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

Matrix representation of operators

  1. Jan 2, 2016 #1
    Let the operators ##\hat{A}## and ##\hat{B}## be ##-i\hbar\frac{\partial}{\partial x}## and ##x## respectively.

    Representing these linear operators by matrices, and a wave function ##\Psi(x)## by a column vector u, by the associativity of matrix multiplication, we have

    ##\hat{A}(\hat{B}##u##)##=##(\hat{A}\hat{B})##u.

    By the definitions of ##\hat{A}## and ##\hat{B}##, we have

    LHS ##= -i\hbar\frac{\partial}{\partial x}\big(x\ \Psi(x)\big) = -i\hbar\big(x\frac{\partial\Psi(x)}{\partial x} + \Psi(x)\big)##

    RHS ##= -i\hbar\Psi(x)##

    LHS ##\neq## RHS, a contradiction. Where is the mistake?

    The extract from a textbook below only talks about the matrix representation of a Hermitian operator. But is it true that all linear operators can be represented by matrices, not just those that are Hermitian?

    Screen Shot 2016-01-03 at 6.48.28 am.png
     
  2. jcsd
  3. Jan 2, 2016 #2

    andrewkirk

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    The mistake is in the RHS calc. It arises from the abuse of notation of writing ##\hat B## as ##x##. THe correct representation of ##\hat B## is ##\Psi(x)\mapsto x\Psi(x)##. That is, it is the operator that, given a function of ##x##, returns a new function that is the original function, multiplied by x. So we get:
    ##\hat A\hat B=-i\hbar\partial_x \Big(\Psi(x)\mapsto x\Psi(x)\Big)=
    \Psi(x)\mapsto -i\hbar \Big(\Psi(x)\partial_x( x)+x\partial_x\Psi(x)\Big)## which will match the LHS when applied to a specific ket.
    Yes. The matrix element given in formula 5.156 is just as meaningful for non-Hermitian operators. It will just be (in most cases) a non-Hermitian matrix.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook