Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Lie bracket question

  1. Feb 15, 2010 #1

    I was doing an assignment in quantum mechanics and came upon the following fact I cannot explain to me.

    I hope someone of you can and will be willing to :)

    Consider the creation and annihilation operators: a^+ and a and also the momentum and position operators p and x:

    [tex]x=\frac{1}{\sqrt 2 c}(a+a^{\dagger})[/tex]
    [tex]p=\frac{\hbar c}{\sqrt 2 i}(a-a^{\dagger})[/tex]
    [tex]a=\frac{1}{\sqrt 2}(cx+\frac{i}{c\hbar}p)[/tex]
    [tex]a^{\dagger}=\frac{1}{\sqrt 2}(cx-\frac{i}{c\hbar}p)[/tex]


    and the canonical commucator relation: [tex][x,p]=i\hbar 1[/tex], where 1 is the identity operator

    It follows immediately from the canonical commutator relation between x and p that

    Now, observe what happens when I take the adjoint of this equation:


    which is peculiar since I thought that the Identity is hermitian: [tex]1^{\dagger}=1[/tex], which apperantly doesn't hold here..

    Can anyone tell me why this is so?

    thanks in advance,

  2. jcsd
  3. Feb 15, 2010 #2
    I think I found your mistake here
    Hi Martin,

    Using the fact that transposing the product of two operator swaps them and gives the product of their transposes

    [tex] (A B)^\dagger = B^\dagger A^\dagger [/tex]

    we have,

    [tex] ([a,a^{\dagger}])^{\dagger}=(aa^{\dagger}-a^{\dagger}a)^{\dagger}=a^{\dagger\dagger}a^\dagger - a^\dagger a^{\dagger\dagger} = a a^\dagger - a^\dagger a = 1 [/tex]
    Last edited: Feb 15, 2010
  4. Feb 15, 2010 #3
    yes, of course!

    sorry for asking...

    and thanks once again!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook