# A question about the linear product

1. Jul 31, 2004

### somy

A question about the "linear product"

Hi everyone;
As you know, we assume the answer of the dot product as a complex number: <a|b>

Also, we have the property: <a|b>=<b|a>*

I just want to know how can we say this, or is it just a definition???
Thanks a lot.
Somy

2. Jul 31, 2004

### HallsofIvy

First, of course, it IS a definition, but there is a reason for that specific definition.

The absolute value of a real number can be defined in two ways:
"algebraically" as |x|= sqrt(x*x) or
"geometrically" as the distance from x to 0 on the number line ("distance", of course, is always positive. The distance from 4 to 0 and the distance from -4 to 0 are both 4).

Extending to complex number, if we think of the number z= x+ iy as the point in the "complex plane" (x,y), then the distance from (x,y) to (0, 0) is sqrt(x2+ y2). This is NOT sqrt(z.z) but is sqrt(z.z*)= sqrt(<z, z>)

3. Jul 31, 2004

### somy

Thank HallsofIvy!!!
the answer was very usefull. but;
just see the exact equality:

<a|b>=<b|a>*
the * sign is out of the dot product.
IT is the thing that I can't understand.
Somy

4. Aug 2, 2004

### Tom Mattson

Staff Emeritus
The inner product is defined the way it is because it is an algebraic abstraction of overlap integrals of wavefunctions.

Look at the inner product in terms of functions:

$\newcommand{\mean}[1]{{<\!\!{#1}\!\!>}}\newcommand{\braket}[2]{{<\!\!{#1|#2}\!\!>}}\newcommand{\braketop}[3]{{<\!\!{#1|\hat{#2}|#3}\!\!>}}\braket{\phi}{\psi} \equiv \int \phi^*(x) \psi(x)\,dx$

Take the complex conjugate of both sides, and it should be clear why <φ|ψ>*=<&psi;|&phi;>

Last edited: Aug 2, 2004
5. Aug 2, 2004

### somy

Thanks Tom;
YOU did it!!!