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Sakurai page 54: Is this a Taylor expansion?

  1. Apr 27, 2013 #1
    From page 54 of 'Modern Quantum Mechanics, revised edition" by J. J. Sakurai.

    Obtaining equation (1.7.15),[tex]
    \begin{eqnarray}
    \left(1- \frac{ip\Delta x'}{\hbar} \right) \mid \alpha \rangle & = & \int dx' \mathcal{T} ( \Delta x' ) \mid x' \rangle \langle x' \mid \alpha \rangle \\
    & = & \int dx' \mid x' + \Delta x' \rangle \langle x' \mid \alpha \rangle \\
    & = & \int dx' \mid x' \rangle \langle x' - \Delta x' \mid \alpha \rangle \\
    & = & \int dx' \mid x' \rangle \left( \langle x' \mid \alpha \rangle - \Delta x' \frac{\partial}{\partial x'} \langle x' \mid \alpha \rangle \right)
    \end{eqnarray}
    [/tex]How does the last line come about? Is it a Taylor expansion?
     
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  3. Apr 27, 2013 #2

    mfb

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    It looks like a Taylor expansion, indeed.
     
  4. Apr 27, 2013 #3
    Can a Bra be Taylor expanded like this? I guess it is like Taylor expanding a vector? I know that Kets are elements of a vector space, and Bras are the elements of the corresponding dual space.
     
  5. Apr 27, 2013 #4

    mfb

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    If the result is differentiable in some variable, and behaves nicely (insert strict mathematical formulation here): why not?
     
  6. Apr 27, 2013 #5

    fzero

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    It's not quite a bra that is being expanded here. ##\langle x | \alpha \rangle = \psi_\alpha(x)## is the position space wavefunction corresponding to the state ##|\alpha\rangle##. What appears in the expression above is ##\psi_\alpha(x'-\Delta x')##, which is usually a nice enough function to have a well-defined Taylor expansion. It may be the case that you can also manipulate the states themselves in some way, but I expect that if the procedure is to be well-defined, there must be a way to express the same manipulations in terms of appropriate wavefunctions.
     
  7. Apr 27, 2013 #6
    Oh, OK. That's easier to understand.

    I was about to ask since all operators [itex]A[/itex] should be [itex]A^{\dagger}[/itex] and all constants [itex]c[/itex] should be [itex]c^*[/itex] in the Bra space, why is it,[tex]
    \int dx' \mid x' \rangle \langle x' - \Delta x' \mid \alpha \rangle = \int dx' \mid x' \rangle \left( \langle x' \mid \alpha \rangle - \Delta x' \frac{\partial}{\partial x'} \langle x' \mid \alpha \rangle \right)
    [/tex] and not, [tex]
    \int dx' \mid x' \rangle \langle x' - \Delta x' \mid \alpha \rangle = \int dx' \mid x' \rangle \left( \langle x' \mid \alpha \rangle - \Delta (x')^* {\frac{\partial}{\partial x'}}^{\dagger} \langle x' \mid \alpha \rangle \right)
    [/tex]?
     
  8. Apr 27, 2013 #7
    I mean, I understand that what's given in the book easily follows if you think of it as a wavefunction. But if you think of it as Taylor expanding a Bra, why don't the complex and hermitian conjugates don't come in to play?
     
  9. Apr 28, 2013 #8
    Because position is a real variable?
     
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