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Translation of a ket

  1. Jan 22, 2014 #1
    2z4gy9u.png

    I'm not sure how they got equation 4.5

    Starting from ##\psi'> = U_{(\vec{a})} |\psi>## where ##U_{(\vec{a})} = e^{\frac{-i\vec {a} \cdot \vec {p}}{\hbar}}##

    Differentiating both sides with respect to ##a_x## and setting ##a = 0##:

    [tex]\frac{\partial |\psi'>}{\partial a_x} = \frac{-ip_x}{\hbar} U_0 |\psi> + U_0\frac{\partial |\psi>}{\partial a_x} [/tex]

    Using the fact that ##U_0 = 1##,

    [tex]\frac{\partial |\psi'>}{\partial a_x} = \frac{-ip_x}{\hbar} |\psi> + \frac{\partial |\psi>}{\partial a_x} [/tex]

    It's wildly different from what they have..
     
  2. jcsd
  3. Jan 22, 2014 #2

    dextercioby

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    Obviously not both kets (the original one and the transformed one) depend on a, so that 1 derivative is 0, right ?
     
  4. Jan 22, 2014 #3
    So that means ## \frac{\partial |\psi>}{\partial a_x} = 0 ##, which doesn't really fit
     
  5. Jan 22, 2014 #4

    ChrisVer

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    why would ψ depend on α?
    The alpha shows you how you do the translation. So the dependence of alpha is only within the operator U and not in your initial state...
     
  6. Jan 22, 2014 #5

    So it means that:
    [tex]\frac{\partial |\psi'>}{\partial a_x} = \frac{-ip_x}{\hbar} |\psi> [/tex]

    Not sure how this leads to the final result..

    Unless ##|\psi'> = |\psi>## since ##\vec{a} = 0##
     
  7. Jan 22, 2014 #6

    ChrisVer

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    1st, reread what your image says above equation 4.5
    2nd, you can also check in 4.3 what happens for a=0
    finally, even without the words of a book, if the vector tha gives you how much you translate something is equal to zero, it means that you didn't translate it at all- so it still remains in the initial state.

    both 3 ways are equivalent, and work XD
     
  8. Jan 22, 2014 #7
    Ok I got it, but how did they change the ##\partial a_x## to ##-\partial x##?

    This would mean that ##\frac{\partial x}{\partial a_x} = -1## from chain rule.

    Therefore ##x = -a_x##.

    This would mean that the centre of mass was at the origin in state ##|\psi>##?
     
  9. Jan 22, 2014 #8

    ChrisVer

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    first of all, you can see the derivative over x by letting the momentum to act as an operator on your state...

    But also there is a physical insight again. Suppose you have a body at point A. And you want to somehow move it right to a point B.
    One way to do that, is to take the point particle and move it along (that corresponds to alpha variable).
    The equivalent way, is to move the whole space so that B will come and coincide on A where your body is positioned.

    So in fact, what changes you can do to alpha are equivalent to changes you'd do to your x with a minus...

    for a quantitive example-
    if you have a particle at x=0 and want to move it at x=1.
    You can either move it from 0->1 by a vector alpha.
    or move the whole axis 1 unit left, so that what was before 0, will become 1...

    (it's like rotations- you can either rotate your body by an angle θ or rotate the whole axis by an angle -θ)
     
    Last edited: Jan 22, 2014
  10. Jan 23, 2014 #9
    Yeah that makes sense, thanks!

    Passive vs. Active Transformation
     
    Last edited: Jan 23, 2014
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