A question about commutaiton between operator

[hgnk708113]

Hamiltonian operator commutes with the linear momentum operator
and for a particle in the box its wavefunction is Nsin(nπx/a) , N is the normalization constant
But I found this wavefuntion is not a eigenfuntion for the momentum operator, why? Isn't the two operators commut with each other?

 

[Orodruin]

They only commute if the potential is not a function of x, i.e., if it is translationally invariant. In the case of an infinite square well, this is not the case.

 

[hgnk708113]

Thanks for your help.
But I still have a question:
Does "translationally invariant" mean the general solution of Hamiltonian operator for free particle used and no boundary condition?
I have substituted the solution Ae^(ikx)+Be^(-ikx) but it is still not an eigenfunction of mometum operator.

 

[Orodruin]

Of course it not an eigenfunction, the problem in itself is not translationally invariant due to the boundary conditions. The eigenfunctions of momentum would be the exp(ikx) functions. These are not in the appropriate Hilbert space (they do not satisfy the boundary conditions). Momentum is not conserved in this problem, which is not particular for the quantum description. Even if you solve the problem classically you get violation of momentum conservation when the particle bounces on the potential walls.

 

[hgnk708113]

I mean Nsin(nπx/a) is the result of boundary conditions
but Ae^(ikx)+Be^(-ikx) has not been subjected to boundary conditions yet.
It's a free particle, and its potential is not a function of x .
And you said they only commute if the potential is not a function of x.
So I ask second time and I don't know why in this stage they do not commute. (Ae^(ikx)+Be^(-ikx) is not momoentum's eigenfunction.)
I knew momoentum's eigenfunction is exp(ikx) functions, I just want to know where I misunderstand.

 

[Orodruin]

That there exists a common set of eigenfunctions does not mean all eigenfunctions of one operator is an eigenfunction to the other. This is only true when the spectrum is non-degenerate, which is not the case here (both the positive and negative exponentials have the same eigenvalue for the Hamiltonian).

Compare with the finite dimensional case of a unit matrix and any other matrix. All vectors are eigenvectors of the unit matrix, but no necessrily of the other one.