A question about commutaiton between operator

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Discussion Overview

The discussion revolves around the commutation relationship between the Hamiltonian operator and the linear momentum operator in quantum mechanics, particularly in the context of a particle in an infinite square well. Participants explore the implications of boundary conditions on the eigenfunctions of these operators and the conditions under which they commute.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant states that the Hamiltonian operator commutes with the momentum operator, but questions why the wavefunction for a particle in a box is not an eigenfunction of the momentum operator.
  • Another participant clarifies that the two operators only commute if the potential is translationally invariant, which is not the case for an infinite square well.
  • A participant seeks clarification on whether "translationally invariant" implies using the general solution of the Hamiltonian for a free particle without boundary conditions, noting that their substitution does not yield an eigenfunction of the momentum operator.
  • It is pointed out that the wavefunction is not an eigenfunction due to the boundary conditions, and that momentum is not conserved in this scenario, even classically.
  • A participant emphasizes that the wavefunction Nsin(nπx/a) results from boundary conditions, while the form Ae^(ikx)+Be^(-ikx) does not account for these conditions, leading to confusion about the commutation of the operators.
  • Another participant explains that the existence of a common set of eigenfunctions does not imply that all eigenfunctions of one operator are eigenfunctions of the other, especially in cases of degenerate spectra.

Areas of Agreement / Disagreement

Participants express differing views on the implications of boundary conditions and translational invariance on the commutation of the Hamiltonian and momentum operators. The discussion remains unresolved regarding the specific conditions under which the operators commute.

Contextual Notes

The discussion highlights the limitations of applying general solutions to specific boundary conditions and the nuances of operator commutation in quantum mechanics.

hgnk708113
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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?
 
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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.
 
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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.
 
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.
 
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.
 
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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.
 
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