[Q]Eigenfunction of inverse opreator and another question.

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Discussion Overview

The discussion revolves around the properties of eigenfunctions of inverse operators, specifically addressing the relationship between an operator \(\hat{A}\) and its inverse \(\hat{A^{-1}}\). Participants explore the implications of the commutator theorem and the application of these operators to functions, examining the validity of certain mathematical assertions.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant asks for a proof that if \(\hat{A}\varphi = a\varphi\), then \(\hat{A^{-1}}\varphi = \frac{1}{a}\varphi\), suggesting that the eigenfunction of the inverse operator is the same as that of the original operator.
  • Another participant suggests that operating with \(\hat{A^{-1}}\) directly leads to the conclusion without needing to commute \(\hat{A}\) with its inverse.
  • There is a clarification regarding the notation, with one participant noting that the last term should be \(\phi = a \hat{A^{-1}} \phi\).
  • One participant expresses confusion about the application of the product of operators, questioning whether \(AA^{-1}\varphi = \varphi\) holds true in all contexts, particularly when considering functions as matrices.
  • Another participant asserts that the law of association holds for matrices, indicating that the expressions are indeed equal.
  • A further clarification is made that the equality of operators means they must yield the same result when applied to any function, reinforcing the concept of operator equality.

Areas of Agreement / Disagreement

Participants exhibit disagreement regarding the application of operators and the implications of operator equality. While some assert that the equality holds universally, others raise concerns about specific cases involving functions as matrices.

Contextual Notes

There are unresolved assumptions regarding the nature of the operators and functions involved, particularly in the context of matrix representations and operator application. The discussion does not resolve whether the commutator theorem applies universally in this scenario.

good_phy
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Hi.

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

textbook said eigenfunction of inverse operator A is the same as [itex]\varphi[/itex]

which eigenvalue is [itex]\frac{1}{a}[/itex]

Can you prove that?

And is it really that [itex][A,A^{-1}] = 0[/itex] so both opreatator have a common
eigenfunction if eigenvalue is not degenerate, this theorem is called commutator theorem?
 
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Just operate with A^-1 so you get [tex]A^{-1}A\phi =\phi =a A^{-1}\phi[/tex].
No need to commute A with its inverse.
Then proving the commutator=0 follows.
 
Last edited by a moderator:
The latex doesn't seem to work. The last term is \phi=a A^{-1} \phi.
 
I don't understand why [itex]AA^{-1}\varphi = \varphi[/itex] It is absolute true that

[itex]AA^{-1} = 1[/itex] But applying [itex]AA^{-1}[/itex] to some function is

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

[itex]A(A^{-1}f)[/itex] is not [itex](AA^{-1})f[/itex] right?
 
Hm, yes, the law of association holds for matrices, so those two last expressions are equal.
 
good_phy said:
It is absolute true that

[itex]AA^{-1} = 1[/itex] But applying [itex]AA^{-1}[/itex] 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.
 

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