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!

Commute an operator with a constant

  1. Aug 10, 2011 #1
    1. The problem statement, all variables and given/known data
    Suppose we had an operator A that multiplied a vector by it's norm:
    [itex]A \mid \psi \rangle = \langle \psi \mid \psi \rangle \mid \psi \rangle[/itex]
    I wanted to know what it's commutator with a constant would be.


    2. Relevant equations
    [itex]\left[A,B\right] = AB - BA[/itex]

    3. The attempt at a solution
    Suppose b is a real number greater than 1, then
    [itex]\left[A,b\right] \mid \psi \rangle =(Ab-bA)\mid \psi \rangle [/itex]
    [itex]=A (b\mid \psi \rangle ) - b (A \mid \psi \rangle ) = (b^2 - b)\langle \psi \mid \psi \rangle \mid \psi \rangle[/itex]

    I'm think the second equality is wrong since I read that the commutator of an operator with a constant should be 0 and the operator A looks nonlinear. But what should be done instead? Thanks
     
  2. jcsd
  3. Aug 10, 2011 #2
    Just looking quickly at this, shouldn't you have got

    [itex]\left( b^2-1 \right) \langle \psi | \psi \rangle \left( b | \psi \rangle \right)[/itex]?
     
  4. Aug 10, 2011 #3
    oh you're right, but it still doesn't equal 0 though unless I'm missing something again
     
  5. Aug 10, 2011 #4

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    How did you get the b^2 ? It shouldn't be there.
     
  6. Aug 10, 2011 #5
    Why should it equal to zero? A is non-linear, so there is no guarantee that [itex]A (b|\psi\rangle) = b A |\psi\rangle[/itex].
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Commute an operator with a constant
  1. Commuting operators (Replies: 2)

Loading...