Why is the inner product conjugate symmetric?

[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

 

[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>)

 

[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.
Thamks in advance.
Somy

 

[quantumdude]

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

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;>

 

[somy]

Thanks Tom;
YOU did it!