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Sakurai Equation (1.6.26)

  1. Nov 4, 2009 #1
    1. The problem statement, all variables and given/known data

    This isn't a homework problem. I am reading Sakurai (Modern Quantum Mechanics) and came upon this:

     
  2. jcsd
  3. Nov 5, 2009 #2
    The only way I can see how that work is if the following is true:

    [tex]\textbf{Kx}-\textbf{xK}=\left[\textbf{K},\textbf{x}\right]=-i[/tex]

    Any insight on this?
     
  4. Nov 5, 2009 #3
    Are x and dx' vectors? If so, it also seems to require that dx' and x are in the same direction?
     
  5. Nov 5, 2009 #4
    According to Sakurai, x has elements x, y, and z.
     
  6. Nov 5, 2009 #5

    gabbagabbahey

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    Equation 1.6.25 was derived by considering the effect of the commutator on an arbitrary position eigenket, 1.6.26 is derived from this result by using the definition of the translation operator eq. 1.6.20.
     
  7. Nov 5, 2009 #6
    I know that.

    [tex]\hat{\textbf{T}}\left(d\textbf{x}'\right)=1-i\textbf{K}\cdot d\textbf{x}'[/tex] (1.6.20)

    I plugged that into (1.6.25) to work out the commutator. Sakurai claims it is [tex]d\textbf{x}'[/tex], but as you can see, when I work it out, I do not understand how that claim is true.
     
  8. Nov 5, 2009 #7

    gabbagabbahey

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    It's true because of eqs. 1.6.23 and 1.6.24...I don't understand the source of your confusion here. If you agree that eq. 1.6.25 is true, and also that 1.6.20 is true, then 1.6.26 must also be true...it is basic logic.
     
  9. Nov 5, 2009 #8
    My confusion is in working out the commutator; using (1.6.26) to verify that (1.6.25) is true.

    But I guess (1.6.26) wasn't meant to be used to verify (1.6.25).

    How about the next part? The text says:

    I do not know how they come up with (1.6.27) either.

    If I choose dx' in the direction of [itex]\hat{\textbf{x}}_j[/itex] , I get:

    [tex]dx\hat{\textbf{x}}_j[/tex]

    If I form the scalar product with [itex]\hat{\textbf{x}}_i[/itex], I get:

    [tex]\langle dx\hat{\textbf{x}}_j|\hat{\textbf{x}}_i \rangle = dx\delta_{ij}[/tex]

    How do I get (1.6.27) from that?
     
  10. Nov 5, 2009 #9

    gabbagabbahey

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    1.6.27 comes from 1.6.26...

    [tex]-i\textbf{xK}\cdot d\textbf{x}'+i\textbf{K}\cdot d\textbf{x}'\textbf{x}=d\textbf{x}'[/tex]

    Using the Einstein summation convention, [itex]\textbf{K}\cdot d\textbf{x}'=K_jdx'_j[/itex] and your equation becomes

    [tex]-i\textbf{x}K_jdx'_j+iK_jdx'_j\textbf{x}=d\textbf{x}'[/tex]

    So, each component satisfies,

    [tex]-ix_kK_jdx'_j+iK_jdx'_jx_k=-ix_kK_jdx'_j+iK_jx_kdx'_j=dx'_k[/tex]

    (the second step is because [itex][x_k,dx'_j]=0[/itex]) and you should be able to take i from here.
     
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