1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: Properties of Hermitian operators in complex vector spaces

  1. Sep 15, 2011 #1
    1. The problem statement, all variables and given/known data

    Given a Hermitian operator A = [itex]\sum \left|a\right\rangle a \left\langle a\right|[/itex] and B any operator (in general, not Hermitian) such that [itex]\left[A,B\right][/itex] = [itex]\lambda[/itex]B show that B[itex]\left|a\right\rangle[/itex] = const. [itex]\left|a\right\rangle[/itex]

    2. Relevant equations

    Basically those listed above plus possibly the Hermitian condition and eigenvalue definition which I will not list since they are well known.

    3. The attempt at a solution

    I have tried expanding it out in terms of the commutator, but this seems like the wrong approach. I am not sure that there is a way to calculate it directly. I do not think the proof is very involved but I am approaching it the wrong way and cant seem to get anywhere. This is listed as a fundamental property of a Hermitian operator in a complex vector space in my text, though it does not prove it and other references seem unconcerned. I am also not very comfortable using this kind of spectral decomposition, which is a bit of a problem.
  2. jcsd
  3. Sep 15, 2011 #2
    I should point out that the end of the above problem is to show that B[itex]\left|a\right\rangle[/itex] = const. [itex]\left|a +\lambda\right\rangle[/itex]
  4. Sep 16, 2011 #3


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    These two statements are inconsistent. In both cases, you're applying B to |a>, but you're getting different answers.

    You can prove the second statement using the commutation relation.
  5. Sep 16, 2011 #4
    They are inconsistent because the statement in the first post is incorrect. The correct relation to prove is the second.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook