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Important Info regarding the quiz:
Major topics:
We reviewed a little from the chapter 3 material. We had several questions and spent some time on them.
Highlights from Chapter 3:
Part 1. Inner Product
“<math>\int_a^b f(x)*g(x)\,\mathrm dx = <f|g></math>”
1 & 0 & 0 & …
0 & 1 & 0 & …
0 & 0 & 1 & …
. & . & .
. & . & .
. & . & . \end{bmatrix}=I</math>, unit matrix.
<math><f|g>=\sum_{i} <f|e_i><e_i|g></math>, as <math><f|e_i>=f_i^*</math>, <math><e_i|g>=g_i</math>, so <math><f|g>=\sum_{i} f_i^*g_i</math>
Q_1_1 & Q_1_2 & Q_1_3 & …
Q_2_1 & Q_2_2 & Q_2_3 & …
Q_3_1 & Q_3_2 & Q_3_3 & …
. & . & .
. & . & .
. & . & . \end{bmatrix}</math>
<math><e_i|\hat{Q}|e_j>=Q_i_j</math>
<math><f|\hat{Q}|g>=\sum_{i,j} <f|e_i><e_i|\hat{Q}|e_j><e_j|g> = \sum_{i,j} f_i^* Q_i_j g_j </math>
Part 2. Hermitian Operator
<f|Âg> = <Âf|g> for all f(x) and all g(x)
where  is a hermitian operator and f and g are functions of x.
Hermitian Operators represent observables.
Âf = af
Eigenvalues are numbers only. (i.e., they are NOT operators or functions)
prove: A, B are two Hermitian operators, so <math>A^+=A</math>, <math>B^+=B</math> (see Problem 3.5)
then <math>(AB)^+=B^+A^+=BA</math> (see Problem 3.5)
Only if BA=AB can we get that <math>(AB)^+=BA=AB</math>. AB is Hermitian. QED
Part 3. Orthonormality for eigenstates
This can be generalized to all Hermitian operators. (Theorem 2 in textbook) Eigenfunctions belonging to distinct eigenvalues are orthogonal.
<math><\psi_n|\psi_m>=<n|m>=\int \psi_n^* \psi_m\, dx =\delta_n_m</math>
<math>|\psi>=\sum C_n |\psi_n> </math>
To be continued :)
Good luck on the quiz!
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