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A question about physical chemistry

  1. Oct 14, 2015 #1
    • Please post this type of questions in HW section using the template.
    For the problem of particle in a 1D box
    V(x)=0 for 0≦x≦a
    ∞ for anywhere outside the box
    I know that Hamiltonian operator commutes with momentum operator
    so they should have smae eigenfunction but it's obvious that the wavefunction
    (2/a)^1/2sin(nπx/a) is not a eigenfunction of linear momentum operator
    what is the reason for the contradiction ?
     
    Last edited: Oct 14, 2015
  2. jcsd
  3. Oct 14, 2015 #2

    blue_leaf77

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    The momentum operator in infinite well potential is not well defined to begin with, this is because momentum operator is supposed to act as a generator of translation, while such a translation in infinite well potential cannot take place. Moreover, even after you define a "momentum-like" operator ##\hat{p}##, it does not commute with the global Hamiltonian.
     
  4. Oct 14, 2015 #3
    Thanks for your answer.
     
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