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Shankar's QM 2nd Edition

  1. Jun 19, 2015 #1
    Hi Folks,

    How do we arrive at equation 1.8.7 page 33.

    We have ##<i|\Omega-\omega I|V>=0##, given ##I=\Sigma_{j=1}|i><i|## we can write

    ##<i|\Omega-\omega \Sigma_{j=1}|i><i||V>=0##

    not sure how to proceed from here...
     
  2. jcsd
  3. Jun 19, 2015 #2
    You can decompose
    You can write your vector V as a linear combination of the basis vectors

    ##|V>=Sigmavj|j>##

    and then using the fact that the basis vector are orthogonal with each other you can evaluate the bracket, taking into account that ωij are the matrix element of Ω in the |i> basis (<i|Ω|j>=ωij)
     
    Last edited: Jun 19, 2015
  4. Jun 20, 2015 #3
    I can get as far as

    ##<I| \Omega - \omega \Sigma_{I} |I><I| \Sigma_{j} v_{j} |j>=0##

    ##\Sigma_{j} \Omega - \omega \Sigma_{I} |I><I| v_{j} \delta_{ij}=0##

    How is the identity dealt with?
     
  5. Jun 20, 2015 #4
    Don't put the identity in the first place, and remember that <i|Ω|j>=Ωij (eq 1.6.1) (I wrote it wrong in my last post)
     
  6. Jun 21, 2015 #5
    I do not follow regarding the identity. I do not know how to remove it. Can u clarify?
     
  7. Jun 21, 2015 #6
    I could show you the whole process but I don't know how to type it.
     
  8. Jun 21, 2015 #7
    Ok here it is

    ##\left\langle i\right|\Omega-\omega\left|v\right\rangle = \left\langle i\right|\Omega-\omega(\sum_{j}v_{j}\left|j\right\rangle )##
    ## = \sum_{j}v_{j}\left\langle i\right|\Omega\left|j\right\rangle -\omega\sum_{j}v_{j}\left\langle i\right.\left|j\right\rangle ##
    ## =\sum_{j}v_{j} \Omega_{ij}-\omega\sum_{j}v_{j}\delta_{ij}##
    ## = \sum_{j}v_{j}(\Omega_{ij}-v_{j}\delta_{ij})##
     
    Last edited: Jun 21, 2015
  9. Jun 22, 2015 #8
    Thanks for your efforts, just some questions:

    1) How do you justify omitting the identity I?
    Are you treating this equal to 1 because the action of the identity operator on a ket is just the same ket?
    Additionally, Shankar specifically instructed to apply the identity operator to the left of |V>

    2) How did you arrive with a 2nd bra <i| on the second line? There is only one.

    regards
     
  10. Jun 22, 2015 #9
    1) Yeah, the identity is implied, it action of a ket and a bra is to leave it as they are. (reading Shankar I don't know why one would introduce the representation of the identity though)

    2)Well, the actual equation is
    ##\left\langle i\right|(\Omega-\omega\left)|v\right\rangle=0 ##

    Look at 1.8.3 and multiply by the left by the i bra.
     
  11. Jun 22, 2015 #10
    To clarify, You can't really have

    ##(\left\langle i\right|\Omega)-(\omega\left|v\right\rangle) ##

    because that would be adding a bra to a ket and you can't do that since they belong to different vector spaces.
     
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