The discussion revolves around the evaluation of a particle in a box using momentum representation, specifically addressing the challenges associated with the momentum operator in the context of infinite and finite potential wells. Participants explore the implications of boundary conditions on the momentum operator and the feasibility of working in momentum space versus position space.
Participants express differing views on the feasibility of using momentum representation for a particle in a box, with some advocating for its use while others emphasize the limitations imposed by boundary conditions. The discussion remains unresolved regarding the practicality of momentum representation in this context.
Participants note that the momentum operator's self-adjointness is critical for its proper definition, and the discussion reveals a reliance on boundary conditions that complicate the application of momentum representation. The mathematical steps and assumptions involved in transitioning between representations are acknowledged as complex and potentially problematic.
Actually the existence of the well defined eigenfunctions is not enough. They have to form a complete set and satisfy the boundary conditions. The enginefunctions of momentum operator in this problem don't form a complete set. You may want to read post #4.Adam Landos said:Oh, I missed that. So, by not being self-adjoint, it does not have proper eigenfunctions. But, this is valid for this particular case. Because for the free particle for example, the momentum operator has well-defined eigenfuctions(plane waves). So, what's up with that?
P.S. Your blog seems useful! Thanks
Please correct me if I am wrong, but from the paper that you provided a link to, I concluded that the self-adjointness of the momentum operator depends on the problem that we examine, right? Because the momentum operator is hermitian(and this is independent of the problem-again correct me if I am wrong) and for it to be self-adjoint it's domain must be the same as the domain of its hermitian conjugate. In the case of the free particle, their domains are the same, so the momentum operator is also self-adjoint, right?vanhees71 said:
Yes, I understand that, but in the case of the free particle, all these conditions are satisfied, so the momentum operator is self-adjoint. So, am I right to conclude that the self-adjointness of an operator depends on the given problem?Ravi Mohan said:Actually the existence of the well defined eigenfunctions is not enough. They have to form a complete set and satisfy the boundary conditions. The enginefunctions of momentum operator in this problem don't form a complete set. You may want to read post #4.
Ah, I am sorry. I didn't read the "free particle". In that case you are correct. Also, as you said, the nature of an operator depends on the given physical system.Adam Landos said:Yes, I understand that, but in the case of the free particle, all these conditions are satisfied, so the momentum operator is self-adjoint. So, am I right to conclude that the self-adjointness of an operator depends on the given problem?
If you have some extra time, could you please show me how to conclude if the momentum operator is self-adjoint or not in the case of the harmonic oscillator? Is it easy to show if it's self-adjoint? I am guessing that it is since there are no boundary conditions, so again the domains of P and P+(dagger) are the same, right?Ravi Mohan said:Ah, I am sorry. I didn't read the "free particle". In that case you are correct. Also, as you said, the nature of an operator depends on the given physical system.
Great, I think I got it! Thanks!Ravi Mohan said:I will have to think and work it out. But essentially, as you say, the vectors in the domain of ##\hat{P}^\dagger## and ##\hat{P}## need not satisfy any extra boundary condition (save square integrability and vanishing at infinity), their domain should turn out same.