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

Confusion about position operator in QM

  1. Nov 21, 2014 #1

    ShayanJ

    User Avatar
    Gold Member

    In quantum mechanics, the position operator(for a single particle moving in one dimension) is defined as [itex] Q(\psi)(x)=x\psi(x) [/itex], with the domain [itex] D(Q)=\{\psi \epsilon L^2(\mathbb R) | Q\psi\epsilon L^2 (\mathbb R) \} [/itex]. But this means no square-integrable function in the domain becomes non-square-integrable after being acted by this operator which, in turn, means there exist no function in the domain for which you can't have a M that satisfies [itex] ||Q \psi|| \leq M ||\psi|| [/itex] and so this operator should be bounded. But people say its not bounded!
    I'm really confused. What's the point here?
    Thanks
     
  2. jcsd
  3. Nov 22, 2014 #2

    rubi

    User Avatar
    Science Advisor

    The point is that there needs to be an ##M## such that this inequality holds for all ##\psi##. However, for the position operator, you can easily check that there cannot be such an ##M##, for example by picking a sequence ##\psi_n## of normed, centered Gaussians with increasing standard deviation ##\sigma_n = n## (since ##\lVert Q\psi_n\rVert = \sigma^2##, so you would need an ##M## such that ##\forall n\in\mathbb N:M \geq n^2##, which doesn't exist).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Confusion about position operator in QM
  1. Positive operator? (Replies: 1)

  2. Positive Operator (Replies: 3)

Loading...