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I Kets and vectors

  1. Jul 16, 2016 #1

    dyn

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    Hi. I have read that when working in 3-D the following kets | x > and | p > are not vectors in 3-D. If that is correct what are they ? I know | ψ > is an abstract vector but I thought | x > and | p > would be 3-D vectors in the position and momentum representation ?
    Thanks
     
  2. jcsd
  3. Jul 16, 2016 #2

    andrewkirk

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    They are vectors in an infinite-dimensional space. It is natural to feel that ##|x\rangle## should be in a 3d space because ##x##, without the enclosing symbols, is a coordinate in 3D space. But ##|x\rangle## is not a coordinate or any other type of number. In the position representation it is the Dirac delta function ##\delta_x##, which is an element of a space of functions, which - like most function spaces - is infinite-dimensional.
     
  4. Aug 3, 2016 #3
    <a is conjugate of a>
    <a|b> kind of dot(a.conjugate , b)
    b><b|a> projection into other direction like b*dot(b.conjugate , a)

    v2=pol><pol|v> is equivalent with this pseudo code
    amp=dot(v,pol)
    v2.x = pol.x*amp
    v2.y = pol.y*amp

    conjugate of V is V.imaginary=-V.imaginary
     
  5. Aug 3, 2016 #4
    The Pauli matrices transform the 3d direction into 2d complex vector like spin.
     
  6. Aug 3, 2016 #5
    Why do we need conjugation?
    because there is a minus sign in real part of multiplication of complex numbers.

    complex operator *(complex c)
    {
    complex e;
    e.real = this->real*c.real - this->img*c.img;
    e.img = this->img*c.real + this->real*c.img;
    return e;
    }
     
  7. Aug 3, 2016 #6

    A. Neumaier

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    They are distributions, elements in the dual space of a Schwartz space of smooth functions. Only the labels are vectors in 3D.
     
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