[Q]Eigenfunction of inverse opreator and another question.

[good_phy]

Hi.

Do you know eigenfunction of inverse operator, for instance \hat{A^{-1}} given that \hat{A}\varphi = a\varphi

textbook said eigenfunction of inverse operator A is the same as \varphi

which eigenvalue is \frac{1}{a}

Can you prove that?

And is it really that [A,A^{-1}] = 0 so both opreatator have a common
eigenfunction if eigenvalue is not degenerate, this theorem is called commutator theorem?

 

[Meir Achuz]

Just operate with A^-1 so you get A^{-1}A\phi =\phi =a A^{-1}\phi.
No need to commute A with its inverse.
Then proving the commutator=0 follows.

 

[Meir Achuz]

The latex doesn't seem to work. The last term is \phi=a A^{-1} \phi.

 

[good_phy]

I don't understand why AA^{-1}\varphi = \varphi It is absolute true that

AA^{-1} = 1 But applying AA^{-1} to some function is

different matter. for example we assume A and its inverse can be matrix, function f is also matrix.

A(A^{-1}f) is not (AA^{-1})f right?

 

[Manilzin]

Hm, yes, the law of association holds for matrices, so those two last expressions are equal.

 

[Fredrik]

It is absolute true that

AA^{-1} = 1 But applying AA^{-1} to some function is

different matter.

No, it can't be a different matter. To say that two operators X and Y are equal means that Xf=Yf for all functions f. This is no different from saying that two functions f and g are equal if f(x)=g(x) for all x.