Commutation and Eigenfunctions

  • #1
My first question is, does any operator commute with itself? If this is the case, is there a simple proof to show so? If not, what would be a counter-example or a "counter-proof"?

My second question has to do with the properties of an eigenvalue problem. If you have an operator Q such that [tex]\hat{Q}[/tex][tex]\psi[/tex]=[tex]\lambda\psi[/tex], then is the following always true?

[tex]\hat{Q}^{2}[/tex][tex]\psi[/tex]=[tex]\lambda^{2}\psi[/tex]

Since [tex]\hat{Q}^{2}[/tex][tex]\psi[/tex]=[tex]\hat{Q}\hat{Q}[/tex][tex]\psi[/tex]=[tex]\hat{Q}\lambda\psi[/tex]=[tex]\lambda\hat{Q}\psi[/tex]=[tex]\lambda\lambda\psi[/tex]=[tex]\lambda^{2}\psi[/tex].

And is there a more general way of saying this? Like,

[tex]\hat{Q}[/tex][tex]_{i}[/tex][tex]\hat{Q}[/tex][tex]_{j}[/tex][tex]\psi[/tex]=[tex]\hat{Q}[/tex][tex]_{i}[/tex][tex]\lambda[/tex][tex]_{j}[/tex][tex]\psi[/tex]=[tex]\lambda[/tex][tex]_{j}[/tex][tex]\hat{Q}[/tex][tex]_{i}[/tex][tex]\psi[/tex]=[tex]\lambda[/tex][tex]_{i}[/tex][tex]\lambda[/tex][tex]_{j}[/tex][tex]\psi[/tex]

Which should always be true even if the operators don't commute because the lambdas are just scalars right?
 
Last edited:

Answers and Replies

  • #2
213
8


Of course an operator commutes with itself [Q,Q] is always 0 from the definition of the commutator

For the last part you wouldn't be able to write it like that if the operators don't commute. Commuting operators have simultaneous eigenfunctions, non-commuting ones do not
 

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