# [Q]Eigenfunction of inverse opreator and another question.

1. Nov 7, 2008

### 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?

2. Nov 7, 2008

### clem

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.

Last edited: Nov 7, 2008
3. Nov 7, 2008

### clem

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

4. Nov 7, 2008

### 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?

5. Nov 8, 2008

### Manilzin

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

6. Nov 8, 2008

### Fredrik

Staff Emeritus
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.