1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Sakurai 1.27 - Transformation Operators

  1. Dec 17, 2008 #1
    Problem
    Suppose that [tex]f(A)[/tex] is a function of a Hermitian operator [tex]A[/tex] with the property [tex]A|a'\rangle = a'|a'\rangle[/tex]. Evaluate [tex]\langle b''|f(A)|b'\rangle[/tex] when the transformation matrix from the [tex]a'[/tex] basis to the [tex]b'[/tex] basis is known.


    The attempt at a solution
    Here's what I have... I'm not sure if the last step is correct, but apart from that, I'm sure everything else is right. So,

    [tex]\langle b''|A|b'\rangle = \sum_{a'} \sum_{a''} \langle b'' | a'' \rangle \langle a'' | A | a' \rangle \langle a' | b' \rangle[/tex]
    [tex]= \sum_{a'} \sum_{a''} \delta(a'-a'') a' \langle b'' | a'' \rangle \langle a' | b' \rangle[/tex]
    [tex]= \sum_{a'} a' \langle b'' | a' \rangle \langle a' | b' \rangle[/tex]

    And then I said that the above was

    [tex]= a' \langle b'' | b' \rangle = a' \delta(b' - b'') \Leftrightarrow \langle b'' | f(A) | b' \rangle = f(a') \delta(b' - b'')[/tex]

    where [tex]a'[/tex] was the eigenvalue of corresponding eigenket to [tex]b'[/tex].

    Is the last step correct? That was the only part I was a bit shaky on... there is an [tex]a'[/tex] index inside the summation...
     
  2. jcsd
  3. Dec 19, 2008 #2

    turin

    User Avatar
    Homework Helper

    I would say that, while your first step is correct, you can do better than that. I suggest that you think of an even better first step which will obviate your suspect final step.

    BTW, your problem statement did not specify that the inner product is the Dirac delta. Is this stated somewhere else?
     
  4. Dec 19, 2008 #3
    It's the Dirac delta because they're eigenkets of some operator (okay well to make the argument more rigorous, the eigenkets a and b are eigenkets of some Hermitian operator where there is no degeneracy).
     
  5. Dec 19, 2008 #4

    turin

    User Avatar
    Homework Helper

    This remains insufficient to determine that [tex]\langle{}a'|a''\rangle=\delta\left(a'-a''\right)[/tex], etc.. But, at any rate, you can avoid this issue entirely (see my previous post).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Sakurai 1.27 - Transformation Operators
  1. Sakurai 1.27 (Replies: 24)

  2. Sakurai 1.7 (Replies: 24)

  3. Sakurai 1.15 (Replies: 3)

  4. Sakurai 1.17 (Replies: 1)

Loading...