Commuting Operators and Eigenfunctions in One Dimension

  • Thread starter Thread starter LagrangeEuler
  • Start date Start date
  • Tags Tags
    Commutator
LagrangeEuler
Messages
711
Reaction score
22
For every operator ##A##, ##[A,A^n]=0##. And if operators commute they have complete eigen- spectrum the same. But if I look for ##p## and ##p^2## in one dimension ##sin kx## is eigen- function of ##p^2##, but it isn't eigen-function of ##p##.
p^2 \sin kx=number \sin kx
p\sin kx \neq number \sin kx
where
p=-i\hbar\frac{d}{dx}
p^2=-\hbar^2\frac{d^2}{dx^2}
 
Physics news on Phys.org
When two operators commute with each other this doesn't mean that they have the same set of eigenstates. This means that you can alway find a set of eigenstates common to both the operators.
In you particular case the set of common eigenstates must be composed of some other functions.
 
Einj said:
When two operators commute with each other this doesn't mean that they have the same set of eigenstates. This means that you can alway find a set of eigenstates common to both the operators.
In you particular case the set of common eigenstates must be composed of some other functions.

If I understand your post correctly, you are saying that if two operators commute, then it is wrong to conclude that every eigenstate of one operator is necessarily an eigenstate of the other operator. However it is right to conclude that at least one set of eigenstates of one is a set of eigenstates of the other.

And so the set of eigenstates the original poster found for p^2 just happened to be one of those sets that isn't a set of eigenstates for p.

Is that right?
 
As a simple example of what is going on here, consider two matrices: diag(1, 1) and diag(1, -1). These matrices commute (after all, one is the identity). Yet the column vector (1 1) is a eigenvalue of the first matrix, but not of the second. The fact that the matrices commute only tells us that there is some basis of common eigenvectors; in this case, the vectors (1, 0) and (0, 1).
 
DocZaius said:
And so the set of eigenstates the original poster found for p^2 just happened to be one of those sets that isn't a set of eigenstates for p.

Is that right?

That's correct :biggrin:
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. Towards the end of the first lecture for the Qiskit Global Summer School 2025, Foundations of Quantum Mechanics, Olivia Lanes (Global Lead, Content and Education IBM) stated... Source: https://www.physicsforums.com/insights/quantum-entanglement-is-a-kinematic-fact-not-a-dynamical-effect/ by @RUTA
If we release an electron around a positively charged sphere, the initial state of electron is a linear combination of Hydrogen-like states. According to quantum mechanics, evolution of time would not change this initial state because the potential is time independent. However, classically we expect the electron to collide with the sphere. So, it seems that the quantum and classics predict different behaviours!
Back
Top