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A question about the linear product

  1. Jul 31, 2004 #1
    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 :smile:
  2. jcsd
  3. Jul 31, 2004 #2


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    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>)
  4. Jul 31, 2004 #3
    Thank HallsofIvy!!!
    the answer was very usefull. but;
    just see the exact equality:

    the * sign is out of the dot product.
    IT is the thing that I can't understand.
    Thamks in advance.
  5. Aug 2, 2004 #4

    Tom Mattson

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    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:

    [itex]\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[/itex]

    Take the complex conjugate of both sides, and it should be clear why <φ|ψ>*=<&psi;|&phi;>
    Last edited: Aug 2, 2004
  6. Aug 2, 2004 #5
    Thanks Tom;
    YOU did it!!!
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