A question about physical chemistry

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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 ?
 
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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.
 
  • #3
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Thanks for your answer.
 

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