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A Question while i study Modern Quantum physics by Sakurai

  1. Mar 3, 2017 #1
    In the book of Modern Quantum physics by Sakurai
    upload_2017-3-3_22-33-37.png

    I wonder how 1.6.26 can be 1.6.27.
     
  2. jcsd
  3. Mar 3, 2017 #2

    PeroK

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    Actually, I'll put hats on the operators (something that perhaps Sakurai ought to have thought about):

    If you set ##\mathbf{dx'} = (dx_1, 0, 0)## then ##\mathbf{\hat{K}} \cdot \mathbf{dx'} = \hat{K_1} dx_1## and (1.6.26) becomes:

    ##-i \mathbf{\hat{x}}\hat{K_1} dx_1 + i\hat{K_1}\mathbf{\hat{x}}dx_1 = (dx_1, 0, 0)##

    Where ##dx_1## is an "infinitesimal" number and can be cancelled, and ##\mathbf{\hat{x}} = (\hat{x_1}, \hat{x_2}, \hat{x_3})##; giving:

    ##-i (\hat{x_1}\hat{K_1} - \hat{K_1}\hat{x_1}, \ \hat{x_2}\hat{K_1} - \hat{K_1}\hat{x_2},\ \hat{x_3}\hat{K_1} - \hat{K_1}\hat{x_3}) = (1, 0, 0)##

    Which gives the result for ##\hat{K_1}##. Repeat for ##\mathbf{dx'} = (0, dx_2, 0)## etc.
     
  4. Mar 3, 2017 #3

    BvU

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    Thanks, @PeroK. Out of curiosity: How did you know ##{\bf x K} \cdot d{\bf x'}## was to be interpreted as ##{\bf x} \left ({\bf K} \cdot d{\bf x'} \right)## and not as ##\left ({\bf x K} \right ) \cdot d{\bf x'}## ?
     
  5. Mar 3, 2017 #4

    PeroK

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    I looked at my notes and, unlike Sakurai, I tend to distinguish between operators and numbers/vectors. You have to do that at the beginning of the section/proof and keep track of it.
     
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