ways.
y=\sum_{i=1}^m y_i v_i
y=Tx=T\bigl( \sum_{j=1}^n x_j u_j\bigr) =\sum_{j=1}^n x_j Tu_j=\sum_{j=1}^n x_j \sum_{i=1}^{m} [Tu_j]_i v_i=\sum_{i=1}^m \bigl( \sum_{j=1}^n [Tu_j]_i x_j \bigr) v_i
I hope the notation is easy enough to understand.
Since basis vectors are linearly independent, we must have
y_i=\sum_{j=1}^n [Tu_j]_i x_j
and this can be interpreted as a matrix equation:
\begin{pmatrix}y_1\\ \vdots\\ y_m\end{pmatrix}=<br />
<br />
\begin{pmatrix}[Tu_1]_1& \dots& [Tu_n]_1\\<br />
\vdots& \ddots& \vdots& \\<br />
[Tu_1]_m& \dots& [Tu_n]_m\end{pmatrix}<br />
<br />
\begin{pmatrix}x_1\\ \vdots\\ x_n\end{pmatrix}
The column matrices "represent" the vectors x and y, and the m×n matrix "represents" the operator T.
The term "linear operator" is usually reserved for the special case U=V. (This is why I called T a linear
transformation).
In quantum mechanics we're dealing with linear operators that map a Hilbert space H into itself (
onto itself, in the case of observables). The bases are orthonormal. Because of this, we can use the inner product to write the matrix equation in a different way:
\begin{pmatrix}\langle u_1|y\rangle\\ \vdots\\ \langle u_n|y\rangle\end{pmatrix}=<br />
<br />
\begin{pmatrix}\langle u_1|Tu_1\rangle& \dots& \langle u_1|Tu_n\rangle\\<br />
\vdots& \ddots& \vdots& \\<br />
\langle u_n|Tu_1\rangle& \dots& \langle u_n|Tu_n\rangle\end{pmatrix}<br />
<br />
\begin{pmatrix}\langle u_1|x\rangle\\ \vdots\\ \langle u_n|x\rangle\end{pmatrix}
