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Sakurai: Page 46

  1. Apr 21, 2013 #1
    From page 46 of "Modern Quantum Mechanics, revised edition", by J.J. Sakurai.

    In equation (1.6.24),
    [tex] \left[\mathbf{x}, \mathcal{T}(d\mathbf{x'}) \right] = d \mathbf{x'} \mid \mathbf{x'} + d \mathbf{x'} \rangle \approx d \mathbf{x'} \mid \mathbf{x'} \rangle [/tex]
    It is written: "where the error made in writing the last line of (1.6.24) is of second order in [itex]d \mathbf{x'}[/itex]". How does that happen? From a Taylor expansion of [itex]\mid \mathbf{x'} + d \mathbf{x'} \rangle [/itex] ? If so, how do you Taylor expand a ket?
     
    Last edited: Apr 21, 2013
  2. jcsd
  3. Apr 21, 2013 #2
    You may just use the definition of the transformation operator [itex]\mathcal{T}[/itex].
    [tex] \mathcal{T}(d\mathbf{x'})\mid \mathbf{x'} \rangle = \mid \mathbf{x'} +d\mathbf{x'} \rangle [/tex]
    [tex] \left[\mathbf{x}, \mathcal{T}(d\mathbf{x'}) \right]\mid \mathbf{x'} \rangle = d \mathbf{x'} \mid \mathbf{x'} + d \mathbf{x'} \rangle = d \mathbf{x'}\mathcal{T}(d\mathbf{x'})\mid\mathbf{x'} \rangle = d \mathbf{x'} \left(1 - i\mathbf{K} \cdot d\mathbf{x'}\right) \mid\mathbf{x'} \rangle [/tex]

    You may notice that [itex]1 - i\mathbf{K} \cdot d\mathbf{x'}[/itex] does look like the first two terms of a Taylor expansion.
     
    Last edited: Apr 21, 2013
  4. Apr 21, 2013 #3
    Oh, yeah. That never occurred to me. Thanks!

    I also found out just now that I had written my equation wrong. I just corrected it. Sorry about that and thanks for looking at Sakurai to get through that.
     
  5. Apr 10, 2014 #4
    You can’t use [itex](1-\mathbf{K}\cdot dx')[/itex] because you need to prove this later. I believe ket can be expanded as [tex]|x'+dx'\rangle=|x'\rangle+\frac{|x'\rangle}{dx'}dx'+\dots[/tex] and then ignore [itex]O(dx')^2[/itex]. What do you think?
     
    Last edited: Apr 10, 2014
  6. Apr 10, 2014 #5

    ChrisVer

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    Gold Member

    The problem is how you can expand the ket?
    I guess a good way would be to use this:
    [itex] <f|x+dx>= f(x+dx) \approx f(x)+ f'(x) dx = <f|x>+ \frac{<f| x+dx>-<f|x>}{dx} dx= <f| (|x>+\frac{| x+dx>-<f|x>}{dx} dx)[/itex]
    or
    [itex] |x+dx> \approx |x>+\frac{| x+dx>-|x>}{dx} dx [/itex]

    so multiplying with dx:
    [itex] dx |x+dx> \approx dx |x>+\frac{| x+dx>-|x>}{dx} (dx)^{2} [/itex]

    Am I somewhere wrong?
     
  7. Apr 10, 2014 #6
    Exactly ... the second term [tex]\frac{|x+dx\rangle-|x\rangle}{dx}[/tex] is nothing but [itex]d|x\rangle/dx[/itex] since [itex]dx[/itex] is infinitesimal. It is totally legal to apply dervative to ket. Remember SHE [itex]\hat{H} |\psi\rangle=i\hbar d|\psi\rangle/dt[/itex]. So you dont need to project the ket into function.
     
    Last edited: Apr 10, 2014
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