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CanIExplore

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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?

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?

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