Commutators, operators and eigenvalues

[dyn]

Hi
I just wanted to check my understanding of something which has come up when first studying path integrals in QM. If x and p_x are operators then [ x , p_x ] = iħ but if x and p_x operate on states to produce eigenvalues then the eigenvalues x and p_x commute because they are just numbers. Is that correct ?
Thanks

 

[Hill]

Operators $x$ and $p_x$ don't have eigenstates in common.

 

[dyn]

Yes . i know that but x can act on a position eigenstate and p can act on a momentum eigenstate. I just want to check that once the operators act on states to produce eigenvalues , that eigenvalues commute because they are just numbers ?

 

[PeterDonis]

the eigenvalues x and p_x commute because they are just numbers. Is that correct ?

No, it makes no sense, because you don't multiply eigenvalues in QM. You could of course measure the position of one quantum system and the momentum of another and then multiply the results you get, but what sense would it make?

 

[PeterDonis]

something which has come up when first studying path integrals in QM

I'm struggling to see how the question you are asking could arise in that context. Can you give a reference?

 

[anuttarasammyak]

. I just want to check that once the operators act on states to produce eigenvalues , that eigenvalues commute because they are just numbers ?

Numbers and operators commute. It does not matter from where the numbers come. 2024 commutes with operators. We do not care whether it is a measured value of position of some state or new calendar year. Happy new year!

 

[dyn]

I'm struggling to see how the question you are asking could arise in that context. Can you give a reference?

The derivation of the path integral in QM on P211-212 of "QFT for the gifted amateur" by Lancaster & Blundell

 

[Hill]

The derivation of the path integral in QM on P211-212 of "QFT for the gifted amateur" by Lancaster & Blundell

Yes, that is correct. All the $p$'s and $q_n$'s here,

are just real numbers and of course commute.