The discussion centers on the challenges of evaluating the momentum operator in the context of a particle in an infinite potential well. The momentum operator cannot be defined as a self-adjoint operator due to boundary conditions, which restricts the use of momentum representation. The Hamiltonian operator, however, is well-defined and can be used effectively in position representation. The conversation highlights the limitations of momentum representation for infinite potential wells and suggests that while Fourier transforms can yield momentum states, they lack physical meaning in this context.
PREREQUISITESQuantum mechanics students, physicists working on quantum systems, and researchers interested in the mathematical foundations of quantum theory, particularly in relation to potential wells and operator theory.
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.