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

I Products of operators : products of matrices

  1. Apr 26, 2017 #1

    dyn

    User Avatar

    Hi.
    If I have an operator in matrix form eg. < i | x | j > then the matrix of the operator x2 is given by the square of the former matrix. This seems like common sense but how would i prove this using Dirac notation ?
    Thanks
     
  2. jcsd
  3. Apr 26, 2017 #2

    Charles Link

    User Avatar
    Homework Helper

    When you square the matrices, i.e. multiply one matrix by the same matrix, you wind up doing a summation over ## k ## in the following way (to get the ij element of the x matrix multiplied by the x matrix) : ## <i|x|k><k|x|j> ##. in the Dirac notation ## |k><k| ## summed over all states is the identity operator and can be removed.
     
  4. Apr 27, 2017 #3

    dyn

    User Avatar

    Thanks for that
     
  5. Apr 28, 2017 #4

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    The full calculation is
    $$\langle i|\hat{x}^2|j \rangle=\sum_{k} \langle i|\hat{x} k \rangle \langle k|\hat{x} j \rangle,$$
    i.e., in matrix notation
    $$(x^2)_{ij}=\sum_k x_{ik} x_{kj}.$$
    That's like matrix multiplication in finite-dimensional vector spaces, only that the matrix here is "infinite dimensional".
     
  6. Apr 28, 2017 #5

    Charles Link

    User Avatar
    Homework Helper

    It should be noted that there are many other operators for which it is simply a matrix multiplication in a finite dimensional space.
     
  7. Apr 29, 2017 #6

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    Yes, e.g., for spins.
     
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: Products of operators : products of matrices
Loading...