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Complex wavefunction.

  1. Feb 16, 2005 #1
    From what I heard, the wavefunction is made up of both real and imaginary parts. How do I prove this? Also, what is the physical interpretation of complex numbers? How does a complex wavefunction fit into physical reality?
     
  2. jcsd
  3. Feb 16, 2005 #2

    dextercioby

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    Use the definition,because it is specified there.

    They're a "necessary evil"...They don't have "physical interpretation".They're just VERY USEFUL mathematical tools...


    By Born's statistical interpretation of Schroedinger's wave function.
    "The probability density to find the particle in the point [itex] \vec{r} [/itex] at the moment "t" is:
    [tex] \mathcal{P}(\vec{r},t)=|\Psi(\vec{r},t)|^{2} [/tex]

    Daniel.
     
  4. Feb 16, 2005 #3
    A complex wavefunction arises because we need a way to describe the wave nature of particles, without having them actually be a disturbance in some medium. So we introduce a complex scalar field and have it oscillate in the complex dimensions.

    So when two wavefunctions add we can have the wave type of destructive/constructive intereference in the complex dimensions. When we need to be brought back to physical reality, we find the length of the complex number (i.e. it's magnitude).

    So we use the complex number description because it is a very easy way to describe reality, even though mathematicians invented complex numbers thinking that they would have no real physical counterpart. And it all works out quite well. Isn't physics amazing?!

    Masud.
     
  5. Feb 16, 2005 #4

    dextercioby

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    The reason for the introduction of COMPLEX scalar fields into classical and quantum field theory is because the "COMPLEX" attribute allows for a correct description of ELECTRIC charge,as the the field of complex numbers is the field of scalars over which the fields determine an associative algebra with involution,the involution in this case being the complex conjugation...
    No connection with any disturbance,whatsoever...:wink:

    Daniel.
     
  6. Feb 16, 2005 #5
    1. The wavefunction is stipulated to be complex, therefore not only can't you prove it, you aren't supposed to.

    2. There is no physical interpretation of a complex number.

    3. Complex wavefunction doesn't fit into reality well at all, that's why you see everyone trying to convert back to real functions of a real variable.

    Regards,

    Guru
     
  7. Feb 16, 2005 #6
    Daniel,

    I wasn't talking about either classical or quantum field theory (an area where my knowledge only partially extends into); instead I was talking about non-relativistic quantum mechanics, where we give our quantum particles both wave and particulate natures.
     
  8. Feb 16, 2005 #7

    dextercioby

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    That's weid,your words
    don't seem to say it...:wink:You should try to be more careful with the terminology,coz people may not understand what u really wanna say...

    Daniel.
     
  9. Feb 16, 2005 #8
    When I say scalar field, what I mean is every point in space is associated with a scalar number, as opposed to a vector field where every point in space is associated with a vector (like the electric field).

    I am aware that, even within maths/physics, the word "field" has several meanings (for example we have the definition of a field as two Abelian groups of the same set, with distributive properties over the operations), and I should have made that clear. But what I meant was that every point in space is associated with a complex number.

    Masud.
     
  10. Feb 16, 2005 #9

    dextercioby

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    I know i'm nit-picking you,but the [itex] \Psi(\vec{r},t) [/tex] is a VECTOR...:wink:

    Daniel.
     
  11. Feb 16, 2005 #10
    Yes you're right :smile: , although I'm pretty sure we're both correct, because we have a complex valued function (which assigns a complex number to each point in spacetime) which also inhabits a vector space. So the function itself is a vector, but not part of a vector field.

    And it's good to nit-pick, the more physicists nit-pick the closer we get to being proper mathematicians.
     
  12. Feb 16, 2005 #11

    dextercioby

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    I resent that...No physicist wants to be a mathematician...

    Daniel.
     
  13. Feb 16, 2005 #12
    Ok, that is true, but I hate it when physicists are mathematically inaccurate, or when they are sloppy, or take shortcuts etc. e.g. Newton's calculus was not rigorous; it took mathematicians such as Cauchy and Euler to put calculus on rigorous foundations and hence founded Analysis.
     
  14. Feb 16, 2005 #13

    dextercioby

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    I resent that.It surely was,for that time .It's unfair to judge the past with the mind of a contemporary indivdual.Mathematics was not rigurous then...After Leibniz it began to be...

    Daniel.
     
  15. Feb 16, 2005 #14

    selfAdjoint

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    Euler and Cauchy weren't rigorous in the modern sense either; it took Weierstrass to put THEIR work on a solid foundation. Newton's ultimate ratio can be mapped into the modern limit concept easily. He wasn't always complete in his demonstrations but they can be made complete without violating his thought.

    As I have quoted before, sufficient unto the day is the rigor thereof.
     
  16. Feb 16, 2005 #15
    Hehehe... "infinitesimals".... dx wandering around on its own... ehehehe... :biggrin:
     
  17. Feb 17, 2005 #16

    Galileo

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    **AHHHUUUMMMMM**!!!
     
  18. Feb 17, 2005 #17

    dextercioby

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    Galileo,please sustain your point,an honomatopeical interjection will not suffice...:wink:

    Daniel.
     
  19. Feb 17, 2005 #18

    Galileo

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    What's there to say? I`m a counterexample to your statement.
    I love both physics and mathematics and I believe physics should strive to be as rigorous as mathematics. Or in any case, every small step that is made by physical arguments must be justified and thoroughly analyzed.
    This is not done in the material presented in the college lectures.

    I`m one the those persons who actually checks if interchanging limits, differentiating delta functions, assuming completeness, taking fourier transforms of arbitrary vector fields etc., treating dy and dx as fractions to your fancy, etc, is allowed. (And get a weird from my physics professors at the same time :uhh: )
     
  20. Feb 17, 2005 #19

    dextercioby

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    I may be old fashioned and it's probably not the kinda idea i should be taking in a (hopefully successful) theorist carrier,but i still think that FOR A THEORETICAL PHYSICIST,MATHEMATICS IS A TOOL AND NOT A PURPOSE...

    Daniel.
     
  21. Feb 17, 2005 #20

    Galileo

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    That's what I hear from every physicist. 'Mathematics is just a tool', 'this or that is a purely mathematical result, there is nothing physical about it'.
    I disagree to quite an extend with this view. We honestly cannot do physics without mathematics (practically and probably theoretically).
    I`m not saying physics is a branch of mathematics, it is not. The point was that physicist aren't always mathematically rigorous. Fact is: we use mathematics to describe nature and to make our results quantitative, so even if you consider it a tool, you are using it so make sure that what you are doing is mathematically justified. That much seems obvious to me.

    Anyway, I stand by it. Mathematics is more than just a tool, it's essential.

    Philosophy is written in this grand book, the universe, which stands continually open to our gaze. But the book cannot be understood unless one first learns to comprehend the language and read the characters in which it is written. It is written in the language of mathematics, and its characters are triangles, circles, and other geometric figures without which it is humanly impossible to understand a single word of it; without these one is wandering in a dark labyrinth.

    - Galileo
     
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