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

A Transformation of position operator under rotations

  1. Nov 27, 2017 #1
    In the momentum representation, the position operator acts on the wavefunction as

    1) ##X_i = i\frac{\partial}{\partial p_i}##

    Now we want under rotations $U(R)$ the position operator to transform as

    ##U(R)^{-1}\mathbf{X}U(R) = R\mathbf{X}##

    How does one show that the position operator as represented in 1) indeed transforms like this?
     
  2. jcsd
  3. Nov 27, 2017 #2

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    In non-relativistic QT you can separate the angular-momentum operator into orbital and spin angular momentum. The spin part commutes. So you are left with the orbital part. Then from the Heisenberg commutator relations you get
    $$[x_k,L_l]=[x_k,\epsilon_{lab} x_a p_b]=x_a \epsilon_{lab} [x_k,p_b]=x_a \epsilon_{lab} \mathrm{i} \delta_{kb}=\mathrm{i} \epsilon_{lak} x_a=-\mathrm{i} \epsilon_{lka} x_a,$$
    i.e., the infinitesimal rotation is correctly represented on the position operators. Exponentiation yields that indeed ##\vec{x}## are vector operators under rotation.

    As you see you don't need a cumbersome explicit representation but you can deal with the abstract operator algebra only!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Transformation of position operator under rotations
  1. The Position Operator (Replies: 12)

  2. Position operator (Replies: 1)

  3. Position operator (Replies: 14)

Loading...