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Momentum operator identity

  1. Aug 22, 2007 #1

    malawi_glenn

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    1. The problem statement, all variables and given/known data
    I wanna show:

    [tex] \langle x' - \triangle x' \vert \alpha \rangle = \langle x' \vert \alpha \rangle - \triangle x' \dfrac{\partial}{\partial x'}\langle x' \vert \alpha \rangle [/tex]

    2. Relevant equations

    [tex] \vert \alpha \rangle [/tex] is a state.

    3. The attempt at a solution

    i have no clue, can anyone give me a hint? Is the bra being splitted in to two, or what?

    This is a part of eq (1.7.15) in Sakurai mordern quantum mechanics, 2ed.
     
    Last edited: Aug 22, 2007
  2. jcsd
  3. Aug 22, 2007 #2

    Dick

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    It's just the first term in a power series expansion. Like f(x-a)=f(x)-a*df(x)/dx.
     
  4. Aug 22, 2007 #3

    malawi_glenn

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    aha okay I'll try that one, thanks alot!
     
  5. Jan 4, 2009 #4
    But I think the first term of the series expansion must be a constant but not a function of x.

    So I'm wondering..
     
  6. Jan 4, 2009 #5

    Dick

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    In f(x-a)=f(x)-a*df(x)/dx I'm thinking of 'x' as the constant (point to expand around) and 'a' as the expansion variable.
     
  7. Jan 5, 2009 #6
    But f is differentiated by 'x' in f(x-a)=f(x)-a*df(x)/dx.

    I think if 'a' is the variable the series expansion must be

    [tex]

    f(x-a)=f(x)-a*\frac{df(x-a)}{da} |_{a=0}

    [/tex]

    I'm studying the part of QM and have difficulty in understanding the eq (1.7.15).

    I wanna verify that eq,

    but I think there may be some mistakes in the explanation by series expansion.
     
  8. Jan 5, 2009 #7

    Dick

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    Now that has the wrong sign in the derivative if you are taking it d/da. It will give you -f'(x). If you want to be super precise how about (to first order in a),

    [tex]
    f(x-a)=f(x)-a \frac{df(z)}{dz} |_{z=x}
    [/tex]

    the variable in the differentiation symbol is really just a dummy, not x or a. Or just f(x-a)=f(x)-a*f'(x) is also clear.
     
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