# A question about physical chemistry

1. Oct 14, 2015

### hgnk708113

• 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. Oct 14, 2015

### blue_leaf77

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. Oct 14, 2015